PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 5 1

Physics 513, Quantum Field Theory

Homework 5
Due Tuesday, 7th October 2003
Jacob Lewis Bourjaily
1. We are to verify the identity
[γ µ , S ρσ ] = (J ρσ )µν γ ν .
It will be helpful to first have a good representation of (J ρσ )µν . This can be obtained by raising
one of the indices of (J ρσ )λν which is defined in Peskin and Schroeder’s equation 3.18.
(J ρσ )µν = g µλ (J ρσ )λν = ig µλ (δλρ δνσ − δνρ δλσ ),
= i(g µρ δνσ − g µσ δνρ ).
We will use this expression for (J ρσ )µν in the last line of our derivation below. We will proceed
by direct computation.
i
[γ µ , S ρσ ] = ([γ µ , γ ρ γ σ ] − [γ µ , γ σ γ ρ ]) ,
4
i
= ({γ µ , γ ρ }γ σ − γ ρ {γ µ , γ σ } − {γ µ , γ σ }γ ρ + γ σ {γ µ , γ ρ }) ,
4
i
= (g µρ γ σ − γ ρ g µσ − g µσ γ ρ + γ σ g µρ ) ,
2
= i (g µρ γ σ − g µσ γ ρ ) ,
= i (g µρ δνσ γ ν − g µσ δνρ γ ν ) ,
= i (g µρ δνσ − g µσ δνρ ) γ ν ,
∴ [γ µ , S ρσ ] = (J ρσ )µν γ ν .
‘ ’
óπ²ρ ²́δ²ι δ²ιξαι
2. All of the required identities will be computed by directly.
a) γµ γ µ = 4
γµ γ µ = (γ 0 )2 + (γ 1 )2 + (γ 2 )2 + (γ 3 )2 = 4.
b) γµ6 kγ µ = −26 k
γµ6 kγ µ = γµ γν k ν γ µ ,
= (2gµν − γν γµ )k ν γ µ ,
= 2kµ γ µ − γν k ν γµ γ µ ,
µ
∴ γµ6 kγ = −26 k
c) γµ6 p6 qγ µ = 4p · q
γµ6 p6 qγ µ = γµ γν pν qρ γ ρ γ µ ,
= (2gµν − γν γµ )pν qρ (2g ρµ − γ µρ ),
= (2pµ −6 pγµ )(2q µ −6 qγ µ ),
= 4p · q − 26 p6 q − 26 p6 q + 46 p6 q,
µ
∴ γµ6 p6 qγ = 4p · q.
µ
d) γµ6 k6 p6 qγ = −26 p6 q6 k
By repeated use of the identity γ µ γ ν = 2g µν − γ ν γ µ ,
γµ6 k6 p6 qγ µ = γµ γ ν kν γ ρ pρ γ σ qσ γ µ ,
= 2γµ6 k6 pqσ g σµ − 2γµ6 kpρ g ρµ6 q + 2γµ kν g νµ6 p6 q − 46 k6 p6 q,
= 26 q6 k6 p − 26 p6 k6 q − 26 k6 p6 q,
= 46 qk · p − 26 q6 p6 k − 4p · k6 q,
µ
∴ γµ6 k6 p6 qγ = −26 p6 q6 k.
‘ ’
óπ²ρ ²́δ²ι δ²ιξαι
2 JACOB LEWIS BOURJAILY

3. We are to prove the Gordon identity,

· ¸
0 µ (p0 + p)µ iσ µν qν
0
ū(p )γ u(p) = ū(p ) + u(p).
2m 2m
Explicitly writing out each term in the brackets and recalling the anticommutation relations of
γ µ , the right hand side becomes,
· 0 ¸ · ¸
0 (p + p)µ iσ µν qν 0 1 0µ µ µ ν 0 ν µ 0
ū(p ) + u(p) = ū(p ) (p + p − ½γ γ (p − p )ν + ½γ γ (p − p )ν ) u(p),
2m 2m 2m
· ¸
0 1 0µ µ µ ν 0 νµ 0 µ ν 0
= ū(p ) (p + p − ½γ γ (p − p )ν + g (p − p )ν − ½γ γ (p − p )ν ) u(p),
2m
· ¸
0 1 0µ µ ν 0
= ū(p ) (2p − γ γ (p − p )ν ) u(p),
2m
· ¸
0 1 0µ µ µ 0
= ū(p ) (2p − γ 6 p − γ 6 p ) u(p).
2m

Now, recall that the Dirac equation for u(p) is

6 pu(p) = mu(p).
0 0
Converting this for ū(p )6 p , one obtains
ū(p0 )6 p0 = mū(p0 ).
Applying both of these equations where we left of, we see that
· 0 ¸
0 (p + p)µ iσ µν qν p0µ
ū(p ) + u(p) = ū(p0 ) u(p).
2m 2m m
Looking again at the Dirac equation, mū(p0 ) = ū(p0 )6 p0 = ū(p0 )γ µ p0µ , it is clear that
· 0 ¸
(p + p)µ iσ µν qν
ū(p0 )γ µ u(p) = ū(p0 ) + u(p).
2m 2m
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óπ²ρ ²́δ²ι δ²ιξαι
4. a) To demonstrate that γ 5 ≡ iγ 0 γ 1 γ 2 γ 3 anticommutes each of the γ µ , it will be helpful to
remember that whenever µ 6= ν, γ µ γ ν = −γ ν γ µ by the anticommutation relations. There-
fore, any odd permutation in the order of some γ 0 s will change the sign of the expression.
It should therefore be quite clear that
γ 5 γ 0 = iγ 0 γ 1 γ 2 γ 3 γ 0 = −iγ 1 γ 2 γ 3 = −iγ 0 γ 0 γ 1 γ 2 γ 3 = −γ 0 γ 5 ;
γ 5 γ 1 = iγ 0 γ 1 γ 2 γ 3 γ 1 = iγ 0 γ 2 γ 3 = −iγ 1 γ 0 γ 1 γ 2 γ 3 = −γ 1 γ 5 ;
γ 5 γ 2 = iγ 0 γ 1 γ 2 γ 3 γ 2 = −iγ 0 γ 1 γ 3 = −iγ 2 γ 0 γ 1 γ 2 γ 3 = −γ 2 γ 5 ;
γ 5 γ 3 = iγ 0 γ 1 γ 2 γ 3 γ 3 = iγ 0 γ 1 γ 2 = −iγ 3 γ 0 γ 1 γ 2 γ 3 = −γ 3 γ 5 ;
∴ {γ 5 , γ µ } = 0.
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óπ²ρ ²́δ²ι δ²ιξαι
b) We will first show that γ 5 is hermitian. Note that the derivation relies on the fact that
(γ 0 )† = γ 0 and (γ i )† = −γ i . These facts are inherent in our chosen representation of the γ
matrices.
(γ 5 )† = −i(γ 0 γ 1 γ 2 γ 3 )† ,
= −i(γ 3 )† (γ 2 )† (γ 1 )† (γ 0 )† ,
= iγ 3 γ 2 γ 1 γ 0 ,
= −iγ 2 γ 1 γ 0 γ 3 ,
= −iγ 1 γ 0 γ 2 γ 3 ,
= iγ 0 γ 1 γ 2 γ 3 ,
= γ5.
 PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 5 3

Let us now show that (γ 5 )2 = 1.

(γ 5 )2 = −iγ3 γ2 γ1 γ0 iγ 0 γ 1 γ 2 γ 3 ,
= γ3 γ2 γ1 γ0 γ 0 γ 1 γ 2 γ 3 ,
= γ3 γ2 γ1 γ 1 γ 2 γ 3 ,
= γ3 γ2 γ 2 γ 3 ,
= γ3 γ 3 ,
= 1.
c) Note that ²κλµν is only nonzero when κ 6= λ 6= µ 6= ν which leaves exactly 4! = 24 nonzero
terms from the 24 possible permutations. Also note that γ κ γ λ γ µ γ ν , like ²κλµν , is totally
antisymmetric–any odd permutation of indices changes the sign of the argument. Therefore,
they change sign exactly together, ²κλµν γ κ γ λ γ µ γ ν does not change sign. That is to say that
each of the 24 nonzero terms of ²κλµν γ κ γ λ γ µ γ ν is identical to ²0123 γ 0 γ 1 γ 2 γ 3 . So
24 5
²κλµν γ κ γ λ γ µ γ ν = 24²0123 γ 0 γ 1 γ 2 γ 3 = − γ ,
i
i
∴ γ5 = − ²κλµν γ κ γ λ γ µ γ ν .
24
This implies that
γ 5 = −i²κλµν γ [κ γ λ γ µ γ ν] ,
∴ γ [κ γ λ γ µ γ ν] = −i²κλµν γ 5 .

5. We will begin by simply directly computing the form of ξ± from the eigenvalue equation
(p̂ · ½~σ ) ξ± (p̂) = ±½ξ± (p̂).
Let us begin to expand the left hand side of the eigenvalue equation,
µ ¶ µ ¶ µ ¶
1 0 sin θ cos φ 1 0 −i sin θ sin φ 1 cos θ 0
(p̂ · ½~σ ) = + + ,
2 sin θ cos φ 0 2 i sin θ sin φ 0 2 0 − cos θ
µ ¶
1 cos θ sin θe−iφ
∴ (p̂ · ½~σ ) = .
2 sin θeiφ − cos θ
Note that we can see here that because this matrix has determinant −1 and trace 0, the eigen-
values must be are ±1. Therefore, we may write the eigenvalue equation as the system of
equations, µ ¶µ 1 ¶ µ 1 ¶
1 cos θ sin θe−iφ ξ± 1 ξ±
2 =± 2 .
2 sin θeiφ − cos θ ξ± 2 ξ±
These two equations are equivalent; I will use the first row of equations. This becomes
1 1
±ξ± = cos θξ± + sin θe−iφ ξ±
2
.
Therefore,
1 sin θe−iφ ξ+
2
sin θe−iφ ξ−
2
ξ+ = = e−iφ tan(θ/2)ξ+
2
and 1
ξ− =− = −e−iφ tan(θ/2)ξ−
2
1 − cos θ 1 + cos θ
So that
µ −iφ 2
¶ µ ¶
e cot(θ/2)ξ+ −e−iφ tan(θ/2)ξ−
2
ξ+ = 2 and ξ− = 2 .
ξ+ ξ−
†
To find the normalization, we must apply the normalization conditions ξ± ξ± = 1. By direct
calculation,
† 2 2
ξ+ ξ+ = 1 = (ξ+ ) (cot2 (θ/2) + 1),
2 2
(ξ+ )
= 2 ,
sin (θ/2)
+
2
∴ ξ+ = eiη sin(θ/2).
4 JACOB LEWIS BOURJAILY

Likewise for ξ− ,
† 2 2
ξ− ξ− = 1 = (ξ− ) (tan2 (θ/2) + 1),
2 2
(ξ− )
= ,
cos2 (θ/2)
−
2
∴ ξ− = eiη cos(θ/2).
Notice that if ξ+ satisfies ξ † ξ = 1 then so does ξ 0 = eiη ξ. So in solving the normalization
equations, we necessarily introduced an arbitrary phase η. Noting, this, spinors become
µ −iφ ¶ µ ¶
+ e cos(θ/2) − −e−iφ sin(θ/2)
ξ+ = eiη and ξ− = eiη .
sin(θ/2) cos(θ/2)
Lastly, we would like to set the phase η so that when the particle is moving in the +z−direction,
they reduce to the usual spin-up/spin-down forms. It should be quite obvious that η − = 0
satisfies this condition for ξ− . For ξ + , we will set the phase to η + = φ so that we may lose the
e−iφ term when θ = 0. So we may write our final spinors as
µ ¶ µ ¶
cos(θ/2) −e−iφ sin(θ/2)
ξ+ = and ξ− = .
eiφ sin(θ/2) cos(θ/2)
‘ ’
óπ²ρ ²́δ²ι δ²ιξαι