Show for the non-relativistic quantum mechanical problem of $n$ electrons in a static homogenous magnetic field $\bf B$ and ignoring spin that the probability current density is gauge independent.

Solution for $n=1$

I would be happy with a derivation analogous to the 1 particle case, thus I start of with this (also to introduce my notations and units):

The (spin-free) $1$-electron probability current is defined as $$ j = {m_e}^{-1}\psi^* \hat\pi \psi$$ with the one-electron wave function $\psi$, the kinematic (mechanical) momentum operator $$\hat\pi = \hat p - \frac{q}{c}\bf{A}$$ with charge $q$ and magnetic vector potential $\bf A$ with $\nabla\times\bf A=\bf B$ and speed of light $c$.

Now its well known that a gauge transformation of the magnetic vector potential: \begin{align} {\bf A \to A ' = A + \nabla} f (\vec{r}) \end{align}

is compensated by a mere phase change of the wave function in the time dependent Schrödinger equation (read "is equivalent to")

\begin{align} \psi \to \psi ' = \psi e^{-i\frac{q}{c\hbar} f(\vec{r})} \end{align}

To show that $j$ is gauge independent one needs only to evaluate

$$j\to j'$$ under $\bf A\to A'$:

\begin{align} (j'/m_e) = & \psi'^* \hat\pi ' \psi'\\ = & (\psi e^{-i\frac{q}{c\hbar} f(\vec{r})})^* \{\hat p - \frac{q}{c}({\bf A}+\nabla f (\vec{r}))\} \psi e^{-i\frac{q}{c\hbar} f(\vec{r})} \\ & \dots \\ = & \psi^* \hat\pi \psi + (\frac{q}{c}\nabla f (\vec{r}))-\frac{q}{c}\nabla f (\vec{r})))\psi^*\psi \\ = & \psi^* \hat\pi \psi \\ = & j/m_e \end{align}

The $n>1$ electron case

In case of $n$-electrons the probability current density is defined as (an effective one-particle current density:

$$J=\frac{1}{m_e}\Re{\{\hat\pi P[\vec{r},\vec{s}]\}_{\vec{s}=\vec{r}}}$$

with the probability density matrix $$P[\vec{r};\vec{s}]=\int\Psi^*(\vec{r},\vec{r}_2,\dots,\vec{r}_n) \Psi(\vec{s},\vec{r}_2,\dots,\vec{r}_n)d\vec{r}_2 \dots d\vec{r}_n$$

(note: $J$ thus reduces for $n=1$ to $j$)

Where $\Psi$ is an n-electron wave function and the Hamiltonian of the system is $$\hat H = \frac{1}{2 m_e}\sum_{j=1}^n \big( \hat p_j - \frac{q}{c}{\bf A}(\vec{r}_j) \big)^2 + \hat V(\vec{r}_1,\vec{r}_2,\dots,\vec{r}_n)$$

which is not simply the sum of n one particle problems, due to the $V$ term, that might contain interelectronic repulsions such that $$ V(\vec{r}_1,\vec{r}_2,\dots,\vec{r}_n)\ne \sum_j V(\vec{r} _j). $$

My idea would be to look at the effect of the gauge transformation $\bf A\to A'$ on $P[\vec{r};\vec{s}]$ in the hope that simply

$$ P'[\vec{r};\vec{s}] = P[\vec{r};\vec{s}] e^{i\frac{q}{c\hbar} f(\vec{r})} e^{-i\frac{q}{c\hbar} f(\vec{s})} $$

would hold, but I fail to see if and how one could show that. The rest would be completely equivalent to the $1$-electron case.


1 Answer 1


Regardless of the specifics of the situation, if you have a transformation that acts as $\psi(r) \mapsto a(r)\psi(r)$ on one-particle wavefunctions for some function $a(r)$, then it acts as $\psi(r_1,\dots,r_n)\mapsto a(r_1)\dots a(r_n)\psi(r_1,\dots,r_n)$ on an $n$-particle wavefunction.

This is because the separable $n$-particle wavefunctions $\psi_\text{sep}(r_1,\dots,r_n) = \psi_1(r_1)\dots \psi_n(r_n)$ (i.e. products of 1-particle wavefunctions) form a basis of the space of $n$-particle wavefunctions. (Abstractly, the $n$-particle space is the $n$-fold tensor product of the 1-particle space, and tensor products are spanned by simple tensors rather by definition. Concretely, $L^2(\mathbb{R}^3n)\cong L^2(\mathbb{R}^3)^{\otimes n}$ as Hilbert spaces.)

It is straightforward to see that the separable functions transform in this way, so the transformation behaviour extends to all $n$-particle wavefunctions by linearity.

  • $\begingroup$ This is a great answer, much better than I had hoped was possible. I was thinking along those lines but I completely miss the math basis required to express it in that way. Thanks a lot! $\endgroup$ Commented Sep 16, 2019 at 17:32
  • $\begingroup$ Btw. would it change anything if antisymmetry w. r. t. electron permutation is required for all wave functions? $\endgroup$ Commented Sep 16, 2019 at 17:37
  • 1
    $\begingroup$ @Rudi_Birnbaum No, the symmetric and anti-symmetric wavefunctions are simply subspaces of the space of all n-particle wavefunctions (spanned by (anti-)symmetric sums of separable tensors), so the argument carries through the same way. $\endgroup$
    – ACuriousMind
    Commented Sep 16, 2019 at 17:41
  • $\begingroup$ Your argument thus yields: $$ \Psi'=\Psi \Pi_j e^{-i\frac{q}{c\hbar}f(\vec{r}_j)}$$ which gives with the intergration for $P'[r, s]$ exactly what I have conjured. $\endgroup$ Commented Sep 16, 2019 at 17:44

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