Temporaly uncertainty principle

[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.