6
$\begingroup$

There is a question in SE about the fact that the Hamiltonian isn't invariant under the EM gauge transformations. I wanted to ask about its consequences here.

I know that in general, the Hamiltonian and wave functions would transform under the gauge transformation in such a way that the Schrödinger equation remains true. Specifically, if under a gauge transformation $H$ goes to $\tilde{H}$ and $\psi$ goes to $\tilde{\psi}$, We still have: $$ \tilde{H} \tilde{\psi}=i\hbar\partial_t\tilde{\psi}$$

Also, I know under a time-independent gauge transformation, if $\psi$ is a eigenstate of Hamiltonian with energy $E$, the $\tilde{\psi}$ would be the eigenstate of $\tilde{H}$ with the same energy. So the energy spectrum of Hamiltonian is invariant under the time-independent gauge transformations. But what confuses me, are the time-dependent gauge transformations. It can be seen that under the gauge transformation $$(\phi,\mathbf{A})\to(\phi+\partial_t \lambda,\mathbf{A}-\nabla \lambda)\qquad (1),$$ the eigenvalue equation for the Hamiltonian transforms to: $$\tilde{H}\tilde{\psi}=E\tilde{\psi}+\partial_t\lambda(x,t) \tilde{\psi}\qquad (2)$$ The derivation of this relation is appended at the end of the question. So as it can be seen $\tilde{\psi}$ isn't anymore the eigenstate of the Hamiltonian and it seems that the energy spectrum would also change. I mean that even the energy differences would change under this gauge transformation.

This somehow looks odd to me. Take for example the hydrogen atom. On one hand, we know that the $13.6$ eV can be measured in experiment, or the energy differences between various level can be observed with spectroscopy techniques, but on the other hand they are somehow dependent on the gauge we choose. Take into account that gauge transformations doesn't represents a physical change in the system like changing the observer, but it's just a reformulation of the same problem. It seems that even the orbital shapes would change under time-dependent gauge transformations.

Is the above reasoning correct? Is energy deference an observable physical quantity or not? or maybe only the eigenvalues of time-independent Hamiltonians represents the energy levels of the system and we should look at time-dependent Hamiltonians as just the time evolution generator and do not relate its spectrum to energies of the system. Can someone clarify the situation?


Proof of Eq. 2:(I have set $\hbar=e=1$ for simplicity)

Suppose that $H=\frac{(p-A)^2}{2m}+\phi$ is the systems Hamiltonian in the original gauge and $\psi$ is one of its eigenvectors with energy $E$, so in the position representation: $$H\psi=[\frac{(-i\nabla-A)^2}{2m}+\phi]\psi(x)=E\psi(x),\qquad (3)$$ Now consider the gauge transformation which is given by Eq. (1), under which the Hamiltonian transforms to: $$\tilde H=\frac{(p-A+\nabla \lambda)^2}{2m}+\phi+\partial_t \lambda$$ and also the $\psi$ wave function would go to: $$\tilde \psi(x,t)=\exp{[-i\lambda(x,t)]}\psi(x,t)$$ It is easy to see that: $$(-i\nabla-A+\nabla \lambda)\exp{[-i\lambda(x,t)]}\psi(x,t)=\exp{[-i\lambda(x,t)]}(-i\nabla-A)\psi(x,t),\qquad (4)$$ If we operate the $\tilde H$ on the $\tilde \psi$ form left and use Eq. 4 twice: $$\tilde H \tilde \psi =\exp[-i\lambda(x,t)]\Big(\frac{(-i\nabla-A)^2}{2m}+\phi\Big)\psi(x,t) +\partial_t \lambda\exp[-i\lambda(x,t)]\psi(x,t)$$ but the expression in the big parentheses is just the original Hamiltonian $H$ and according to Eq. 3, it can be replace by $E$, So: $$\tilde H\tilde \psi=E\tilde\psi +(\partial_t\lambda)\tilde\psi $$ which completes the proof.

$\endgroup$
  • 2
    $\begingroup$ Don't forget that the wave-function has a phase dependent in time, so the time derivative compensates the shift in scalar potential. $\endgroup$ – Adam Jul 24 '16 at 9:39
  • $\begingroup$ Sure, As I said I understand that the Schrodinger Equation is still correct, My question is about the eigenvalues of Hamiltonian. And it isn't just a shift. It is a function of space. $\endgroup$ – seyed Jul 24 '16 at 17:58
  • $\begingroup$ Not sure why you're so sure, since your second SE is not correct... $\endgroup$ – Adam Jul 25 '16 at 7:11
  • $\begingroup$ It isn't SE, but only the action of new Hamiltonian on the new wave function. Some QM books call it the time-independent SE. I edited the question to clarify the issue and also added its proof at the end of the question. If you still think it is not correct please let me know. $\endgroup$ – seyed Jul 26 '16 at 7:18
6
$\begingroup$

The time-independent Schrodinger equation $\hat{H} \psi = E \psi$ only holds when the Hamiltonian does not depend explicitly on time. If you start with a time-independent Hamiltonian and make a time-dependent gauge transformation, then the new Hamiltonian will depend explicitly on time, and there is no reason to expect that the (time-dependent) eigenvalues of $\tilde{H}$ will correspond to any kind of physical energy.

$\endgroup$
  • $\begingroup$ OK, now I understand the point, If I understood correctly, this is a question in classical mechanics more than quantum mechanics. I will add my thoughts bellow for the future use. $\endgroup$ – seyed Jul 29 '16 at 15:36
1
$\begingroup$

The gauge transformed Hamiltonian doesn't represent the energy of the system in classical mechanics and therefore the quantized version of it doesn't represent the energy operator and its eigenvalues doesn't represent the energy levels despite the fact that it is still the time evolution generator by construction.

If we are interested in the systems energy we should find the classical representation of energy in terms of transformed gauge fields and quantize it to find the energy operator, say $\hat{E}$ then it is easy to see that the gauge transformed wave function is the eigenstate of this operator with the old $E$ energy:$$\hat E \tilde \psi=E\tilde \psi$$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.