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As is well known, the Heisenberg uncertainty principle states that the position and momentum satisfy an uncertainty relation, which follows from the canonical commutation relation \begin{equation} [\hat{x}_i,\hat{p}_j]=i\hbar\delta_{ij}. \end{equation} There is also the well known energy-time uncertainty principle, with the canonical commutation relation \begin{equation} \Big[\hat{H},\hat{t}\Big]= i\hbar. \end{equation} This is not as well defined, however, because the time operator is not an operator on the Hilbert space, even though this formally follows from the Schrodinger equation. It is known that the energy-time uncertainty principle is not as simple as the position momentum one, but is there a natural operator that does satisfy canonical commutation relations with the Hamiltonian? I am specifically interested in second quantized Hamiltonians, where the Hamiltonian is written in terms of creation and annihilation operators.

From wikipedia, it says that canonically conjugate variables are fourier transforms of one another. What would that mean in this context? Does that generalize to the Hamiltonian? Is there an operator that is the "Fourier transform" of the Hamiltonian.

Qmechanic
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Teddy Baker
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  • Related: http://physics.stackexchange.com/q/6584/2451 , https://physics.stackexchange.com/q/220697/2451 and links therein. – Qmechanic Dec 28 '17 at 23:28

2 Answers2

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One of the true misconceptions people learning QM - perhaps beyond an introductory textbook - encounter is that "According to Pauli's theorem, there is no way one can build a self-adjoint operator conjugate to the Hamiltonian in the same way the momentum operator is conjugate to the coordinate operator". Loosely speaking "there is no time operator in QM". One can even support this claim by quoting Pauli's original argument (p. 7 from here: https://arxiv.org/pdf/quant-ph/0609163.pdf - but the reader be warned, this article has some blatant errors in it, such as the square root of the delta distribution on page 3!) which goes back to the beginning of the 1930s at the same time when Johann von Neumann was writing in German about QM in a fresh mathematically acceptable language.

It is no surprise then that the original Pauli's argument, so easily brought forward in a lot of topics on PhysSE, was not written in the language of Hilbert space functional analysis. This had just been invented by Marshall Harvey Stone in the US and Johann von Neumann in Germany. [As a side remark, I cannot speculate whether (the otherwise great physicist) Pauli did eventually become fluent in von Neumann's functional analysis up to his death in 1958]. Actually, the topic itself of a time operator in Quantum Physics (hence Pauli's analysis) was not a popular one (or seemed to be forgotten), until the advent of operator-based string theory and its endless possibilities.

The only mathematical scrutiny of Pauli's theorem and its shortcomings I recommend is the one by Eric Galapon here: https://arxiv.org/abs/quant-ph/9908033 and some further comments here: https://arxiv.org/abs/quant-ph/0303106. Galapon shows that, under special conditions, there is an operator canonically conjugate to an acceptable self-adjoint Hamiltonian (the time of arrival operator). The overall answer to the question in the title is, therefore "yes". Neglecting this result and still pushing forward Pauli's argument and conclusion are wrong. So, at the end of the day, all textbooks on QM ignoring Pauli's argument were/are right and do not spread this aforementioned misconception.

DanielC
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  • But those are for particular Hamiltonians, right? Pauli's argument does hold for bounded Hamiltonians? – Teddy Baker Jan 18 '18 at 15:27
  • There is no requirement for a Hamiltonian to be bounded as an operator in QM/QFT, but only that its spectrum be bounded from below. – DanielC Jan 18 '18 at 18:09
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Probably not due to Pauli's theorem. I only say "probably" because I don't know how splitting the role of Hamiltonian and energy (the former defining the time translation operator, the latter the operator with a lower bound that defines the ground state) would affect Pauli's theorem.

Sean E. Lake
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  • Thank you, that is an interesting point. I was not familiar with Pauli's theorem. I will do some research on this. If we are in a many-particle system where the energy eigenvalues are semi-continuous, might there be an operator that is "close" to what I am looking for? – Teddy Baker Dec 28 '17 at 23:25
  • @TeddyBaker The problem isn't the continuity of the spectrum of the Hamiltonian, the problem is the lower bound on its spectrum. – Sean E. Lake Dec 28 '17 at 23:26
  • Ok, I marked it correct, even though maybe slightly dissapointed :) – Teddy Baker Dec 29 '17 at 18:34
  • Actually, also I'm wondering why the time operator necessarily has to be self adjoint. That is an assumption of the theorem, but what if it was relaxed? – Teddy Baker Dec 29 '17 at 19:00
  • I'm guessing that the eigenvalues have to be real for the operator to track time. Although imaginary time is interesting for thermodynamics type calculations. – Teddy Baker Dec 29 '17 at 19:06
  • @TeddyBaker the Hermitian requirement is imposed because there is a requirement in QM that all observable correspond to Hermitian operators (though not all Hermitian operators are observables - consider $m\omega X + P$ in a simple harmonic oscillator, for example). Since time is most definitely an observable quantity, we have the conundrum. I do wonder, right now, whether the problem vanishes if time is also bounded from below (as in some pictures of the big bang). – Sean E. Lake Dec 29 '17 at 19:08
  • @TeddyBaker The difference between Hermitian and anti-Hermitian is a factor of $i$, so it would be a trivial change to the commutation relation that I'm pretty sure wouldn't alter the results of the theorem. – Sean E. Lake Dec 29 '17 at 19:09
  • Let us continue this discussion in chat. – Teddy Baker Dec 29 '17 at 19:16