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An argument by Pauli is usually referred to in the literature when it is stated that there cannot be a time operator in quantum mechanics. This argument can be found as a footnote to P63 of W. Pauli, The general principles of quantum mechanics, p. 63. Springer, Berlin 1980. The corresponding page is available from google books (look for P.63):

https://books.google.co.in/books?id=hVjsCAAAQBAJ&printsec=frontcover&dq=pauli+general+principles+of+quantum+mechanics&hl=en&sa=X&ved=0ahUKEwiOwKX1vvXNAhWJMI8KHSAODC0Q6AEIHDAA#v=onepage&q=discrete%20eigen%20values&f=false

But I do know cases, for example a particle on a circle or other cases with periodic boundary conditions, where momentum operator $P$ has only discrete eigen values. But then shouldn't Pauli's argument also say that there is no $X$ operator for such systems, since the commutation relation between $X$ and $P$ is the same as the commutation relation between $T$ and $E$ assumed by Pauli?

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Curiosa
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  • This blog article argues that the argument is invalid: http://www.soulphysics.org/2013/09/paulis-theorem-is-not-a-theorem-not-as-pauli-stated-it/ – Ján Lalinský Jul 15 '16 at 15:01

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I'm not sure what "Pauli's argument" precisely is because he refers to pages in the first edition of Dirac's Principles of Quantum Mechanics which contain nothing of evident relevance in my fourth edition, but the more common thing to say is that it is the boundedness of the energy from below that forbids a naive time operator, not the discreteness.

However, the "particle on a ring" scenario you propose is not so simple: The space of wavefunctions on a circle may be naively written as $H = \{f\in L^2([0,2\pi],\mathrm{d}\phi)\mid f(0) = f(2\pi)\}$, i.e. all the possible square-integrable function with periodic boundary conditions. Multiplying by the "position operator" $\phi$, you must observe that $\phi$ does not preserve the boundary condition: $(\phi f)(0) = (\phi f)(2\pi)$ if and only if $f(0) = 0 = f(2\pi)$. So either you have the unnatural restriction to wavefunctions vanishing at a particular point, or you must relax the boundary conditions for a general wavefunction (none of these have direct physical meaning because the value at the single point does not change the probability density). For more on where exactly the commutation relation is valid, see this answer of mine.

The issue here is that the commutation relations themselves involve unbounded operators and are rather pathological. If they only hold on a subset of the space that is not the same as the dense domain on which the operators are defined, which can happen as in the case of the particle on the ring, then the Stone-von Neumann theorem does not hold and it is possible to construct all sorts of counterexamples.

The formally correct way to state Pauli's theorem is then this:

If the Hamiltonian $H$ is bounded from below or has a discrete part of the spectrum, then there is no operator $T$ such that the Weyl relations $$ U(t)V(e) = \exp(-\mathrm{i}te)V(e)U(t)$$ for $U(t) = \exp(\mathrm{i}Ht),S(e) = \exp(\mathrm{i}Te)$ hold everywhere.

This follows directly from the Stone-von Neumann theorem, since unitarily equivalent operators necessarily have the same spectrum, and the canonical representation of the Weyl relations has both operators continuous and unbounded. It means that although you might be able to construct the infinitesimal version of the time operator in some cases on some part of the Hilbert space, you cannot get the "tame" version of the CCR, the Weyl relations, from them. To see that Weyl relations are really physically necessary for something to be the proper "time operator", note that $U(t)V(e)U(-t) = \exp(-\mathrm{i}te)V(e)$ is really just the statement that the time translation operator $U(-t)$ shifts the value of the time observable by $t$.

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ACuriousMind
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  • I too couldn't figure out where the relevant portion is in the 4th edition. I could not find the first edition even on google books. But Pauli's argument is mentioned in Ballentine, Quantum Mechanics, 1st edition in Section 12.3. But Ballentine mentions only boundedness of energy from below. – Curiosa Jul 19 '16 at 16:34
  • You have mentioned the boundary condition as $f(0)=f(2\pi)$. But I think the boundary condition should be $f(\phi) = f(\phi + 2\pi)$ for any $\phi$. Then, the $\phi$ operator will preserve the boundary condition only if $f(\phi) = 0$ everywhere! What will that mean? Or is there some error? – Curiosa Jul 19 '16 at 16:38
  • @Curiosa: Note that the boundary condition $f(\phi) = f(\phi+2\pi)$ does not make any sense for the space of functions I've chosen, which are functions on $[0,2\pi]$. However, functions on $[0,2\pi]$ with $f(0) = f(2\pi)$ are in bijection to functions on $\mathbb{R}$ with $f(x) = f(x+2\pi)$, so, in essence, I am talking about such function, but the presentation as function on $[0,2\pi]$ makes it a bit easier to talk about the position operator. – ACuriousMind Jul 20 '16 at 13:21
  • I cannot replace S(e) in the formula by V(e) because it is too short. could you do it? tbanks – Naima Mar 28 '18 at 14:56
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Commutation relations between position and momentum operators, in quantum mechanics, are valid only when the actions of either of the two operators is contained in the domain of the remaining one. In particular we have: $$ [x,p]\psi = (xp)\psi-(px)\psi. $$ In order for the above to be defined $\psi$ must be in the domain of definition of both operators and, also, $x\psi$ must be in the domain of $p$ (and viceversa). For the particle on the circle it is just not the case.

Also, commutation rules between time and energy are not actual commutation rules, rather they are indicative statements that only hold in particular cases.

gented
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  • I'm not so sure what you mean by "for the particle on the circle it is just not the case". The domain of the commutator of $\mathrm{i}\partial_\phi$ and $\phi$ on $L^2([0,2\pi],\mathrm{d}\phi)$ is given by ${f\in L^2\mid f'\in L^2\land f(0) = 0 = f(2\pi)}$. But the domain of the commutator being smaller than the whole space is not unusual, I'm not sure this is the reason the argument fails. – ACuriousMind Jul 15 '16 at 13:40
  • @ACuriousMind My argument is that I'm not entirely sure that the commutator is defined at all: is $p\psi$ in the domain of $x$ and viceversa? (I remember one of the two fails with the boundary requirements and this implies the uncertainty principle is no longer true). – gented Jul 15 '16 at 13:49
  • I discuss the domains of the operators involved and a "better" version of the uncertainty principle in this answer. I suspect Pauli's argument is really supposed to be that energy is bounded from below. One can show that there no consistent momentum operators on the half-line, IIRC - but one operator bounded and continuous and the other unbounded below and above and discrete like for the particle on the interval/circle is not problematic. But energy is never unbounded below and continuous, so we're not in that case. – ACuriousMind Jul 15 '16 at 13:58
  • Oh yes, that answer is what I was referring to! – gented Jul 15 '16 at 18:46