Multiple choices of gauge for a single many-body wavefunction in non-relativistic quantum mechanics

Question

It is known that in a single particle quantum mechanics problem with the Hamiltonian, $H = \frac{(\vec p-q\vec A)^2}{2m} + V(\vec r)$, one can perform the following gauge transformation:

$$\vec A \rightarrow \vec A' = \vec A + \vec \nabla \lambda(\vec r),$$

provided we also transform the wave-function, $\Psi(\vec r) \rightarrow \Psi'(\vec r) = e^{iq\lambda(\vec r)/\hbar} \Psi(\vec r) $, so that the Schr$\ddot o$dinger equation remains satisfied.

However, if we have multiple particles with the Hamiltonian being,

$$H = \sum \frac{(\vec p_n-q\vec A_n)^2}{2m} + V(\vec r_n),$$

is one allowed to choose the gauge for each term $\vec A_n$ differently for each particle?

Also, would the corresponding wave-function after gauge transformation look like $$\Psi'(\vec r_1, ..., \vec r_N) = e^{\frac{iq}{\hbar}\sum \lambda_n(\vec r_n) } \Psi(\vec r_1, ..., \vec r_N)~?$$

I tried to verify if this is true and it seems to be so. In that case, this is very strange since a single source of magnetic field will give rise to as many gauge choices as there are particles.

Answer 1

In electromagnetism there is one vector potential, $\vec{A}$. So the correct way to write the Hamiltonian is

\begin{equation} H = \sum_n \frac{(p_n-q_n \vec{A})^2}{2m_n} + V(\vec{r}_1, \vec{r}_2,...\vec{r}_N) \end{equation}

This Hamiltonian is invariant under gauge transformations of the form (apologies if I get a sign wrong) \begin{eqnarray} \Psi_n &\rightarrow& e^{i q_n \lambda(\vec{r}_n)/\hbar} \Psi_n \\ \vec{A} &\rightarrow& \vec{A} + \nabla \lambda \end{eqnarray}

ie, the vector potential transforms the same way each place it appears, and all fields transform with the same $\lambda(\vec{r}_n)$, with the exact phase factor determined by the charge $q_n$. It is not consistent to allow the same field ($\vec{A}$) to transform in different ways if it appears in multiple places in the Hamiltonian.

(Note: in an earlier version I wrote $x$ instead of $\vec{r}_n$ as the argument to $\lambda$, but changed this due to a comment)

Aside

Based on the comments, what follows from this point is not the situation the OP is interested in. But, for completeness, I'll note we can consider a situation with $N$ vector potentials (where $N$ is the number of particles). We can represent this situation with a Hamiltonian

\begin{equation} H = \sum_n \frac{(p_n-q_n \vec{A}_n)^2}{2m_n} + V(\vec{r}_1, \vec{r}_2,...\vec{r}_N) \end{equation}

which would be invariant under $N$ different gauge symmetries (the group is then $U(1)^N$). In this situation, we can transform each field separately. Let me use $j$ to label the fields, to try to make it obvious there is a difference with the situation above, in that here only two fields are transforming, $\Psi_j$ and $\vec{A}_j$, as opposed to $N+1$ in the case of electromagnetism. \begin{eqnarray} \Psi_j &\rightarrow& e^{i q_j \lambda_j(\vec{r}_j)/\hbar} \Psi_j \\ \vec{A}_j &\rightarrow& \vec{A}_j + \nabla \lambda_j \end{eqnarray} In this situation, since there are $N$ different vector potentials, it is consistent for them each to transform in a different way.

Answer 2

You have a single electromagnetic field described by single potentials $\vec{A}(\vec{r},t)$ and $V(\vec{r},t)$. These potentials extend over the whole space (all $\vec{r}$).

And you have $N$ particles represented by the wave function $\Psi(\vec{r}_1,\dots,\vec{r}_N,t)$.

Then the Hamiltonian is $$H=\sum_{n=1}^N\left (\frac{1}{2m}(\vec{p}_n-q\vec{A}(\vec{r}_n,t))^2 +qV(\vec{r}_n,t)\right).$$ Notice that you still have only one electromagnetic field ($\vec{A}$ and $V$). However, only the field values at positions $\vec{r}_n$ are relevant, because the electromagnetic interaction takes place only where the particles are.

Then Schrödinger's equation is $$\sum_{n=1}^N\left (\frac{1}{2m}(\vec{\nabla}_n-q\vec{A}(\vec{r}_n,t))^2 +qV(\vec{r}_n,t)\right)\Psi(\vec{r}_1,\dots,\vec{r}_N,t) =i\hbar\frac{\partial}{\partial t}\Psi(\vec{r}_1,\dots,\vec{r}_N,t)$$ where $\vec{\nabla}_n$ is with respect to $\vec{r}_n$.

It is easy to verify that this Schrödinger equation is invariant to the following gauge transformation: $$\begin{align} \vec{A}'(\vec{r},t)&=\vec{A}(\vec{r},t)+\vec{\nabla}\lambda(\vec{r},t) \\ V'(\vec{r},t)&=V(\vec{r},t)-\frac{\partial}{\partial t}\lambda(\vec{r},t) \\ \Psi'(\vec{r}_1,\dots,\vec{r}_N,t)&= e^{iq\lambda(\vec{r}_1,t)/\hbar}...e^{iq\lambda(\vec{r}_N,t)/\hbar} \Psi(\vec{r}_1,\dots,\vec{r}_N,t) \end{align}$$ Notice that you have one common function $\lambda$. Otherwise gauge invariance would not hold.