{abstract}
As you know, there are still problems with Quantum-Field theory in flat spacetime -- the renormalization of even Quantum Electrodynamics is not completed yet. One hopes to remove all divergences (in first place the ``cosmological constant problem") in the hypothetical Theory of Everything. However, one can try to make the Quantum Mechanics more self-consistent.
Indeed, the commutators are wrong in the very first line. But that is not all and it is further quite incorrect. For example, the uncertainty relation is a property that holds for arbitrary states. It is meaningless to use it, as you do, to limit the possible forms of the Hamiltonian.
"Indeed, the commutators are wrong in the very first line. "
How such sht could happen? I am respectful researcher with papers in Physical Review!
@Friends, do not rush with decisions, you are not Mr. Trump. :-)
Look on the first page the commutator [E, t] acts on the wave function psi. That is why is written [E, t] psi. And that is calculated, comes: [E, t] psi = (\hbar/2) psi.
Dmitri,
you wrote as commutators [Ê, t] and [Ĥ,t]. However, t is not an operator.
With kind regards,
Sofia
Dmitri> And that is calculated, comes: [E, t] psi = (\hbar/2) psi.
No, [iħ∂t, t] ψ = iħψ, as you must be able to calculate in your head.
I repeat, t is not an operator in QM, unless a time-operator is explicitly constructed.
I see one more wrong statement
"Then, momentum eigenvalue, which can be detected, is px = p̑xψ/ψ is function of x for general form of V(x)."
The value obtained for the momentum is one of the eigenvalues picked from the range of values of the momentum. This range is obtained by doing the Fourier transform of ψ, let's denote it by Fψ(px). The function |Fψ(px)|2 gives the probability of each value of px. So, |Fψ(px)|2 is a functional dependent on ψ, and thereby on V(x). In consequence, NOT the value obtained for px is a function of x or of V(x). What depends on V(x) is but the probability distribution, in particular, the probability that a value of px appears in a trial of the experiment.
In no case does the value of px depend on x. The operators of linear momentum and of position don't commute.
It's similar with the fact that the value of the spin detected along a direction a , does not depend on the value which would have been obtained if the spin were tested along a direction b. Ŝa and Ŝb don't commute.
@Dmitri Martila
"do not rush with decisions, you are not Mr. Trump"
Neither all here assume themselves as Mr. Trump...You asked a query and they are just answering your query...
Once you had written "How such sht could happen? ", you lose your respect...It does matter in which journal you have published your work...
"Once you had written "How such sht could happen? ", you lose your respect."
I have done no more wrong, than Trump is doing trying to mess in Syria government business.
@friends, why there can not be time operator \hat t = t? Show me couple simple formula, which contradict this choice. Do not pretend, what there is all perfect and shiny in Quantum Physics. Remember the huge prediction for vacuum energy density! The cosmological constant problem. That prediction is called "the most bad prediction in Science".
Once thought thinking works like arbitrary moments in sense - but every little thing influences time flow or time itself!
@Sofia: "I repeat, t is not an operator in QM, unless a time-operator is explicitly constructed."
Nevertheless, you can compute [iħ∂t, t] = iħ as Kåre Olaussen correctly remarks. But [E,t] =ħ /2 is wrong, if E is either the energy or the Hamiltonian. t is a number and commutes with any other number. I suppose it also commutes with H, because H is composed only from operators that do not act on time.
Ah, Klaus,
What is the connection between my statement (that you cite) and your comment in continuation. In my original country, Romania, there was a popular saying
"I ask you where is my sheep, and you answer me that the rain begins".
Dmitri asked the users - see his last comment - why cannot the variable t be also the time operator. I explained him, but in vain, that this is a wrong question. Here is the reason.
For testing whether the variable t can be also an operator, one should check whether this assumption does not contradict some experiment in QM. I can wish him a long life. Anyway, I doubt that somebody would volunteer to do this search for him.
- @Sofia @klaus @Dmitri / Hi, If the energy AND time operators are considered in an extended QPSR [Quantum Phase Space Representation] . For example: Eop = i hbar D(_ , t) +E/2 , Top = -i hbar D(_ , E) +T/2 , then considering these energy AND time operators, still [Eop,Top]= i hbar works, as can be verified by direct calculation. But this is quite expected, as the Physics of these extended phase space variables E and T , must be the same regardless of what representation the quantum system is framed in. So while various representations may more clearly show various mathematical connections to the physics, still the Physics is all the same, regardless of 'pictures' [i.e., representations]/
- Best to y'all/
- tony from Kentucky
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