I have been treating this Hamiltonian:


where $\vec{\sigma}=(\sigma_x , \sigma_y)$. It is relevant for 2D graphene quantum dots with some vector potential $\vec{A}$, also 2D, applied inside the dot while outside the dot there is no vector potential. Now, I am stuck with an exercise that my tutor gave me:

I had solved the above Hamiltonian for the case that $\vec{A}=(A_x, 0)$ and basically found there are no states for $E=0$. He told me to now convince myself that the same should be true for an arbitrary vector potential in the $\sigma_{x,y}$-plane, namely


However, he does not want me to calculate the whole thing again but rather use rotational invariance. It seems logic to me that I previously really had done the calculation for a special case of the Hamiltonian considered now and that changing the choice of the direction of the vector potential should really not have any effect on the solution but I don't know how to approach it.

I guess I have to proof that $H'\psi=0 \Leftrightarrow H\psi=0$ somehow, $H'$ being the Hamiltonian I had treated and $H$ being the more general Hamiltonian.

Edit: Using the comment, I have got: $$H(r,\theta) \psi = \begin{pmatrix} 0 & e^{-i\theta} \\ e^{i\theta} & 0\\ \end{pmatrix} (vp_r + A_r) \begin{pmatrix} \psi_1 \\ \psi_2\\ \end{pmatrix} $$ whist $$H(r, 0)\psi = \begin{pmatrix} 0 & 1 \\ 1 & 0\\ \end{pmatrix} (vp_r + A_r) \begin{pmatrix} \psi_1 \\ \psi_2\\ \end{pmatrix}$$

I guess the interpretation is that the eigenvalues of the two Hamiltonians are the same while their eigenvectors differ by a phasefactor which means they ultimately represent the same states. However, how could I show this mathematically. I feel stupid but I cant figure out the final step in the calculus where the relation between the eigenvectors becomes obvious.

  • $\begingroup$ If you use $A\equiv A_r(\cos \theta, \sin \theta)$ and the same basis for $p\equiv p_r (\cos \theta, \sin \theta)$ it is straight forward to derive a relation between eigenvectors and eigenvalues of $H(r,\theta=0)$ and $H(r,\theta)$. Hint: remember Euler's formula. $\endgroup$ – M. J. Steil Feb 27 '17 at 17:36
  • $\begingroup$ Hi there, thanks for the comment, I used your hint to go on, as you can check out above. I am still not 100% sure about the interpretation of it though. Apart from that, I tried to derive the form of $p_r$ by writing $\vec{p}$ in polar coordinates but I just wasnt able to bring it to a form of $p=p_r (cos\theta, sin \theta)$. Any ideas how $p_r$ actually looks like? $\endgroup$ – Marsl Mar 2 '17 at 11:26
  • $\begingroup$ Well you do not have to guess about the eigensystems of those two Hamiltonians: you can (or in the case of (r,0) have) solved the eigenvalue problem. The hamiltonians are isospectral; they have the same eigenvalues $\lambda_\pm=\pm(v p_r +A_r)$. The eigenvectors are $\left\{(-1,1),(1,1)\right\}$ and $\left\{(-e^{-i\theta},1),(e^{-i\theta},1)\right\}$. So the only difference between the two systems is a complex phase in the x-direction. $\endgroup$ – M. J. Steil Mar 2 '17 at 13:42
  • $\begingroup$ Thanks, I get it now. However, I still struggle with $p_r$. If $\vec{p}=-i\hbar \nabla$, then it should look like this in polar coordinates: $\vec{p}=-i\hbar (\partial_r \hat{e}_r + \frac{1}{r} \partial_{\theta} \hat{e}_{\theta})$. The second term however is not of the form $ (\cos \theta, \sin \theta)$ and I wonder how one could even transfer $\vec{p}$ to that polar base. $\endgroup$ – Marsl Mar 2 '17 at 14:32

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