Temporaly uncertainty principle

[Fredrik]

In the case of x and p, the |p> states themselves are a problem. But it's also the case that if [x,p]=i, as we've been taught (where the right-hand side must be interpreted as i times the identity operator, since nothing else has been indicated), then the commutator [x,p] can clearly be sandwiched between any two states that the identity operator can be sandwiched between.

 

[tom.stoer]

I still don't understand.

I can use your proof from post #13 with x, p and [x,p]=i and show that everything works fine; I get a continuum of |p-a> states and I know that p is not bounded from below. OK.

Why does this proof not work for two other symbols T and H? They are just symbols b/c I don't care about any specific H.The only thing I avoid is sandwiching as I now that I may run into trouble and that using sandwiching may spoil my argument (your argument :-)

So that's why I say that you shouldnot use <E|...|E> as you cannot be sure (w/o knowing H) that this is allowed.

In order to use the argument with x and p I do not even have to use |p> states. Any function of x is fine (if I use the x-representation and p=-id/dx)

 

[Fredrik]

Let me first state the argument regarding T and H in the clearest possible way:

Suppose that [T,H]=iI, and that H has an eigenvector. Then

1=\langle E|E\rangle=-i\langle E|iI|E\rangle=-i\langle E|[T,H]|E\rangle=-iE\langle E|(T-T)|E\rangle=0

There's nothing wrong with this calculation, so there must be something wrong with the assumptions that went into it. (It can't be wrong to sandwich the commutator like that if it's equal to a constant times the identity operator). The two assumptions can't both be true, so either H doesn't have an eigenvector, or there's no T such that [T,H]=iI.

If it's true that [x,p]=iI, then this argument implies that neither x nor p have eigenvectors. (Apply it once with x=T and p=H, and then again with x=H and p=T). However, I don't think it makes sense to say that [x,p]=iI, because both operators are defined on dense proper subsets of the Hilbert space. So I think the commutator should be [x,p]=iJ, where J is the restriction of the identity operator to the union of the domains of x and p. This means that the argument about H and T can't be used to argue that neither x nor p have eigenvectors. (They don't have eigenvectors, but we have to use some other method to prove it).

I'm not sure I explained anything here that you didn't already understand, but I think this is the best I can do right now. I still know very little about unbounded operators.

 

[naima]

If we write <E,(HT -TH)E> we get <E,HTE> - E<E,TE>
We have the paradox if we use (HT)* = T*H* which is a property of bounded operators. http://en.wikipedia.org/wiki/Unbounded_operator" says it not correct for unbounded operator adjoints.

 

[Fredrik]

Hm, that's interesting, but is it really an issue here? We only need to take the adjoint of H. If we write the eigenvalue equation as He=Ee (because now I don't want to use bra-ket notation),

<e,THe>=<e,TEe>=E<e,Te>
<e,HTe>=<He,Te>=E*<e,Te>=E<e,Te>

 

[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:

<br /> (TH - HT)\, \psi(t) ~=~ i \psi(t) ~,<br />

he notes that this is valid for

<br /> \psi(t) \in D_c ~=~ \left\{ \psi(t) \in D(H) : \psi(0) = \psi(2\pi) = 0 \right\}<br />

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"