Why do ##t## and ##-i\hbar\partial_t## not satisfy the definition of a linear map/operator in Hilbert space?

[Dr_Nate]

TL;DR Summary

It is common to say that $t$ and $-i\hbar\partial_t$ are not operators in quantum mechanics. But an unambiguous mathematical justification seems lacking.

It is common to say that $t$ and $-i\hbar\partial_t$ are not operators in quantum mechanics. But I haven't seen a satisfying justification.

As an example of the precision of our discourse, someone has said that $-i\hbar\partial_t$ satisfies the definition of Hermicity, but it is not an operator in quantum mechanics. That seems wrong to me because Hermicity requires the above expression to be a linear map/operator.

Alternatively, some say that time is a parameter and not a variable in Hilbert space, so it can't be an operator. However, when I look at the definition of a linear map, I don't see the words \emph{parameter} or \emph{variable} used, so there seems to be a gap in the justification.

Interestingly, no one directly answered this post in that same thread linked above.

Applying $t$ or $-i\hbar\partial_t$ to kets, I don't see a case where additivity and scalar multiplication are not preserved. I don't see how they violate the requirements of a linear mapping back to Hilbert space. Note, though, that I'm an experimentalist, so I can't tell you the difference between a vector space and a field in these definitions (wikipedia link).

Question: How does the application of $t$ and $-i\hbar\partial_t$ to a ket not satisfy the mathematical definition of a linear map/operator in Hilbert space as used in the mathematical formalism of quantum mechanics?

 

[PeterDonis]

no one directly answered this post in that same thread linked above.

Yes, there was an answer given, in post #20 of that thread--which was before the one you linked to here (post #30).

 

[Demystifier]

The Hilbert space in QM is the space of functions of $x$, not a space of functions of $x$ and $t$. The time derivative operator is therefore not an operator in this Hilbert space. The functions in the Hilbert space must have a finite norm, if they were functions of both $x$ and $t$ then the computation of norm would involve integration over both $x$ and $t$, but the integral over $t$ would not converge because the wave functions satisfying Schrodinger equation do not vanish at $t\to\pm\infty$. Of course, the wave functions do depend on both $x$ and $t$, but only the dependence on $x$ is associated with a Hilbert space structure. The dependence on $t$ is also associated with some space of functions, but this space is not a Hilbert space, because the integration over $t$ does not lead to a finite norm.

 

[Dr_Nate]

Yes, there was an answer given, in post #20 of that thread--which was before the one you linked to here (post #30).

The question I am asking is pretty much about straight math. That answer invokes observables, so isn't what I am looking.

 

[Dr_Nate]

The Hilbert space in QM is the space of functions of $x$, not a space of functions of $x$ and $t$. The time derivative operator is therefore not an operator in this Hilbert space. The functions in the Hilbert space must have a finite norm, if they were functions of both $x$ and $t$ then the computation of norm would involve integration over both $x$ and $t$, but the integral over $t$ would not converge because the wave functions satisfying Schrodinger equation do not vanish at $t\to\pm\infty$. Of course, the wave functions do depend on both $x$ and $t$, but only the dependence on $x$ is associated with a Hilbert space structure. The dependence on $t$ is also associated with some space of functions, but this space is not a Hilbert space, because the integration over $t$ does not lead to a finite norm.

Thank you. This is the answer I am looking for. If I recall correctly something like this was said in the linked thread, but now it is clear to me.

I have read many lists of postulates and have never noticed before in those lists that references to Hilbert space were at fixed time. Now, that I know, I can see it's clear on Wikipedia. I'm curious to see how often it's clearly stated in other more formal places.