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Hossein
Hossein
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In the calculation of the so-called Caroli-de Gennes energy states in a type II superconductor the following Bogoliubov-de Gennes Hamiltonian is used: $$ H=- (\frac{1}{2m}(\nabla+ie\mathbf{A}\sigma_z)^2+\mu)\sigma_z+\Delta(r)e^{i\theta} \sigma_+ + \Delta(r)e^{-i\theta}\sigma_- $$ to solve the eigenvalue problem $H\psi=E\psi$ where $\psi=(u,v)^T$ is the wavefunction in particle-hole space and $\Delta(r)$ is real. Then, to get rid of the phase in front of order parameter a transformation of the form $\psi\rightarrow e^{-i\sigma_z\theta/2}\psi$ is applied and one will have the following transformed Hamiltonian: $$H=-(\frac{1}{2m}(\nabla+ie\mathbf{A}\sigma_z+i\sigma_z \frac{\hat{\theta}}{2r})^2+\mu)\sigma_z+\Delta(r)\sigma_x$$ Both the original paper (Caroli,de Gennes,Matricon, 1964) and de Gennes' book they proceed to neglect the $\mathbf{A}$ term since it is smaller than $\hat{\theta}$ term and they say that we can neglect all of the magnetic field effects and they solve this Hamiltonian: $$H=-(\frac{1}{2m}(\nabla+i\sigma_z \frac{\hat{\theta}}{2r})^2+\mu)\sigma_z+\Delta(r)\sigma_x$$

Now, my problem is that apparently we cannot neglect all of the effects of $\mathbf{A}$ as I describe below: The magnetic flux quantization condition says that: $$\oint d\varphi=-2e \oint \mathbf{A}.d\mathbf{r}$$ Where $\varphi$ is the phase of order parameter. At arbitrarily long distances from the vortex the magnetic field is zero and hence $\mathbf{A}$ is a pure gauge but this does not mean that it is zero. Using the flux condition we can see that for our vortex where $\varphi(\mathbf{r})=\theta$ we have: $$\mathbf{A}=-\frac{1}{2er}\hat{\theta}$$ And therefore the original Hamiltonian after the transformation $\psi\rightarrow e^{-i\sigma_z\theta/2}\psi$ takes the form: $$H=-(\frac{\nabla^2}{2m}+\mu)\sigma_z+\Delta(r)\sigma_x$$ which is much simpler than the one mentioned in the original paper. The problem now, is that without the $\hat{\theta}$ term the spectrum will be different if one follows the method which the authors use in the paper.

So, apparently I have made a mistake, but everything in my statement seems clear to me.

In the calculation of the so-called Caroli-de Gennes energy states in a type II superconductor the following Bogoliubov-de Gennes Hamiltonian is used: $$ H=- (\frac{1}{2m}(\nabla+ie\mathbf{A}\sigma_z)^2+\mu)\sigma_z+\Delta(r)e^{i\theta} \sigma_+ + \Delta(r)e^{-i\theta}\sigma_- $$ to solve the eigenvalue problem $H\psi=E\psi$ where $\psi=(u,v)^T$ is the wavefunction in particle-hole space and $\Delta(r)$ is real. Then, to get rid of the phase in front of order parameter transformation of the form $\psi\rightarrow e^{-i\sigma_z\theta/2}\psi$ is applied and one will have the following transformed Hamiltonian: $$H=-(\frac{1}{2m}(\nabla+ie\mathbf{A}\sigma_z+i\sigma_z \frac{\hat{\theta}}{2r})^2+\mu)\sigma_z+\Delta(r)\sigma_x$$ Both the original paper (Caroli,de Gennes,Matricon, 1964) and de Gennes' book they proceed to neglect the $\mathbf{A}$ term since it is smaller than $\hat{\theta}$ term and they say that we can neglect all of the magnetic field effects and they solve this Hamiltonian: $$H=-(\frac{1}{2m}(\nabla+i\sigma_z \frac{\hat{\theta}}{2r})^2+\mu)\sigma_z+\Delta(r)\sigma_x$$

Now, my problem is that apparently we cannot neglect all of the effects of $\mathbf{A}$ as I describe below: The magnetic flux quantization condition says that: $$\oint d\varphi=-2e \oint \mathbf{A}.d\mathbf{r}$$ Where $\varphi$ is the phase of order parameter. At arbitrarily long distances from the vortex the magnetic field is zero and hence $\mathbf{A}$ is a pure gauge but this does not mean that it is zero. Using the flux condition we can see that for our vortex where $\varphi(\mathbf{r})=\theta$ we have: $$\mathbf{A}=-\frac{1}{2er}\hat{\theta}$$ And therefore the original Hamiltonian after the transformation $\psi\rightarrow e^{-i\sigma_z\theta/2}\psi$ takes the form: $$H=-(\frac{\nabla^2}{2m}+\mu)\sigma_z+\Delta(r)\sigma_x$$ which is much simpler than the one mentioned in the original paper. The problem now, is that without the $\hat{\theta}$ term the spectrum will be different if one follows the method which the authors use in the paper.

So, apparently I have made a mistake, but everything in my statement seems clear to me.

In the calculation of the so-called Caroli-de Gennes energy states in a type II superconductor the following Bogoliubov-de Gennes Hamiltonian is used: $$ H=- (\frac{1}{2m}(\nabla+ie\mathbf{A}\sigma_z)^2+\mu)\sigma_z+\Delta(r)e^{i\theta} \sigma_+ + \Delta(r)e^{-i\theta}\sigma_- $$ to solve the eigenvalue problem $H\psi=E\psi$ where $\psi=(u,v)^T$ is the wavefunction in particle-hole space and $\Delta(r)$ is real. Then, to get rid of the phase in front of order parameter a transformation of the form $\psi\rightarrow e^{-i\sigma_z\theta/2}\psi$ is applied and one will have the following transformed Hamiltonian: $$H=-(\frac{1}{2m}(\nabla+ie\mathbf{A}\sigma_z+i\sigma_z \frac{\hat{\theta}}{2r})^2+\mu)\sigma_z+\Delta(r)\sigma_x$$ Both the original paper (Caroli,de Gennes,Matricon, 1964) and de Gennes' book proceed to neglect the $\mathbf{A}$ term since it is smaller than $\hat{\theta}$ term and they say that we can neglect all of the magnetic field effects and they solve this Hamiltonian: $$H=-(\frac{1}{2m}(\nabla+i\sigma_z \frac{\hat{\theta}}{2r})^2+\mu)\sigma_z+\Delta(r)\sigma_x$$

Now, my problem is that apparently we cannot neglect all of the effects of $\mathbf{A}$ as I describe below: The magnetic flux quantization condition says that: $$\oint d\varphi=-2e \oint \mathbf{A}.d\mathbf{r}$$ Where $\varphi$ is the phase of order parameter. At arbitrarily long distances from the vortex the magnetic field is zero and hence $\mathbf{A}$ is a pure gauge but this does not mean that it is zero. Using the flux condition we can see that for our vortex where $\varphi(\mathbf{r})=\theta$ we have: $$\mathbf{A}=-\frac{1}{2er}\hat{\theta}$$ And therefore the original Hamiltonian after the transformation $\psi\rightarrow e^{-i\sigma_z\theta/2}\psi$ takes the form: $$H=-(\frac{\nabla^2}{2m}+\mu)\sigma_z+\Delta(r)\sigma_x$$ which is much simpler than the one mentioned in the original paper. The problem now, is that without the $\hat{\theta}$ term the spectrum will be different if one follows the method which the authors use in the paper.

So, apparently I have made a mistake, but everything in my statement seems clear to me.

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Hossein
Hossein
  • 1.4k
  • 10
  • 21

The Energy Spectrum of Bound states in the Vortex Core of Type II Superconductors

In the calculation of the so-called Caroli-de Gennes energy states in a type II superconductor the following Bogoliubov-de Gennes Hamiltonian is used: $$ H=- (\frac{1}{2m}(\nabla+ie\mathbf{A}\sigma_z)^2+\mu)\sigma_z+\Delta(r)e^{i\theta} \sigma_+ + \Delta(r)e^{-i\theta}\sigma_- $$ to solve the eigenvalue problem $H\psi=E\psi$ where $\psi=(u,v)^T$ is the wavefunction in particle-hole space and $\Delta(r)$ is real. Then, to get rid of the phase in front of order parameter transformation of the form $\psi\rightarrow e^{-i\sigma_z\theta/2}\psi$ is applied and one will have the following transformed Hamiltonian: $$H=-(\frac{1}{2m}(\nabla+ie\mathbf{A}\sigma_z+i\sigma_z \frac{\hat{\theta}}{2r})^2+\mu)\sigma_z+\Delta(r)\sigma_x$$ Both the original paper (Caroli,de Gennes,Matricon, 1964) and de Gennes' book they proceed to neglect the $\mathbf{A}$ term since it is smaller than $\hat{\theta}$ term and they say that we can neglect all of the magnetic field effects and they solve this Hamiltonian: $$H=-(\frac{1}{2m}(\nabla+i\sigma_z \frac{\hat{\theta}}{2r})^2+\mu)\sigma_z+\Delta(r)\sigma_x$$

Now, my problem is that apparently we cannot neglect all of the effects of $\mathbf{A}$ as I describe below: The magnetic flux quantization condition says that: $$\oint d\varphi=-2e \oint \mathbf{A}.d\mathbf{r}$$ Where $\varphi$ is the phase of order parameter. At arbitrarily long distances from the vortex the magnetic field is zero and hence $\mathbf{A}$ is a pure gauge but this does not mean that it is zero. Using the flux condition we can see that for our vortex where $\varphi(\mathbf{r})=\theta$ we have: $$\mathbf{A}=-\frac{1}{2er}\hat{\theta}$$ And therefore the original Hamiltonian after the transformation $\psi\rightarrow e^{-i\sigma_z\theta/2}\psi$ takes the form: $$H=-(\frac{\nabla^2}{2m}+\mu)\sigma_z+\Delta(r)\sigma_x$$ which is much simpler than the one mentioned in the original paper. The problem now, is that without the $\hat{\theta}$ term the spectrum will be different if one follows the method which the authors use in the paper.

So, apparently I have made a mistake, but everything in my statement seems clear to me.