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I'm given an spinor $\Psi$ which is solution of the Free Dirac equation, such that is an eigenfunction of $\hat{\vec{p}}$ and has positive energy. Then I'm asked to calculate the expectation value of the operator $\beta$. I normalize the functions inside a box of volume $V$

So, is this correct?

$$\langle \hat{\beta} \rangle = \displaystyle\int d^3r (\Psi^{\dagger}\hat{\beta}\Psi )$$

For instance, let me begin:

$$\Psi(\vec{r},t)=\displaystyle\frac{1}{(2\pi\hbar)^{3/2}}\exp \left( \frac{\vec{p}\cdot\vec{r}-Et}{\hbar} \right) \psi(\vec{p)}$$


$$\psi(\vec{p})=\left(\begin{matrix}1 \\ 0 \\ c\frac{\vec{\sigma}\cdot\vec{p}}{E_p+mc^2} \\ 0 \\ \end{matrix}\right)$$

In the matrix $\psi(\vec{p})$, did I miss half of the solutions? I mean, there is a pair of two element linearly independent matrices that satisfies the equation with the given constraints (one for each value of spin) but can I take, for simplicity the above matrix instead of a more general

$$\psi(\vec{p})=\left(\begin{matrix}c_1\phi_1+c_2\phi_2 \\ (c_1\phi_1+c_2\phi_2)c\frac{\vec{\sigma}\cdot\vec{p}}{E_p+mc^2} \\ \end{matrix}\right)$$

with $c_1,c_2$ constants and $\phi_1,\phi_2$ arbitrary two elements matrices (the case I choose is a particular case of this one)

Now, in the actual calculation, is this correct?

$$\langle \hat{\beta} \rangle = \frac{1}{(2\pi \hbar)^3 V}\displaystyle\int d^3r \left(1-\frac{c^2(\vec{\sigma}\cdot\vec{p})^2}{(E_p+mc^2)^2}\right) $$


The solution shold be $\langle \hat{\beta} \rangle = \gamma^{-1}$ but first of all I don't know how to cancel the $\hbar$ and the rest of terms in the denominator, and the calculation is more than probably wrong for other reasons apart of that. I think the $\hbar$ stuff comes as an arbitrary multiplicative constant for a plane wave over the entire space time, wich is not normalizable. Now, for normalizable wave functions, I think it should dissapear, but then the other factor

$$\left(1-\frac{c^2(\vec{\sigma}\cdot\vec{p})^2}{(E_p+mc^2)^2}\right) (1)$$

does not seem $\gamma^{-1}$. I know that $(\vec{\sigma}\cdot{\vec{p}})^2=p^2$ but I can't find the final answer.

Thanks for your time.

Note: I'm using Pauli representation of the $\gamma$ matrices.

As I said in the comment, the above is wrong, but I'm going to leave it unaltered and continue with I think it's better but not correct.

The wave function should be

$$ \Psi(\vec{r},t)=\frac{\exp(\frac{i(\vec{p}\cdot\vec{r}-Et)}{\hbar})}{\sqrt{V}}\psi(\vec{p}) $$

choosing the solution as $c_2=0, c_1=1, \phi_1=\left(\begin{matrix} 1 \\ 0\end{matrix}\right)$

First, the action of $\hat{\beta}$ on $\Psi$ is

$$\left( \begin{matrix} \mathbb{I} & 0 \\ 0 & -\mathbb{I} \end{matrix} \right)\frac{\exp(\frac{i(\vec{p}\cdot\vec{r}-Et)}{\hbar})}{\sqrt{V}}\left(\begin{matrix}1 \\ 0 \\ c\frac{\vec{\sigma}\cdot\vec{p}}{E+mc^2} \\ 0 \\ \end{matrix}\right) = \frac{\exp(\frac{i(\vec{p}\cdot\vec{r}-Et)}{\hbar})}{\sqrt{V}}\left(\begin{matrix}1 \\ 0 \\ -c\frac{\vec{\sigma}\cdot\vec{p}}{E+mc^2} \\ 0 \\ \end{matrix}\right) $$

So the integrand $\Psi^{\dagger}\hat{\beta}\Psi$ is

$$ \frac{1}{V}\left(1-c^2\frac{(\vec{\sigma}\cdot\vec{p})^2}{(E+mc^2)^2}\right) $$

The second factor is constant and the integral is $V$ which cancels with the $V$ in the denominator and then

$$\langle \hat{\beta} \rangle = 1-c^2\frac{(\vec{\sigma}\cdot\vec{p})^2}{(E+mc^2)^2}=1-\left(\frac{cp}{E+mc^2}\right)^2$$

is my result, which is not $\gamma^{-1}$.

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Thanks for the corrections. I'm reading this and noticed that $\Psi$ is actually quite bad written. It should be something like $\Psi=\frac{1}{\sqrt{V}}\exp(\frac{\vec{p}\cdot\vec{r}-Et}{h})\psi(\vec{p})$ but the answer I will get it's (1) for the expectation value. – Jorge Apr 20 '12 at 19:30
I would answer this question, but your user name is making me wary... – Ron Maimon Apr 21 '12 at 4:55
I'm sorry about that, however I would like to point out that Burzum is the name of the band, and not the name of the composer :) (I enjoy the music, even if Vikernes is a criminal for more than one reason) – Jorge Apr 21 '12 at 11:46
No worries. I just got scared for a second that you would burn down some church if my answer was less than insightful! :) – Ron Maimon Apr 22 '12 at 4:53

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