In non-relativistic QM, on one hand we have the following relations:
⟨x|P|ψ⟩ = −iℏ∂∂xψ(x),
⟨p|X|ψ⟩ = iℏ∂∂pψ(p).
On the other hand, despite the similarities, the relations cannot be directly applied to energy and time:
⟨t|H|ψ⟩ = iℏ∂∂tψ(t),
⟨E|T|ψ⟩ = −iℏ∂∂Eψ(E).
Just wondering, how can one mathematically prove that the "classical time" (which means no QFT or relativity involved), unlike its close relative "position", is not a Hermitian operator?
I ask your pardon if you feel the question clumsy or scattered. But to be honest, if I can clearly point out where the core issue of the problem is, I may have already answered it by myself :/
Answer
Time is not a variable in Quantum Mechanics (QM), it's a parameter — much in the same way as it is in Classical (Newtonian) Mechanics.
So, if you have a Hamiltonian, e.g., for the harmonic oscillator, you have ω as a parameter, as well as the masses of the particle(s) involved, say m, and you also have time — even though it's not something that shows up explicitly in the Hamiltonian (remember explicit time dependency from Classical Mechanics: Poisson Brackets, Canonical Transformations, etc — in fact, you could get your answer straight from these kinds of arguments).
In this sense, just like you don't have a 'transformation pair' between m and ω, you also don't have one between time and Energy.
What do you say to convince yourself that ω≠−i∂m? Why can't you use this same argument to justify E≠−i∂t? ;-)
I think Roger Penrose makes a nice illustration of how this whole framework works in his book The Road to Reality: A Complete Guide to the Laws of the Universe: check chapter 17.
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