Temporaly uncertainty principle

[naima]

another thing.
When we use inner product states must belong to the HS.
HE> does but is it true for TE> ?

 

[Fredrik]

When we use inner product states must belong to the HS.
HE> does but is it true for TE> ?

Yes. The range of an unbounded operator is a subset of the Hilbert space.

 

[tom.stoer]

@Fredrik:

I hope I can explain the problem with x and p:

$$1=\langle p|p\rangle=-i\langle p|iI|p\rangle=-i\langle p|[x,p]|p\rangle=-ip\langle p|(x-x)|p\rangle=0$$

Suppose that [x,p]=iI, and that p has an eigenvector.

...

There's nothing wrong with this calculation, so there must be something wrong with the assumptions that went into it. ... The two assumptions can't both be true, so either p doesn't have an eigenvector, or there's no x such that [x,p]=iI.

The problem is obviously the first equation

$$1=\langle p|p\rangle$$

You want to show that something is wrong with the second assumption [x,p] but you plug this into an expression that is already ill-defined from the very beginning.

The result of your reasoning is not that [x,p]=iI is wrong, but that p does not have a normalizable eigenvector. For [T,H]=iI this means that you have not shown that T does not exist, but that something may be wrong with |E> being normalizable. That's the reason I am insisting in not sandwiching the operators if you can't be sure that both H and T have discrete spectra.

[you can check this using x, p=-id/dx and plane wave states; the problem is not the commutator but the normalization of the plane wave states]

 

[Fredrik]

All eigenvectors, and all other members of a Hilbert space, have finite norm by definition of "Hilbert space". So there's nothing wrong with the first equality. Just about everything else is wrong though. p doesn't have an eigenvector, so we wouldn't start by assuming that it does. And neither x nor p is defined on the entire Hilbert space (what I said about the operator I called Q in #27 proves it for x), so it's not possible that [x,p] is equal to something that is.

If we work with a rigged Hilbert space instead, then p does have an eigenvector, but I think this problem is complicated enough in a Hilbert space. I know a lot less about rigged Hilbert spaces than I do about Hilbert spaces. I'm not even sure if those "eigenvectors" are even called eigenvectors.

 

[tom.stoer]

Fredrik,

if you restrict to finite-norm states in L² you can solve the square well and the harmonic oscillator, but nothing else - not even the free particle case!

 

[Fredrik]

I don't see that as a problem. Rigged Hilbert spaces may be easier to work with once you know them well, but since we don't, I think we should stick to Hilbert spaces for now.

 

[dextercioby]

Fredrik,

if you restrict to finite-norm states in L² you can solve the square well and the harmonic oscillator, but nothing else - not even the free particle case!

It's not on whether we need RHS for QM (the answer is yes), but rather if we can live without them for quantum systems in which particular assumptions are made. There's a huge literature on the CCR in a single Hilbert space, that's for sure. Very few treatments of CCR in RHS, however.

What I know of it is quite easy to rememeber: If A and B satisfy the CCR on some complex, separable Hilbert space, then at least of the 2 operators is unbounded. Futhermore, depending on the system under analysis, then both A and B, wether time op and Hamiltonian operator, wether position and momentum, can be rendered self-adjoint, thus describing real measurable quantities. This is rather widely known.

What I've been trying to say all along is the think that if A and B are self-adj, linear operators in some Hilbert space and they satisfy the CCR on some subset of vector of the Hilbert space, then a semibounded B doesn't contradict the existence of some bounded A.

A and B can be time or Hamiltonian, can be position and momentum. The Hilbert space depends of course on the physical problem.

 

[dextercioby]

@Fredrik:

I hope I can explain the problem with x and p:

$$1=\langle p|p\rangle=-i\langle p|iI|p\rangle=-i\langle p|[x,p]|p\rangle=-ip\langle p|(x-x)|p\rangle=0$$

Suppose that [x,p]=iI, and that p has an eigenvector.

...

There's nothing wrong with this calculation, so there must be something wrong with the assumptions that went into it. ... The two assumptions can't both be true, so either p doesn't have an eigenvector, or there's no x such that [x,p]=iI.

The problem is obviously the first equation

$$1=\langle p|p\rangle$$

You want to show that something is wrong with the second assumption [x,p] but you plug this into an expression that is already ill-defined from the very beginning.

The result of your reasoning is not that [x,p]=iI is wrong, but that p does not have a normalizable eigenvector. For [T,H]=iI this means that you have not shown that T does not exist, but that something may be wrong with |E> being normalizable. That's the reason I am insisting in not sandwiching the operators if you can't be sure that both H and T have discrete spectra.

[you can check this using x, p=-id/dx and plane wave states; the problem is not the commutator but the normalization of the plane wave states]

It's not <p,p>=1 that is wrong. Under certain conditions, it's right. What's wrong in the flow of equalities invoked by Fredrik is the fact that he sandwiched the commutator between the eigenvectors of p. The eigenvectors- when memebers of the H space (RHS left aside)- are not in the domain of the commutator. That's all.

 

[tom.stoer]

Regardig Pauli's theorem, counter examples and attempts to construct T please refer to
http://arxiv.org/PS_cache/quant-ph/pdf/9908/9908033v4.pdf
http://arxiv.org/PS_cache/quant-ph/pdf/0410/0410070v2.pdf
http://arxiv.org/PS_cache/quant-ph/pdf/0702/0702111v1.pdf
http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.3312v1.pdf
http://arxiv.org/PS_cache/arxiv/pdf/1005/1005.2870v1.pdf
and references therein.

I begin to understand where the problem is, but it will take some time go go through all the details.

 

[tom.stoer]

It's not <p,p>=1 that is wrong. Under certain conditions, it's right. ... The eigenvectors - when members of the H space - are not in the domain of the commutator.

Can you show me an example where <p|p>=1 holds but where <p|[x,p]p> is ill-defined?

 

[Fredrik]

It's not on whether we need RHS for QM (the answer is yes),

I would say that the answer is clearly no. If we need it, then the theory based on RHS must be experimentally distinguishable from the one based on Hilbert spaces. (I would take that as the definition of "need"). The RHS just gives us a different way to express the same ideas.

It's not <p,p>=1 that is wrong. Under certain conditions, it's right. What's wrong in the flow of equalities invoked by Fredrik is the fact that he sandwiched the commutator between the eigenvectors of p. The eigenvectors- when memebers of the H space (RHS left aside)- are not in the domain of the commutator.

As I said in at least one of my previous posts, it can't be wrong to "sandwich" the identity operator between two state vectors, because the identity operator is defined on the entire Hilbert space. If the commutator isn't, then it isn't a constant times the identity operator.

 

[dextercioby]

Can you show me an example where <p|p>=1 holds but where <p|[x,p]p> is ill-defined?

Since you mentioned and provided links to my supporting documentation, then an example is to be found in the article http://arxiv.org/PS_cache/quant-ph/pdf/9908/9908033v4.pdf Section 5, Page 460 from the Journal.

 

[tom.stoer]

As I said in at least one of my previous posts, it can't be wrong to "sandwich" the identity operator between two state vectors, because the identity operator is defined on the entire Hilbert space.

I am referring to not normalizable states. The identity is defined on the entire Hilbert space, but the scalar product <.|.> isn't. In that case <.|1|.> isn't defined, either.

 

[dextercioby]

I would say that the answer is clearly no. If we need it, then the theory based on RHS must be experimentally distinguishable from the one based on Hilbert spaces. (I would take that as the definition of "need").

I disagree. We need it for correctness purposes. Without RHS, Dirac's formalism would be full of hand-waving arguments. And we know that Dirac's abstract theory is experimentally indistinguishable from a theory with Schroedinger wave functions based on a Hilbert space, and not to an extension of it.

The RHS just gives us a different way to express the same ideas.

It gives the proper way to express all possible ideas pertaining to the mathematical formulation of the theory.

As I said in at least one of my previous posts, it can't be wrong to "sandwich" the identity operator between two state vectors, because the identity operator is defined on the entire Hilbert space. If the commutator isn't, then it isn't a constant times the identity operator.

The 1 in the i1 is indeed not the 1 on all the H space, but only a restriction of it to the domain of the commutator. The operatorial equality expressed by the CCR takes place only in the domain of the commutator. You can 'sandwich' your commutator between any <state vectors> you like, just as long as they're in the domain of the commutator. But the eigenvectors of H in all cases ARE NOT.

 

[dextercioby]

I am referring to not normalizable states. The identity is defined on the entire Hilbert space, but the scalar product <.|.> isn't. In that case <.|1|.> isn't defined, either.

I don't agree with the wording but I got your point. The scalar product is defined on all the Hilbert space. For any vector p in H, <p,1p> is defined and can be made equal to the number 1. You must have meant that p is not normalizable, therefore lies outside the Hilbert space. So your last statement is true only if the 'sandwich' is made with vectors outside the Hilbert space, which would be nonsensical, of course.

 

[tom.stoer]

I don't agree with the wording but I got your point. The scalar product is defined on all the Hilbert space. For any vector p in H, <p,1p> is defined and can be made equal to the number 1. You must have meant that p is not normalizable, therefore lies outside the Hilbert space. So your last statement is true only if the 'sandwich' is made with vectors outside the Hilbert space, which would be nonsensical, of course.

I think we really have a problem with the wording. Let's make a specific example: plane wave states on the real line. Clearly <p|p> does NOT exist; it does not makes sense, even in a Gelfand triple. But of course plane waves are physically relevant.

 

[Fredrik]

I disagree. We need it for correctness purposes. Without RHS, Dirac's formalism would be full of hand-waving arguments.

So? We don't need to use Dirac's formalism. It's convenient, not necessary.

The 1 in the i1 is indeed not the 1 on all the H space, but only a restriction of it to the domain of the commutator.

I know, but the assumptions that went into the "proof" of 1=0 didn't specify that, so in that calculation it was the actual identity operator.

 

[Fredrik]

The identity is defined on the entire Hilbert space, but the scalar product <.|.> isn't.

The inner product on a Hilbert space H is a function from H×H into ℂ that satisfies certain properties.

Let's make a specific example: plane wave states on the real line. Clearly <p|p> does NOT exist; it does not makes sense, even in a Gelfand triple.

I don't know rigged Hilbert spaces well enough to know if this is correct. (I'm guessing that it is). This is why I wanted to keep the discussion about Hilbert spaces. In a Hilbert space, there's no such thing as |p>. In a RHS, |p> is (I think) an antilinear unbounded functional on the subspace of states that are in the domain of x and p, and is closed under a finite number of applications of x and/or p. (For example, if |u> is in that subspace, then pⁿx^mp^k|u> is too, for all integers n,m,k).

 

[tom.stoer]

OK, I understand why you want to restrict to Hilbert spaces; as far as I can see even in that case (where you don't have problems with the inner product) one can bypass Pauli's theorem: http://arxiv.org/PS_cache/quant-ph/pdf/9908/9908033v4.pdf section 5 as bigubau mentioned.

Let me explain why I didn't like sandwiching: in Pauli's theorem one starts with one energy eigenket |E> and then shows that given T one can construct |E-a> with arbitrary real a. That means that H would have a continuous spectrum. But for a continuous spectrum the expresssion <E|E> is ill-defined, one would have to use <E'|E> instead.

The most simplest example with x and p on the real line is

$$\langle p^\prime|p\rangle \sim \int_{-\infty}^{-\infty} dk \delta(k-p^\prime) \delta(k-p) = \delta(p-p^\prime)$$

which is certainly not defined setting p = p'. In order to avoid that I proposed not to used sandwiching.

But I see that already in the Hilbert space case there are problems with Pauli's theorems. The example mentioned above discusses something like p (or E) eigenstates on L²[0,1].

 

[strangerep]

[...] an example is to be found in the article http://arxiv.org/PS_cache/quant-ph/pdf/9908/9908033v4.pdf Section 5, Page 460 from the Journal.

(For other readers, Galapon's example uses the Hilbert space $$L^2[0, 2\pi]$$ .)

Just after his equation 5.2, which is:

$$(TH - HT)\, \psi(t) ~=~ i \psi(t) ~,$$

he notes that this is valid for

$$\psi(t) \in D_c ~=~ \left\{ \psi(t) \in D(H) : \psi(0) = \psi(2\pi) = 0 \right\}$$

but then he notes that the eigenvectors of H lie outside this domain (my italics).
Further, although this domain is invariant under the action of T, it is not invariant
under the action of H.

Now let's consider what all this implies: the $$\psi(t)$$ in $$D_c$$ cannot
be sensibly expanded as a linear combination of H eigenvectors because none of
the latter are in that domain. IMHO this is quite a serious problem because such
superpositions are kinda fundamental to QM. This makes me think that Galapon's
example might be mathematically correct, but physically irrelevant.

 

[tom.stoer]

This makes me think that Galapon's
example might be mathematically correct, but physically irrelevant.

This is something that came up to my mind as well. In all cases you restrict H to something one could ask if H is too small or too big in order to solve a certain problem.

 

[dextercioby]

Now let's consider what all this implies: the $$\psi(t)$$ in $$D_c$$ cannot
be sensibly expanded as a linear combination of H eigenvectors because none of
the latter are in that domain. IMHO this is quite a serious problem because such
superpositions are kinda fundamental to QM. This makes me think that Galapon's
example might be mathematically correct, but physically irrelevant.

I don't find it disturbing that the vectors in D_c are not physically relevant. The relevant ones, at least in the common framework quite widely accepted, are the eigenvectors of H.

 

[strangerep]

I don't find it disturbing that the vectors in D_c are not physically relevant. The relevant ones, at least in the common framework quite widely accepted, are the eigenvectors of H.

Yes, but the time-energy commutation relation is only valid on $$D_c$$ (iiuc).

However, a cursory read of the early parts of Galapon's paper conveys the impression
that he's going to establish a physically meaningful/useful context for the
time-energy commutation relation, and hence a physically sensible time operator.
But this impression is misleading -- unless I've missed something.

 

[dextercioby]

He uses strong mathematics to destroy a myth. He then tries to give a reasonable physical meaning to his mathematical objects. Actually that accounts for extending a theory which hasn't had significant changes in the past 80 years. Accepting and widely-recognizing his proposals would mean a significant change to the overall mechanism, I think.

The least value one can give to his paper(s) is that he generated a solution to a problem very much neglected, probably because it was considered to be obvious. In 80 years nobody took the task of questioning Pauli's result seriously enough, even though he had the mathematical tools and means to do it since the 1930's.

As a side note, it's interesting that no textbook mentions Pauli's result, which is very much ok, because it's actually false. Not even von Neumann or Prugovecki didn't tackle this result in their famous books.

Anyways, the majority imposes what's to be taught in schools. Pauli's result and Galapon's counter-examples are worth spreading around, so that they could "hit" the next serious mathematical study of this quite controversial theory.

 

[unusualname]

This thread is very good, its title could be better, to aid future searches, eg "Time Operator conjugate to a Hamiltonian"

 

[tom.stoer]

I tried to answer a related question here, but I would like to come back to it.

Looking at x and p it's trivial to define T an H. We use the energy representation with

$$H|E\rangle = H|E\rangle$$

and

$$\psi(E) = \lange E|\psi\rangle$$

Then we can define

$$T = i\frac{\partial}{\partial E}$$

All this can be constructed using well-known relations for x and p. The problem is that we want to relate H to some function on phase space, that means H = H[x,p].

So the question is: for which H[x,p] can one define T using the E-representation? Is this question reasonable? (what about the fact that this seems to fail for discrete E?)

 

[dextercioby]

I tried to answer a related question here, but I would like to come back to it.

Looking at x and p it's trivial to define T an H. We use the energy representation with

$$H|E\rangle = H|E\rangle$$

and

$$\psi(E) = \lange E|\psi\rangle$$

Then we can define

$$T = i\frac{\partial}{\partial E}$$

All this can be constructed using well-known relations for x and p. The problem is that we want to relate H to some function on phase space, that means H = H[x,p].

So the question is: for which H[x,p] can one define T using the E-representation? Is this question reasonable? (what about the fact that this seems to fail for discrete E?)

To me what you did is "sounds" like substituting the linear operators X and P with T and H and nothing more. You copied the results you know to be valid for X and P and assign them to the 2 new operators. This I understand. What I don't understand is why you'd want to express the H as a function of 'x' and 'p' ? You can have it as an operator in the 't' representation (t can be thought as a labeling of the spectrum of T) or in the 'E' representation (E plays the same role for H as t for T). In the <normal> scenario, P and Q are fundamental operators, now the fundamental ones are H and T. So one should ask for the equations such as Q(T,H) and P(T,H), i.o. H(P,Q) and T(Q,P).

As per Galapon, it's easier to use the 't' representation, so that H, P, Q eventually become operators acting on functions of time.

 

[tom.stoer]

I think you got my idea.

Why I want to express H in terms of x and p? Because this is standard I am QM! I want to relate the "new picture" with T and H to the normal picture using x and p simply b/c if that's not possible I don't see how this could be relevant for physics.

I guess in the very end we have four Hilbert spaces (let's forget about rigged Hilbert spaces for a moment), namely the x-, p-, E- and t- representation. Now all these separable Hilbert spaces are isomorphic, so it must be possible to construct a map between the state vectors and the observables.

 

[strangerep]

Tom,

I don't understand the notation in your post #61. (Typos perhaps?)

[...] We use the energy representation with

$$H|E\rangle = H|E\rangle$$

Did you mean $$H|E\rangle = E|E\rangle ~~$$ ?

and

$$\psi(E) = \lange E|\psi\rangle$$

There was a typo in your latex. I guess you meant $$\psi(E) = \langle E|\psi\rangle$$?


Also, was it your intent that the energy spectrum is bounded below?

 

[tom.stoer]

Sh...

Yeah, typos; I am sorry. Unfortunately I can't fix it, the post is already locked against editing.

My intention was not to to use the t-representation b/c this is the new ingredient which should be understood based on something that is already known. So my idea was to write down some general formulae in the E- representation.

Then the idea is that a specific H is never given in terms of the t-rep. but in terms of x an p. So the idea should be to check whether for a given H(x,p) a derivation of an operator T in the E-rep. is possible (or if it's not possible, why the derivation fails). That seems to be a natural step b/c it starts with something that is well-known - both theoretically and experimentally.

I'll add the corrected formulas:

$$H|E\rangle = E|E\rangle$$

$$\psi(E) = \langle E|\psi\rangle$$

$$T = i\frac{\partial}{\partial E}$$

 

[strangerep]

My intention was not to to use the t-representation b/c this is the new ingredient which should be understood based on something that is already known. So my idea was to write down some general formulae in the E- representation.

Then the idea is that a specific H is never given in terms of the t-rep. but in terms of x an p. So the idea should be to check whether for a given H(x,p) a derivation of an operator T in the E-rep. is possible (or if it's not possible, why the derivation fails). That seems to be a natural step b/c it starts with something that is well-known - both theoretically and experimentally.

I'll add the corrected formulas:

$$H|E\rangle = E|E\rangle$$

$$\psi(E) = \langle E|\psi\rangle$$

$$T = i\frac{\partial}{\partial E}$$

You didn't answer the last question in my post, i.e.,

[...] was it your intent that the energy spectrum is bounded below?

If the answer is "yes", then I think the conclusion to be drawn from
Galapon's work is that one can only construct a T operator satisfying
a CCR with H such that the CCR does not hold on a domain spanned
by H's eigenstates.

 

[tom.stoer]

You didn't answer the last question in my post ...

Sorry; I wasn't clear about that: my intention was to check whether for a given H(x,p) a derivation of an operator T in the E-rep. is possible (or if it's not possible, why the derivation fails); of course for a reasonable given H(x,p) that means that this H indeed is bounded from below.

 

[A. Neumaier]

Sh...

Yeah, typos; I am sorry. Unfortunately I can't fix it, the post is already locked against editing.

My intention was not to to use the t-representation b/c this is the new ingredient which should be understood based on something that is already known. So my idea was to write down some general formulae in the E- representation.

Then the idea is that a specific H is never given in terms of the t-rep. but in terms of x an p. So the idea should be to check whether for a given H(x,p) a derivation of an operator T in the E-rep. is possible (or if it's not possible, why the derivation fails). That seems to be a natural step b/c it starts with something that is well-known - both theoretically and experimentally.

I'll add the corrected formulas:

$$H|E\rangle = E|E\rangle$$

$$\psi(E) = \langle E|\psi\rangle$$

$$T = i\frac{\partial}{\partial E}$$

This works only if the spectrum of H is purely continuous and there are no additional quantum numbers, so that the eigenstate (in the rigged Hilbert space) is uniquely determined by E.

Moreover, if H is bounded below (as for all physical Hamiltonians) then T is not self-adjoint, hence not a good observable.

 

[tom.stoer]

This works only if the spectrum of H is purely continuous ...

I agree,additionalwork is required.

... and there are no additional quantum numbers, so that the eigenstateis uniquely determined by E.

Of course there's more work tobe done for additonal quantum numbers

Moreover, if H is bounded below (as for all physical Hamiltonians) then T is not self-adjoint, hence not a good observable.

Having an observable T on a non-dense set would be some progress. But there are physicalexamples w/o where an "observable" is not self-adjoint but only symmetric, i.e. for a particle on R⁺ with a boundary condition for u(x) at x=0: u(0)=0. The momentum operator -id/dx is only symmetric (as it has index 1), nevertheless I would say it's an observable.

 

[A. Neumaier]

there are physical examples w/o where an "observable" is not self-adjoint but only symmetric, i.e. for a particle on R⁺ with a boundary condition for u(x) at x=0: u(0)=0. The momentum operator -id/dx is only symmetric (as it has index 1), nevertheless I would say it's an observable.

According to the traditional terminology, it isn't, since its spectrum consists of all complex numbers. But an observable must not only be symmetric but may have only real spectrum; this is equivalent to being self-adjoint.

In your example, there are infinitely many observables corresponding to -id/dx, one each corresponding to each self-adjoint extension of the operator.
[Correction: Since the deficiency indices don't match, there is no self-adjoint extension, and therefore also no observable corresponding to it.
See Example 2.5.10;3 in Vol. 3 of Thirring, A course in mathematical physics.]

In fact, if the spectrum of H is R^+ and the centralizer of H consists of functions of H only (so that there are no other quantum numbers apart from the energy) then your construction is isomorphic to the example you just mentioned, with x denoting the
energy.