Temporaly uncertainty principle

[IRobot]

Hi, I am currently having problems solving a an exercise:
Let's make the assumption of the existence of an operator H such as [T,H]=i \hbar I.
by examining the state: |\psi\rangle}=He^{i \alpha T}|E\rangle} with H|E\rangle}=E|E\rangle}
show that the spetrum of H is not bounded below.

 

[Fredrik]

The idea is to show that for all \alpha, e^{i\alpha T}|E\rangle is an eigenstate of H with an eigenvalue that can be given any value you want by an appropriate choice of \alpha. To do this, you need to find the commutator [H,e^{i\alpha T}].

 

[IRobot]

Thank you Fredrik, it works well ;)

 

[Demystifier]

See also Sec. 3 of
http://xxx.lanl.gov/pdf/quant-ph/0609163v2
especially Eqs. (9)-(13).

 

[tom.stoer]

I suspect that such an operator T does not exist in general

 

[Fredrik]

I suspect that such an operator T does not exist in general

I think that's the point of this exercise. If it does exist, the Hamiltonian isn't bounded from below. This would mean e.g. that atoms don't have a ground state.

 

[naima]

The KG hamiltonian is not bounded from below but I never read one such T with it.

 

[tom.stoer]

The KG hamiltonian is not bounded from below but I never read one such T with it.

:-)

As far as I understand the idea is to show that given T with [T, H] = i one can show that H is not bounded from below; not the other way round that given H is not bounded from below one should construct T.

As a simple exercise one could try to construct T for the simple harminic oscillator

H = a^\dagger a + \frac{1}{2}

and check where this construction fails

 

[Fredrik]

The KG hamiltonian is not bounded from below but I never read one such T with it.

Isn't the vacuum state a minimum energy state?

 

[naima]

The Klein Gordon hamiltonian expressed with annihilation and creation operators
contains a term proportional to Dirac delta function.
After ignoring it (changing the Hamiltonian) one can find a ground state.
Refer to Peskin and Schroeder on page 21-22.
I hope i understand id correctly.

 

[dextercioby]

Hi, Naima, you are absolutely correct. The <natural quantization of the KG field> delivers a Hamiltonian operator not bounded from below. We then use the trick of <normal ordering> to get the physical boundedness from below of the spectrum of H.

 

[dextercioby]

I think that's the point of this exercise. If it does exist, the Hamiltonian isn't bounded from below. This would mean e.g. that atoms don't have a ground state.

Hi, Fredrik, Hrvoje, Tom and the rest of the crew. There was another thread last week briefly touching Pauli's theorem of 1926. I know I'm kind of repeating myself, but I say once more, probably this time more precisely and naming Eric Galapon to be responsible with the juicy math stuff.

Given the operator relation [T,H]=i1 valid on some properly chosen subset of a complex separable Hilbert space (or the antidual of this set wrt to some topology, if one uses RHS), H being a properly defined self-adjoint Hamiltonian with a spectrum bounded from below, one cannot conclude that there does not exist a self-adjoint T operator (let's call it <time operator>, because its spectral values are measured in seconds in the SI units) 'canonically conjugate to the Hamiltonian'. Rephrasing, T exists as a self-adjoint linear operator.

Such a statement is proved by simply building 2 operators -H and T- with the aforementioned properties for a quantum system.

Case closed.

 

[Fredrik]

Hi, Fredrik, Hrvoje, Tom and the rest of the crew. There was another thread last week briefly touching Pauli's theorem of 1926. I know I'm kind of repeating myself, but I say once more, probably this time more precisely and naming Eric Galapon to be responsible with the juicy math stuff.

Given the operator relation [T,H]=i1 valid on some properly chosen subset of a complex separable Hilbert space (or the antidual of this set wrt to some topology, if one uses RHS), H being a properly defined self-adjoint Hamiltonian with a spectrum bounded from below, one cannot conclude that there does not exist a self-adjoint T operator (let's call it <time operator>, because its spectral values are measured in seconds in the SI units) 'canonically conjugate to the Hamiltonian'. Rephrasing, T exists as a self-adjoint linear operator.

Such a statement is proved by simply building 2 operators -H and T- with the aforementioned properties for a quantum system.

Case closed.

That's interesting. Thanks for the information, and for spelling my name right. That only happens about 5% of the time, even though it's right there on the screen.

Do you understand why this exercise seems to prove that such a T can't exist? I mean, do you know what parts of the reckless proof are invalid? I mean this reckless proof:

The commutation relation [T,H]=i implies [T^n,H]=niT^{n-1}. From this we get

[H,e^{iaT}]=\left[H,\sum_{n=0}^\infty\frac{(iaT)^n}{n!}\right]=\sum_{n=1}^\infty\frac{(ia)^n}{n!}[H,T^n]=a\sum_{n=1}^\infty\frac{(ia)^{n-1}}{(n-1)!}T^{n-1}=ae^{iaT}

He^{iaT}|E\rangle=(e^{iaT}H+[H,e^{iaT}])|E\rangle=(E+a)e^{iaT}|E\rangle

 

[dextercioby]

Well, it's incredibly easy to see what's wrong in the <proof> in the post 13. The domain of the commutator between T and H is equal to the domain of the commutator between H and exp (iaT) and does not contain the 'eigenvectors' of the Hamiltonian. Thus, under these circumstances, the last line of what Fredrik wrote makes no sense.

 

[tom.stoer]

The domain of the commutator between T and H is equal to the domain of the commutator between H and exp (iaT) and does not contain the 'eigenvectors' of the Hamiltonian.

Why? Can you elaborate?
In this case already the equation [T, H] = i would be wrong; it would have to contain a projector on the r.h.s. As it's written here it is defined globally.

What I see immediately is that the spectrum must not be discrete as otherwise the new state |E+a> would no longer has the chance to be an eigenstate. Looking at the same calculation for the operators x and p the shift |p> to |p+a> is valid; that means that p is not bounded from below; what goes wrong with this argument when using H?

 

[dextercioby]

[T,H]=i1 is not a global relation, it holds only on some subset of a Hilbert space. And the Hilbert space needn't be the L^2 on whole of R, but only on a subset of it. This means that T and H may be conjugate to one another, but they needn't be a system of imprimitivities on the real line, as the Q and P operator for the free particle are.

It actually turns out that the commutation relation does not imply that the spectrum of the Hamiltonian is necessarily continuous, so a thorough analysis can be made without resorting to rigged Hilbert spaces. The same commutation relation doesn't imply that both operators be unbounded, one of them can be bounded.

For the maths I refer you to the article by E. Galapon published in Proc.Roy.Soc. Lond. A, volume 458, page 2671 and the theorem 4.1 on the page 2687.

 

[tom.stoer]

Why? Can you give me an example where it fails (I mean on the original Hilbert space, not on its dual)?

 

[dextercioby]

Why? Can you give me an example where it fails (I mean on the original Hilbert space, not on its dual)?

Sorry, I've changed my post. It took a lot to write it and the system logged me off. A thorough analysis has been included in the article I've sent reference to. Actually there's a series of articles which should serve as a back up, all of them written by Eric Galapon.

 

[tom.stoer]

I don't think we need a thorough analysis and a lot of math.

The original post talks about T, H and [T, H]=i. Of course H should become a Hamiltonian eventually, but currently we do not need this interpretation; we do neither need properties like self-adjointness.

I would rather turn the whole story round and ask about properties of H that are enforced or preculed by the relation [T, H]=i

Looking at x and p the situation is rather trivial. One can use

x \to x
p \to -i\partial_x
[x,p]=i \to [x, -i\partial_x]\psi(x) = -i(x\partial_x - \partial_x x)\psi(x) = i(\partial_x x)\psi(x) = i\psi(x)

In that sense there is no problem with the argument; all one needs is the definition of the two operators. Now one can check whether the relation [T, H]=i enforces properties on H which preclude H from being a reasonable Hamiltonian.

 

[Fredrik]

I think this is also relevant here:

Now suppose that two hermitian operators have a commutator that's proportional to the identity operator: [A,B]=cI, and eigenvectors satisfying A|a>=a|a> and B|b>=b|b>.

1=\frac c c\langle a|a\rangle=\frac 1 c\langle a|cI|a\rangle=\frac 1 c\langle a|[A,B]|a\rangle=\frac 1 c\langle a|(aB-Ba)|a\rangle=0

(Thanks to George Jones for posting this in some other thread).

The same thing happens if we use |b> instead of |a>. What this means is that two hermitan operators that satisfy such a commutation relation (like x and p) can't have eigenvectors. Bounded hermitian operators always do, so this is one way to see that x and p must be unbounded.

 

[dextercioby]

What's wrong in that line that determines you to conclude that 1=0 ?

 

[tom.stoer]

I think this is also relevant here:

Of course you have to avoid sandwiching!

 

[dextercioby]

I don't think we need a thorough analysis and a lot of math.

The original post talks about T, H and [T, H]=i. Of course H should become a Hamiltonian eventually, but currently we do not need this interpretation; we do neither need properties like self-adjointness.

I would rather turn the whole story round and ask about properties of H that are enforced or preculed by the relation [T, H]=i

[...]

In that sense there is no problem with the argument; all one needs is the definition of the two operators. Now one can check whether the relation [T, H]=i enforces properties on H which preclude H from being a reasonable Hamiltonian.

What are the properties which would "preclude H from being a reasonable Hamiltonian" ? Could you, please, spell them out ? Thank you

 

[tom.stoer]

H should be self adjoint or at least symmetric; H should allow for "generalized eigenvectors"; H should be bounded from below; in QFT H should be "finite after normal ordering / regularization".

 

[dextercioby]

Do you agree then with the claim that assuming [T,H]=i1 on some subset of a Hilbert space/RHS, with T a s-adj operator and H satisfying all your requirements, doesn't lead to any inconsistencies, both mathematically and physically ?

 

[tom.stoer]

Do you agree then with the claim that assuming [T,H]=i1 on some subset of a Hilbert space/RHS, with T a s-adj operator and H satisfying all your requirements, doesn't lead to any inconsistencies, both mathematically and physically ?

a) I do not understand why you are talking about a subset
b) I guess that - coming back to the original post - one can show that the requirement of H being bounded from below is in conflict with [T,H]=i1 (of course I would keep H and forget about T)
So except for the subset in a) and ruling out the existence of T accoring to b) I do not have a problem

 

[Fredrik]

What's wrong in that line that determines you to conclude that 1=0 ?

As you can see in the text after that line, I was thinking that the problem wasn't in that line, but in the assumptions before it, in particular the assumption that two operators with a commutator like that can both have eigenvectors. But now I'm thinking that since |b> doesn't appear in that line, that explanation can't be right. Based on what you said before, I guess it has to be that if A has an eigenvector, it can't be in the domain of B.

a) I do not understand why you are talking about a subset

Unbounded operators are usually (always?) defined on a dense subset of the Hilbert space, not on the entire space. Suppose e.g. that Q is defined by Qf=xf for all functions f in L²(ℝ). Then Qf isn't square integrable for each choice of f. To ensure that the range of this linear operator doesn't contain any functions that are outside of L²(ℝ), we must choose its domain to be a proper subset of L²(ℝ).

b) I guess that - coming back to the original post - one can show that the requirement of H being bounded from below is in conflict with [T,H]=i1 (of course I would keep H and forget about T)
So except for the subset in a) and ruling out the existence of T accoring to b) I do not have a problem

The proof assumes that |E> is in the domain of T, which implies that 1=0. I'd call that a problem.

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

Hey, wait a second, this also implies that [T,H]=iI is impossible. It might make sense if we replace the identity operator with its restriction to the domain of T, but it doesn't make sense as it stands.

 

[tom.stoer]

Unbounded operators are usually (always?) defined on a dense subset of the Hilbert space, not on the entire space. Suppose e.g. that Q is defined by Qf=xf for all functions f in L²(ℝ). Then Qf isn't square integrable for each choice of f. To ensure that the range of this linear operator doesn't contain any functions that are outside of L²(ℝ), we must choose its domain to be a proper subset of L²(ℝ).

The problem is to use L²(ℝ)

 

[Fredrik]

The problem is to use L²(ℝ)

Since all infinite-dimensional complex separable Hilbert spaces are isomorphic to L²(ℝ), I don't think you can solve any problems of this magnitude by using a different Hilbert space.

 

[tom.stoer]

Hey, wait a second, this also implies that [T,H]=iI is impossible. It might make sense if we replace the identity operator with its restriction to the domain of T, but it doesn't make sense as it stands.

But the problem is not |E> but sandwiching H and T between |E>, or simply writing down <E|!

Please do the same for x and p; there is no problem with x and p and there is no problem with [x,p] acting on a wave function; the problem is sandwiching [x,p] between |p> states!

 

[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.