The Hamiltonian for an electron of mass $m$ and charge $e$ in an exterior electromagnetic field is $$H=\frac{1}{2m}(p-(e/c)A)^2+e\varphi.$$ The corresponding (via canonical quantization) quantum mechanical Hamiltonian is not invariant under Gauge transformations (for $A$ and $\varphi$). What's the physical meaning of this? And what is the physical meaning of the fact that the classical Hamiltonian is not invariant under gauge transformations?

  • 4
    $\begingroup$ The classical Hamiltonian is not gauge invariant, either. If I shift $\phi$ by 5 V, then Hamiltonian shifts by 5 eV. Perhaps it's better to ask why Hamiltonians in general (classical/quantum) are gauge variant? $\endgroup$
    – Nanite
    Jan 21, 2014 at 21:40
  • $\begingroup$ I agree on this $\endgroup$
    – user37252
    Jan 21, 2014 at 22:01
  • $\begingroup$ The classical Hamiltonian isn't gauge invariant. But the equations of motion are. On the quantum level, the Hamiltonian isn't gauge invariant either. But the Schodinger equaiton is invariant under a gauge transformation, provided you also change the wavefunction by a phase $e^{ie\Lambda / \hbar c }$. A google search shows some useful reference, like this. $\endgroup$
    – anecdote
    Jan 21, 2014 at 22:26
  • $\begingroup$ @anecdote: thanks for the reference, looks promising. but in the first three lines you are pointing out the obvious (=things which I already know): I was asking about the physical reason for it, not whether it is so or not $\endgroup$
    – user37252
    Jan 21, 2014 at 22:28
  • 1
    $\begingroup$ @KarlKäfer, the physical observables has to be gauge invariant, but in classical mechanics, the Lagrangian itself is not unique. We can change it by a total derivative without affecting the EOM. The problem arises when we are doing a path integral where the action is on the exponent e^{iS}. I've evaluated the path integral for this Hamiltonian for a loop before, and surprisingly, the gauge invariant condition leads to the quantization of flux. $\endgroup$
    – anecdote
    Jan 21, 2014 at 22:47

2 Answers 2


I will try to slightly elaborate on @VladimirKalitvianski answer.

From Maxwell's equations, we can derive that the following combination of gauge transformations on $\mathbf{A}$ and $\Phi$ leave both $\mathbf{B}$ and $\mathbf{E}$ invariant: \begin{align} & \mathbf{A}'=\mathbf{A}-\mathbf{\nabla} \alpha \\& \Phi'=\Phi+\frac{\partial \alpha}{\partial t} \end{align} where $\alpha=\alpha(\mathbf{x},t)$. This means that all the field configurations of $\mathbf{B}$ and $\mathbf{E}$ related by a gauge transformation are physically equivalent. Note that this has nothing to do with the Hamiltonian operator in QM.

Now in QM, we know that a wave function can always be multiplied by a phase factor: $$ \psi'=e^{-iq\alpha}\psi, $$ where $\alpha \neq \alpha(\mathbf{x},t)$, because the probability of finding the particle at a particular position is unaffected by the above transformation, and also the Schrodinger equation and the probability current are unaffected by the above transformation. If we now demand that the above also holds for when $\alpha = \alpha(\mathbf{x},t)$ (i.e. a gauge transformation), then the Schrodinger equation must be made gauge invariant: \begin{equation} i\frac{\partial\psi}{\partial t}=-\frac{1}{2m}(\mathbf{\nabla}-iq\mathbf{A})^2\psi+(V+q\Phi)\psi \end{equation} such that the Schrodinger equation is invariant under the simultaneous gauge transformations: \begin{align} &\mathbf{A}'=\mathbf{A}-\mathbf{\nabla} \alpha \\& \Phi'=\Phi+\frac{\partial \alpha}{\partial t} \\& \psi'=e^{-iq\alpha}\psi \tag{1} \end{align} Note that we can say that we have adjusted the "normal" Hamiltonian by replacing the ordinary (partial) derivatives by: \begin{equation} \begin{array}{cc} \displaystyle \mathbf{\nabla} \rightarrow \mathbf{D}\equiv \mathbf{\nabla} - i q \mathbf{A} \; ,& \displaystyle \frac{\partial}{\partial t} \rightarrow D^0 \equiv\frac{\partial}{\partial t} + iq \end{array} \end{equation} To sum up, by demanding that our theory is invariant under the gauge transformation expressed by equation $(1)$, we are forced to change the Hamiltonian operator as we have done above. However, by doing this, the new Hamiltonian describes a particle interacting with the potentials $\mathbf{A}$ and $\Phi$. If you're not convinced by this argument, I strongly recommend you to read up on the Aharonov–Bohm effect (http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect).

Furthermore, note that we require that a gauge transformation does not affect any observables. This means that we must demand that the probability current is also unaffected. You can show (although it is quite tedious) that the current is made gauge invariant by making the replacement: $\displaystyle \mathbf{\nabla} \rightarrow \mathbf{D}$.


What is physical meaning of gauge non-invariance of potentials $\mathbf{A}$ and $\phi$? Following their definitions, only their differences matter, not the absolute values, to be short.

The distance between two points is also invariant, but the liberty in choosing the frame reference position is translated into non invariance of a single point position.

  • $\begingroup$ I am not sure whether I understand your first paragraph: I was asking about the non-invariance of the Hamiltonian $H$. So you are basically saying $H$ itself doesn't matter, just differences? $\endgroup$
    – user37252
    Jan 21, 2014 at 23:24
  • 2
    $\begingroup$ Yes, the Hamiltonian serves for deriving equations, equations serve for obtaining solutions, solutions serve for composing observables. The latter are gauge invariant. $\endgroup$ Jan 22, 2014 at 9:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy