Let's say we have a system of $n$ stationary charges interacting via Coulomb potential. Let's ignore possible external electromagnetic fields. Moreover the system is quantum, and its wave function is $\psi(R_1, R_2, ...,R_n)$ where $R_k$ denotes the coordinate of charge $k$-th. The question is the following: given a gauge transformation with a phase $\alpha(R_1,R_2, ...,R_n)$, is it correct to write the wave function transformation as \begin{equation} \psi'(R_1, R_2, ...,R_n) = e^{iQ\alpha(R_1,R_2, ...,R_n)} \psi(R_1, R_2, ...,R_n) \, , \end{equation} where $Q$ is the total charge, i.e $Q=\sum_{i} Q_i$. How do you derive the total charge conservation then?

What about the following relation: \begin{equation} \psi'(R_1, R_2, ...,R_n) = e^{iQ_1g_1(R_1)} e^{iQ_2g_2(R_2)} \cdots \psi(R_1, R_2, ...,R_n) \, , \end{equation} where $g_l$ are some functions. I am thinking of phases which change at each charge position independently from each other. Can it be still considered a gauge transformation?

This case probably holds only insofar $\psi$ may be written as a product, i.e., $\psi_1(R_1)\psi_2(R_2)\cdots \psi_n(R_n)$. Does this imply that the charges are non-interacting at all? Please let me know what do you think.

  • $\begingroup$ How can something be stationary on quantum mechanics? $\endgroup$
    – Jun Seo-He
    Oct 21, 2021 at 12:09
  • 1
    $\begingroup$ @JunSeo-He A system of charges cannot be stationary even in classical mechanics: Earnshaw's theorem $\endgroup$ Oct 21, 2021 at 12:25
  • $\begingroup$ Even in the ground state at 0K a particle has zero point energy which exists in order for the particle not to violate the Heisenberg uncertainty principle. $\endgroup$
    – Jun Seo-He
    Oct 21, 2021 at 12:32

1 Answer 1


In quantum mechanics coupled to the electromagnetic field, a gauge transformation is a single function $g : \mathbb{R}^3\to\mathbb{R}$ and every one-particle wave-function $\psi(x)$ transforms as $\psi\mapsto\mathrm{e}^{\mathrm{i}qg(x)}\psi(x)$.

Since the $n$-particle space is just the (properly (anti-)symmetrized) tensor product of the 1-particle space, this already defines the action on an $n$-particle wavefunction as $$ \psi(x_1,\dots,x_n)\mapsto \prod_{i=1}^n\mathrm{e}^{\mathrm{i}q_i g(x_i)}\psi(x_1,\dots,x_n)$$ You cannot call any of the two transformation you have written down a gauge transformation, since you only defined the transformation of one particular state with one particular particle number. It does not extend to states with more or less particles in them, and is hence not a proper transformation on the space of state, and in particular not a gauge transformation.

  • $\begingroup$ Thank you very much for your reply. If I wish to write down the Hamiltonian of this system, this would be clearly not gauge invariant, isn't it? Meanwhile the relative Schrodinger should be gauge invariant as the physics does not change after a gauge transformation. Am I wrong? $\endgroup$
    – Caute
    Mar 31, 2016 at 17:05

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