## Temporaly uncertainty principle

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.

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 Mentor 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 $[H,e^{i\alpha T}]$.
 Thank you Fredrik, it works well ;)

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## Temporaly uncertainty principle

http://xxx.lanl.gov/pdf/quant-ph/0609163v2
especially Eqs. (9)-(13).

 Recognitions: Science Advisor I suspect that such an operator T does not exist in general

Mentor
 Quote by tom.stoer 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.

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

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 Quote by naima 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

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 Quote by naima 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?

 Recognitions: Gold Member 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.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Hi, Naima, you are absolutely correct. The delivers a Hamiltonian operator not bounded from below. We then use the trick of to get the physical boundedness from below of the spectrum of H.

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Homework Help
 Quote by 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.
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.

Mentor
 Quote by bigubau 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
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

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

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

 Blog Entries: 9 Recognitions: Homework Help Science Advisor Well, it's incredibly easy to see what's wrong in the 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.

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