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This is a seemingly odd question, but it's come up in a few different areas related to the supercurrent from GL theory. I am happy that one can vary wrt to the vector potential to option the equation for the supercurrent, but a lot of references and literature refer to the supercurrent in an imaginary expression: \begin{align} \vec{J}_s = \text{Im}\left( \psi^* (\nabla - i \vec{A}) \psi \right ) \end{align} This can also be written in terms of a real component, as is written at the bottom of the first page of this paper. The confusion comes mainly from this paper by Sadovskyy, where they switch between the common representation, where one can write something like \begin{align} \vec{J}_s = \frac{1}{2i}\left( \psi^* \nabla \psi - \psi \nabla \psi^* \right ) - |{\psi}|^2 \vec{A} \end{align}

What I don't understand is how these two are related? I'm sure it's just a simple mathematical trick, but I can't see it at the moment. Any help or guidance would be greatly appreciated. Thank you.

EDIT: The latter expression for the supercurrent may have differing factors between sources; this is just an example of a broader idea.

Closed: Thanks to all for the help. The key idea is remembering basic complex numbers, as in the ticked answer; the imaginary part of a complex number can be calculated as \begin{align} \text{Im}[z] = \frac{1}{2i} (z - z^*). \end{align} Hopefully this will help someone else having a brainfart!

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  • $\begingroup$ The last equation you wrote can be rearranged as $\vec{J}_{s}^{} = \frac{1}{2 i}\left[\psi_{}^{*}\left(\vec{\nabla} _{}^{} - i \vec{A} _{}^{}\right) \psi_{}^{} - \psi_{}^{}\left(\vec{\nabla} _{}^{}+ i \vec{A} _{}^{}\right) \psi_{}^{*}\right]$. $\endgroup$
    – Sunyam
    Commented May 1, 2020 at 8:07
  • $\begingroup$ @Sunyam thanks for the comment; I've tried to write it out, but I'm not sure how that helps. I'm currently having a play around with the phase of the order parameter and seeing if I can work something out that way. $\endgroup$
    – user262693
    Commented May 1, 2020 at 8:09
  • $\begingroup$ It works fine on my mobile app (you have to use render mathjax option). You dont need to split order parameter into phase and amplitude to see the equivalence, observe that second term in the above comment is the complex conjugate of first. $\endgroup$
    – Sunyam
    Commented May 1, 2020 at 8:12
  • $\begingroup$ It's taken me a shockingly long time to remember that $Im[z] = \frac{1}{2i} (z - z^*)$... I had that rearrangement but just forgot some 6th form calculus... A Google quickly helped that. Thanks for pointing that out! $\endgroup$
    – user262693
    Commented May 1, 2020 at 8:18

1 Answer 1

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Any imaginary number can be expressed as:

$$z=a+ib$$

Therefore, the imaginary part will be expressed as:

$$z-z^*=2i Im(z)$$

Thus, if we take what is inside Im as z, we can express:

$$Im(z)=\frac{1}{2i}(\psi^* \nabla \psi - i\psi^* \vec{A} \psi-\psi \nabla \psi^* - i\psi \vec{A} \psi^*)$$

And this is what you get in your last expression.

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