Let $p_\mu$ be a generic non-null 4-vector, and let $m=\sqrt{p_\mu p^\mu}\neq 0$ (let us choose $\operatorname{Im}m\ge0$). Let also $\gamma^\mu$ be the Dirac gamma matrices.

Prove that the eigenvalues of $\not p=\gamma^\mu p_\mu$ are $\pm m$, and they are both 2-degenerate.

I know that must be true because of the explicit solutions of the equations $(\not p\pm m)u=0$ when I use a given representation of the gamma matrices. But is there a way to prove it independently of the representation? (I assume anyway that $\gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0$.)

I also proved the first part of the statement, by using $\not p \not p = m^2$, but I can't think of any way to determine the multiplicity of the two eigenvalues. I know the only possibilities for the multiplicities of $m$ and $-m$ can be $(1,3)$, $(2,2)$, $(3,1)$, but I can't rule out the $1,3$ possibilities.

EDIT: My answer below relies on the assumption that the multiplicity of the eigenvalues $m$ and $-m$ sum to $4$, which is equivalent to say that $\not p$ is diagonalizable. But since $\not p^\dagger=\gamma^0\not p \gamma ^0\neq \not p$, this seems to me anything but trivial! How can I prove that $\not p$ is diagonalizable?

I have found that the whole statement is equivalent to $$\exists S\:\colon\:\not p=S^\dagger \gamma^0DS,S^\dagger\gamma^0=S^{-1},D=\operatorname {diag}(m,m,-m,-m)\,,$$ but I can't manage to prove it.


After struggling for hours, the answer appeared clearly in my head just after having posted.

Since $\not p \not p = m^2$, then, for each eigenvector $u$, corresponding to the eigenvalue $\lambda$, $$\not p u = \lambda u \qquad \implies \qquad \not p \not p u = \lambda ^2 u \qquad \implies \qquad \lambda ^2 = m^2\,,$$ which proves that $\pm m$ are the only possible eigenvalues.

Answer to edit of OP (part 1):

Since $(m^{-1}\not p)^2=1$, then $m^{-1} \not p$ is diagonalizable, and so is $\not p$. As a consequence, the sum of the multiplicities of $m$ and $-m$ is $4$. Note that one of the two multiplicities may be zero at this point.

Let then $n$ be the multiplicity of the eigenvalue $m$. Then $4-n$ is the multiplicity of the eigenvalue $-m$. $$0=\operatorname{tr} \not p=nm+(4-n)(-m)=(2n-4)m\qquad \implies \qquad n=2\,.$$

This proves that $\exists S$ such that $\not p = S^{-1}DS$.

Answer to edit of OP (part 2):

As for the condition $S^{-1}=S^\dagger \gamma^0$ it is not provable as in general it is false; however, it is possible to choose such an $S$, since $$\begin{aligned}\not pu & =mu\\ \not pv & =-mv \end{aligned} \quad\implies\quad\bar{v}u=v^{\dagger}\gamma^{0}u=0\,.$$


As written, the statement seems to be incorrect: if $p_\mu$ is a space-like vector, $p_\mu p^\mu=-m^2$ (provided the signature is $\{1,-1,-1,-1\}$). It is not assumed in the wording of the statement that the Dirac equation is satisfied. Moreover, the sign of the magnitude || in the statement suggests that $p_\mu$ can indeed be space-like.

  • $\begingroup$ Can you be a bit clearer? No, I am not assuming the Dirac equation is satisfied; yes, I did use some notation that is used while handling the Dirac equation, because answering this question makes the Dirac equation be solved very very easily. Also, I don't understand what you are saying to be false; if the statement is false, can you provide a counterexample to it? $\endgroup$ – renyhp Apr 8 at 20:31
  • $\begingroup$ @renyhp : If $p_\mu$ is space-like, then $p_\mu p^\mu=-m^2$, so $(p_\mu\gamma^\mu)^2=p_\mu p^\mu=-m^2$, so $p_\mu\gamma^\mu$ cannot have eigenvalues $\pm m$, as $(\pm m)^2\neq -m^2$. $\endgroup$ – akhmeteli Apr 8 at 21:43
  • $\begingroup$ Oh! Sorry, my own writing made me keep thinking of $p$ as a four-momentum, and so as a time-like vector. It seems to me that removing the absolute value inside the square root in the OP (and choosing the principal determination in case $p$ is space-like) should make the statement true. Am I right? I will edit the post if you confirm my reasoning is not wrong. Thanks! $\endgroup$ – renyhp Apr 8 at 23:15
  • $\begingroup$ @renyhp : Sorry, I don't feel comfortable guessing what question you would like to ask. I cannot guarantee that the adjustments you are going to make will make the statement correct. $\endgroup$ – akhmeteli Apr 9 at 0:27
  • $\begingroup$ That's fair enough. I will edit the OP and wait for a possible comment of yours if the statement is wrong again. $\endgroup$ – renyhp Apr 9 at 0:37

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