Non-negativity of the eigenvalues of the Dirac operator

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  • #1
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Main Question or Discussion Point

How can I prove that the eigenvalues of the operator ## i\gamma^\mu \partial_\mu ## are non-negative?
I've tried using the ansatz ## \psi=u(p) e^{ip_\nu x^\nu} ## but it didn't help.
I've also tried playing with the equation using the properties of gamma matrices but that doesn't seem to lead anywhere too.
Thanks
 

Answers and Replies

  • #2
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Why would it not have negative EVs? [I am sorry for asking this but I am trying to see the motivation]
I think its because the Dirac equation is ## i\gamma^\mu\partial_\mu\psi=m \psi ## and so the eigenvalue of that operator is the mass of the Dirac field, so it should be non-negative.

This was given to one of my friends to prove and she asked me for help, so I don't know what was the motivation.
 
  • #3
ChrisVer
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So OK:
[itex] i \gamma^\mu \partial_\mu \psi = a \psi[/itex]
[itex] [i \gamma^\mu \partial_\mu - a] \psi =0[/itex]
Now the eigenvalues can be found by solving:
[itex] det[i \gamma^\mu \partial_\mu - a \textbf{1}] =0[/itex]
?
I have never tried that...
But I think the expression is obtained much more easily by the ansatzes taken while deriving the Dirac equation.
 
  • #4
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That's a differential operator so you can't just cancel the Dirac spinor.
 
  • #5
ChrisVer
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That's a differential operator so you can't just cancel the Dirac spinor.
try just write it as the momentum...
 
  • #6
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try just write it as the momentum...
As I said in the OP, I tried the ansatz ## \psi=u(p) e^{i p_\nu x^\nu} ##, but it doesn't help because you'll get a quadratic equation in m which has both negative and positive solutions.
 
  • #7
ChrisVer
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how do you get a quadratic term by differentiating once the exponential?
 
  • #8
ChrisVer
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[itex] i \gamma^\mu \partial_\mu u(p) e^{i p x} = a u(p)e^{ipx}[/itex]
[itex] - \gamma^\mu p_\mu u(p) e^{ipx} = a u(p) e^{ipx}[/itex]
[itex] (a + \slash{p} ) u(p) e^{ipx} = 0[/itex]
so just try the determinant of the matrix on the left?
 
  • #9
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so just try the determinant of the matrix on the left?
And that's exactly what gives you the quadratic equation! (Actually its fourth order, but you can solve it to get a quadratic equation.)
 
  • #10
ChrisVer
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In fact the more I think of it, the more I tend to say that by construction there is no reason to have positive EVs...
It's again coming from the fact that negative masses wouldn't make much sense and would violate causality...
And then, the rest comes from the way we derive the Dirac equation, by using the [itex]E^2= p^2+m^2[/itex] where [itex]m^2>0[/itex].
 
  • #11
Paul Colby
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I guess I would expect the operator, ##i\gamma^\mu\partial_\mu## to have a sign symmetric spectrum. Is there a reason to think otherwise? Take the special case ##v_ne^{-i\omega t}## where ##v_n## is an eigen vector of ##\gamma_t##. Both signs occur.
 

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