Question regarding Dirac matrices

  • Thread starter peperone
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  • #1
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Hey there.

In an exercise I was trying to show that every solution of the Dirac equation also solves the Klein-Gordon equation.

Now I have two very simple questions:

Is [tex]\gamma_\nu \gamma^\mu = \gamma^\nu \gamma_\mu[/tex] ?

And is [tex]\partial_\nu \partial^\mu = \partial^\nu \partial_\mu[/tex] ?

Thanks for the help.
 

Answers and Replies

  • #2
dextercioby
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Not really.U should pay attention with this "Lorentz covariance" and always try to see these expressions as tensor products of 4-vectors.

Daniel.
 
  • #3
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I kinda felt it couldn't be that simple. :p

Thanks anyway.
 
  • #4
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Can anybody verify this?

[tex]\gamma^\nu\partial_\nu\gamma_\mu\partial^\mu = \partial_\mu\partial^\mu[/tex]

Just as clarification: I am trying to show that every solution of the Dirac equation also solves the Klein-Gordon equation. So I started with the Dirac equation:

[tex](i\gamma_\mu\partial^\mu - m)\Psi(x) = 0[/tex]

Then, I multiplied it with the operator [tex]-(i\gamma^\nu\partial_\nu + m)[/tex] (from left) which gave me:

[tex](\gamma^\nu\partial_\nu\gamma_\mu\partial^\mu + m^2)\Psi(x) = 0[/tex]

This surely looks somewhat like the Klein-Gordan equation if the above assumption is correct. I just fail to see why it is true.

Any hints appreciated!
 
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  • #5
dextercioby
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Yes,it can be shown easily once you manipulate the Lorentz indices and the anticommutation relation between the gamma's.
[tex] \gamma_{\nu}\partial^{\nu}\gamma_{\mu}\partial^{\mu}=\gamma_{\nu}\gamma_{\mu}\partial^{\nu}\partial^{\mu}=\frac{1}{2}\gamma_{(\nu}\gamma_{\mu)}\partial^{\nu}\partial^{\mu}=g_{\mu\nu}\partial^{\nu}\partial^{\mu}=d'alembertian [/tex]

Daniel.

P.S.Did u follow the reasoning...?
 
  • #6
dextercioby
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Why's the KG equation:

[tex] (\partial_{\mu}\partial^{\mu}-m^{2})\Psi (\vec{r},t) =0 [/tex]

,instead of:

[tex] -(\partial_{\mu}\partial^{\mu}+m^{2})\Psi (\vec{r},t)=0 [/tex]

What metric are u using...?

Daniel.
 
  • #7
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dextercioby said:
Why's the KG equation:

[tex] (\partial_{\mu}\partial^{\mu}-m^{2})\Psi (\vec{r},t) =0 [/tex]

,instead of:

[tex] -(\partial_{\mu}\partial^{\mu}+m^{2})\Psi (\vec{r},t)=0 [/tex]
Sorry, it was just a typo. I edited my post. Thanks for all the help. I think I got it now...

Start with the Dirac equation:

[tex](i\gamma_\mu\partial^\mu - m)\Psi(x) = 0[/tex]

Multiply the operator [tex]-(i\gamma_\mu\partial^\mu + m)[/tex] (from left):

[tex](\gamma_\mu\partial^\mu\gamma_\mu\partial^\mu + m^2)\Psi(x) = (\gamma_\mu\gamma_\mu\partial^\mu\partial^\mu + m^2)\Psi(x) = 0[/tex]

Because of the anticommutation relation [tex]\{\gamma^\mu,\gamma^\nu\} = 2 g^{\mu\nu}[/tex] it is possible to simplify [tex]\gamma_\mu\gamma_\mu[/tex] to [tex]g_{\mu\mu}[/tex] which proves [tex]\gamma_\mu\gamma_\mu\partial^\mu\partial^\mu = \partial^\mu\partial_\mu[/tex].

Thanks again.
 
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  • #8
dextercioby
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Note that u have 4 identical (Lorentz) indices in the same tensor equation and that's not good,not good at all.
Pay attention with these indices,as u might confuse them.

Daniel.
 
  • #9
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Hm, that's true. Do you know a better way? I'm highly interested in alternative solutions.
 
  • #10
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dextercioby said:
Yes,it can be shown easily once you manipulate the Lorentz indices and the anticommutation relation between the gamma's.
[tex] \gamma_{\nu}\partial^{\nu}\gamma_{\mu}\partial^{\mu}=\gamma_{\nu}\gamma_{\mu}\partial^{\nu}\partial^{\mu}=\frac{1}{2}\gamma_{(\nu}\gamma_{\mu)}\partial^{\nu}\partial^{\mu}=g_{\mu\nu}\partial^{\nu}\partial^{\mu}=d'alembertian [/tex]
To be honest, I don't understand what you're doing here (step 2 -> step 3)
 
  • #11
dextercioby
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You multiply with the same operator,but with index "nu" (instead of "mu") and then use the trick i've used in post #5.It should come out nicely.


Daniel.
 
  • #12
dextercioby
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peperone said:
To be honest, I don't understand what you're doing here (step 2 -> step 3)
[tex] \gamma_{\nu}\gamma_{\mu}T^{\nu\mu}=\frac{1}{2}(\gamma_{\nu}\gamma_{\mu}+\gamma_{\mu}\gamma_{\nu})T^{\mu\nu} [/tex]

where T is a symmetric second rank tensor and to get the second term from the bracket i turned "mu"--->"nu" and "nu"---->"mu" and to factor the same tensor i used the fact that under this renotation,the tensor T is symmetric.

Daniel.
 
  • #13
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This surely looks slick! :) But why is [tex]\partial^\nu\partial^\mu[/tex] a symmetric second rank tensor?
 
  • #14
dextercioby
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Because wrt bosonic coordinates,the partial derivatives commute.

Daniel.
 

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