'Diracology'-some simple identities

1. Sep 26, 2015

CAF123

1. The problem statement, all variables and given/known data

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

2. Relevant equations
As given above (defintion of $\tilde a$)

3. 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!

2. Sep 26, 2015

Orodruin

Staff Emeritus
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.

You may want to reexamine this.

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

3. Sep 26, 2015

CAF123

Hi Orodruin,
I see, ok thanks.
Do you mean to say there is a mistake in the equality you quoted? I don't see one yet.
Ah ok, I tried \slashed before and it didn't render. Thanks.

4. Sep 26, 2015

Orodruin

Staff Emeritus
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.

5. Sep 26, 2015

CAF123

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?

6. Sep 28, 2015

CAF123

@Orodruin Any further comments here? Thanks.

7. Sep 28, 2015

Orodruin

Staff Emeritus
It is, so where did the 4 come from?

8. Sep 28, 2015

CAF123

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 $g_{\mu \beta}\gamma^{\beta}\gamma^{\mu} = \gamma^{\mu}\gamma_{\mu} = 4$.

9. Sep 28, 2015

Orodruin

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

10. Sep 28, 2015

CAF123

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.

11. Sep 30, 2015

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.

12. Sep 30, 2015

Orodruin

Staff Emeritus
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.

13. Oct 5, 2015

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!

Last edited: Oct 5, 2015
14. Oct 13, 2015

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

15. Oct 13, 2015

Orodruin

Staff Emeritus
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 realise 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.

16. Oct 14, 2015

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.

17. Oct 17, 2015

samalkhaiat

See this

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18. Feb 2, 2016

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!

Last edited: Feb 2, 2016
19. Feb 2, 2016

Orodruin

Staff Emeritus
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.