4Spinors

III Quantum Field Theory

4.2 The Clifford algebra and the spin representation

It turns out there is a useful way of finding a representation of the Lorentz

algebra which gives rise to nice properties of the representation. This is via the

Clifford algebra.

We will make use of the following notation:

Notation (Anticommutator). We write

{A, B} = AB + BA

for the anticommutator of two matrices/linear maps.

Definition

(Clifford algebra)

.

The Clifford algebra is the algebra generated by

γ

0

, γ

1

, γ

2

, γ

3

subject to the relations

{γ

µ

, γ

ν

} = γ

µ

γ

ν

+ γ

ν

γ

µ

= 2η

µν

1.

More explicitly, we have

γ

µ

γ

ν

= −γ

ν

γ

µ

for µ 6= ν.

and

(γ

0

)

2

= 1, (γ

i

)

2

= −1.

A representation of the Clifford algebra is then a collection of matrices (or linear

maps) satisfying the relations listed above.

Note that if we have a representation of the Clifford algebra, we will also

denote the matrices of the representation by γ

µ

, instead of, say, R(γ

µ

).

Example

(Chiral representation)

.

The simplest representation (in the sense

that it is easy to write out, rather than in any mathematical sense) is the

4-dimensional representation called the chiral representation. In block matrix

form, this is given by

γ

0

=

0 1

1 0

, γ

i

=

0 σ

i

−σ

i

0

,

where the σ

i

are the Pauli matrices given by

σ

1

=

0 1

1 0

, σ

2

=

0 −i

i 0

, σ

3

=

1 0

0 −1

,

which satisfy

{σ

i

, σ

j

} = 2δ

ij

1.

Of course, this is not the only representation. Given any 4

×

4 matrix

U

, we

can construct a new representation of the Clifford algebra by transforming by

γ

µ

7→ Uγ

µ

U

−1

.

It turns out any 4-dimensional representation of the Clifford algebra comes from

applying some similarity transformation to the chiral representation, but we will

not prove that.

We now claim that every representation of the Clifford algebra gives rise to

a representation of the Lorentz algebra.

Proposition.

Suppose

γ

µ

is a representation of the Clifford algebra. Then the

matrices given by

S

ρσ

=

1

4

[γ

ρ

, γ

σ

] =

(

0 ρ = σ

1

2

γ

ρ

γ

σ

ρ 6= σ

=

1

2

γ

ρ

γ

σ

−

1

2

η

ρσ

define a representation of the Lorentz algebra.

This is known as the spin representation.

We will need the following technical result to prove the proposition:

Lemma.

[S

µν

, γ

ρ

] = γ

µ

η

νρ

− γ

ν

η

ρµ

.

Proof.

[S

µν

, γ

ρ

] =

1

2

γ

µ

γ

ν

−

1

2

η

µν

, γ

ρ

=

1

2

γ

µ

γ

ν

γ

ρ

−

1

2

γ

ρ

γ

µ

γ

ν

= γ

µ

(η

νρ

− γ

ρ

γ

ν

) − (η

µρ

− γ

µ

γ

ρ

)γ

ν

= γ

µ

η

νρ

− γ

ν

η

ρµ

.

Now we can prove our claim.

Proof of proposition. We have to show that

[S

µν

, S

ρσ

] = η

νρ

S

µσ

− η

µρ

S

νσ

+ η

µσ

S

νρ

− η

νσ

S

µρ

.

The proof involves, again, writing everything out. Using the fact that

η

ρσ

commutes with everything, we know

[S

µν

, S

ρσ

] =

1

2

[S

µν

, γ

ρ

γ

σ

]

=

1

2

[S

µν

, γ

ρ

]γ

σ

+ γ

ρ

[S

µν

, γ

σ

]

=

1

2

γ

µ

η

νρ

γ

σ

− γ

ν

η

µρ

γ

σ

+ γ

ρ

γ

µ

η

νσ

− γ

ρ

γ

ν

η

µσ

.

Then the result follows form the fact that

γ

µ

γ

σ

= 2S

µσ

+ η

µσ

.

So we now have a representation of the Lorentz algebra. Does this give us a

representation of the (restricted) Lorentz group? Given any Λ

∈ SO

+

(1

,

3), we

can write

Λ = exp

1

2

Ω

ρσ

M

ρσ

.

We now try to define

S[Λ] = exp

1

2

Ω

ρσ

S

ρσ

.

Is this well-defined?

We can try using an example we have previously computed. Recall that given

any rotation parameter φ = (φ

1

, φ

2

, φ

3

), we can pick

Ω

ij

= −ε

ijk

φ

k

to obtain a rotation denoted by

φ

. What does this give when we exponentiate

with S

ρσ

? We use the chiral representation, so that

S

ij

=

1

4

0 σ

i

−σ

i

0

,

0 σ

j

−σ

j

0

= −

i

2

ε

ijk

σ

k

0

0 σ

k

Then we obtain

Proposition. Let φ = (φ

1

, φ

2

, φ

3

), and define

Ω

ij

= −ε

ijk

φ

k

.

Then in the chiral representation of S, writing σ = (σ

1

, σ

2

, σ

3

), we have

S[Λ] = exp

1

2

Ω

ρσ

S

ρσ

=

e

iφ·σ/2

0

0 e

iφ·σ/2

.

In particular, we can pick

φ

= (0

,

0

,

2

π

). Then the corresponding Lorentz

transformation is the identity, but

S[1] =

e

iπσ

3

0

0 e

iπσ

3

!

= −1 6= 1.

So this does not give a well-defined representation of the Lorentz group, because

different ways of representing the element

1

in the Lorentz group will give

different values of

S

[

1

]. Before we continue on to discuss this, we take note of

the corresponding formula for boosts.

Note that we have

S

0i

=

1

2

0 1

1 0

0 σ

i

−σ

i

0

=

1

2

−σ

i

0

0 σ

i

.

Then the corresponding result is

Proposition. Write χ = (χ

1

, χ

2

, χ

3

). Then if

Ω

0i

= −Ω

i0

= −χ

i

,

then

Λ = exp

1

2

Ω

ρσ

M

ρσ

is the boost in the χ direction, and

S[Λ] = exp

1

2

Ω

ρσ

S

ρσ

=

e

χ·σ/2

0

0 e

−χ·σ/2

.

Now what is this

S

[Λ] we have constructed? To begin with, writing

S

[Λ]

is a very bad notation, because

S

[Λ] doesn’t only depend on what Λ is, but

also additional information on how we “got to” Λ, i.e. the values of Ω

ρσ

. So

going from

1

to

1

by rotating by 2

π

is different from going from

1

to

1

by doing

nothing. However, physicists like to be confusing and this will be the notation

used in the course.

Yet,

S

[Λ] is indeed a representation of something. Each point of this “some-

thing” consists of an element of the Lorentz group, and also information on

how we got there (formally, a (homotopy class of) paths from the identity to Λ).

Mathematicians have come up with a fancy name of this, called the universal

cover, and it can be naturally given the structure of a Lie group. It turns out

in this case, this universal cover is a double cover, which means that for each

Lorentz transformation Λ, there are only two non-equivalent ways of getting to

Λ.

Note that the previous statement depends crucially on what we mean by

ways of getting to Λ being “equivalent”. We previously saw that for a rotation,

we can always add 2

nπ

to the rotation angle to get a different Ω

ρσ

. However,

it is a fact that whenever we add 4

π

to the angle, we will always get back the

same

S

[Λ] for any representation

S

of the Lorentz algebra. So the only two

non-equivalent ways are the original one, and the original one plus 2π.

(The actual reason is backwards. We know for geometric reasons that

adding 4

π

will give an equivalent path, and thus it must be the case that any

representation must not change when we add 4

π

. Trying to supply further details

and justification for what is going on would bring us deep into the beautiful

world of algebraic topology)

We give a name to this group.

Definition

(Spin group)

.

The spin group

Spin

(1

,

3) is the universal cover of

SO

+

(1

,

3). This comes with a canonical surjection to

SO

+

(1

,

3), sending “Λ and

a path to Λ” to Λ.

As mentioned, physicists like to be confusing, and we will in the future keep

talking about “representation of the Lorentz group”, when we actually mean the

representation of the spin group.

So we have the following maps of (Lie) groups:

Spin(1, 3) SO

+

(1, 3) O(1, 3) .

This diagram pretty much summarizes the obstructions to lifting a representation

of the Lorentz algebra to the Lorentz group. What a representation of the Lorentz

algebra gives us is actually a representation of the group

Spin

(1

,

3), and we want

to turn this into a representation of O(1, 3).

If we have a representation

D

of O(1

,

3), then we can easily produce a

representation of

Spin

(1

,

3) — given an element

M ∈ Spin

(1

,

3), we need to

produce some matrix. We simply apply the above map to obtain an element of

O(1, 3), and then the representation D gives us the matrix we wanted.

However, going the other way round is harder. If we have a representation of

Spin

(1

,

3), then for that to give a representation of

SO

+

(1

,

3), we need to make

sure that different ways of getting to a Λ don’t give different matrices in the

representation. If this is true, then we have found ourselves a representation

of

SO

+

(1

,

3). After that, we then need to decide what we want to do with the

elements of O(1, 3) not in SO

+

(1, 3), and this again involves some work.