'Diracology'-some simple identities

[CAF123]

1. Homework Statement
I can't get the latex to render so I use $\tilde a$ to mean the slashed notation commonly seen in the Dirac equation, so $\tilde a = \gamma^{\nu}a_{\nu}$ which is also $a_{\nu}\gamma^{\nu}$ I think.

Prove the following:
$\gamma^{\mu} \tilde{a} \gamma_{\mu} = -2 \tilde a$
$\gamma^{\mu} \tilde a \tilde b \gamma_{\mu} = 4ab$

Homework Equations

As given above (defintion of $\tilde a$)

The Attempt at a Solution

$$\gamma^{\mu} \tilde{a} \gamma_{\mu} = \gamma^{\mu}\gamma^{\nu}a_{\nu}\gamma_{\mu} = \gamma^{\mu}a_{\nu} \gamma^{\nu} \gamma_{\mu} = g_{\mu \beta}\gamma^{\mu}a_{\nu}\gamma^{\nu}\gamma^{\beta} = g_{\mu \beta}\gamma^{\mu} a_{\nu}(2 g^{\beta \nu} - \gamma^{\beta}\gamma^{\nu})$$ which is then $$2 \gamma^{\mu}a_{\mu} - 4a_{\nu}\gamma^{\nu} = -2\tilde a$$ using that $\gamma^{\mu}\gamma_{\mu} = 4$ and the Clifford algebra. Does that seem ok?

Then
$$\gamma^{\mu} \tilde{a}\tilde{b}\gamma_{\mu} = g_{\mu \beta}\gamma^{\mu}a_{\nu}\gamma^{\nu} b_{\sigma}\gamma^{\sigma}\gamma^{\beta} = g_{\mu \beta} \gamma^{\mu} a_{\nu} \gamma^{\nu}b_{\sigma}(2g^{\sigma \beta} - \gamma^{\beta} \gamma^{\sigma}) = 2 \gamma^{\mu}a_{\nu} \gamma^{\nu} b_{\mu} - 4 a_{\nu} \gamma^{\nu}b_{\sigma}\gamma^{\sigma}$$ which is $$2(2g^{\mu \nu} - \gamma^{\nu}\gamma^{\mu})a_{\nu}b_{\mu} - 4 \tilde a \tilde b = 4 ab - 6 \tilde a \tilde b?$$
Where did I go wrong?
Thanks!

 

[Orodruin]

I suggest you instead work with $\gamma^\mu \gamma^\sigma \gamma^\nu \gamma_\mu$ and forget about the $a_\mu$ and $b_\mu$ for the time being. This will save you some terms which are just extra luggage in your equations.

$$ g_{\mu \beta} \gamma^{\mu} a_{\nu} \gamma^{\nu}b_{\sigma}(2g^{\sigma \beta} - \gamma^{\beta} \gamma^{\sigma}) = 2 \gamma^{\mu}a_{\nu} \gamma^{\nu} b_{\mu} - 4 a_{\nu} \gamma^{\nu}b_{\sigma}\gamma^{\sigma}$$

You may want to reexamine this.

Also, this is the easiest way to do slash notation in LaTeX: "\not a" = $\not a$

 

[CAF123]

Hi Orodruin,

I suggest you instead work with $\gamma^\mu \gamma^\sigma \gamma^\nu \gamma_\mu$ and forget about the $a_\mu$ and $b_\mu$ for the time being. This will save you some terms which are just extra luggage in your equations.

I see, ok thanks.

You may want to reexamine this.

Do you mean to say there is a mistake in the equality you quoted? I don't see one yet.

Also, this is the easiest way to do slash notation in LaTeX: "\not a" = $\not a$

Ah ok, I tried \slashed before and it didn't render. Thanks.

 

[Orodruin]

Do you mean to say there is a mistake in the equality you quoted? I don't see one yet.

Yes. It is not even clear to me which identity you are trying to apply. You should be applying whay you showed in the first problem.

 

[CAF123]

Yes. It is not even clear to me which identity you are trying to apply. You should be applying whay you showed in the first problem.

Ah ok, I wasn't trying to apply any identity, just use properties of the metric: $$g_{\mu \beta} \gamma^{\mu} a_{\nu} \gamma^{\nu} b_{\sigma} (2g^{\sigma \beta} - \gamma^{\beta} \gamma^{\sigma}) = 2g_{\mu \beta}g^{\sigma \beta}\gamma^{\mu}a_{\nu}\gamma^{\nu}b_{\sigma} - g_{\mu \beta}\gamma^{\mu}a_{\nu}\gamma^{\nu}b_{\sigma}\gamma^{\beta}\gamma^{\sigma}$$ which is simplified to the form I gave using $g_{\mu \beta}g^{\sigma \beta} = g_{\mu}^{\sigma}$, is it not?

 

[CAF123]

@Orodruin Any further comments here? Thanks.

 

[Orodruin]

Ah ok, I wasn't trying to apply any identity, just use properties of the metric: $$g_{\mu \beta} \gamma^{\mu} a_{\nu} \gamma^{\nu} b_{\sigma} (2g^{\sigma \beta} - \gamma^{\beta} \gamma^{\sigma}) = 2g_{\mu \beta}g^{\sigma \beta}\gamma^{\mu}a_{\nu}\gamma^{\nu}b_{\sigma} - g_{\mu \beta}\gamma^{\mu}a_{\nu}\gamma^{\nu}b_{\sigma}\gamma^{\beta}\gamma^{\sigma}$$ which is simplified to the form I gave using $g_{\mu \beta}g^{\sigma \beta} = g_{\mu}^{\sigma}$, is it not?

It is, so where did the 4 come from?

 

[CAF123]

It is, so where did the 4 come from?

The expression I am working with is $$g_{\mu \beta} \gamma^{\mu} a_{\nu}\gamma^{\nu}b_{\sigma}(2g^{\sigma \beta} - \gamma^{\beta}\gamma^{\sigma})$$ Multiplying out gives $$2\gamma^{\mu}a_{\nu}\gamma^{\nu}b_{\mu} - 4a_{\nu}\gamma^{\nu}b_{\sigma}\gamma^{\sigma},$$ where in the first term I made use of $g_{\mu \beta}g^{\sigma \beta}=g_{\mu}^{\sigma}$ and in the second term the fact that $g_{\mu \beta}\gamma^{\beta}\gamma^{\mu} = \gamma^{\mu}\gamma_{\mu} = 4$.

 

[Orodruin]

Multiplying out gives

No it doesnt. The second term should have four gamma matrices.

 

[CAF123]

No it doesnt. The second term should have four gamma matrices.

Ah ok, many thanks, I see what I did wrong. The second term should be $-\gamma_{\beta}a_{\nu}\gamma^{\nu}b_{\sigma}\gamma^{\beta}\gamma^{\sigma}$. I don't think any of these objects commute (apart from the sum $a_{\nu}\gamma^{\nu} = \gamma^{\nu}a_{\nu}$) so what would you suggest I try next?
Thanks.

 

[CAF123]

@Orodruin any suggestions on how to progress? I'm starting to have doubts that $\gamma_{\nu}a^{\nu} = a^{\nu}\gamma_{\nu}$ is even correct because the $\gamma^{\mu}$'s are matrices and $a_{\nu}$ a vector.

 

[Orodruin]

@Orodruin any suggestions on how to progress? I'm starting to have doubts that $\gamma_{\nu}a^{\nu} = a^{\nu}\gamma_{\nu}$ is even correct because the $\gamma^{\mu}$'s are matrices and $a_{\nu}$ a vector.

The $a_\nu$ are just numbers which happen to be the components of a vector. They commute with everything. Again, I suggest you work without the as and bs and contract with them only in a last step if you find it necessary. All they do is to clog up your equations.

Regarding your current expression, apply what you have already shown ...

 

[CAF123]

Thanks Orodriun, I was able to obtain the result and some results for similar identities. Can I ask about some other ones?

Prove that a) Tr$(\not a \not b) = 4ab$ and b) Tr$(\gamma^5 \not a \not b) = 0$

Attempt:

a) $$\text{Tr}(\not a \not b) = \text{Tr}(a^{\mu}\gamma_{\mu}b^{\nu}\gamma_{\nu}) = a_{\mu}b_{\nu}\text{Tr}(\gamma^{\mu} \gamma^{\nu}) = \frac{1}{2} a_{\mu}b_{\nu} \text{Tr}(2 g^{\mu \nu}) = \text{Tr}(ab)$$ using the fact that since tr(AB)=tr(BA) and tr(A+B) = tr(A)+tr(B) then Tr(AB) = Tr(AB+BA)/2 and using the Clifford algebra to bring in the metric. I think I am nearly there but I don't know how the factor of 4 comes in?

Thanks!

 

[CAF123]

Before the last equality, I thought I could write the terms like $g_{\mu \sigma} a^{\sigma} b_{\nu} \text{Tr}(g^{\mu \nu}) = a^{\sigma}b_{\nu} \text{Tr}(g^{\mu \nu}g_{\mu \sigma}) = a^{\sigma}b_{\nu}\text{Tr}(\delta^{\nu}_{\sigma})$ Then I said that the delta there constrains $\nu = \sigma$ so this is the same as $a^{\sigma}b_{\sigma} \delta^{\nu}_{\nu} = 4ab$ where $\text{Tr}(\delta^{\nu}_{\sigma}) = \delta^{\nu}_{\nu} = 4$. Is that ok?

And if so would you be able to give me a hint on how to start proving $\text{Tr}(\gamma^5 \not a \not b) = 0$?

Thanks

 

[Orodruin]

Careful here, the metric and the Kronecker delta have absolutely nothing to do with the trace. For each fixed value of the indices, you just have numbers, which may be moved out of the tracw due to linearity. The main thing to realize is that the metric component resulting from the anticommutation relation is not just a metric component, but also multiplies a 4x4 identity matrix in spinor space. It is in this space that you take the trace.

 

[CAF123]

Thanks I see what you are saying. Would you be able to give me a hint on how to begin proving $\text{Tr}(\gamma^5 \not a \not b) = 0$?
I have tried writing $$\text{Tr}(\gamma^5 \gamma^{\mu} \gamma^{\nu}) = \frac{1}{2} \text{Tr}(\left\{\gamma^0, \gamma^1 \gamma^2 \gamma^3 \gamma^{\mu}\gamma^{\nu}\right\}),$$ making use of the explicit form for $\gamma^5$. I played around with shifting the ordering of the terms there (and therefore introducing minus signs) to try and get a result that gave me that Tr = - Tr and thus infer it was zero but I have not managed.

 

[samalkhaiat]

Thanks I see what you are saying. Would you be able to give me a hint on how to begin proving $\text{Tr}(\gamma^5 \not a \not b) = 0$?
I have tried writing $$\text{Tr}(\gamma^5 \gamma^{\mu} \gamma^{\nu}) = \frac{1}{2} \text{Tr}(\left\{\gamma^0, \gamma^1 \gamma^2 \gamma^3 \gamma^{\mu}\gamma^{\nu}\right\}),$$ making use of the explicit form for $\gamma^5$. I played around with shifting the ordering of the terms there (and therefore introducing minus signs) to try and get a result that gave me that Tr = - Tr and thus infer it was zero but I have not managed.

See this

 

[CAF123]

Not really following on from this thread but thought it fits fine with the title 'diracology' in the title: I am just trying to get the dirac equation for spinor $\psi$ given that for $\bar{\psi}$. My steps are as follows and just wondering if they are ok:

For $\bar{\psi}$ have, $$\bar{\psi} (i \not{\partial}^{\leftarrow} + m + q\not{A}) = 0$$ Then insert $\bar{\psi} = \psi^{\dagger}\gamma_o$ and removed the gammas from the slash: $$\psi^{\dagger} \gamma_0 \gamma^{\mu} (i \partial_{\mu}^{\leftarrow} + qA_{\mu}) + \psi^{\dagger} \gamma_o m = 0.$$ Then conjugate: $$(-i\partial_{\mu} + qA_{\mu}) (\psi^{\dagger} \gamma^0 \gamma^{\mu})^{\dagger} + \gamma^0 m \psi = 0$$ Insert $\gamma_o^2 = 1$ and use the fact that $\gamma_o (\gamma^{\mu})^{\dagger}\gamma_o = \gamma^{\mu}$ gives $$(-i\partial_{\mu} + qA_{\mu}) \gamma_o \gamma^{\mu} \psi + m \gamma_o \psi = 0$$ Hit by $\gamma_o^{-1}$ on the left and manipulate gives $(i\not{D}-m)\psi = 0$

I know the final equation is correct, however, I want to check all manipulations above are valid.

Thanks!

 

[Orodruin]

Looks fine, but instead of inserting $1 = \gamma_0^2$ and then multiplying by $\gamma_0$ in the end, it would be simpler to just multiply by $\gamma_0$ to start with.