IB Quantum Mechanics (Full)

Part IB — Quantum Mechanics

Based on lectures by J. M. Evans

Notes taken by Dexter Chua

Michaelmas 2015

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Physical background

Photo electric effect. Electrons in atoms and line spectra. Particle diffraction. [1]

Schr¨odinger equation and solutions

De Broglie waves. Schr¨odinger equation. Superposition principle. Probability interpre-

tation, density and current. [2]

Stationary states. Free particle, Gaussian wave packet. Motion in 1-dimensional

potentials, parity. Potential step, square well and barrier. Harmonic oscillator. [4]

Observables and expectation values

Position and momentum op erators and expectation values. Canonical commutation

relations. Uncertainty principle. [2]

Observables and Hermitian operators. Eigenvalues and eigenfunctions. Formula for

expectation value. [2]

Hydrogen atom

Spherically sy mmetric wave functions for spherical well and hydrogen atom.

Orbital angular momentum operators. General solution of hydrogen atom. [5]

Contents

0 Introduction

0.1 Light quanta

0.2 Bohr model of the atom

0.3 Matter waves

1 Wavefunctions and the Schr¨odinger equation

1.1 Particle state and probability

1.2 Operators

1.3 Time evolution of wavefunctions

2 Some examples in one dimension

2.1 Introduction

2.2 Infinite well — particle in a box

2.3 Parity

2.4 Potential well

2.5 The harmonic oscillator

3 Expectation and uncertainty

3.1 Inner products and expectation values

3.2 Ehrenfest’s theorem

3.3 Heisenberg’s uncertainty principle

4 More results in one dimensions

4.1 Gaussian wavepackets

4.2 Scattering

4.3 Potential step

4.4 Potential barrier

4.5 General features of stationary states

5 Axioms for quantum mechanics

5.1 States and observables

5.2 Measurements

5.3 Evolution in time

5.4 Discrete and continuous spectra

5.5 Degeneracy and simultaneous measurements

6 Quantum mechanics in three dimensions

6.1 Introduction

6.2 Separable eigenstate solutions

6.3 Angular momentum

6.4 Joint eigenstates for a spherically symmetric potential

7 The hydrogen atom

7.1 Introduction

7.2 General solution

7.3 Comments

0 Introduction

Quantum mechanics (QM) is a radical generalization of classical physics. Pro-

found new features of quantum mechanics include

(i)

Quantisation — Quantities such as energy are often restricted to a discrete

set of values, or appear in definite amounts , called quanta.

(ii)

Wave-particle duality — Classical concepts of particles and waves become

merged in quantum mechanics. They are different aspects of a single entity.

So an electron will no longer be thought of a “particle” but an entity that

has properties of both particles and waves.

(iii)

Probability and uncertainty — Predictions in quantum mechanics involve

probability in a fundamental way. This probability does not arise from our

lack of knowledge of the system, but is a genuine uncertainty in reality. In

particular, there are limits to what we can ask about a physical system,

even in principle. For example, the Heisenberg uncertainty principle entails

that we cannot accurately know both the position and momentum of a

particle.

Quantum mechanics also involves a new fundamental constant

2π

. The

dimension of this is

[h] = ML

−1

= [energy] × [time] = [position] ×[momentum].

We can think of this constant as representing the “strength” of quantum effects.

Despite having these new profound features, we expect to recover classical physics

when we take the limit ~ → 0.

Historically, there are a few experiments that led to the development of

quantum mechanics.

0.1 Light quanta

In quantum mechanics, light (or electromagnetic waves) consists of quanta called

photons. We can think of them as waves that come in discrete “packets” that

behave like particles.

In particular, photons behave like particles with energy

hν

~ω

, where

is the frequency and

= 2

πν

is the angular frequency. However, we usually

don’t care about ν and just call ω the frequency.

Similarly, the momentum is given by

h/λ

, where

is the wavelength

and k = 2π/λ is the wave number.

For electromagnetic waves, the speed is

ω/k

νλ

. This is consistent

with the fact that photons are massless particles, since we have

E = cp,

as entailed by special relativity.

Historically, the concept of quanta was introduced by Planck. At that time,

the classical laws of physics did not accurately depict how the spectrum of black-

body radiation will behave. In particular, it predicted that a black body will emit

an infinite amount of energy through radiation, which is clearly nonsense. Using

the concept of quanta and the energy-frequency relation, Planck was to derive the

spectrum of “black-body” radiation that is consistent with experimental results.

However, they were not yet sure that light indeed come in quanta physically. It

could have just been a mathematical trick to derive the desired result.

The physical reality of photons was clarified by Einstein in explaining the

photo-electric effect.

When we shine some light (or electromagnetic radiation

) of frequency

onto certain metals, this can cause an emission of electrons (

). We can measure

the maximum kinetic energy K of these electrons.

Experiments show that

(i)

The number of electrons emitted depends on the intensity (brightness) of

the light, but not the frequency.

(ii)

The kinetic energy

depends only (linearly) on the frequency but not the

intensity.

(iii) For ω < ω

(for some critical value ω

), no electrons are emitted at all.

This is hard to understand classically, but is exactly as expected if each electron

emitted is due to the impact with a single photon. If

is the energy required

to liberate an electron, then we would expect

~ω − W

by the conservation

of energy. We will have no emission if ω < ω

= W/~.

0.2 Bohr model of the atom

When we heat atoms up to make them emit light; or shine light at atoms so that

they absorb light, we will find that light is emitted and absorbed at very specific

frequencies, known as the emission and absorption spectra. This suggests that

the inner structure of atoms is discrete.

However, this is not the case in the classical model. In the classical model,

the simplest atom, the hydrogen, consists of an electron with charge

−e

and

mass m, orbiting a proton of charge +e and mass m

 m fixed at the origin.

The potential energy is

V (r) = −

4πε

and the dynamics of the electron is governed by Newton’s laws of motions, just as

we derived the orbits of planets under the gravitational potential in IA Dynamics

and Relativity. This model implies that the angular momentum

is constant,

and so is the energy E =

+ V (r).

This is not a very satisfactory model for the atom. First of all, it cannot

explain the discrete emission and absorption spectra. More importantly, while

this model seems like a mini solar system, electromagnetism behaves differently

from gravitation. To maintain a circular orbit, an acceleration has to be applied

onto the electron. Indeed, the force is given by

F =

4πε

Accelerating particles emit radiation and lose energy. So according to classical

electrodynamics, the electron will just decay into the proton and atoms will

implode.

The solution to this problem is to simply declare that this cannot happen.

Bohr proposed the Bohr quantization conditions that restricts the classical orbits

by saying that the angular momentum can only take values

L = mrv = n~

for

= 1

, ···

. Using these, together with the force equation, we can solve

and v completely for each n and obtain

4πε

= −



4πε



Now we assume that the electron can make transitions between different energy

levels

and

m > n

, accompanied by emission or absorption of a photon of

frequency ω given by

E = ~ω = E

− E



4πε





−



This model explains a vast amount of experimental data. This also gives an

estimate of the size of a hydrogen atom:



4πε



≈ 5.29 × 10

−11

known as the Bohr radius.

While the model fits experiments very well, it does not provide a good

explanation for why the radius/angular momentum should be quantized. It

simply asserts this fact and then magically produces the desired results. Thus,

we would like a better understanding of why angular momentum should be

quantized.

0.3 Matter waves

The relations

E = hν = ~ω

p =

= ~k

are used to associate particle properties (energy and momentum) to waves. They

can also be used the other way round — to associate wave properties (frequency

and wave number) to particles. Moreover, these apply to non-relativistic particles

such as electrons (as well as relativistic photons). This

is known as the de

Broglie wavelength.

Of course, nothing prevents us from assigning arbitrary numbers to our

particles. So an immediate question to ask is — is there any physical significance

to the “frequency” and “wavenumber” of particles? Or maybe particles in fact

are waves?

Recall that the quantization of the Bohr model requires that

L = rp = n~.

Using the relations above, this is equivalent to requiring that

nλ = 2πr.

This is exactly the statement that the circumference of the orbit is an integer

multiple of the wavelength. This is the condition we need for a standing wave

to form on the circumference. This looks promising as an explanation for the

quantization relation.

But in reality, do electrons actually behave like waves? If electrons really are

waves, then they should exhibit the usual behaviour of waves, such as diffraction

and interference.

We can repeat our favorite double-slit experiment on electrons. We have a

sinusoidal wave incident on some barrier with narrow openings as shown:

wave

density of

electrons

At different points, depending on the difference

in path length, we may

have constructive interference (large amplitude) or destructive interference (no

amplitude). In particular, constructive interference occurs if

nλ

, and

destructive if δ = (n +

)λ.

Not only does this experiment allow us to verify if something is a wave. We

can also figure out its wavelength λ by experiment.

Practically, the actual experiment for electrons is slightly more complicated.

Since the wavelength of an electron is rather small, to obtain the diffraction

pattern, we cannot just poke holes in sheets. Instead, we need to use crystals as

our diffraction grating. Nevertheless, this shows that electrons do diffract, and

the wavelength is the de Broglie wavelength.

This also has a conceptual importance. For regular waves, diffraction is

something we can make sense of. However, here we are talking about electrons.

We know that if we fire many many electrons, the distribution will follow the

pattern described above. But what if we just fire a single electron? On average,

it should still follow the distribution. However, for this individual electron, we

cannot know where it will actually land. We can only provide a probability

distribution of where it will end up. In quantum mechanics, everything is

inherently probabilistic.

As we have seen, quantum mechanics is vastly different from classical mechan-

ics. This is unlike special relativity, where we are just making adjustments to

Newtonian mechanics. In fact, in IA Dynamics and Relativity, we just “derived”

special relativity by assuming the principle of relativity and that the speed

of light is independent of the observer. This is not something we can do for

quantum mechanics — what we are going to do is just come up with some theory

and then show (or claim) that they agree with experiment.

1 Wavefunctions and the Schr¨odinger equation

The time evolution of particles in quantum mechanics is governed by the

Schr¨odinger equation, which we will come to shortly. In general, this is a

difficult equation to solve, and there aren’t many interesting cases where we are

able to provide a full solution.

Hence, to begin with, we will concentrate on quantum mechanics in one

(spatial) dimension only, as the maths is much simpler and diagrams are easier

to draw, and we can focus more on the physical content.

1.1 Particle state and probability

Classically, a point particle in 1 dimension has a definitive position

(and

momentum

) at each time. To completely specify a particle, it suffices to write

down these two numbers. In quantum mechanics, this is much more complicated.

Instead, a particle has a state at each time, specified by a complex-valued

wavefunction ψ(x).

The physical content of the wavefunction is as follows: if

is appropriately

normalized, then when we measure the position of a particle, we get a result

with probability density function

|ψ

(

)

, i.e. the probability that the position

is found in [

x, x

δx

] (for small

δx

) is given by

|ψ

(

)

δx

. Alternatively, the

probability of finding it in an interval [a, b] is given by

P(particle position in [a, b]) =

|ψ(x)|

dx.

What do we mean by “appropriately normalized”? From our equation above, we

see that we require

∞

−∞

|ψ(x)|

dx = 1,

since this is the total probability of finding the particle anywhere at all. This

the normalization condition required.

Example (Gaussian wavefunction). We define

ψ(x) = Ce

−

(x−c)

2α

where c is real and C could be complex.

We have

∞

−∞

|ψ(x)|

dx = |C|

∞

−∞

−

(x−c)

dx = |C|

(απ)

= 1.

So for normalization, we need to pick C = (1/απ)

1/4

(up to a multiple of e

iθ

is small, then we have a sharp peak around

. If

is large, it is

more spread out.

While it is nice to have a normalized wavefunction, it is often inconvenient

to deal exclusively with normalized wavefunctions, or else we will have a lot of

ugly-looking constants floating all around. As a matter of fact, we can always

restore normalization at the end of the calculation. So we often don’t bother.

If we do not care about normalization, then for any (non-zero)

(

) and

λψ

(

) represent the same quantum state (since they give the same probabilities).

In practice, we usually refer to either of these as “the state”. If we like fancy

words, we can thus think of the states as equivalence classes of wavefunctions

under the equivalence relation ψ ∼ φ if φ = λψ for some non-zero λ.

What we do require, then, is not that the wavefunction is normalized, but

normalizable, i.e.

∞

−∞

|ψ(x)|

dx < ∞.

We will very soon encounter wavefunctions that are not normalizable. Mathe-

matically, these are useful things to have, but we have to be more careful when

interpreting these things physically.

A characteristic property of quantum mechanics is that if

(

) and

(

) are

wavefunctions for a particle, then

(

) +

(

) is also a possible particle state

(ignoring normalization), provided the result is non-zero. This is the principle of

superposition, and arises from the fact that the equations of quantum mechanics

are linear.

Example (Superposition of Gaussian wavefunctions). Take

ψ(x) = B



exp



−(x − c)

2α



+ exp



−

2β



Then the resultant distribution would be something like

We choose

so that

in a normalized wavefunction for a single particle. Note

that this is not two particles at two different positions. It is one particle that is

“spread out” at two different positions.

It is possible that in some cases, the particles in the configuration space

may be restricted. For example, we might require

−

≤ x ≤

with some

boundary conditions at the edges. Then the normalization condition would not

be integrating over (−∞, ∞), but [−

1.2 Operators

We know that the square of the wavefunction gives the probability distribution of

the position of the particle. How about other information such as the momentum

and energy? It turns out that all the information about the particle is contained

in the wavefunction (which is why we call it the “state” of the particle).

We call each property of the particle which we can measure an observable.

Each observable is represented by an operator acting on

(

). For example, the

position is represented by the operator

ˆx

. This means that (

ˆxψ

)(

) =

xψ

(

We can list a few other operators:

position ˆx = x ˆxψ = xψ(x)

momentum ˆp = −i~

∂

∂x

ˆpψ = −i~ψ

(x)

energy H =

ˆp

+ V (ˆx) Hψ = −

(x) + V (x)ψ(x)

The final

is called the Hamiltonian, where

is the mass and

is the potential.

We see that the Hamiltonian is just the kinetic energy

and the potential

energy

. There will be more insight into why the operators are defined like

this in IIC Classical Dynamics and IID Principles of Quantum Mechanics.

Note that we put hats on

ˆx

and

ˆp

to make it explicit that these are operators,

as opposed to the classical quantities position and momentum. Otherwise, the

definition ˆx = x would look silly.

How do these operators relate to the actual physical properties? In general,

when we measure an observable, the result is not certain. They are randomly

distributed according to some probability distribution, which we will go into full

details later.

However, a definite result is obtained if and only if

is an eigenstate, or

eigenfunction, of the operator. In this case, results of the measurements are the

eigenvalue associated. For example, we have

ˆpψ = pψ

if and only if ψ is a state with definite momentum p. Similarly,

Hψ = Eψ

if and only if ψ has definite energy E.

Here we are starting to see why quantization occurs in quantum mechanics.

Since the only possible values of

and

are the eigenvalues, if the operators

have a discrete set of eigenvalues, then we can only have discrete values of

and

Example. Let

ψ(x) = Ce

ikx

This has a wavelength of

= 2

π/k

. This is a momentum eigenstate, since we

have

ˆpψ = −~ψ

= (~k)ψ.

So we know that the momentum eigenvalue is p = ~k. This looks encouraging!

Note that if there is no potential, i.e. V = 0, then

Hψ =

ˆp

ψ = −

ψ.

So the energy eigenvalue is

E =

Note, however, that our wavefunction has

|ψ

(

)

|C|

, which is a constant.

So this wavefunction is not normalizable on the whole line. However, if we

restrict ourselves to some finite domain

−

≤ x ≤

, then we can normalize by

picking C =

√

Example. Consider the Gaussian distribution

ψ(x) = C exp



−

2α



We get

ˆpψ(x) = −i~ψ

(x) 6= pψ(x)

for any number p. So this is not an eigenfunction of the momentum.

However, if we consider the harmonic oscillator with potential

V (x) =

then this

(

) is an eigenfunction of the Hamiltonian operator, provided we

picked the right α. We have

Hψ = −

ψ = Eψ

when

. Then the energy is

. This is to be verified on the

example sheet.

Despite being a simple system, the harmonic oscillator is incredibly useful in

theoretical physics. We will hence solve this completely later.

Definition

(Time-independent Schr¨odinger equation)

The time-independent

Schr¨odinger equation is the energy eigenvalue equation

Hψ = Eψ,

−

+ V (x)ψ = Eψ.

This is in general what determines what the system behaves. In particular,

the eigenvalues E are precisely the allowed energy values.

1.3 Time evolution of wavefunctions

So far, everything is instantaneous. The wavefunction specifies the state at a

particular time, and the eigenvalues are the properties of the system at that

particular time. However, this is quantum mechanics, or quantum dynamics. We

should be looking at how things change. We want to know how the wavefunction

changes with time. This is what we will get to now.

Time-dependent Schr¨odinger equation

We will write Ψ instead of

to indicate that we are looking at the time-dependent

wavefunction. The evolution of this Ψ(

x, t

) is described by the time-dependent

Schr¨odinger equation.

Definition

(Time-dependent Schr¨odinger equation)

For a time-dependent

wavefunction Ψ(x, t), the time-dependent Schr¨odinger equation is

∂Ψ

∂t

= HΨ. (∗)

For a particle in a potential V (x), this can reads

∂Ψ

∂t

= −

∂

∂x

+ V (x)Ψ.

While this looks rather scary, it isn’t really that bad. First of all, it is linear. So

the sums and multiples of solutions are also solutions. It is also first-order in

time. So if we know the wavefunction Ψ(

x, t

) at a particular time

, then this

determines the whole function Ψ(x, t).

This is similar to classical dynamics, where knowing the potential

(and

hence the Hamiltonian

) completely specifies how the system evolves with

time. However, this is in some ways different from classical dynamics. Newton’s

second law is second-order in time, while this is first-order in time. This is

significant since when our equation is first-order in time, then the current state

of the wavefunction completely specifies the evolution of the wavefunction in

time.

Yet, this difference is just an illusion. The wavefunction is the state of the

particle, and not just the “position”. Instead, we can think of it as capturing the

position and momentum. Indeed, if we write the equations of classical dynamics

in terms of position and momentum, it will be first order in time.

Stationary states

It is not the coincidence that the time-independent Schr¨odinger equation and

the time-dependent Schr¨odinger equation are named so similarly (and it is also

not an attempt to confuse students).

We perform separation of variables, and consider a special class of solutions

Ψ(

x, t

) =

(

)

(

), where Ψ(

0) =

(

) (i.e.

(0) = 1). If

satisfies the

time-independent Schr¨odinger equation

Hψ = Eψ,

then since

does not involve time derivatives, we know Ψ is an energy eigenstate

at each fixed t, i.e.

HΨ = EΨ.

So if we want this Ψ to satisfy the Schr¨odinger equation, we must have

T = ET.

The solution is obvious:

T (t) = exp



−

iEt



We can write our full solution as

Ψ(x, t) = ψ(x) exp



−

iEt



Note that the frequency is

. So we recover the Energy-frequency relation

we’ve previously had.

Definition (Stationary state). A stationary state is a state of the form

Ψ(x, t) = ψ(x) exp



−

iEt



where

(

) is an eigenfunction of the Hamiltonian with eigenvalue

. This term

is also sometimes applied to ψ instead.

While the stationary states seem to be a rather peculiar class of solutions

that would rarely correspond to an actual physical state in reality, they are in

fact very important in quantum mechanics. The reason is that the stationary

states form a basis of the state space. In other words, every possible state can be

written as a (possibly infinite) linear combination of stationary states. Hence, by

understanding the stationary states, we can understand a lot about a quantum

system.

Conservation of probability

Note that for a stationary state, we have

|Ψ(x, t)|

= |ψ(x)|

which is independent of time. In general, this is true in most cases.

Consider a general Ψ(

x, t

) obeying the time-dependent Schr¨odinger equation.

Proposition. The probability density

P (x, t) = |Ψ(x, t)|

obeys a conservation equation

∂P

∂t

= −

∂j

∂x

where

j(x, t) = −



∗

dΨ

−

dΨ

∗



is the probability current.

Since Ψ

∗

is the complex conjugate of Ψ

0∗

Ψ, we know that Ψ

∗

−

0∗

Ψ is

imaginary. So multiplying by

ensures that

(

x, t

) is real, which is a good thing

since P is also real.

Proof.

This is straightforward from the Schr¨odinger equation and its complex

conjugate. We have

∂P

∂t

= Ψ

∗

∂Ψ

∂t

∂Ψ∗

∂t

= Ψ

∗

−

00∗

where the two V terms cancel each other out, assuming V is real

= −

∂j

∂x

The important thing here is not the specific form of

, but that

∂P

∂t

can be

written as the space derivative of some quantity. This implies that the probability

that we find the particle in [a, b] at fixed time t changes as

|Ψ(x, t)|

dx =

−

∂j

∂x

(x, t) dx = j(a, t) − j(b, t).

We can think of the final term as the probability current getting in and out of

the interval at the boundary.

In particular, consider a normalizable state with Ψ

, j →

0 as

x → ±∞

for fixed t. Taking a → −∞ and b → +∞, we have

∞

−∞

|Ψ(x, t)|

dx = 0.

What does this tell us? This tells us that if Ψ(

0) is normalized, Ψ(

x, t

) is

normalized for all

. Hence we know that for each fixed

Ψ(

x, t

)

is a probability

distribution. So what this really says is that the probability interpretation is

consistent with the time evolution.

2 Some examples in one dimension

2.1 Introduction

In general, we are going to consider the energy eigenvalue problem for a particle

in 1 dimension in a potential V (x), i.e.

Hψ = −

+ V (x)ψ = Eψ.

In other words, we want to find the allowed energy eigenvalues.

This is a hard problem in general. However, we can consider the really easy

case where

(

) =

, where

is a constant. Then we can easily write down

solutions.

If U > E, then the Schr¨odinger equation is equivalent to

− κ

ψ = 0,

where κ is such that U − E =

. We take wlog κ > 0. The solution is then

ψ = Ae

κx

+ Be

−κx

On the other hand, if U < E, then the Schr¨odinger equation says

ψ + k

ψ = 0,

where k is picked such that E − U =

. The solutions are

ψ = Ae

ikx

+ Be

ikx

Note that these new constants are merely there to simplify our equations. They

generally need not have physical meanings.

Now why are we interested in cases where the potential is constant? Wouldn’t

that be just equivalent to a free particle? This is indeed true. However, knowing

these solutions allow us to to study piecewise flat potentials such as steps, wells

and barriers.

Here a finite discontinuity in

is allowed. In this case, we can have

ψ, ψ

continuous and

discontinuous. Then the discontinuity of

cancels that of

V , and the Schr¨odinger equation holds everywhere.

In this chapter, we will seek normalizable solutions with

∞

−∞

|ψ(x)|

dx.

This requires that

(

)

→

0 as

x → ±∞

. We see that for the segments and

the end, we want to have decaying exponentials

−κx

instead of oscillating

exponentials e

−ikx

2.2 Infinite well — particle in a box

The simplest case to consider is the infinite well. Here the potential is infinite

outside the region [

−a, a

], and we have much less to think about. For

|x| > a

we must have ψ(x) = 0, or else V (x)ψ(x) would be infinite.

−a

V (x) =

(

0 |x| ≤ a

∞ |x| > a.

We require

= 0 for

|x| > a

and

continuous at

±a

. Within

|x| < a

, the

Schr¨odinger equation is

−

= Eψ.

We simplify this to become

+ k

ψ = 0,

where

E =

Here, instead of working with the complex exponentials, we use

sin

and

cos

since

we know well when these vanish. The general solution is thus

ψ = A cos kx + B sin kx.

Our boundary conditions require that ψ vanishes at x = ±a. So we need

A cos ka ±B sin ka = 0.

In other words, we require

A cos ka = B sin ka = 0.

Since

sin ka

and

cos ka

cannot be simultaneously 0, either

= 0 or

= 0. So

the two possibilities are

(i) B = 0 and ka = nπ/2 with n = 1, 3, ···

(ii) A = 0 and ka = nπ/2 with n = 2, 4, ···

Hence the allowed energy levels are

8ma

where n = 1, 2, ···, and the wavefunctions are

(x) =





(

cos

nπx

n odd

sin

nπx

n even

These are normalized with

−a

|ψ

(x)|

dx.

−a −a

This was a rather simple and nice example. We have an infinite well, and the

particle is well-contained inside the box. The solutions just look like standing

waves on a string with two fixed end points — something we (hopefully) are

familiar with.

Note that

(

−x

) = (

−

n+1

(

). We will see that this is a general feature

of energy eigenfunctions of a symmetric potential. This is known as parity.

2.3 Parity

Consider the Schr¨odinger equation for a particle of mass m

Hψ = −

+ V (x)ψ = Eψ.

with potential

V (x) = V (−x).

By changing variables

x → −x

, we see that

(

) is an eigenfunction of

with

energy

if and only if

(

−x

) is an eigenfunction of

with energy

. There

are two possibilities:

(i)

(

) and

(

−x

) represent the same quantum state, this can only happen

(

−x

) =

ηψ

(

) for some constant

. Since this is true for all

, we can

do this twice and get

ψ(x) = ηψ(−x) = η

ψ(x).

So we get that

1 and

(

−x

) =

±ψ

(

). We call

the parity, and say

ψ has even/odd parity if η is +1/ −1 respectively.

For example, in our particle in a box, our states ψ

have parity (−1)

n+1

(ii)

(

) and

(

−x

) represent different quantum states, then we can still

take linear combinations

(x) = α(ψ(x) ±ψ(−x)),

and these are also eigenstates with energy eigenvalue

, where

is for

normalization. Then by construction,

(

−x

) =

±ψ

(

) and have parity

η = ±1.

Hence, if we are given a potential with reflective symmetry

(

−x

) =

(

), then

we can restrict our attention and just look for solutions with definite parity.

2.4 Potential well

We will consider a potential that looks like this:

−a

−U

The potential is given by

V (x) =

(

−U |x| < a

0 |x| ≥ a

for some constant

U >

0. Classically, this is not very interesting. If the energy

E <

0, then the particle is contained in the well. Otherwise it is free to move

around. However, in quantum mechanics, this is much more interesting.

We want to seek energy levels for a particle of mass

, defined by the

Schr¨odinger equation

Hψ = −

+ V (x)ψ = Eψ.

For energies in the range

−U < E < 0,

we set

U + E =

> 0, E = −

where

k, κ >

0 are new real constants. Note that these coefficients are not

independent, since U is given and fixed. So they must satisfy

+ κ

2mU

Using these constants, the Schr¨odinger equation becomes

(

+ k

ψ = 0 |x| < a

− κ

ψ = 0 |x| > a.

As we previously said, we want the Schr¨odinger equation to hold even at the

discontinuities. So we need ψ and ψ

to be continuous at x = ±a.

We first consider the even parity solutions

(

−x

) =

(

). We can write our

solution as

ψ =

(

A cos kx |x| < a

−κ|x|

|x| > a

We match ψ and ψ

at x = a. So we need

A cos ka = Be

−κa

−Ak sin ka = −κBe

−κa

By parity, there is no additional information from x = −a.

We can divide the equations to obtain

k tan ka = κ.

this is still not something we can solve easily. To find when solutions exist, it is

convenient to introduce

ξ = ak, η = aκ,

where these two constants are dimensionless and positive. Note that this

has

nothing to do with parity. It’s just that we have run out of letters to use. Hence

the solution we need are solutions to

η = ξ tan ξ.

Also, our initial conditions on k and κ require

+ η

2ma

We can look for solutions by plotting these two equations. We first plot the

curve η = ξ tan ξ:

1π

3π

5π

7π

9π

11π

The other equation is the equation of a circle. Depending on the size of the

constant 2ma

U/~

, there will be a different number of points of intersections.

So there will be a different number of solutions depending on the value of

2ma

U/~

. In particular, if

(n − 1)π <



2mUa



1/2

< nπ,

then we have exactly n even parity solutions (for n ≥ 1).

We can do exactly the same thing for odd parity eigenstates. . . on example

sheet 1.

For

E >

0 or

E < −U

, we will end up finding non-normalizable solutions.

What is more interesting, though is to look at the solutions we have now. We

can compare what we’ve got with what we would expect classically.

Classically, any value of

in the range

−U < E <

0 is allowed, and the

motion is deeply uninteresting. The particle just goes back and forth inside the

well, and is strictly confined in −a ≤ x ≤ a.

Quantum mechanically, there is just a discrete, finite set of allowed energies.

What is more surprising is that while

decays exponentially outside the well, it

is non-zero! This means there is in theory a non-zero probability of finding the

particle outside the well! We call these particles bound in the potential, but in

fact there is a non-zero probability of finding the particle outside the well.

2.5 The harmonic oscillator

So far in our examples, the quantization (mathematically) comes from us requiring

continuity at the boundaries. In the harmonic oscillator, it arises in a different

way.

This is a harmonic oscillator of mass m with

V (x) =

mω

Classically, this has a motion of

A cos ω

(

t − t

), which is something we

(hopefully) know well too.

This is a really important example. First of all, we can solve it, which is a

good thing. More importantly, any smooth potential can be approximated by a

harmonic oscillator near an equilibrium x

, since

V (x) = V (x

) +

)(x − x

)

+ ··· .

Systems with many degrees like crystals can also be treated as collections of

independent oscillators by considering the normal modes. If we apply this to

the electromagnetic field, we get photons! So it is very important to understand

the quantum mechanical oscillator.

We are going to seek all normalizable solutions to the time-independent

Schr¨odinger equation

Hψ = −

mω

ψ = Eψ.

So simplify constants, we define

y =



mω



x, E =

~ω

both of which is dimensionless. Then we are left with

−

+ y

ψ = Eψ.

We can consider the behaviour for

 E

. For large

, the

term will

be large, and so we want the

term to offset it. We might want to try the

Gaussian

−y

, and when we differentiate it twice, we would have brought

down a factor of y

. So we can wlog set

ψ = f(y)e

−

Then the Schr¨odinger equation gives

− 2y

+ (E − 1) = 0.

This is known as Hermite’s equation. We try a series solution

f(y) =

r≥0

and substitute in to get

r≥0



(r + 2)(r + 1)a

n+2

+ (E − 1 − 2r)a



= 0.

This holds if and only if

r+2

2r + 1 − E

(r + 2)(r + 1)

, r ≥ 0.

We can choose

and

independently, and can get two linearly independent

solutions. Each solution involves either all even or all odd powers.

However, we have a problem. We want normalizable solutions. So we want

to make sure our function does not explode at large

. Note that it is okay if

(

) is quite large, since our

is suppressed by the

−

terms, but we cannot

grow too big.

We look at these two solutions individually. To examine the behaviour of

f(y) when y is large, observe that unless the coefficients vanish, we get

p−2

∼

This matches the coefficients of

for some power

(e.g.

p≥0

). This

is bad, since our ψ will then grow as e

, and cannot be normalized.

Hence, we get normalizable

if and only if the series for

terminates to

give a polynomial. This occurs iff

= 2

+ 1 for some

. Note that for each

only one of the two independent solutions is normalizable. So for each

, we get

exactly one solution.

So for n even, we have

r+2

2r − 2n

(r + 2)(r + 1)

for r even, and a

= 0 for r odd, and the other way round when n is odd.

The solutions are thus

(

) =

(

), where

is a polynomial of degree

with h

(−y) = (−1)

(y).

For example, we have

(y) = a

(1 − 2y

)

(y) = a



y −



These are known as the Hermite polynomials. We have now solved our harmonic

oscillator. With the constant restored, the possible energy eigenvalues are

= ~ω



n +



for n = 0, 1, 2, ···.

The wavefunctions are

(x) = h





mω





exp



−

mω



where normalization fixes a

and a

As we said in the beginning, harmonic oscillators are everywhere. It turns

out quantised electromagnetic fields correspond to sums of quantised harmonic

oscillators, with

− E

= n~ω

This is equivalent to saying the

th state contains

photons, each of energy

~ω

3 Expectation and uncertainty

So far, all we’ve been doing is solving Schr¨odinger’s equation, which is arguably

not very fun. In this chapter, we will look at some theoretical foundations of

quantum mechanics. Most importantly, we will learn how to extract information

from the wavefunction

. Two important concepts will be the expectation

and uncertainty. We will then prove two important theorems about these —

Ehrenfest’s theorem and the Heisenberg uncertainty principle.

3.1 Inner products and expectation values

Definitions

Definition

(Inner product)

Let

(

) and

(

) be normalizable wavefunctions

at some fixed time (not necessarily stationary states). We define the complex

inner product by

(φ, ψ) =

∞

−∞

φ(x)

∗

ψ(x) dx.

Note that for any complex number α, we have

(φ, αψ) = α(φ, ψ) = (α

∗

φ, ψ).

Also, we have

(φ, ψ) = (ψ, φ)

∗

These are just the usual properties of an inner product.

Definition

(Norm)

The norm of a wavefunction

, written,

kψk

is defined by

kψk

= (ψ, ψ) =

∞

−∞

|ψ(x)|

dx.

This ensures the norm is real and positive.

Suppose we have a normalized state

, i.e.

kψk

= 1, we define the expectation

values of observables as

Definition

(Expectation value)

The expectation value of any observable

the state ψ is

hHi

= (ψ, Hψ).

For example, for the position, we have

hˆxi

= (ψ, ˆxψ) =

∞

−∞

x|ψ(x)|

dx.

Similarly, for the momentum, we have

hˆpi

= (ψ, ˆpψ) =

∞

−∞

∗

(−i~ψ

) dx.

How are we supposed to interpret this thing? So far, all we have said about

operators is that if you are an eigenstate, then measuring that property will give

a definite value. However, the point of quantum mechanics is that things are

waves. We can add them together to get superpositions. Then the sum of two

eigenstates will not be an eigenstate, and does not have definite, say, momentum.

This formula tells us what the average value of any state is.

This is our new assumption of quantum mechanics — the expectation value

is the mean or average of results obtained by measuring the observable many

times, with the system prepared in state ψ before each measurement.

Note that this is valid for any operator. In particular, we can take any

function of our existing operators. One important example is the uncertainty:

Definition

(Uncertainty)

The uncertainty in position (∆

)

and momentum

(∆p)

are defined by

(∆x)

= h(ˆx − hˆxi

)

= hˆx

− hˆxi

with exactly the same expression for momentum:

(∆p)

= h(ˆp − hˆpi

)

= hˆp

− hˆpi

We will later show that these quantities (∆

)

and (∆

)

are indeed real

and positive, so that this actually makes sense.

Hermitian operators

The expectation values defined can be shown to be real for

ˆx

and

ˆp

specifically,

by manually fiddling with stuff. We can generalize this result to a large class of

operators known as Hermitian operators.

Definition

(Hermitian operator)

An operator

is Hermitian iff for all nor-

malizable φ, ψ, we have

(φ, Qψ) = (Qφ, ψ).

In other words, we have

∗

Qψ dx =

(Qφ)

∗

ψ dx.

In particular, this implies that

(ψ, Qψ) = (Qψ, ψ) = (ψ, Qψ)

∗

So (ψ, Qψ) is real, i.e. hQi

is real.

Proposition.

The operators

ˆx

ˆp

and

are all Hermitian (for real potentials).

Proof.

We do

ˆx

first: we want to show (

φ, ˆxψ

) = (

ˆxφ, ψ

). This statement is

equivalent to

∞

−∞

φ(x)

∗

xψ(x) dx =

∞

−∞

(xφ(x))

∗

ψ(x) dx.

Since position is real, this is true.

To show that

ˆp

is Hermitian, we want to show (

φ, ˆpψ

) = (

ˆpφ, ψ

). This is

equivalent to saying

∞

−∞

∗

(−i~ψ

) dx =

∞

−∞

(i~φ

)

∗

ψ dx.

This works by integrating by parts: the difference of the two terms is

−i~[φ

∗

ψ]

∞

−∞

= 0

since φ, ψ are normalizable.

To show that H is Hermitian, we want to show (φ, Hψ) = (Hφ, ψ), where

H = −

+ V (x).

To show this, it suffices to consider the kinetic and potential terms separately.

For the kinetic energy, we just need to show that (

φ, ψ

) = (

, ψ

). This is true

since we can integrate by parts twice to obtain

(φ, ψ

) = −(φ

, ψ

) = (φ

, ψ).

For the potential term, we have

(φ, V (ˆx)ψ) = (φ, V (x)ψ) = (V (x)φ, ψ) = (V (ˆx)φ, ψ).

So H is Hermitian, as claimed.

Thus we know that hxi

, hˆpi

, hHi

are all real.

Furthermore, observe that

X = ˆx − α, P = ˆp − β

are (similarly) Hermitian for any real α, β. Hence

(ψ, X

ψ) = (ψ, X(Xψ)) = (Xψ, Xψ) = kXψk

≥ 0.

Similarly, we have

(ψ, P

ψ) = (ψ, P (P ψ)) = (P ψ, P ψ) = kP ψk

≥ 0.

If we choose

hˆxi

and

hˆpi

, the expressions above say that (∆

)

and

(∆p)

are indeed real and positive.

Cauchy-Schwarz inequality

We are going to end with a small section on a technical result that will come

handy later on.

Proposition

(Cauchy-Schwarz inequality)

and

are any normalizable

states, then

kψkkφk ≥ |(ψ, φ)|.

Proof. Consider

kψ + λφk

= (ψ + λφ, ψ + λφ)

= (ψ, ψ) + λ(ψ, φ) + λ

∗

(φ, ψ) + |λ|

(φ, φ) ≥ 0.

This is true for any complex λ. Set

λ = −

(φ, ψ)

kφk

which is always well-defined since

is normalizable, and then the above equation

becomes

kψk

−

|(ψ, φ)|

kφk

≥ 0.

So done.

3.2 Ehrenfest’s theorem

We will show that, in fact, quantum mechanics is like classical mechanics.

Again, consider a normalizable state Ψ(

x, t

) satisfying the time-dependent

Schr¨odinger equation, i.e.

Ψ = HΨ =



ˆp

+ V (ˆx)



Ψ.

Classically, we are used to

and

changing in time. However, here

ˆx

and

ˆp

are

fixed in time, while the states change with time. However, what does change

with time is the expectations. The expectation values

hˆxi

= (Ψ, ˆxΨ), hˆpi

= (Ψ, ˆpΨ)

depend on t through Ψ. Ehrenfest’s theorem states the following:

Theorem (Ehrenfest’s theorem).

hˆxi

hˆpi

= −hV

(ˆx)i

These are the quantum counterparts to the classical equations of motion.

Proof. We have

hˆxi

= (

Ψ, ˆxΨ) + (Ψ, ˆx

Ψ)



HΨ, ˆxΨ





Ψ, ˆx





Since H is Hermitian, we can move it around and get

= −

(Ψ, H(ˆxΨ)) +

(Ψ, ˆx(HΨ))

(Ψ, (ˆxH − H ˆx)Ψ).

But we know

(ˆxH − H ˆx)Ψ = −

(xΨ

− (xΨ)

) + (xV Ψ − V xΨ) = −

ˆpΨ.

So done.

The second part is similar. We have

hˆpi

= (

Ψ, ˆpΨ) + (Ψ, ˆp

Ψ)



HΨ, ˆpΨ





Ψ, ˆp





Since H is Hermitian, we can move it around and get

= −

(Ψ, H(ˆpΨ)) +

(Ψ, ˆp(HΨ))

(Ψ, (ˆpH − H ˆp)Ψ).

Again, we can compute

(ˆpH − H ˆp)Ψ = −i~



−~



((Ψ

)

− (Ψ

)

) − i~((V (x)Ψ)

− V (x)Ψ

)

= −i~V

(x)Ψ.

So done.

Note that in general, quantum mechanics can be portrayed in different

“pictures”. In this course, we will be using the Schr¨odinger picture all the time,

in which the operators are time-independent, and the states evolve in time. An

alternative picture is the Heisenberg picture, in which states are fixed in time,

and all the time dependence lie in the operators. When written in this way,

quantum mechanics is even more like classical mechanics. This will be explored

more in depth in IID Principles of Quantum Mechanics.

3.3 Heisenberg’s uncertainty principle

We will show that, in fact, quantum mechanics is not like classical mechanics.

Statement

The statement of the uncertainty principle (or relation) is

Theorem

(Heisenberg’s uncertainty principle)

is any normalized state (at

any fixed time), then

(∆x)

(∆p)

≥

Example. Consider the normalized Gaussian

ψ(x) =



απ



−x

/α

We find that

hˆxi

= hˆpi

= 0,

and also

(∆x)

, (∆p)

2α

So we get

(∆x)

(∆p)

We see that a small

corresponds to

sharply peaked around

= 0, i.e. it has

a rather definite position, but has a large uncertainty in momentum. On the

other hand, if

is large,

is spread out in position but has a small uncertainty

in momentum.

Recall that for

mω

, this is the lowest energy eigenstate for harmonic

oscillator with

H =

ˆp +

mω

ˆx

with eigenvalue

~ω

. We can use the uncertainty principle to understand why

we have a minimum energy of

~ω

instead of 0. If we had a really small energy,

then we would just sit at the bottom of the potential well and do nothing, with

both a small (and definite) momentum and position. Hence for uncertainty to

occur, we need a non-zero ground state energy.

To understand where the uncertainty principle come from, we have to under-

stand commutation relations.

Commutation relations

We are going to define a weird thing called the commutator. At first sight, this

seems like a weird definition, but this turns out to be a really important concept

in quantum mechanics.

Definition

(Commutator)

Let

and

be operators. Then the commutator

is denoted and defined by

[Q, S] = QS − SQ.

This is a measure of the lack of commutativity of the two operators.

In particular, the commutator of position and momentum is

[ˆx, ˆp] = ˆxˆp − ˆpˆx = i~.

This relation results from a simple application of the product rule:

(ˆxˆp − ˆpˆx)ψ = −i~ψ

− (−i~(xψ)

= i~ψ.

Note that if α and β are any real constants, then the operators

X = ˆx − α, P = ˆp − β

also obey

[X, P ] = i~.

This is something we will use when proving the uncertainty principle.

Recall that when we proved Ehrenfest’s theorem, the last step was to calculate

the commutator:

[ˆx, H] = ˆxH − H ˆx =

ˆp

[ˆp, H] = ˆpH − H ˆp = −i~V

(ˆx).

Since

is defined in terms of

ˆp

and

ˆx

, we can indeed find these relations just

using the commutator relations between

ˆx

and

ˆp

(plus a few more basic facts

about commutators).

Commutator relations are important in quantum mechanics. When we first

defined the momentum operator

ˆp

−i~

∂

∂x

, you might have wondered where

this weird definition came from. This definition naturally comes up if we require

that

ˆx

is “multiply by

” (so that the delta function

(

x − x

) is the eigenstate

with definition position

), and that

ˆx

and

ˆp

satisfies this commutator relation.

With this requirements, ˆp must be defined as that derivative.

Then one might ask, why would we want

ˆx

and

ˆp

to satisfy this commutator

relation? It turns out that in classical dynamics, there is something similar

known as the Poisson bracket

{·, ·}

, where we have

{x, p}

= 1. To get from

classical dynamics to quantum mechanics, we just have to promote our Poisson

brackets to commutator brackets, and multiply by i~.

Proof of uncertainty principle

Proof of uncertainty principle. Choose α = hˆxi

and β = hˆpi

, and define

X = ˆx − α, P = ˆp − β.

Then we have

(∆x)

= (ψ, X

ψ) = (Xψ, Xψ) = kXψk

(∆p)

= (ψ, P

ψ) = (P ψ, P ψ) = kP ψk

Then we have

(∆x)

(∆p)

= kXψkkPψk

≥ |(Xψ, P ψ)|

≥ |im(Xψ, P ψ)|

≥



(Xψ, P ψ) − (P ψ, Xψ)



(ψ, XP ψ) − (ψ, P Xψ)



(ψ, [X, P ]ψ)



(ψ, ψ)



So done.

4 More results in one dimensions

4.1 Gaussian wavepackets

When we solve Schr¨odinger’s equation, what we get is a “wave” that represents

the probability of finding our thing at each position. However, in real life,

we don’t think of particles as being randomly distributed over different places.

Instead, particles are localized to some small regions of space.

These would be represented by wavefunctions in which most of the distribution

is concentrated in some small region:

These are known as wavepackets.

Definition

(Wavepacket)

A wavefunction localised in space (about some point,

on some scale) is usually called a wavepacket.

This is a rather loose definition, and can refer to anything that is localized in

space. Often, we would like to consider a particular kind of wavepacket, known

as a Gaussian wavepacket.

Definition (Gaussian wavepacket). A Gaussian wavepacket is given by

(x, t) =





1/4

γ(t)

1/2

−x

/2γ(t)

for some γ(t).

These are particularly nice wavefunctions. For example, we can show that for

a Gaussian wavepacket, (∆

)

(∆

)

exactly, and uncertainty is minimized.

The Gaussian wavepacket is a solution of the time-dependent Schr¨odinger

equation (with V = 0) for

γ(t) = α +

Substituting this

(

) into our equation, we find that the probability density is

(x, t) = |Ψ

(x, t)|





1/2

|γ(t)|

−αx

/|γ(t)|

which is peaked at

= 0. This corresponds to a particle at rest at the origin,

spreading out with time.

A related solution to the time-dependent Schr¨odinger equation with

= 0 is

a moving particle:

(x, t) = Ψ

(x − ut) exp





exp



−i



The probability density resulting from this solution is

(x, t) = |Ψ

(x, t)|

= P

(x − ut, t).

So this corresponds to a particle moving with velocity u. Furthermore, we get

hˆpi

= mu.

This corresponds with the classical momentum, mass × velocity.

We see that wavepackets do indeed behave like particles, in the sense that we

can set them moving and the quantum momentum of these objects do indeed

behave like the classical momentum. In fact, we will soon attempt to send them

to a brick wall and see what happens.

In the limit

α → ∞

, our particle becomes more and more spread out in

space. The uncertainty in the position becomes larger and larger, while the

momentum becomes more and more definite. Then the wavefunction above

resembles something like

Ψ(x, t) = Ce

ikx

−iEt/~

which is a momentum eigenstate with

and energy

Note, however, that this is not normalizable.

4.2 Scattering

Consider the time-dependent Schr¨odinger equation with a potential barrier. We

would like to send a wavepacket towards the barrier and see what happens.

Classically, we would expect the particle to either pass through the barrier or get

reflected. However, in quantum mechanics, we would expect it to “partly” pass

through and “partly” get reflected. So the resultant wave is something like this:

AΨ

ref

BΨ

Here Ψ, Ψ

ref

and Ψ

are normalized wavefunctions, and

ref

= |A|

, P

= |B|

are the probabilities of reflection and transmission respectively.

This is generally hard to solve. Scattering problems are much simpler to

solve for momentum eigenstates of the form

ikx

. However, these are not

normalizable wavefunctions, and despite being mathematically convenient, we

are not allowed to use them directly, since they do not make sense physically.

These, in some sense, represent particles that are “infinitely spread out” and

can appear anywhere in the universe with equal probability, which doesn’t really

make sense.

There are two ways we can get around this problem. We know that we can

construct normalized momentum eigenstates for a single particle confined in a

box −

≤ x ≤

, namely

ψ(x) =

√

ikx

where the periodic boundary conditions require

(

) =

(

), i.e.

2πn

for some integer

. After calculations have been done, the box can be removed

by taking the limit ` → ∞.

Identical results are obtained more conveniently by allowing Ψ(

x, t

) to repre-

sent beams of infinitely many particles, with

Ψ(

x, t

)

being the density of the

number of particles (per unit length) at

x, t

. When we do this, instead of having

one particle and watching it evolve, we constantly send in particles so that the

system does not appear to change with time. This allows us to find steady

states. Mathematically, this corresponds to finding solutions to the Schr¨odinger

equation that do not change with time. To determine, say, the probability of

reflection, roughly speaking, we look at the proportion of particles moving left

compared to the proportion of particles moving right in this steady state.

In principle, this interpretation is obtained by considering a constant stream

of wavepackets and using some limiting/averaging procedure, but we usually

don’t care about these formalities.

For these particle beams, Ψ(

x, t

) is bounded, but no longer normalizable.

Recall that for a single particle, the probability current was defined as

j(x, t) = −

(Ψ

∗

− ΨΨ

0∗

If we have a particle beam instead of a particle, and Ψ is the particle density

instead of the probability distribution, j now represents the flux of particles at

x, t, i.e. the number of particles passing the point x in unit time.

Recall that a stationary state of energy

is of the form Ψ(

x, t

) =

(

)

iEt/~

We have

|Ψ(x, t)|

= |ψ(x)|

and

j(x, t) = −

(ψ

∗

− ψψ

0∗

Often, when solving a scattering problem, the solution will involve sums of

momentum eigenstates. So it helps to understand these better.

Our momentum eigenstates are

ψ(x) = Ce

ikx

which are solutions to the time-independent Schr¨odinger equation with

= 0

with E =

Applying the momentum operator, we find that

is the momentum of

each particle in the beam, and

|ψ

(

)

|C|

is the density of particles in the

beam. We can also evaluate the current to be

j =

|C|

This makes sense.

is the velocity of the particles, and

|C|

is how many

particles we have. So this still roughly corresponds to what we used to have

classically.

In scattering problems, we will seek the transmitted and reflected flux

ref

in terms of the incident flux

inc

, and the probabilities for transmission and

reflection are then given by

inc

, P

ref

inc

4.3 Potential step

Consider the time-independent Schr¨odinger equation for a step potential

V (x) =

(

0 x ≤ 0

U x > 0

where

U >

0 is a constant. The Schr¨odinger equation is, in case you’ve forgotten,

−~

+ V (x)ψ = Eψ.

V (x)

We require ψ and ψ

to be continuous at x = 0.

We can consider two different cases:

(i)

< E < U

: We apply the standard method, introducing constants

k, κ >

such that

E =

, U −E =

Then the Schr¨odinger equation becomes

(

+ k

ψ = 0 x < 0

− κ

ψ = 0 x > 0

The solutions are

ikx

−ikx

for

x <

0, and

−κx

for

x >

(since ψ has to be bounded).

Since ψ and ψ

are continuous at x = 0, we have the equations

(

I + R = C

ikI − ikR = −κC

So we have

R =

k − iκ

k + iκ

I, C =

k + iκ

x <

(

) is a superposition of beams (momentum eigenstates) with

|I|

particles per unit length in the incident part, and

|R|

particles per

unit length in the reflected part, with p = ±~k. The current is

j = j

inc

+ j

ref

= |I|

− |R|

The probability of reflection is

ref

inc

|R|

|I|

= 1,

which makes sense.

On the right hand side, we have

= 0. So

= 0. However,

|ψ

(

)

= 0

in this classically forbidden region.

(ii) E > U: This time, we set

E =

, E − U =

with k, κ > 0. Then the Schr¨odinger equation becomes

(

+ k

ψ = 0 x < 0

+ κ

ψ = 0 x > 0

Then we find

ikx

−ikx

x <

0, with

T e

ikx

x >

0. Note

that it is in principle possible to get an

−ikx

term on

x >

0, but this

would correspond to sending in a particle from the right. We, by choice,

assume there is no such term.

We now match ψ and ψ

at x = 0. Then we get the equations

(

I + R = T

ikI − ikR = ikT.

We can solve these to obtain

R =

k − κ

k + κ

I, T =

k + κ

Our flux on the left is now

j = j

inc

+ j

ref

= |I|

− |R|

while the flux on the right is

j = j

|T |

~κ

The probability of reflection is

ref

inc

|R|

|I|



k − κ

k + κ



while the probability of transmission is

inc

|T |

|I|

4kκ

(k + κ)

Note that P

ref

+ P

= 1.

Classically, we would expect all particles to be transmitted, since they all

have sufficient energy. However, quantum mechanically, there is still a

probability of being reflected.

4.4 Potential barrier

So far, the results were slightly weird. We just showed that however high energy

we have, there is still a non-zero probability of being reflected by the potential

step. However, things are really weird when we have a potential barrier.

Consider the following potential:

We can write this as

V (x) =











0 x ≤ 0

U 0 < x < a

0 x ≥ a

We consider a stationary state with energy

with 0

< E < U

. We set the

constants

E =

, U −E =

Then the Schr¨odinger equations become

+ k

ψ = 0 x < 0

− κ

ψ = 0 0 < x < a

+ k

ψ = 0 x > a

So we get

ψ = Ie

ikx

+ Re

−ikx

x < 0

ψ = Ae

κx

+ Be

−κx

0 < x < a

ψ = T e

ikx

x > a

Matching ψ and ψ

at x = 0 and a gives the equations

I + R = A + B

ik(I − R) = κ(A − B)

κa

+ Be

−κa

= T e

ika

κ(Ae

κa

− Be

−κa

) = ikTe

ika

We can solve these to obtain

I +

κ − ik

κ + ik

R = T e

ika

−κa

I +

κ + ik

κ − ik

R = T e

ika

κa

After lots of some algebra, we obtain

T = Ie

−ika



cosh κa − i

− κ

2kκ

sinh κa



−1

To interpret this, we use the currents

j = j

inc

+ j

ref

= (|I|

− |R|

)

for x < 0. On the other hand, we have

j = j

= |T |

for

x > a

. We can use these to find the transmission probability, and it turns

out to be

trj

inc

|T |

|I|



1 +

4E(U −E)

sinh

κa



−1

This demonstrates quantum tunneling. There is a non-zero probability that the

particles can pass through the potential barrier even though it classically does

not have enough energy. In particular, for

κa 

1, the probability of tunneling

decays as

−2κa

. This is important, since it allows certain reactions with high

potential barrier to occur in practice even if the reactants do not classically have

enough energy to overcome it.

4.5 General features of stationary states

We are going to end the chapter by looking at the difference between bound

states and scattering states in general.

Consider the time-independent Schr¨odinger equation for a particle of mass

Hψ = −

+ V (x)ψ = Eψ,

with the potential

(

)

→

0 as

x → ±∞

. This is a second order ordinary

differential equation for a complex function

, and hence there are two complex

constants in general solution. However, since this is a linear equation, this

implies that 1 complex constant corresponds to changing

ψ 7→ λψ

, which gives

no change in physical state. So we just have one constant to mess with.

As |x| → ∞, our equation simply becomes

−

= Eψ.

So we get

ψ ∼

(

ikx

+ Be

−ikx

E =

> 0

κx

+ Be

−κx

E = −

< 0.

So we get two kinds of stationary states depending on the sign of

. These

correspond to bound states and scattering states.

Bound state solutions, E < 0

If we want ψ to be normalizable, then there are 2 boundary conditions for ψ:

ψ ∼

(

κx

x → −∞

−κx

x → +∞

This is an overdetermined system, since we have too many boundary conditions.

Solutions exist only when we are lucky, and only for certain values of

. So

bound state energy levels are quantized. We may find several bound states or

none.

Definition

(Ground and excited states)

The lowest energy eigenstate is called

the ground state. Eigenstates with higher energies are called excited states.

Scattering state solutions, E > 0

Now

is not normalized but bounded. We can view this as particle beams, and

the boundary conditions determines the direction of the incoming beam. So we

have

ψ ∼

(

ikx

+ Re

−ikx

x → −∞

T e

ikx

x → +∞

This is no longer overdetermined since we have more free constants. The solution

for any E > 0 (imposing condition on one complex constant) gives

j ∼

(

inc

+ j

ref

(

|I|

− |R|

x → −∞

|T |

x → +∞

We also get the reflection and transmission probabilities

ref

= |A

ref

inc

= |A

inc

where

ref

(k) =

are the reflection and transmission amplitudes. In quantum mechanics, “ampli-

tude” general refers to things that give probabilities when squared.

5 Axioms for quantum mechanics

We are going to sum up everything we’ve had so far into a big picture. To do

so, we will state a few axioms of quantum mechanics and see how they relate to

what we have done so far.

Our axioms (or postulates) will be marked by bullet points in this chapter,

as we go through them one by one.

5.1 States and observables

–

States of a quantum system correspond to non-zero elements of a complex

vector space

(which has nothing to do with the potential), with

and

αψ physically equivalent for all α ∈ C \ {0}.

Furthermore, there is a complex inner product (

φ, ψ

) defined on

satisfying

(φ, α

+ α

) = α

(φ, ψ

) + α

(φ, ψ

)

(β

+ β

, ψ) = β

∗

(φ

, ψ) + β

∗

(φ

, ψ)

(φ, ψ) = (ψ, φ)

∗

kψk

= (ψ, ψ) ≥ 0

kψk = 0 if and only if ψ = 0.

Note that this is just the linear algebra definition of what a complex inner

product should satisfy.

An operator A is a linear map V → V satisfying

A(αψ + βφ) = αAψ + βAφ.

For any operator

, the Hermitian conjugate or adjoint, denoted

†

defined to be the unique operator satisfying

(φ, A

†

ψ) = (Aφ, ψ).

An operator Q is called Hermitian or self-adjoint if

†

= Q.

A state χ 6= 0 is an eigenstate of Q with eigenvalue λ if

Qχ = λχ.

The set of all eigenvalues of Q is called the spectrum of Q.

A measurable quantity, or observable, in a quantum system corresponds to

a Hermitian operator.

So far, we have worked with the vector space of functions (that are sufficiently

smooth), and used the integral as the inner product. However, in general, we can

work with arbitrary complex vector spaces and arbitrary inner products. This

will become necessary when we, say, study the spin of electrons in IID Principles

of Quantum Mechanics.

Even in the case of the vector space of (sufficiently smooth) functions, we will

only work with this formalism informally. When we try to make things precise,

we will encounter a lot of subtleties such as operators not being well-defined

everywhere.

5.2 Measurements

Key results for Hermitian operators

Our next set of axioms relate Hermitian operators with physical measurements.

Before we state them, we first look at some purely mathematical properties of

Hermitian operators. We first prove some purely mathematical results about

Hermitian operators.

Proposition. Let Q be Hermitian (an observable), i.e. Q

†

= Q. Then

(i) Eigenvalues of Q are real.

(ii)

Eigenstates of

with different eigenvalues are orthogonal (with respect to

the complex inner product).

(iii)

Any state can be written as a (possibly infinite) linear combination of

eigenstates of

, i.e. eigenstates of

provide a basis for

. Alternatively,

the set of eigenstates is complete.

Note that the last property is not actually necessarily true. For example,

the position and momentum operators do not have eigenstates at all, since the

“eigenstates” are not normalizable. However, for many of the operators we care

about, this is indeed true, and everything we say in this section only applies to

operators for which this holds.

Proof.

(i) Since Q is Hermitian, we have, by definition,

(χ, Qχ) = (Qχ, χ).

Let χ be an eigenvector with eigenvalue λ, i.e. Qχ = λχ. Then we have

(χ, λχ) = (λχ, χ).

So we get

λ(χ, χ) = λ

∗

(χ, χ).

Since (χ, χ) 6= 0, we must have λ = λ

∗

. So λ is real.

(ii) Let Qχ = λχ and Qφ = µφ. Then we have

(φ, Qχ) = (Qφ, χ).

So we have

(φ, λχ) = (µφ, χ).

In other words,

λ(φ, χ) = µ

∗

(φ, χ) = µ(φ, χ).

Since λ 6= µ by assumption, we know that (φ, χ) = 0.

(iii) We will not attempt to justify this, or discuss issues of convergence.

Measurement axioms

Consider an observable

with discrete spectrum (i.e. eigenvalues) and normalized

eigenstates. From the results we just had, any given state can be written

ψ =

with Qχ

= λ

, and (χ

, χ

) = δ

. This means we can calculate α

= (χ

, ψ).

We can assume also that

= 0 since we can just pretend the terms with

coefficient 0 do not exist, and that the corresponding

are distinct by choosing

an appropriate basis. Then

– The outcome of a measurement is some eigenvalue of Q.

– The probability of obtaining λ

= |α

where α

= (χ

, ψ) is the amplitude.

–

The measurement is instantaneous and forces the system into the state

This is the new state immediately after the measurement is made.

The last statement is rather weird. However, if you think hard, this must be

the case. If we measure the state of the system, and get, say, 3, then if we

measure it immediately afterwards, we would expect to get the result 3 again with

certainty, instead of being randomly distributed like the original state. So after

a measurement, the system must be forced into the corresponding eigenstate.

Note that these axioms are consistent in the following sense. If

is normalized,

then

(ψ, ψ) =





m,n

∗

(χ

, χ

)

m,n

∗

|α

= 1

So if the state is normalized, then the sum of probabilities is indeed 1.

Example.

Consider the harmonic oscillator, and consider the operator

with eigenfunctions χ

= ψ

and eigenvalues

= E

= ~ω



n +



Suppose we have prepared our system in the state

ψ =

√

(ψ

+ 2ψ

− iψ

Then the coefficients are

√

, α

√

, α

−i

√

This is normalized since

kψk

= |α

+ |α

= 1.

Measuring the energy gives the following probabilities:

Energy Probability

~ω P

If a measurement gives

, then

is the new state immediately after measure-

ment.

Expressions for expectation and uncertainty

We can use these probability amplitudes to express the expectation and uncer-

tainty of a state in a more familiar way.

Proposition.

(i) The expectation value of Q in state ψ is

hQi

= (ψ, Qψ) =

with notation as in the previous part.

(ii) The uncertainty (δQ)

is given by

(∆Q)

= h(Q − hQi

)

= hQ

− hQi

(λ

− hQi

)

From this, we see that

is an eigenstate of

with eigenvalue

if and only

if hQi

= λ and (∆Q)

= 0.

Proof.

(i) Let ψ =

. Then

Qψ =

So we have

(ψ, Qψ) =

n,m

(α

, χ

, α

) =

∗

(ii) This is a direct check using the first expression.

Example.

Consider the harmonic oscillator as we had in the previous example.

Then the expectation value is

hHi





~ω =

~ω.

Note that the measurement axiom tells us that after measurement, the

system is then forced into the eigenstate. So when we said that we interpret the

expectation value as the “average result for many measurements”, we do not

mean measuring a single system many many times. Instead, we prepare a lot of

copies of the system in state ψ, and measure each of them once.

5.3 Evolution in time

– The state of a quantum system Ψ(t) obeys the Schr¨odinger equation

Ψ = HΨ,

where

is a Hermitian operator, the Hamiltonian; this holds at all times

except at the instant a measurement is made.

In principle, this is all there is in quantum mechanics. However, for practical

purposes, it is helpful to note the following:

Stationary states

Consider the energy eigenstates with

Hψ

= E

, (ψ

, ψ

) = δ

Then we have certain simple solutions of the Schr¨odinger equation of the form

= ψ

−iE

t/~

In general, given an initial state

Ψ(0) =

since the Schr¨odinger equation is linear, we can get the following solution for all

time:

Ψ(t) =

−iE

t/~

Example. Consider again the harmonic oscillator with initial state

Ψ(0) =

√

(ψ

+ 2ψ

− iψ

Then Ψ(t) is given by

Ψ(t) =

√

(ψ

−iωt/2

+ 2ψ

−3iωt/2

− iψ

−9iωt/2

Ehrenfest’s theorem (general form)

Theorem

(Ehrenfest’s theorem)

is any operator with no explicit time

dependence, then

hQi

= h[Q, H]i

where

[Q, H] = QH − HQ

is the commutator.

Proof. If Q does not have time dependence, then

(Ψ, QΨ) = (−i~

Ψ, QΨ) + (Ψ, Qi~

Ψ)

= (−HΨ, QΨ) + (Ψ, QHΨ)

= (Ψ, (QH − HQ)Ψ)

= (Ψ, [Q, H]Ψ).

has explicit time dependence, then we have an extra term on the right,

and have

hQi

= h[Q, H]i

+ i~h

These general versions correspond to classical equations of motion in Hamiltonian

formalism you will (hopefully) study in IIC Classical Dynamics.

5.4 Discrete and continuous spectra

In stating the measurement axioms, we have assumed that our spectrum of

eigenvalues of

was discrete, and we got nice results about measurements.

However, for certain systems, the spectra of

ˆp

and

may be continuous. This

is the same problem when we are faced with non-normalizable momentum

wavefunctions.

To solve this problem, we can make the spectra discrete by a technical device:

we put the system in a “box” of length

with suitable boundary conditions of

(

). We can then take

` → ∞

at the end of the calculation. We’ve discussed

this in some extent for momentum eigenstates. We shall revisit that scenario in

the more formal framework we’ve developed.

Example.

Consider

(

) with periodic boundary conditions

(

) =

(

So we can restrict to

−

≤ x ≤

We compare with the general axioms, where

Q = ˆp = −i~

The eigenstates are

(x) =

√

, k

2πn

Now we have discrete eigenvalues as before given by

= ~k

We know that the states are orthonormal on −

≤ x ≤

, i.e.

(χ

, χ

) =

−

(x)

∗

(x) dx = δ

We can thus expand our solution in terms of the eigenstates to get a complex

Fourier series

ψ(x) =

(x),

where the amplitudes are given by

= (χ

, ψ).

When we take the limit as

n → ∞

, the Fourier series becomes a Fourier integral.

There is another approach to this problem (which is non-examinable). We

can also extend from discrete to continuous spectra as follows. We replace the

discrete label n with some continuous label ξ. Then we have the equation

Qχ

= λ

These are eigenstates with an orthonormality conditions

(χ

, χ

) = δ(ξ − η),

where we replaced our old

with

(

ξ − η

), the Dirac delta function. To

perform the expansion in eigenstates, the discrete sum becomes an integral. We

have

ψ =

dξ,

where the coefficients are

= (χ

, ψ).

In the discrete case,

|α

is the probability mass function. The obvious gener-

alization here would be to let

|α

be our probability density function. More

precisely,

|α

dξ = probability that the result corresponds to a ≤ ξ ≤ b.

Example.

Consider the particle in one dimension with position as our operator.

We will see that this is just our previous interpretation of the wavefunction.

Let our operator be Q = ˆx. Then our eigenstates are

(x) = δ(x −ξ).

The corresponding eigenvalue is

= ξ.

This is true since

ˆxχ

(x) = xδ(x −ξ) = ξδ(x − ξ) = ξχ

(x),

since δ(x − ξ) is non-zero only when x = ξ.

With this eigenstates, we can expand this is

ψ(x) =

(x) dξ =

δ(x −ξ) dξ = α

So our coefficients are given by α

= ψ(ξ). So

|ψ(ξ)|

dξ

is indeed the probability of measuring the particle to be in

a ≤ ξ ≤ b

. So we

recover our original interpretation of the wavefunction.

We see that as long as we are happy to work with generalized functions like

delta functions, things become much nicer and clearer. Of course, we have to be

more careful and study distributions properly if we want to do this.

5.5 Degeneracy and simultaneous measurements

Degeneracy

Definition

(Degeneracy)

For any observable

, the number of linearly inde-

pendent eigenstates with eigenvalue

is the degeneracy of the eigenvalue. In

other words, the degeneracy is the dimension of the eigenspace

= {ψ : Qψ = λψ}.

An eigenvalue is non-degenerate if the degeneracy is exactly 1, and is degenerate

if the degeneracy is more than 1.

We say two states are degenerate if they have the same eigenvalue.

In one dimension, the energy bound states are always non-degenerate, as we

have seen in example sheet 2. However, in three dimensions, energy bound states

may be degenerate. If

is degenerate, then there is a large freedom in choosing

an orthonormal basis for

. Physically, we cannot distinguish degenerate states

by measuring

alone. So we would like to measure something else as well to

distinguish these eigenstates. When can we do this? We have previously seen

that we cannot simultaneously measure position and momentum. It turns out

the criterion for whether we can do this is simple.

Commuting observables

Recall that after performing a measurement, the state is forced into the corre-

sponding eigenstate. Hence, to simultaneously measure two observables

and

, we need the state to be a simultaneously an eigenstate of

and

. In other

words, simultaneous measurement is possible if and only if there is a basis for

consisting of simultaneous or joint eigenstates χ

with

Aχ

= λ

, Bχ

= µ

Our measurement axioms imply that if the state is in

, then measuring

in rapid succession, in any order, will give definite results

and

respectively

(assuming the time interval between each pair of measurements is short enough

that we can neglect the evolution of state in time).

From IB Linear Algebra, a necessary and sufficient condition for

and

be simultaneously measurable (i.e. simultaneously diagonalizable) is for

and

B to commute, i.e.

[A, B] = AB − BA = 0.

This is consistent with a “generalized uncertainty relation”

(∆A)

(∆B)

≥

|h[A, B]i

since if we if have a state that is simultaneously an eigenstate for

and

, then

the uncertainties on the left would vanish. So

[

A, B

]

= 0. The proof of this

relation is on example sheet 3.

This will be a technique we will use to tackle degeneracy in general. If we have

a system where, say, the energy is degenerate, we will attempt to find another

operator that commutes with

, and try to further classify the underlying states.

When dealing with the hydrogen atom later, we will use the angular momentum

to separate the degenerate energy eigenstates.

6 Quantum mechanics in three dimensions

Finally, we will move on and discuss quantum systems in three dimensions, since

we (probably) live in a three-dimensional world. These problems are usually

substantially more difficult, since we have to deal with partial derivatives. Also,

unlike the case of one dimension, we will run into degenerate states, where

multiple eigenstates have the same energy. We thus have to find some other

observables (that commute with the Hamiltonian) in order to fully classify the

eigenstates.

6.1 Introduction

To begin with, we translate everything we’ve had for the one-dimensional world

into the three-dimensional setting.

A quantum state of a particle in three dimensions is given by a wavefunction

(

) at fixed time, or Ψ(

x, t

) for a state evolving in time. The inner product is

defined as

(ϕ, ψ) =

ϕ(x)

∗

ψ(x) d

We adopt the convention that no limits on the integral means integrating over

all space. If ψ is normalized, i.e.

kψk

= (ψ, ψ) =

|ψ(x)|

x = 1,

then the probability of measuring the particle to be inside a small volume

δV

(containing x) is

|ψ(x)|

δV.

The position and momentum are Hermitian operators

x = (ˆx

, ˆx

), ˆx

ψ = x

and

p = (ˆp

, ˆp

) = −i~∇ = −i~



∂

∂x

∂

∂x

∂

∂x



We have the canonical commutation relations

[ˆx

, ˆp

] = −i~δ

, [ˆx

, ˆx

] = [ˆp

, ˆp

] = 0.

We see that position and momentum in different directions don’t come into

conflict. We can have a definite position in one direction and a definite momentum

in another direction. The uncertainty principle only kicks in when we are in the

same direction. In particular, we have

(∆x

)(∆p

) ≥

Similar to what we did in classical mechanics, we assume our particles are

structureless.

Definition

(Structureless particle)

A structureless particle is one for which all

observables can be written in terms of position and momentum.

In reality, many particles are not structureless, and (at least) possess an

additional quantity known as “spin”. We will not study these in this course, but

only mention it briefly near the end. Instead, we will just pretend all particles

are structureless for simplicity.

The Hamiltonian for a structureless particle in a potential V is

H =

+ V (

x) = −

∇

+ V (x).

The time-dependent Schr¨odinger equation is

∂Ψ

∂t

= HΨ = −

∇

Ψ + V (x)Ψ.

The probability current is defined as

j = −

(Ψ

∗

∇Ψ − Ψ∇Ψ

∗

This probability current obeys the conservation equation

∂

∂t

|Ψ(x, t)|

= −∇ · j.

This implies that for any fixed volume V ,

|Ψ(x, t)|

x = −

∇ · j d

x = −

∂V

j · dS,

and this is true for any fixed volume V with boundary ∂V . So if

|Ψ(x, t)| → 0

sufficiently rapidly as |x| → ∞, then the boundary term disappears and

|Ψ(x, t)|

x = 0.

This is the conservation of probability (or normalization).

6.2 Separable eigenstate solutions

How do we solve the Schr¨odinger equation in three dimensions? Things are

more complicated, since we have partial derivatives. Often, those we can solve

are those we can reduce to one-dimensional problems, using the symmetry of

the system. For example, we will solve the hydrogen atom exploiting the fact

that the potential is spherically symmetric. To do so, we will use the method of

separation of variables, as we’ve seen in the IB Methods course.

Consider the simpler case where we only have two dimensions. The time-

independent Schr¨odinger equation then gives

Hψ = −



∂

∂x

∂

∂x



ψ + V (x

, x

)ψ = Eψ.

We are going to consider potentials of the form

V (x

, x

) = U

) + U

The Hamiltonian then splits into

H = H

+ H

where

= −

∂

∂x

+ U

We look for separable solutions of the form

ψ = χ

)χ

The Schr¨odinger equation then gives (upon division by ψ = χ

) gives



−

+ U





−

+ U



= E.

Since each term is independent of x

and x

respectively, we have

= E

, H

= E

with

+ E

= E.

This is the usual separation of variables, but here we can interpret this physically

— in this scenario, the two dimensions are de-coupled, and we can treat them

separately. The individual

and

are just the contributions from each

component to the energy. Thus the process of separation variables can be seen

as looking for joint or simultaneous eigenstates for

and

, noting the fact

that [H

, H

] = 0.

Example. Consider the harmonic oscillator with

) =

This corresponds to saying

V =

mω

kxk

mω

+ x

Now we have

= H

(ˆx

, ˆp

with H

the usual harmonic oscillator Hamiltonian.

Using the previous results for the one-dimensional harmonic oscillator, we

have

= ψ

), χ

= ψ

where

is the

th eigenstate of the one-dimensional harmonic oscillator, and

, n

= 0, 1, 2, ···.

The corresponding energies are

= ~ω





The energy eigenvalues of the two-dimensional oscillator are thus

E = ~ω (n

+ n

+ 1)

for

ψ(x

, x

) = ψ

)ψ

We have the following energies and states:

State Energy Possible states

Ground state E = ~ω ψ = ψ

)ψ

)

1st excited state E = 2~ω

ψ = ψ

)ψ

)

ψ = ψ

)ψ

)

We see there is a degeneracy of 2 for the first excited state.

Separable solutions also arise naturally when

(i.e. the potential

) has

some symmetry. For example, if the potential is spherically symmetric, we can

find solutions

ψ(x) = R(r)Y (θ, φ),

where r, θ, φ are the spherical polars.

6.3 Angular momentum

Recall that in IA Dynamics and Relativity, we also solved a spherically symmetric

system in classical dynamics. We obtained beautiful, exact solutions for the

orbits of particle in a gravitational potential in terms of conic sections.

To do so, we came up with a conserved quantity known as the angular

momentum. Using the fact that this is conserved, we were able to make a lot of

calculations and derive many results. We can also use this conserved quantity to

understand, say, why the Earth doesn’t just fall into the sun directly.

To understand spherically symmetric potentials in quantum mechanics, it is

also helpful to understand the angular momentum.

Definitions

Definition

(Angular momentum)

The angular momentum is a vector of oper-

ators

L =

x ∧

p = −i~x ∧ ∇.

In components, this is given by

= ε

ijk

ˆx

ˆp

= −i~ε

ijk

∂

∂x

For example, we have

= ˆx

ˆp

− ˆx

ˆp

= −i~



∂

∂x

− x

∂

∂x



Note that this is just the same definition as in classical dynamics, but with

everything promoted to operators.

These operators are Hermitian, i.e.

†

= L,

since

ˆx

and

ˆp

are themselves Hermitian, and noting the fact that

ˆx

and

ˆp

commute whenever i 6= j.

Each component of

is the angular momentum in one direction. We can

also consider the length of L, which is the total angular momentum.

Definition

(Total angular momentum)

The total angular momentum operator

= L

+ L

Again, this is Hermitian, and hence an observable.

Commutation relations

We have the following important commutation relations for angular momentum:

, L

] = i~ε

ijk

. (i)

For example,

, L

] = i~L

Recall that in classical dynamics, an important result is that the angular mo-

mentum is conserved in all directions. However, we know we can’t do this in

quantum mechanics, since the angular momentum operators do not commute,

and we cannot measure all of them. This is why we have this

. It captures the

total angular momentum, and commutes with the angular momentum operators:

, L

] = 0 (ii)

for all i.

Finally, we also have the following commutation relations.

, ˆx

] = i~ε

ijk

ˆx

, [L

, ˆp

] = i~ε

ijk

ˆp

. (iii)

These are rather important results.

and

are observables, but (i) implies we

cannot simultaneously measure, say

and

. The best that can be done is to

measure

and

, say. This is possible by (ii). Finally, (iii) allows computation

of the commutator of L

with any function of ˆx

and ˆp

First, we want to prove these commutation relations.

Proposition.

(i) [L

, L

] = i~ε

ijk

(ii) [L

, L

] = 0.

(iii) [L

, ˆx

] = i~ε

ijk

ˆx

and [L

, ˆp

] = i~ε

ijk

ˆp

Proof.

(i)

We first do a direct approach, and look at specific indices instead of general

i and j. We have

= (−i~)



∂

∂x

− x

∂

∂x



∂

∂x

− x

∂

∂x



= −~



∂

∂x

∂

∂x

− x

∂

∂x

− x

∂

∂x

+ x

∂

∂x



Now note that

∂

∂x

∂

∂x

= x

∂

∂x

+ x

∂

∂x

Similarly, we can compute L

and find

, L

] = −~



∂

∂x

− x

∂

∂x



= i~L

where all the double derivatives cancel. So done.

Alternatively, we can do this in an abstract way. We have

= ε

iar

ˆx

ˆp

jbs

ˆx

ˆp

= ε

iar

jbs

(ˆx

ˆp

ˆx

ˆp

)

= ε

iar

jbs

(ˆx

ˆp

− [ˆp

, ˆx

])ˆp

)

= ε

iar

jbs

(ˆx

ˆx

ˆp

− i~δ

ˆx

ˆp

)

Similarly, we have

= ε

iar

jbs

(ˆx

ˆx

ˆp

− i~δ

ˆx

ˆp

)

Then the commutator is

− L

= −i~ε

iar

jbs

(δ

ˆx

ˆp

− δ

ˆx

ˆp

)

= −i~(ε

iab

jbs

ˆx

ˆp

− ε

iar

jba

ˆx

ˆp

)

= −i~((δ

− δ

)ˆx

ˆp

− (δ

− δ

)ˆx

ˆp

)

= i~(ˆx

ˆp

− ˆx

ˆp

)

= i~ε

ijk

So done.

(ii) This follows from (i) using the Leibnitz property:

[A, BC] = [A, B]C + B[A, C].

This property can be proved by directly expanding both sides, and the

proof is uninteresting.

Using this, we get

, L

] = [L

, L

]

= [L

, L

+ L

, L

]

= i~ε

ijk

+ L

)

= 0

where we get 0 since we are contracting the antisymmetric tensor

ijk

with

the symmetric tensor L

+ L

(iii)

We will use the Leibnitz property again, but this time we have it the other

way round:

[AB, C] = [A, C]B + A[B, C].

This follows immediately from the previous version since [

A, B

] =

−

[

B, A

Then we can compute

, ˆx

] = ε

iab

[ˆx

ˆp

, ˆx

]

= ε

iab

([ˆx

, ˆx

]ˆp

+ ˆx

[ˆp

, ˆx

])

= ε

iab

ˆx

(−i~δ

)

= i~ε

ija

ˆx

as claimed.

We also have

, ˆp

] = ε

iab

[ˆx

ˆp

, ˆp

]

= ε

iab

([ˆx

, ˆp

]ˆp

+ ˆx

[ˆp

, ˆp

])

= ε

iab

(i~δ

ˆp

)

= i~ε

ijb

ˆp

Spherical polars and spherical harmonics

Recall that angular momentum is something coming from rotation. So let’s

work with something with spherical symmetry. We will exploit the symmetry

and work with spherical polar coordinates. To do so, we first need to convert

everything we had so far from Cartesian coordinates to polar coordinates.

We define our spherical polar coordinates (r, θ, ϕ) by the usual relations

= r sin θ cos ϕ

= r sin θ sin ϕ

= r cos θ.

If we express our angular momentum operators as differential operators, then we

can write them entirely in terms of

r, θ

and

using the chain rule. The formula

for

will be rather simple, since

is our axis of rotation. However, those for

and

will be much more complicated. Instead of writing them out directly,

we instead write down the formula for

± iL

. A routine application of

the chain rule gives

= −i~

∂

∂ϕ

= L

± iL

= ±~e

±iϕ



∂

∂θ

± i cot θ

∂

∂ϕ



= −~



sin θ

∂

∂θ

sin θ

∂

∂θ

sin

∂

∂ϕ



Note these operators involve only

and

. Furthermore, the expression for

is something we have all seen before — we have

∇

∂

∂r

r −

Since we know

, L

] = 0,

there are simultaneous eigenfunctions of these operators, which we shall call

(

θ, ϕ

), with

= 0

, ···

and

= 0

, ±

, ··· , ±`

. These have eigenval-

ues ~m for L

and ~

`(` + 1) for L

In general, we have

= const e

imϕ

(cos θ),

where

is the associated Legendre function. For the simplest case

= 0, we

have

= const P

(cos θ),

where P

is the Legendre polynomial.

Note that these details are not important. The important thing to take away

is that there are solutions

(

θ, ϕ

), which we can happily use and be confident

that someone out there understands these well.

6.4 Joint eigenstates for a spherically symmetric potential

Unfortunately, it is completely standard to use

to indicate the eigenvalue of

, as we did above. Hence, here we shall consider a particle of mass

in a

potential V (r) with spherical symmetry. The Hamiltonian is

H =

2µ

ˆp

+ V = −

2µ

∇

+ V (r).

We have seen that we can write H as

H = −

2µ

∂

∂r

r +

2µ

+ V (r).

The first thing we want to check is that

, H] = [L

, H] = 0.

This implies we can use the eigenvalues of

and

to label our solutions

to the equation.

We check this using Cartesian coordinates. The kinetic term is

] = [L

, ˆp

ˆp

]

= [L

, ˆp

]ˆp

+ ˆp

, ˆp

]

= i~ε

ijk

(ˆp

ˆp

+ ˆp

ˆp

)

= 0

since we are contracting an antisymmetric tensor with a symmetric term. We

can also compute the commutator with the potential term

, V (r)] = −i~ε

ijk



∂V

∂x



= −i~ε

ijk

(r)

= 0,

using the fact that

∂r

∂x

Now that

and

are a commuting set of observables, we have the joint

eigenstates

ψ(x) = R(r)Y

(θ, ϕ),

and we have

= ~

`(` + 1)Y

` = 0, 1, 2, ···

= ~mY

m = 0, ±1, ±2, ··· , ±`.

The numbers

and

are usually known as the angular momentum quantum

numbers. Note that

= 0 is the special case where we have a spherically

symmetric solution.

Finally, we solve the Schr¨odinger equation

Hψ = Eψ

to obtain

−

2µ

(rR) +

2µr

`(` + 1)R + V R = ER.

This is now an ordinary differential equation in

. We can interpret the terms

as the radial kinetic energy, angular kinetic energy, the potential energy and

the energy eigenvalue respectively. Note that similar to what we did in classical

dynamics, under a spherically symmetric potential, we can replace the angular

part of the motion with an “effective potential”

2µr

`(` + 1).

We often call

(

) the radial part of the wavefunction, defined on

r ≥

Often, it is convenient to work with

(

) =

(

), which is sometimes called

the radial wavefunction. Multiplying the original Schr¨odinger equation by

, we

obtain

−

2µ

`(` + 1)

2µr

χ + V χ = Eχ.

This is known as the radial Schr¨odinger equation.

This has to obey some boundary conditions. Since we want

to be finite as

r →

0, we must have

= 0 at

= 0. Moreover, the normalization condition is

now

1 =

|ψ(x)|

x =

|R(r)|

(θ, ϕ)|

sin θ dθ dϕ.

Hence, ψ is normalizable if and only if

∞

|R(r)|

dr < ∞.

Alternatively, this requires

∞

|χ(r)|

dr < ∞.

Example

(Three-dimensional well)

We now plot our potential as a function of

−U

This is described by

V (r) =

(

0 r ≥ a

−U r < a

where U > 0 is a constant.

We now look for bound state solutions to the Schr¨odinger equation with

−U < E < 0, with total angular momentum quantum number `.

For r < a, our radial wavefunction χ obeys

−

`(` + 1)

χ + k

χ = 0,

where k is a new constant obeying

U + E =

2µ

For r ≥ a, we have

−

`(` + 1)

χ − κ

χ = 0,

with κ obeying

E = −

2µ

We can solve in each region and match

χ, χ

, with boundary condition

(0) = 0. Note that given this boundary condition, solving this is equivalent to

solving it for the whole R but requiring the solution to be odd.

Solving this for general

is slightly complicated. So we shall look at some

particular examples.

For

= 0, we have no angular term, and we have done this before. The

general solution is

χ(r) =

(

A sin kr r < a

−κr

r > a

Matching the values at x = a determines the values of k, κ and hence E.

For ` = 1, it turns out the solution is just

χ(r) =

(



cos kr −

sin kr



r < a



1 +



−κr

r > a

After matching, the solution is

ψ(r) = R(r)Y

(θ, ϕ) =

χ(r)

(θ, ϕ),

where m can take values m = 0, ±1.

Solution for general

involves spherical Bessel functions, and are studied

more in-depth in the IID Applications of Quantum Mechanics course.

7 The hydrogen atom

Recall that at the beginning of the course, we said one of the main motivations

for quantum mechanics was to study how atoms worked and where the spectrum

of the hydrogen atom came from. This is just an application of what we have

done for general spherically symmetric potentials, but this is so important that

we give it a separate chapter.

7.1 Introduction

Consider an electron moving in a Coulomb potential

V (r) = −

4πε

This potential is due to a proton stationary at

= 0. We follow results from

the last section of the last chapter, and set the mass

, the electron mass.

The joint energy eigenstates of H, L

and L

are of the form

φ(x) = R(r)Y

(θ, ϕ)

for ` = 0, 1, ··· and m = 0, ±1, ··· , ±m.

The radial part of the Schr¨odinger equation can be written

−

`(` + 1)

R +

2λ

R = κ

R, (∗)

with

λ =

4πε

, E =

−~

Note that here we work directly with

instead of

, as this turns out to be

easier later on.

The goal of this chapter is to understand all the (normalizable) solutions to

this equation (∗).

As in the case of the harmonic oscillator, the trick to solve this is see what

happens for large

, and “guess” a common factor of the solutions. In the case of

the harmonic oscillator, we guessed the solution should have

−y

as a factor.

Here, for large r, we get

∼ κ

This implies R ∼ e

−κr

for large R.

For small

, we by assumption know that

is finite, while

and

could

potentially go crazy. So we multiply by

and discard the

and

terms to

get

+ 2rR

− `(` + 1)R ∼ 0.

This gives the solution R ∼ r

We’ll be bold and try a solution of the form

R(r) = Cr

−κr

When we substitute this in, we will get three kinds of terms. The

−κr

terms

match, and so do the terms of the form r

`−2

−κr

. Finally, we see the r

`−1

−κe

terms match if and only if

2(`r

`−1

)(−κe

−κr

) + 2(r

`−1

)(−κe

−κr

) + 2λr

`−1

−κe

= 0.

When we simplify this mess, we see this holds if and only if

(` + 1)κ = λ.

Hence, for any integer

+ 1 = 1

, ···

, there are bound states with

energies

= −



4πε



These are the energy levels of the Bohr model, derived within the framework of

the Bohr model. However, there is a slight difference. In our model, the total

angular momentum eigenvalue is

`(` + 1) = ~

n(n − 1),

which is not what the Bohr model predicted.

Nevertheless, this is not the full solution. For each energy level, this only

gives one possible angular momentum, but we know that there can be many

possible angular momentums for each energy level. So there is more work to be

done.

7.2 General solution

We guessed our solution

−κr

above by looking at the asymptotic behaviour

at large and small

. We then managed to show that this is one solution of the

hydrogen atom. We are greedy, and we want all solutions.

Similar to what we did for the harmonic oscillator, we guess that our general

solution is of the form

R(r) = e

−κr

f(r).

Putting it in, we obtain

−

`(` + 1)

f = 2



κf

+ (κ − λ)



We immediately see an advantage of this substitution — now each side of the

equality is equidimensional, and equidimensionality makes our life much easier

when seeking series solution. This equation is regular singular at

= 0, and

hence we guess a solution of the form

f(r) =

∞

p=0

p+σ

, a

6= 0.

Then substitution gives

p≥0

((p + σ)(p + σ − 1) − `(` + 1))a

p+σ−2

p≥0

2(κ(p + σ + 1) − λ)a

p+σ−1

The lowest term gives us the indicial equation

σ(σ + 1) − `(` + 1) = (σ − `)(σ + ` + 1) = 0.

So either

−

(

+ 1). We discard the

−

(

+ 1) solution since this

would make f and hence R singular at r = 0. So we have σ = `.

Given this, the coefficients are then determined by

2(κ(p + `) − λ)

p(p + 2` + 1)

p−1

, p ≥ 1.

Similar to the harmonic oscillator, we now observe that, unless the series termi-

nates, we have

p−1

∼

2κ

p → ∞

, which matches the behaviour of

2κr

(for some

). So

(

) is

normalizable only if the series terminates. Hence the possible values of λ are

κn = λ

for some

n ≥ `

+ 1. So the resulting energy levels are exactly those we found

before:

= −



4πε



for n = 1, 2, 3, ···. This n is called the principle quantum number.

For any given n, the possible angular momentum quantum numbers are

` = 0, 1, 2, 3, ··· , n − 1

m = 0, ±1, ±2, ··· , ±`.

The simultaneous eigenstates are then

n`m

(x) = R

(r)Y

(θ, ϕ),

with

(r) = r

(r)e

−λr/n

where g

(r) are (proportional to) the associated Laguerre polynomials.

In general, the “shape” of probability distribution for any electron state

depends on

and

θ, ϕ

mostly through

. For

= 0, we have a spherically

symmetric solutions

n00

(x) = g

(r)e

−λr/n

This is very different from the Bohr model, which says the energy levels depend

only on the angular momentum and nothing else. Here we can have many

different angular momentums for each energy level, and can even have no angular

momentum at all.

The degeneracy of each energy level E

n−1

`=0

m=−`

1 =

n−1

`=0

(2` + 1) = n

If you study quantum mechanics more in-depth, you will find that the degeneracy

of energy eigenstates reflects the symmetries in the Coulomb potential. Moreover,

the fact that we have

degenerate states implies that there is a hidden symmetry,

in addition to the obvious

(3) rotational symmetry, since just

(3) itself

should give rise to much fewer degenerate states.

So. We have solved the hydrogen atom.

7.3 Comments

Is this it? Not really. When we solved the hydrogen atom, we made a lot of

simplifying assumptions. It is worth revisiting these assumptions and see if they

are actually significant.

Assumptions in the treatment of the hydrogen atom

One thing we assumed was that the proton is stationary at the origin and the

electron moves around it. We also took the mass to be

. More accurately,

we can consider the motion relative to the center of mass of the system, and we

should take the mass as the reduced mass

µ =

+ m

just as in classical mechanics. However, the proton mass is so much larger and

heavier, and the reduced mass is very close to the electron mass. Hence, what

we’ve got is actually a good approximation. In principle, we can take this into

account and this will change the energy levels very slightly.

What else? The entire treatment of quantum mechanics is non-relativistic.

We can work a bit harder and solve the hydrogen atom relativistically, but the

corrections are also small. These are rather moot problems. There are larger

problems.

Spin

We have always assumed that particles are structureless, namely that we can

completely specify the properties of a particle by its position and momentum.

However, it turns out electrons (and protons and neutrons) have an additional

internal degree of freedom called spin. This is a form of angular momentum, but

with

and

. This cannot be due to orbital motion, since orbital

motion has integer values of

for well-behaved wavefunctions. However, we still

call it angular momentum, since angular momentum is conserved only if we take

these into account as well.

The result is that for each each quantum number

n, `, m

, there are two

possible spin states, and the total degeneracy of level

is then 2

. This agrees

with what we know from chemistry.

Many electron atoms

So far, we have been looking at a hydrogen atom, with just one proton and one

electron. What if we had more electrons? Consider a nucleus at the origin with

charge +

, where

is the atomic number. This has

independent electrons

orbiting it with positions x

for a = 1, ··· , Z.

We can write down the Schr¨odinger equation for these particles, and it looks

rather complicated, since electrons not only interact with the nucleus, but with

other electrons as well.

So, to begin, we first ignore electron-electron interactions. Then the solutions

can be written down immediately:

ψ(x

, x

, ··· , x

) = ψ

)ψ

) ···ψ

where

is any solution for the hydrogen atom, scaled appropriately by

7→ Ze

to accommodate for the larger charge of the nucleus. The energy is then

E = E

+ E

+ ··· + E

We can next add in the electron-electron interactions terms, and find a more

accurate equation for

using perturbation theory, which you will come across

in IID Principles of Quantum Mechanics.

However, there is an additional constraint on this. The Fermi-Dirac statistics

or Pauli exclusion principle states that no two particles can have the same state.

In other words, if we attempt to construct a multi-electron atom, we cannot

put everything into the ground state. We are forced to put some electrons in

higher energy states. This is how chemical reactivity arises, which depends on

occupancy of energy levels.

–

For

= 1, we have 2

= 2 electron states. This is full for

= 2, and this

is helium.

–

For

= 2, we have 2

= 8 electron states. Hence the first two energy

levels are full for Z = 10, and this is neon.

These are rather stable elements, since to give them an additional electron, we

must put it in a higher energy level, which costs a lot of energy.

We also expect reactive atoms when the number of electrons is one more

or less than the full energy levels. These include hydrogen (

= 1), lithium

(Z = 3), fluorine (Z = 9) and sodium (Z = 11).

This is a recognizable sketch of the periodic table. However, for

= 3 and

above, this model does not hold well. At these energy levels, electron-electron

interactions become important, and the world is not so simple.