Temporaly uncertainty principle

[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.

 

[tom.stoer]

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.

I do not fully agree. In my example complex eigenvalues are also excluded, but not by the property of being self-adjoint. It is clear that real spectrum follows from selfajointness, but selfadjointess does not follow from having real spectrum.

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.

That's why I presented this as an example.

 

[A. Neumaier]

I do not fully agree. In my example complex eigenvalues are also excluded, but not by the property of being self-adjoint. It is clear that real spectrum follows from selfajointness, but selfadjointess does not follow from having real spectrum.

Oh, sorry; I meant to say that (with p=-id/dx), p^* has all complex numbers in the spectrum.

An operator A is self-adjoint iff it is symmetric and the spectrum of A^* is real.
See, e.g., Theorem 2.2.5 in Vol. 3 of Thirring, A course in mathematical physics.

On the other hand, p itself has no (generalized) eigenstates at all. This plays havoc with the usual way of interpreting exact measurements. So in which sense would p be an observable?

 

[tom.stoer]

-id/dx has the usual "generalized" eigenvalues with spectrum R⁺

 

[A. Neumaier]

-id/dx has the usual "generalized" eigenvalues with spectrum R⁺

Your space consists of the u(x) with u(0)=0. pu=ku implies u'=iku, hence u(x)=exp(ikx)u(0)=0 identically. Thus there is no generalized eigenvector.

In terms of measurement, it is therefore not clear how you'd prepare a state in which a measurement of p yields the value k.

 

[tom.stoer]

Seems to be convincing, but I remember there was an argument based on H=p²/2m (which has the usual spectrum using sin(kx) instead of exp ikx) and taking the "square root" of H. I have to check this again.

But I think this has not so much to do with ther time operator.