# Eigenvalues of $\not{p}$

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.

• 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? – renyhp Apr 8 at 20:31
• @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 $(\pm m)^2\neq -m^2$. – akhmeteli Apr 8 at 21:43
• 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! – renyhp Apr 8 at 23:15
• @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. – akhmeteli Apr 9 at 0:27
• That's fair enough. I will edit the OP and wait for a possible comment of yours if the statement is wrong again. – renyhp Apr 9 at 0:37