The wave function of a free Dirac particle moving with momentum $\vec{p}=(p,0,0)^T$ is given in the rest frame and the laboratory frame as

$$\Psi_r=N_r\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)e^{-\frac{i}{\hbar}mc^2t}\qquad\Psi_l = N_l\sqrt{\frac{E+mc^2}{2mc^2}}\left(\begin{array}{c}1\\0\\0\\\frac{pc}{E+mc^2}\end{array}\right)e^{-ip_\mu x^\mu/\hbar},$$

where $N_r$ and $N_l$ are normalization constants which are

$$N_i =\left\{\begin{array}{l}1/\sqrt{V},\quad i=r\\\sqrt{\frac{mc^2}{EV}},\quad i=l\end{array}\right.$$

and which I calculated using $\int\mathop{d^3x}\Psi_i^\dagger\Psi_i=1$.

Now, I would like to calculate the spin expectation values $\left<\sigma_z\right> = \frac{1}{N}\int\mathop{d^3x}\Psi^\dagger\Sigma_z\Psi$ in the rest and the laboratory frame, where $\Sigma_z = \frac{\hbar}{2}\left(\begin{array}{cc}\sigma_z & 0 \\ 0 & \sigma_z\end{array}\right)$ is the spin operator and $\sigma_z$ the third Pauli matrix. For the rest frame, I get

$$\left<\sigma_z\right>_r = \frac{\hbar\sqrt{V}}{2}\int\mathop{d^3x}(1,0,0,0)e^{i mc^2t/\hbar}\left(\begin{array}{cc}\sigma_z & 0 \\ 0 & \sigma_z\end{array}\right)\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)e^{-imc^2t/\hbar}=\frac{V^{3/2}\hbar}{2}$$

and analogously in the lab frame

$$\left<\sigma_z\right>_l = \frac{\hbar}{2}\sqrt{\frac{E}{mc^2}}V^{3/2}\frac{mc^2}{E+mc^2}.$$

I'm quite sure that these results cannot be true. Especially in the rest frame, I would expect an expectation value which is independent of the volume $V$. So it would be nice if someone could tell me how to do the calculation correctly.

  • $\begingroup$ I think there might be some ambiguity with what you call „N“. One the one hand every Psi has its own N, but then in the normalization there should be a N^2 on the left-hand side. Similarly, I think there should be a N^2 in the definition of <sigma>. $\endgroup$
    – ersbygre1
    Nov 16, 2017 at 11:17
  • $\begingroup$ @Stephan I edited my post because the line with the normalization constants was nonsense, as you said. In the definition of $<\sigma>$, you mean that it should be $1/N^2$ instead of $1/N$? But then the volumes don't vanish from the results. The definition in my post was taken from an exercise sheet I found on the internet. $\endgroup$
    – MeMeansMe
    Nov 16, 2017 at 11:35

1 Answer 1


Short answer: I believe your "$N$" in $\langle\sigma_z\rangle=\ldots$ refers to the normalization of the spinors, which is (since you've already normalized them) $N=1$.

Long answer: An expectation value is defined as $$ \langle A \rangle = \frac{\langle \Psi |\hat A|\Psi\rangle}{\langle \Psi|\Psi\rangle}.$$ This means, for the spin in $z$-direction, we need to calculate $\langle \Psi |\hat \Sigma_z|\Psi\rangle$ and $\langle \Psi|\Psi\rangle$. Let's do the second one first: \begin{align*} \langle \Psi|\Psi\rangle &= \int\limits_V \text d^3x \Psi^\dagger \Psi\\ &= \frac{1}{V}\int\limits_V \text d^3x \,(1,0,0,0)\begin{pmatrix}1\\0\\0\\0\end{pmatrix}\text e^{-\text i \frac{mc^2t}{\hbar}}e^{+\text i \frac{mc^2t}{\hbar}}\\ &= \frac{1}{V}\int\limits_V \text d^3x \, \cdot 1 = 1. \end{align*} Now onto the first term: \begin{align*} \langle \Psi |\hat \Sigma_z|\Psi\rangle &= \int\limits_V \text d^3x \Psi^\dagger \hat\Sigma_z\Psi\\ &= \frac{1}{V}\frac{\hbar}{2} \int\limits_V \text d^3x \, \underbrace{(1,0,0,0)\begin{pmatrix}1&&&\\&-1&&\\&&1&\\&&&-1\end{pmatrix}\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}_{1}\,\cdot 1\\ &= \frac{1}{V}\frac{\hbar}{2} \int\limits_V \text d^3x \,\cdot 1\\ &= \frac{V}{V}\frac{\hbar}{2} = \frac{\hbar}{2} \end{align*}

So for a positive energy, spin-up spinor, we get the right result, independent of the volume size.

  • $\begingroup$ This makes sense, thank you. In the lab frame, I get $\langle\sigma_z\rangle = \frac{\hbar (mc^2)^2}{E^2+Emc^2}$. Do you think this is right? $\endgroup$
    – MeMeansMe
    Nov 16, 2017 at 14:16
  • $\begingroup$ I get $\langle\Sigma_z\rangle = \frac{\hbar}{2}\frac{mc^2}{E}$. $\endgroup$
    – ersbygre1
    Nov 16, 2017 at 15:24
  • $\begingroup$ You're right, I calculated it again from the very beginning and now it agrees with your result. $\endgroup$
    – MeMeansMe
    Nov 16, 2017 at 16:47

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