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 [tex][T,H]=i \hbar I[/tex].
by examining the state: [tex]|\psi\rangle}=He^{i \alpha T}|E\rangle}[/tex] with [tex]H|E\rangle}=E|E\rangle}[/tex]
show that the spetrum of H is not bounded below.

Fredrik

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

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 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

:-)

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

[tex]H = a^\dagger a + \frac{1}{2}[/tex]

and check where this construction fails

Fredrik

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

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

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

[tex][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}[/tex]

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

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

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)?