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I'm working on an exercise and I'm getting quite stuck.

We define $\sigma$ as the vector of Pauli matrices.The Hamiltonian is formulated as: $ H_1 = -\frac{\hbar^2 k^2}{2 m} + \alpha \left(\sigma \cdot \nabla\right)$

Now, the latter seems to be a weird term that should read $\left(\sigma \cdot \nabla\right) = \nabla \cdot \sigma + \sigma \cdot \nabla$, where the first $\nabla$ is then acting towards the left.

The problem is formulated as

"Consider an electron gas with the given Hamiltonian, where $\alpha$ parameterizes a >model spin-orbit interaction. Compute the eigenvalues and eigenvectors as a function of >wave function $k$ and plot them in the $x$-direction. Interpret the results."

I'm quite confused about this, as $<k'| H_1 |k>$ only has nonzero off-diagonal terms when $k\neq k'$.

At first, I didn't think this $\delta_{k, k'}$ issue was that big of a problem. However, it seems quite weird. When $k=k'$, you just find the regular two options for spin, because $H_{k,k'} = \epsilon_k \delta_{k,k'}$. Actually, there's no reason to assume spin even exists (in the model) at this point, although it is viable to multiply by $I_2$.

But as soon as you go for $k'\neq k$, you get these weird matrix elements that don't seem to fit the problem. Although I note that for $k=k'$, the elements are real and that for $k,k'$ it does seem to be the conjugate of $k',k$, which is good (the matrix is hermitian).

I was expecting an answer along the lines of 'We find eigenvalues $l_+$ and $l_-$ with eigenvectors $u, v$ such that $u\cdot v = 0$, where all of these are functions of $k$.

Thanks for your time!

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  • $\begingroup$ I'm pretty sure it is $\sigma\cdot p$, in other words, $\alpha$ is pure imaginary. $\endgroup$ – Meng Cheng Aug 24 '15 at 23:59
  • $\begingroup$ This is one of the options that presented itself to me, but the professor hinted that it had something to do with the $\nabla \cdot \sigma$ term acting towards the left. $\endgroup$ – Daimonie Aug 25 '15 at 0:11
  • $\begingroup$ Well, the Schrodinger equation is $H\psi=E\psi$, so there is nothing on the left to act towards. $\endgroup$ – Meng Cheng Aug 25 '15 at 0:53
  • $\begingroup$ Then $\alpha = {\rm i} |\alpha|$ seems to be the only option. $\endgroup$ – Daimonie Aug 25 '15 at 6:27
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Maybe you should take into account that $(\sigma_i p_i)^2=p_i p_i$.

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  • $\begingroup$ It might be because it's late here, but I do not see how this applies. Could you explain? $\endgroup$ – Daimonie Aug 25 '15 at 0:41
  • $\begingroup$ @Daimonie: eigenvectors of $\sigma_i p_i$ are also eigenvectors of the Hamiltonian, as far as I can see. $\endgroup$ – akhmeteli Aug 25 '15 at 0:50
  • $\begingroup$ They should be. The question formulation basically asks to look for $<r|k> = e^{i \underline{k}\cdot \underline{r}}$, then solve the matrix system to find the spin-vectors. However, that's the exact problem; the matrix doesn't seem to be hermitian and the professor's hint doesn't seem to make sense. $\endgroup$ – Daimonie Aug 25 '15 at 6:26
  • $\begingroup$ @Daimonie: Maybe you should not bother about hermiticity right now, just find the eigenvectors and the eigenvalues. $p_j$, $\sigma_i p_i$ and $H$ all commute, so look for common eigenvectors of all these operators. For example, can you find eigenvectors of $\sigma_3 p_3$ (no summation)? $\endgroup$ – akhmeteli Aug 25 '15 at 8:19
  • $\begingroup$ Those should just be $<x|k \sigma> = e^{{\rm i} k_x \cdot x} |\chi>$ with $\left\{|\chi>\right\} = \left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \end{pmatrix}\right\}$, right? (And e.v. $\pm k_x {\rm i}$ ?). $\endgroup$ – Daimonie Aug 25 '15 at 10:58
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Just to close shop here. $H$ and $\sigma_i p_i$ do not commute, because $H$ does not necessarily include $\sigma_i p_i$, because $\alpha$ is free. Under the constraint that $\alpha$ is purely imaginary, the term does turn into $\sigma_i p_i$ and the commutation is valid.

So, both of you are right. $H$ and $\sigma \cdot p$ should commute, which they only do if $\alpha$ is purely imaginary.

The exact same thing follows from my concern that $H$ should be hermitian, but I didn't see it yesterday night when I made the post (and was extremely frustrated).

Thanks, both of you.

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  • $\begingroup$ I don't understand this. As far as I can see, $H$ and $\sigma_i p_i$ commute anyway, whether $H$ includes $\sigma_i p_i$ (if $\alpha$ is constant - imaginary or not). $\endgroup$ – akhmeteli Aug 30 '15 at 17:06
  • $\begingroup$ You're probably right. In retrospect this post seems somewhat nosensical. $\endgroup$ – Daimonie Sep 7 '15 at 14:31

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