The most general form of Schrödinger equation is $$i \hbar \frac{d}{d t}\Psi(t) = H\Psi(t) \tag 1,$$ where $\psi(t)$ is an element of a Hilbert space $\mathcal H$ (not necessarily $L^2$), and $H$ is a self-adjoint operator.

Any kind of proof of the Galilean covariance of the Schrödinger equation I have seen so far, ψ is in the coordinate representation and the Hamiltonian operator is taken in its explicit form. Is there a way to deal with Galilean covariance of the Schrödinger equation in the general setting, without picking a specific representation?

  • 4
    $\begingroup$ The general form of the Schrödinger equation isn't guaranteed to be Galilean covariant unless explicitely stated $\endgroup$
    – Slereah
    Jan 27 at 8:01
  • $\begingroup$ OK. My question is whether we can formulate some criteria for the general Schrödinger equation to be Galilean covariant. $\endgroup$
    – mma
    Jan 27 at 8:11
  • $\begingroup$ The Lie algebra involved is not representation-dependent. You failed to write it down. $\endgroup$ Jan 27 at 16:39
  • $\begingroup$ Linked. $\endgroup$ Jan 27 at 16:47
  • $\begingroup$ Does this help? Or this? $\endgroup$ Jan 27 at 17:28

1 Answer 1


Hint: The links provided above address/settle your question, but, as a complementary setup, you might consider restricting to one dimension, starting from the general solution to the general TDSE, namely $$ |\Psi(t)\rangle = e^{-it\hat H/\hbar}|\Psi(0) \rangle\tag 2,$$ and never look back!

Just dot on the left with $\langle x|$, a bra, $\langle x|\Psi(t)\rangle=\psi(x,t)$ ; and monitor the braiding of exponential operators under the action of a Galilean transformation, $$ \hat G= \exp\left ({\frac{iv}{\hbar}(m\hat x-t\hat p)} \right )= e^{-imv^2t/2\hbar}e^{imv\hat x/\hbar}e^{-itv\hat p /\hbar } = e^{imv^2t/2\hbar}e^{-itv\hat p /\hbar } e^{imv\hat x/\hbar} \\ \implies \qquad \hat G^\dagger \hat x \hat G =\hat x-vt, \qquad\qquad\\ \hat G^\dagger \hat p \hat G =\hat p-mv,\\ \hat G^\dagger {\hat{ p} ^2 \over 2m} \hat G = {(\hat{ p}-mv )^2 \over 2m} ,\\ \hat G^\dagger V(\hat x) \hat G = V(\hat x -vt). $$

To get to the expressions you might have, defining $|\Psi'\rangle= \hat G^\dagger |\Psi\rangle$, you have, from above, $$ \psi'(x,t)=\langle x|\Psi'(t)\rangle=\langle x| \hat G^\dagger |\Psi(t)\rangle\\ = e^{-imvx/\hbar+imv^2t/2\hbar}\langle x+vt|\Psi(t)\rangle= e^{-imvx/\hbar+imv^2t/2\hbar} \psi(x+tv,t). $$ $\hat G^\dagger $ acted on the left, the bra. I have used $e^{tv\partial_x} \langle x| = \langle x+vt|$, of course. No fuss, no muss. It’s not the TDSE you care about, it’s its solutions!

  • $\begingroup$ I was stuck already at the beginning. What does $\langle x|$, $\langle x|\Psi(t)\rangle=\psi(x,t)$ mean? $\Psi(t)$ (or $|\Psi(t)\rangle$ if you want) is an element of a Hilbert space, which is not neceserily $L^2$. Then what is $x$? $\endgroup$
    – mma
    Jan 28 at 6:06
  • 1
    $\begingroup$ Dirac bra ket notation. $\endgroup$ Jan 28 at 8:49
  • $\begingroup$ Try the outstanding QM text by Sakurai & Napolitano. $|\Psi\rangle = \int\!\!dx ~~\psi(x) |x\rangle$, so the $\psi(x)$ are the coefficients of the $\cal H$ vector $|\Psi\rangle$ in the $|x\rangle$ basis/representation. $\endgroup$ Jan 28 at 15:13
  • $\begingroup$ The final relation amounts to the coordinate dependent one, (4.17), of Ballentine, through very careful changes of variables. $\endgroup$ Jan 28 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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