# Why doesn't the energy come out right in the Dirac Equation?

1. May 29, 2012

### Xezlec

Hello,

I'm looking at the Dirac Equation, in the form given on Wikipedia, and (foolishly) trying to understand it.

$\left( c \boldsymbol{\alpha}\cdot \mathbf{\hat{p}}+\beta mc^2 \right ) \psi = i\hbar\frac{\partial \psi}{\partial t}\,\!$

So I picture a wavefunction in an eigenstate of the momentum operator in the $e_1$-direction with an eigenvalue of p, and simultaneously an eigenstate of the $\alpha_1$ and $\beta$ operators with an eigenvalue of 1 in both cases. Now obviously, for this case:

$\left( c \boldsymbol{\alpha}\cdot \mathbf{\hat{p}}+\beta mc^2 \right ) \psi = \left( p c +mc^2 \right ) \psi\,\!$

But we know that $E = \sqrt{p^2 c^2 + m^2 c^4}$, so this doesn't seem to give the right eigenvalue for the energy operator on the RHS. We want the hypotenuse of a right triangle with $c p$ and $m c^2$ as its legs, not the length of a line with those two quantities as segments of the line! It seems like it might work out right if somehow they were complex and 90 degrees out of phase, but I can't see any way to get that.

What part of my brain is broken?

Thanks.

2. May 29, 2012

### fzero

3. May 29, 2012

### Xezlec

Wow. I had forgotten those didn't commute. Now I'm really confused. Back to the drawing board.

Thanks!

4. May 30, 2012

### Xezlec

OK, I may have tracked down part of what's confusing me now. I've always been told that if two operators don't commute then they can't have a common eigenstate, but looky here. If

$A = \begin{bmatrix}-1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix}$

and

$B = \begin{bmatrix}0 & 1 & 0 \\1 & 0 & 0 \\0 & 0 & 1 \end{bmatrix}$

then

$AB = \begin{bmatrix}0 & -1 & 0 \\1 & 0 & 0 \\0 & 0 & 1 \end{bmatrix}$

but

$BA = \begin{bmatrix}0 & 1 & 0 \\-1 & 0 & 0 \\0 & 0 & 1 \end{bmatrix}$.

Since these are different, $A$ and $B$ do not commute. However,

$r = \begin{bmatrix}0 \\0 \\1 \end{bmatrix}$

is clearly an eigenvector of both $A$ and $B$. I guess I'm having trouble with the math at a more basic level than I thought.

5. May 30, 2012

### fzero

I was a bit too fast, sorry to have misled you. If the rank of $[A,B]$ is not maximal, there may be some common eigenvectors. The true statement is that, for two Hermitian operators that don't commute, their bases of eigenvectors are not the same. In other words, they are not simultaneously diagonalizable.

Now, to get back to the original problem, it turns out that $\alpha_1$ and $\beta$ still do not have a common eigenvector.