Deduction of the action of this unitary on the wavefunction

[julian]

I have operators satisfying $[\hat{Z} , \hat{E}] = i \hbar$. The operators $\hat{Z}$ and $\hat{E}$ are taken to be Hermitian. You consider the unitary operator

$U_\lambda = \exp (i \lambda \hat{Z} / \hbar)$.

I have proved that

$U_\lambda \hat{E} U_\lambda^\dagger = \hat{E} - \lambda \quad Eq.1$

I know how to apply Eq.1 to any function of $\hat{E}$, $f (\hat{E})$, which may be Taylor expanded. Moreover consider the application of the transformation to some matrix elements

$\int_0^\infty \psi^* (E) f (\hat{E}) \phi (E) dE = \int_0^\infty \psi^* U^\dagger U f (\hat{E}) U^\dagger U \phi dE = \int \psi^* U^\dagger f (\hat{E} - \lambda) U \phi dE$

I'm told that you can deduce from this that (and similarly for $\psi^*$):

$U \phi (E) = \phi (E - \lambda)$.

I not sure I understand this. What am I missing?

Thanks.

 

[blue_leaf77]

The integration should run from $-\infty$ to $\infty$. Then
$$\int_{-\infty}^\infty \psi^* (E) f (E) \phi (E) dE = \int_{-\infty}^\infty \psi^* (E-\lambda) f (\hat{E-\lambda}) \phi (E-\lambda) dE$$
Now compare this with $\int_{-\infty}^\infty (U\psi)^\dagger f (E - \lambda) U \phi dE$

 

[julian]

The integration should run from $-\infty$ to $\infty$. Then
$$\int_{-\infty}^\infty \psi^* (E) f (E) \phi (E) dE = \int_{-\infty}^\infty \psi^* (E-\lambda) f (\hat{E-\lambda}) \phi (E-\lambda) dE$$
Now compare this with $\int_{-\infty}^\infty (U\psi)^\dagger f (E - \lambda) U \phi dE$

Actually $E$ is the Energy and is bounded from below, I've taken the lower bound to be $E=0$, so that there are no energy-eigenstates for $E < 0$. I probably should have mentioned that.

Did a major edit here.

 

[blue_leaf77]

Are you sure $\hat{E}$ is energy and should be bound to zero? What if $E-\lambda$ is negative, $\phi(E-\lambda)$ will then have no defined state?
Actually $U$ operator is called the translation operator, you may remember that there is a pair of well-known operators which satisfy the same commutator relation as our problem here, which are the momentum and position operators. Besides how come you explicitly write that the eigenvalues of $\hat{E}$ is continuous? Perhaps it will help if you tell us in which context/discussion you encountered this problem.

 

[julian]

It is energy - I'm looking into the problems that arise when you attempt defining a putative Hermitian operator conjugate to the Hamiltonian operator (so a putative time operator. I'm aware that time is not a dynamical quantity but I'm ignoring that for the moment). In non-relativistic problems the energy has a lower bound and we may as well take it be 0 as we are allowed to.

By the way I'm not worrying too much about some of the mathematical subtleties right now - just trying to get a first handle on it.

The contradiction I'm leading up to is applying $U$ to an energy eigenstate, $\phi (E) = \delta (E-E_0)$, which only has an amplitude at some particular energy $E_0$, with $\lambda$ is sufficiently big to displace $E_0$ below $E=0$ where, by assumption, there are no energy eigenstates, proving no Hermitian $\hat{Z}$ exists.

In relativistic quantum mechanics there is the gap $-mc^2 < E < mc^2$, so we can apply similar arguments there.

 

[julian]

Come to think of it, my old prof's notes I'm looking at might have a mistake...when you do $\phi(E) \mapsto \phi (E - \lambda)$ what this does is moves the function to the right along the $E-$axis by $\lambda$.

If I want to shift the energy eigenstate $\phi (E) = \delta (E-E_0)$ to the left along the $E-$axis (creating an energy-eigenstate with negative energy and hence a contradiction) don't I need to do $\phi (E) \mapsto \phi (E+\lambda)$?

Obviously $\phi (E+\lambda) = \delta (E+\lambda-E_0)$ only has an amplitude at $E = E_0 - \lambda$!