# Gauge-independence of the “$n$-particle” probability current

Problem

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.

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.
• 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. – Rudi_Birnbaum Sep 16 at 17:44