# Follow-on to: Is the time-derivative Hermitian?

Gold Member
[mentor's note: This thread was forked from https://www.physicsforums.com/threads/is-the-time-derivative-hermitian.791879/ when it looked to to be raising issues beyond the original question. Refer back to that thread for any missing context]

vanhees71, your argument in the first paragraph is mathematically flawed. It's not necessary for both operators to have unbounded spectra.
Can you elaborate on this statement? For position and momentum you find the derivation in many textbooks that, if you have two operators with commutator ##[\hat{x},\hat{p}]= \mathrm{i} \hat{1}##, because momentum is the generator for spatial translations and position the generator for momentum translations. This implies that the spectrum of both operators is the entire real line. So, where is this argument flawed?

The same argument holds true if time and Hamilton operator obey this commutation relation, and this they should, if you consider time as an observable in the sense of QT, because by definition the Hamiltonian is the generator for time translations.

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## Answers and Replies

DrDu
I think this argument by Dirac only rules out that the commutator ##[p_t,t]=i##can be correct, not that p_t can be a self-adjoint operator.

Gold Member
Well, that's indeed true, because ##p_t=H##, and ##H## is an observable (the energy of the system). So, indeed it rules out that ##t## can be an Hermitean operator that satisfies the commutation relation.

dextercioby
Homework Helper
The first thing to notice is that the Wielandt-Wintner theorem (http://de.wikipedia.org/wiki/Satz_von_Wintner-Wielandt) does not force us to take both T and H from [T,H] =1 as unbounded operators. In other words, it makes sense to consider a bounded T and an unbounded H. That also solves the spectrum problem I mentioned in post#3, for if an operator is bounded, its spectrum is also bounded.

What prompted this is the so-called <Pauli's theorem> which can be considered a myth within itself in quantum mechanics. I've yet so see an unique statement in the literature which would force us to consider the pro's and con's for/against it.

EDIT: @DrDu, that's exactly what I said in the now bolded part of this post.

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DrDu
From this wiki article one only learns that two operators fulfilling the Heisenberg commutation relations can t be both bounded.

Gold Member
But the theorem says precisely what Pauli stated in the famous article in the Encyclopedia, i.e., that H and t cannot be bounded if t is an operator with [t,H]=i.

The usual physists' handwaving way to prove that both operators have ##\mathbb{R}## as their spectrum is rougly as follows (I write it for x and p). If ##x## has one generalized eigenvector ##|x_0 \rangle## then ##\exp(-\mathrm{i} \hat{p} \xi) |x_0 \rangle## for ##\xi \in \mathbb{R}## is a generalized eigenvector with eigenvalue ##x_0+\xi##, i.e., if the position operator allows a spectral decomposition at all then its spectrum is entire ##\mathbb{R}##.

In the same way you show with help of the operator ##\exp(\mathrm{i} k \hat{x})## that also ##\hat{p}## has ##\mathbb{R}## as a spectrum.

I guess these arguments can be made mathematically rigorous, e.g., in the rigged-Hilbert space formulation (or in the more conventional formulation a la von Neumann).

dextercioby
Homework Helper
I'm afraid what you say in your first paragraph is not really accurate.

Let me quote from the original Pauli article in the "Handbuch der Physik", Vol.24, 1933, page 140 (Springer Verlag). [1].

<
In der älteren Literatur über Quantenmechanik findet sich an Stelle von ( #176) oft
die Operatorgleichung

Ht - tH = ħ/i I,

die aus (#176) formal durch Einsetzen von t für F entsteht. Es ist indessen im allgemeinen
nicht möglich, einen HERMITEschen Operator (z. B. als Funktion der p und q) zu konstruieren,
der diese Gleichung erfüllt. Dies ergibt sich schon daraus, daß aus der angeschriebenen V.-R.
gefolgert werden kann, daß H kontinuierlich alle Eigenwerte von -oo bis +oo besitzt (vgl.
DIRAC, Quantenmechanik, S. 34 u. 56), während doch andererseits diskrete Eigenwerte von H
vorkommen können. Wir schließen also, daß auf die Einführung eines Operators t grundsätzlich
verzichtet und die Zeit t in der Wellenmechanik notwendig als gewöhnliche Zahl ("c-Zahl")
betrachtet werden muß (vgl. hierzu auch E. SCHRÖDINGER, Berl. Ber. 1931, S. 238).
>

One can make a few comments here:

* There's no mathematical proof of any statement whatsoever
* Pauli sends the reader to Dirac's German translation of his 1st Ed. of <Principles of Quantum Mechanics> where it appears that the argument of full ℝ spectrum of the Hamiltonian is actually given (in this way, without getting a hand on Dirac's book, one might assume that the so-called <theorem of Pauli> is actually <theorem of Dirac>). I don't have Dirac's 1930 book, not its German translation. :(
* He says that it's not possible to build "hermitean" t and H as functions of p,q, but this statement is (not only by 2015 science) definitely mathematically imprecise and moreover not substantiated.
* His famous (bolded by me) conclusion is also reached from relativistic bases by Schrödinger in 1931 (the mentioned article by the Austrian in 'Proceedings of the Prussian Academy of Sciences' is also out of my reach :(), but can't be mathematically substantiated in modern times. It remained (as far as textbook QM and general acceptance) until today as a dogma, just like v.N's collapse.
* Related to your exact wording in the first phrase, there's no mention by Pauli whether t or H as matrices need to be infinite-dimensional and/or 'unbounded' as operators. What you're saying is just an impression based on what we know about the maths of QM nowadays. One must read texts for what they truly contain, not on what we think they contain. ;) :)

Thus my statement above (actually in the original thread from which this derived) that the so-called <Pauli's theorem> is a myth, is justified. Moreover, one should attempt to put all formal derivations of Pauli (and possibly Dirac) present on page 140 on a firm, rigorous mathematical footing, and this is possible, without resorting to RHS-s. (by 1933, the Hilbert space formulation of QM was only 1 year old and von Neumann hadn't addressed the possibility of a time operator/matrix in his book and I can't really speculate on Pauli's degree of familiarity with von Neumann's mathematical discoveries to assess whether Pauli had a note of (rigorous/forma)l proof somewhere of his statements of this famous note 1) to his equation #176).

The reason why I'm using the word 'myth' should be familiar to you, since I asked you for a copy of @Demystifier 's published article in the Found. of Physics [2] where he addresses what he calls <the myth of the time-energy uncertainty relation> exactly by reproducing a mathematically hand-waving argument of what some people believe Pauli would have meant to write, had he been asked to substantiate his claims he'd done in 1933 and quoted by me in this post. So, with all due respect, Hrvoje is/was fighting a myth by invoking and actually propagating another myth (well, that definitely went by refereeing, just like his outragous square root of the paper's eqn. 1).

Notes:
[1] If you're not willing to use google translate or tradukka to go from German to English, read the English official translation which was done by P. Achuthan and K. Venkatesan and was published by Springer Verlag in 1980 under the title <General Principles of Quantum Mechanics>. The exact page is 63. :)

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vanhees71
dextercioby
Homework Helper
[...]

The usual physists' handwaving way to prove that both operators have ##\mathbb{R}## as their spectrum is rougly as follows (I write it for x and p). If ##x## has one generalized eigenvector ##|x_0 \rangle## then ##\exp(-\mathrm{i} \hat{p} \xi) |x_0 \rangle## for ##\xi \in \mathbb{R}## is a generalized eigenvector with eigenvalue ##x_0+\xi##, i.e., if the position operator allows a spectral decomposition at all then its spectrum is entire ##\mathbb{R}##.

In the same way you show with help of the operator ##\exp(\mathrm{i} k \hat{x})## that also ##\hat{p}## has ##\mathbb{R}## as a spectrum.

I guess these arguments can be made mathematically rigorous, e.g., in the rigged-Hilbert space formulation (or in the more conventional formulation a la von Neumann).

Now looking at what I just wrote in post #7 above, I can't comment any further, since you mention <The usual physicists' hand waving way to prove that...>.

There's a subtle trick that one's not willing or not prepared to see: the so-called "canonical commutation relations" set forth by Born and Jordan in 1925, reinterpreted by Weyl in 1927 and "mathematically modernized" by von Neumann (and later by Rellich, Dixmier, Foias, Sz-Nagy, etc. on one hand and Wielandt, Wintner on the other hand) don't force us to work in L2 (ℝ) (one DOF for simplicity). We can consider other Hilbert spaces as well (suited for describing physics, of course). It is actually when one considers another setting than an infinite movement (assumed when bringing ℝ in L2 (ℝ)), that a true (i.e. mathematically pleasing) evasion to any conceivable speculation of the actual content of the 'mythical Pauli's theorem' is present. I have in mind the section 6. of PRSLA, Vol. 458, page 451 by Eric Galapon. (DOI: 10.1098/rspa.2001.0874).

vanhees71
Gold Member
That's very interesting. I'll have a look at the cited paper later. I think this has pretty far-reaching consequences: The usual argument for the foundation of quantum theory rests on the construction of ray representations of the space-time symmetries. Then one constructs with the above mentioned hand-waving physists' argument the usual quantum theory based on the Hilbert space ##L^2(\mathbb{R}^3)##. Glancing over the physics part of the paper (I guess I'll have a hard time to completely understand the hard mathematical part), the additional assumption is that one wants a homogeneous space, reflecting the classical meaning translation invariance.

So does the result of the paper imply in turn that one can construct other space-time symmetry groups already with the usual commutator relations?

strangerep
[...] I have in mind the section 6. of PRSLA, Vol. 458, page 451 by Eric Galapon. (DOI: 10.1098/rspa.2001.0874).
For those following along at home, here is an arXiv copy of Galapon's paper.

dextercioby
Homework Helper
That's very interesting. I'll have a look at the cited paper later. I think this has pretty far-reaching consequences: The usual argument for the foundation of quantum theory rests on the construction of ray representations of the space-time symmetries. Then one constructs with the above mentioned hand-waving physists' argument the usual quantum theory based on the Hilbert space ##L^2(\mathbb{R}^3)##. Glancing over the physics part of the paper (I guess I'll have a hard time to completely understand the hard mathematical part), the additional assumption is that one wants a homogeneous space, reflecting the classical meaning translation invariance.

So does the result of the paper imply in turn that one can construct other space-time symmetry groups already with the usual commutator relations?

I can't make a direct connection of this result with the known theory of symmetry groups, but what I know is that:

When we speak of space-time symmetry, we're already choosing how our space-time looks like, that is finite or not. GR teaches that the shape of space-time is a consequence of its matter content, but not if space-time is finite or not. Imposing limitations to 'infinite space-time' (for example a free spinless particle 'trapped' in a 1D box by putting an nonphysical infinite potential outside the walls) provides means to discover important (mathematical) things in QM, like the role of boundary conditions in the known problem solved by von Neumann: building self-adjoint extensions for symmetric operators.

vanhees71
DrDu
Just a note: When I learned in QM 101 about the commutation relations ##[p,x]=i##, I was told in the same breath that this is a physicists short hand for the Weyl commutation relations ##U_\alpha V_\beta=V_\beta U_\alpha \exp(i\alpha\beta)## where ##U_\alpha=\exp(i\alpha p) ## and ##V_\beta=\exp(i\beta x)## and both unitary. I. e., we don't want the commutation relations to hold for some small domain on which both self adjoint operators are defined but for the unitary operators which span the so called Heisenberg algebra on all of Hilbert space which are part of the Galilei group. I think the original argument by Pauli or Dirac goes through considering the unitary operators instead of their self adjoint generators.
The Heisenberg algebra has but one irreducible representation which means that all Hilbert spaces are equivalent.

vanhees71
Gold Member
Yeah, that's what I was taught and what teach most sufficiently complete textbooks (quantum theory without a proper consideration of the symmetry concepts is pretty incomprehensible if you ask me), but this paper cited by dextercioby shows that the proof that the Heisenberg algebra has one and only one irreducible representation is the one on the (up to equivalence unique) separable Hilbert space (which is equivalent to the beloved Schrödinger realization ##L^2(\mathbb{R}^3,\mathbb{C}^n)##) seems to be flawed. The usual handling of the operators by us physicists is indeed pretty handwaving usually not considering the domains and codomains of the operators carefully as you must do as a mathematician.

More precisely it's not an irreducible representation of the Galilei group but a central extension of its covering group, which implies that there are half-integer realizations of the rotation subgroup, i.e., you represent rotations by the covering group SU(2) of the classical rotation group SO(3), and there's a non-trivial central charge, which is the mass; there are no physically meaning full unitary representations of the covering group of the classical Galilei group.

DrDu
From what I understand of that paper, the problem dissappears if you use ##V_\beta=\exp i\beta H## instead of H in eq. 2.3 and the Weyl commutation relations. There is no domain problem left as unitary operators are defined on the whole Hilbert space, and there are no formal operations involving a Taylor series expansion of the operator U, as the relevant commutator is expressis verbis the Weyl commutator. Galapon acknowledges this but insists on this not being Paulis original theorem, as Pauli used the commutator of the selfadjoint operators H and t.
In the usual sloppy physicist way of Pauli or Dirac it is easy to show why his example of p and x restricted to 0 to 2pi and periodic boundary conditions is irrelevant: The commutator is ##[p,x]=i-i2\pi \delta(x)## as x jumps back from 2pi to 0 at x=0.

aleazk
Gold Member
The Weyl relations are equivalent to a unitary representation of the Heisenberg group, which is the simply connected Lie group associated to the Heisenberg Lie algebra (which corresponds to the usual algebraic commutation relations of unbounded operators we would like to have, but only speaking in the algebraic sense, i.e., the structure constants, not the domains or the representation space; the unitary operators in the Weyl relations are the generators of a Weyl algebra, which is a *-algebra). The principle in QM says that the representation of the Heisenberg algebra in our Hilbert space should come from a unitary representation of the Heisenberg group via Stone's Theorem, i.e., through the infinitesimal generators of the representation of the group (the Garding domain provides an invariant domain of essential self-adjointness for the operators). So, the principle says that we don't want just any representation of the Heisenberg algebra: we want one that, when exponentiated, gives us the Weyl relations. This is not trivial because there are representations of the algebra that do not arise in this way, i.e., if you exponentiate it, it doesn't give you the Weyl relations (take the usual realization but in L2([0,1]) and on the domain of absolutely continuous functions that satisfy obvious consistency conditions related to the form of the operators and the base space [0,1].) In fact, it's the existence of these kind of counterexamples the heart of all of this discussion as far as I can see. The Stone-von Neumann theorem states that, up to unitary equivalence, there's essentialy only one unitary irreducible representation of the (finite dimensional) Heisenberg group. This (through the infinitesimal generators of the representation of the group) gives you, of course, the usual Schrödinger representation in L2(R3), where the position operator is a multiplication operator and its spectrum is the whole real line (the same for momentum, since it's a similar operator in momentum space). The domain is the Schwartz space. The operators are essentially self-adjoint there. Check Folland's 'QFT, a tourist guide for mathematicians' for all this.

Thus, the Stone-von Neumann theorem is not flawed, it explicitly refers to the Weyl relations and not to a general representation of the Heisenberg algebra.

Now, some textbooks usually give a handwaving justification for the transition from the commutation relations to the 'exponential' Weyl relations (usually invoking that it simplifies worries about domains issues). But, as discovered by Mackey, it actually has a very interesting interpretation. Suppose the system is localizable (i.e., there exists a position observable). If you also have a representation, Ux, of the translation group (whose generator is momentum), then Mackey says that this representation acts on the PVM, P, of the position operator in the following way: Ux-1P(E)Ux=P(x.E), where E is a borel element of the real line and x.E is the action of the translation group. This is the 'Mackey imprimitivity condition' and it expresses the homogeneity of space. By a simple calculation using the spectral theorem (which can be found in Jauch's book), you get the Weyl relations between Ux and the one parameter group generated by the position operator. Check Jauch's 'Foundations of QM', Varadarajan's 'Geometry of quantum theory'.

So, in any case, what can be debatable is why the Weyl relations, which goes back to why the imprimitivity condition then.

If time is homogeneous under time translations, then you get the imprimitivity condition, then the Weyl relations and then by the Stone-von Neumann theorem you get that both operators, t and H, have the real line as spectrum. Something which, of course, is at odds with the Hamiltonian being bounded from below. This would be the only way in which Pauli's theorem is true (i.e., not in the general way in which it was stated by Pauli, but with the additional hypothesis of time homogeneity.)

dextercioby and vanhees71
DrDu