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Note:

This question is the same as mine.

After having studied physics decades ago I recently reviewed some basics of symmetry in quantum physics. As a more experimental physicist I never understood deeply, what symmetries have to do with the Hamiltonian operator $H$.

For instance it is always stated, that $H$ is the generator of time evolution, so (I put it down simplified and also set $\hbar=1$)

$$| \psi(t+dt)\rangle = (1- iHdt)\, | \psi(t)\rangle $$

Similarly, $$\psi(x+dx)= \psi(x) + dx \cdot \frac{\partial }{\partial x} \psi(t) = (1- i p \cdot dx) \, \psi(t)$$

with $p$ as the momentum operator.

Why is there a symmetry, when $$[p, H]=0.$$

What has time evolution to do with a shift along the $x$-axis? And what is this symmetry about? I know, that $p$ is a conserved quantity in this case, but how is this derived from the first principles above?

To clarify my question a bit more:

I viewed Susskind's advanced QM lectures and cannot follow his physical arguments in the middle of the video. Unfortunately exactly at this time the video is interrupted and something is missing. He introduced the time evolution operator H and the x-shift operator V and by some "magic" he came to the conclusion, that under a "symmetry" they must commute (see image). My question is from what reasoning this was derived and what symmetry means here in particular?

enter image description here

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If an operator (for example $\hat{p}$) commutes with the Hamiltonian, then the eigenstates of $\hat{H}$ are also eigenstates of $\hat{p}$ (this is a mathematical statement that can be formally proved in the framework of linear algebra). Now consider an eigenstate of $\hat{p}$, $|\psi_p\rangle$ and consider its time evolution through the Schrodinger equation: $$ i \hbar \frac{\partial}{\partial t} |\psi_p\rangle = H |\psi_p\rangle. $$ Since $|\psi_p\rangle$ is an eigenstate of $\hat{H}$, when we apply the Hamiltonian we just get back the same state multiplied by a number, which is the energy $E_p$: $$ i \hbar \frac{\partial}{\partial t} |\psi_p\rangle = E_p |\psi_p\rangle $$ Solving this equation you see that the only possible time evolution for our state is through a phase factor, which does not affect the observables, namely $$ |\psi_p(t)\rangle = e^{i E_p t/\hbar} |\psi_p(0)\rangle. $$ In particular, momentum is conserved over time, because if we measure it, i.e. we take the expectation value $\langle \psi_p(t) | \hat{p} | \psi_p(t) \rangle$, we get: $$ p(t) = \langle \psi_p(t) | \hat{p} | \psi_p(t) \rangle = \langle \psi_p(0)| e^{-iE_pt/\hbar} p e^{iE_pt/\hbar} | \psi_p(0) \rangle = p \langle \psi_p(0) | \psi_p(0) \rangle = p, $$ which is constant.

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  • $\begingroup$ from the viewpoint of known formalism I understand this. The problem is always: do I understand it really or do I understand how to employ the recipe? Anyway, I'm still not aware what is this "symmetry" about. Yes, p is conserved, but what does "symmetry" mean here in particular? To better understand my question I added some details about where my question came from. $\endgroup$
    – MichaelW
    Commented Nov 20, 2023 at 8:15
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    $\begingroup$ @MichaelW Symmetry or rather symmetry transformations do not change the outcome of a measurement. You can apply them at any time you wish and they won't change the outcome. This connects symmetry with time and time evolution and implies that they commute with the Hamiltonian, since the Hamiltonian is the generator of time translation. $\endgroup$
    – Hans Wurst
    Commented Nov 20, 2023 at 8:40
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    $\begingroup$ As always, "symmetry" means that the Hamiltonian is left invariant by a given set of transformations. In the specific case of momentum, the transformations are spatial translations. You can recast $[\hat{H},\hat{p}]=0$ as $\hat{p}^{-1}\hat{H}\hat{p}=\hat{H}$, which clarifies this point: the transformation generated by $\hat{p}$ transforms $\hat{H}$ into itself, which is the usual definition of symmetry. $\endgroup$
    – Matteo
    Commented Nov 20, 2023 at 9:23
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    $\begingroup$ If you want to study more specifically the relation between translation invariance and momentum operator, I recommend this Wikipedia page: en.wikipedia.org/wiki/…. $\endgroup$
    – Matteo
    Commented Nov 20, 2023 at 9:28
  • $\begingroup$ @Matteo: great link :-) $\endgroup$
    – MichaelW
    Commented Nov 20, 2023 at 10:48

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