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:
$$\Psi(\vec{r},t)=\displaystyle\frac{1}{(2\pi\hbar)^{3/2}}\exp \left( \frac{\vec{p}\cdot\vec{r}-Et}{\hbar} \right) \psi(\vec{p)}$$
where
$$\psi(\vec{p})=\left(\begin{matrix}1 \\ 0 \\ c\frac{\vec{\sigma}\cdot\vec{p}}{E_p+mc^2} \\ 0 \\ \end{matrix}\right)$$
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?
$$\langle \hat{\beta} \rangle = \frac{1}{(2\pi \hbar)^3 V}\displaystyle\int d^3r \left(1-\frac{c^2(\vec{\sigma}\cdot\vec{p})^2}{(E_p+mc^2)^2}\right) $$
$$=\frac{1}{(2\pi\hbar)^3}\left(1-\frac{c^2(\vec{\sigma}\cdot\vec{p})^2}{(E_p+mc^2)^2}\right)$$
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 $\gamma^{-1}$.
