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

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.

• Your conclusion is (almost) correct following your premises, but let me check if you intended to ask the question you did. By putting an $n$ label on $\vec{A}_n$, you are saying that you have as many different gauge fields as you have particles. In other words, each particle is coupling to a different field. This means that you aren't really talking about the electromagnetic field, which is the same for all charged particles. If you want to talk about the electromagnetic field, you should just have $\vec{A}$, and you have to simultaneously transform all particles with the same $\lambda(x)$. – Andrew Jul 18 at 0:25
• Sorry for not being clear. There is a single electromagnetic field and I am exploring the possibility whether I can choose different vector potentials $\vec A_n$ (all equivalent under gauge transformation) for different particles. – Abhijeet Melkani Jul 18 at 0:29
• In other words, let's say I choose some gauge and get the expression for $\vec A$. Now 1) if I alter the gauge is the total wave-function transformed by a factor of $~e^{i\lambda}$ or $~e^{iN\lambda}$ where $N$ is the number of particles. 2) Looking at the form of the Hamiltonian can I treat each $\vec A$ term that couples to a $p_n$ term differently? Can I choose the gauge for each of the $\vec A_n$ term differently? – Abhijeet Melkani Jul 18 at 0:35
• Ah ok. In that case the answer is no. You start with one $\vec{A}$ field, and if you do a gauge transformation, then $\vec{A}$ has to transform in the same way in each place it appears in the Hamiltonian. What you are saying amounts to having $N$ different gauge potentials, which is logically fine, but not the situation you are interested in. I'll turn this into an answer. – Andrew Jul 18 at 0:54

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

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

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

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

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.

• In the single vector potential case, when you write $e^{iq_n\lambda(x)/\hbar}$ as the transformation factor, what does $x$ refer to? The wavefunction is a function of the 3N coordinates $\vec r_1, ... \vec r_N$. – Abhijeet Melkani Jul 18 at 1:18
• I'm assuming we have $N$ distinguishable particles, so then I can write the state of each charged particle $\Psi_n$. Then $x$ is short hand for $\vec{r}_n$. – Andrew Jul 18 at 1:21
• Oh okay! And I guess the total factor for the indistinguishable case would be the product of all such exponential factors of one variable. Thanks. – Abhijeet Melkani Jul 18 at 1:23
• Right, you could form a Slater determinant to build the state for distinguishable fermions, for example, but I think it amounts to what you are saying. – Andrew Jul 18 at 1:25

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.