Proving \{ \gamma^5 , \gamma^\mu \} = 0 to Gamma 5 Matrix Verification

[latentcorpse]

How do I verify $\{ \gamma^5 , \gamma^\mu \} = 0$

I have

$\{ \gamma^5 , \gamma^\mu \} = \gamma^5 \gamma^\mu + \gamma^\mu \gamma^5$
$= -i ( \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5 + \gamma^5 \gamma^0 \gamma^1 \gamma^2 \gamma^3 )$
$= -i ( \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5 - \gamma^0 \gamma^5 \gamma^1 \gamma^2 \gamma^3 )$
$= -i ( \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5 + \gamma^0 \gamma^1 \gamma^5 \gamma^2 \gamma^3 )$
$= -i ( \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5 - \gamma^0 \gamma^1 \gamma^2 \gamma^5 \gamma^3 )$
$= -i ( \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5 + \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5 )$

But this is not quite right, because at some point I will have shifted the $\gamma^\mu$ past itself and so I will get an additional term $+2 \eta^{ \mu \mu}$ since $\{ \gamma^\mu , \gamma^\nu \} = 2 \eta^{\mu \nu}$

So I should get three terms:
$= -i ( \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5 + \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5 -2 \eta^{\mu \mu} \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5 )$

and then

$= -i ( (2-2 \eta^{\mu \mu}) \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^5) \neq 0$ since $\eta^{\mu \mu} = 4$, no?

 

[quZz]

So you have to prove that
$$\gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^{\mu} + \gamma^{\mu} \gamma^0 \gamma^1 \gamma^2 \gamma^3 = 0$$

The main idea is to get matrices with identical indexes close to each other.
Note that
1. every time you switch two $\gamma$'s with $\mu \neq \nu$ you get a minus sign
2. if you have to move $\gamma^\mu$ in the first summand k times to the left, you will have to move it in the second summand (3-k) times to the right, thus giving ou additional (-1)

 

[latentcorpse]

So you have to prove that
$$\gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^{\mu} + \gamma^{\mu} \gamma^0 \gamma^1 \gamma^2 \gamma^3 = 0$$

The main idea is to get matrices with identical indexes close to each other.
Note that
1. every time you switch two $\gamma$'s with $\mu \neq \nu$ you get a minus sign
2. if you have to move $\gamma^\mu$ in the first summand k times to the left, you will have to move it in the second summand (3-k) times to the right, thus giving ou additional (-1)

Thanks. Can I also ask, why equation 3.25 is correct? Isn't that the interaction Lagrangian instead of the interaction Hamiltonian? I think you'll need to look at eqn 3.7 as well.

Cheers.

 

[quZz]

em... which book? =)

 

[latentcorpse]

em... which book? =)

sorry. see these online notes:
http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf

 

[quZz]

It comes right from the connection between hamiltonian and lagrangian.

If you have for lagrangian L = L1 + L2 and L2 does not depend on derivatives, then for hamiltonian you have H = H1 + H2, where H1 corresponds to L1 and H2 = -L2.

Even more, if you have infinitesimal addition (that can depend on derivatives) to lagrangian: L = L0 + L', then for hamiltonian H = H0 + H', where H0 corresponds to L0 and H' = -L'.

 

[latentcorpse]

It comes right from the connection between hamiltonian and lagrangian.

If you have for lagrangian L = L1 + L2 and L2 does not depend on derivatives, then for hamiltonian you have H = H1 + H2, where H1 corresponds to L1 and H2 = -L2.

Even more, if you have infinitesimal addition (that can depend on derivatives) to lagrangian: L = L0 + L', then for hamiltonian H = H0 + H', where H0 corresponds to L0 and H' = -L'.

when you say H0 correspsonds to L0, do you mean H0=L0? Or it is just some function of L0?

 

[dextercioby]

H0 can't be equal to L0, because they don't depend on the same (field) variables. For the system without constraints, the Hamiltonian is the Legendre transformation of the Lagrangian wrt the generalized velocities.