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
    Commented Jan 27, 2023 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
    Commented Jan 27, 2023 at 8:11
  • $\begingroup$ The Lie algebra involved is not representation-dependent. You failed to write it down. $\endgroup$ Commented Jan 27, 2023 at 16:39
  • $\begingroup$ Linked. $\endgroup$ Commented Jan 27, 2023 at 16:47
  • $\begingroup$ Does this help? Or this? $\endgroup$ Commented Jan 27, 2023 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
    Commented Jan 28, 2023 at 6:06
  • 1
    $\begingroup$ Dirac bra ket notation. $\endgroup$ Commented Jan 28, 2023 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. $\endgroup$ Commented Jan 28, 2023 at 15:13
  • $\begingroup$ The final relation amounts to the coordinate dependent one, (4.17), of Ballentine, through very careful changes of variables. $\endgroup$ Commented Jan 28, 2023 at 17:53

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