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Are they not supposed to be the same operator according to Schrödinger equation? $$ i\hbar\dfrac{\partial}{\partial t}\psi = H\left(\vec{r},-i\hbar\nabla,t\right)\psi $$

Apparently $[t,i\hbar\dfrac{\partial}{\partial t}] = i\hbar$, while $[t,H] = 0$

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    $\begingroup$ They are not the same operators. Eg. See this SE post. $\endgroup$
    – Physiker
    Jan 25 at 7:35
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    $\begingroup$ Does this answer your question? Why isn't the time-derivative considered an operator in quantum mechanics? And also time is no operator in standard QM (and QFT). $\endgroup$ Jan 25 at 8:35
  • $\begingroup$ I didn't know there was an entire rabbit hole of questions related to this one. They didn't show up on the browser when searching about commutation between t and H. I need to investigate more about relativistic theory since that's were the answer to my question should lie $\endgroup$
    – K. Pull
    Jan 25 at 8:58
  • $\begingroup$ It has nothing to do with relativistics. My comment is completely applicable in non-relativistic QM and is a consequence of the mathematical formalism. $\endgroup$ Jan 25 at 9:01
  • $\begingroup$ @K.Pull Those questions are certainly related to yours. But I think your observation is new and it makes perfect sense. $\endgroup$
    – lcv
    Jan 25 at 9:01

2 Answers 2

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The origin of this confusion stems from the fact that in (standard) quantum mechanics, time is actually not an operator. Rather, time is just a real variable.

The way time enters in QM is, for example in case of a time-independent Hamiltonian, such that the evolution operator is

$$ U(t) = e^{-i t H}.$$

So you see, we never compute the eigenvalues of "time" or things like that.

In QM as we know it now, there is an asymmetry between time and space. An asymmetry that, according to special relativity should not be there.

A lot of efforts are put nowadays in trying to level this asymmetry (quantum gravity, etc.) but this is another story.

Let me now come to the mathy details. Consider a situation where the Hilbert space of a particle is $L^2(\mathbb{R}^3)$. So here $\mathbb{R}^3$ is the allowed (configuration) space. So you see that

$$ \frac{\partial}{\partial t} $$

is not an operator in the Hilbert space. Of course you can consider $t$ as the multiplication operator by $t$ in some other space, and in this space acts also the time derivative. In this space $[t,\partial_t]=1$. But the Hamiltonian is an operator in another space and you cannot even consider the commutator between objects living in different spaces. The way you can make sense of your last equation

$$[t,H]$$

is to consider time simply as a variable. So effectively in the above $t$ stands for $t \mathrm{id}$ where $\mathrm{id}$ is the identity in the Hilbert space (where $H$) lives. Clearly then the result is zero.

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  • $\begingroup$ Could we extend the configuration space to $R^4$? $\endgroup$
    – K. Pull
    Jan 25 at 9:22
  • $\begingroup$ Here we are not trying to unify quantum mechanics with relativity and solve the deepest problem of current physics :). Having said that your observation is reasonable. $\endgroup$
    – lcv
    Jan 25 at 9:32
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Your question is surprisingly multi-faceted. @lcv did a good job addressing the main part. I wanted to touch upon one other aspect that I hope doesn't go unmissed.

Are they not supposed to be the same operator according to Schrödinger equation?

Definitely not.

An example can make this very clear: consider the spatial representation of the Schrodinger equation with no potential. It says

$$ i\hbar\frac{\partial}{\partial t} \psi = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\psi. $$

This equation is true, but from this do we say that the operators $i\hbar\frac{\partial}{\partial t}$ and $-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}$ are the same operator? I think you would agree the answer is very clearly no.

In fact, if the answer were yes, the equation would be completely meaningless, because it'd be tautological.

A non-trivial or meaningful equation takes two different operators $\hat{A}, \hat{B}$ and identifies a class $U$ of functions $u\in U$ for which applying $\hat{A}$ to $u$ and $\hat{B}$ to $u$ gives the same result $v$. We say that the functions for which this happens "solve the differential equation" and the functions for which this doesn't happen "do not solve the differential equation." The set of latter functions should be nonempty or else your equation isn't saying anything because everything solves it.


Regarding the main part of your question, I would challenge you to start with the Schrodinger equation

$$ i\hbar\frac{\partial}{\partial t} |\psi(t)\rangle = \hat{H}|\psi(t)\rangle $$

and try to derive a mathematical contradiction from the facts $[\hat{H}, t] = 0$ and $[\frac{\partial}{\partial t}, t]\ne 0$ (where $t$ is a scalar). For example, try multiplying both sides of the equation by $t$ and see where you go with it. I claim you won't be able to derive any contradiction if you are careful enough, but the attempts nonetheless will illuminate what is going.

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  • $\begingroup$ OK, I will write up the contradiction which I thought of while writing my question. Let $\psi = t\phi$. If $\psi$ is a Schrödinger solution, then $i\hbar(\phi+t\dot{\phi}) = Ht\phi$. Now, if $\phi$ is also a Schrödinger solution, one concludes that $\phi = 0$. Guess that I made a mistake when assuming that $\phi$ was also a Schrödinger solution? I thought for a moment that the Schrödinger equation gave the time evolution for any function whatever. $\endgroup$
    – K. Pull
    Jan 25 at 9:31
  • $\begingroup$ @K.Pull Right. If $\psi$ is a Schrodinger solution, $\phi = \psi/t$ is not necessarily a Schrodinger solution. It's just some other function you defined. $\endgroup$ Jan 25 at 9:41

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