Unitary time evolution operator and the corresponding Hamiltonian

Question

Could someone tell me how to find the eigenvalues $(E)$ of a time independent Hamiltonian $(H)$ if the eigenvalues of the corresponding unitary time evolution operator $U$ $\left(=e^{itH/\hbar}\right)$ are given , at a particular instant of time $t_0$, and given in the form of $e^{i\theta(t_0)}$. That is, I need the relation between $E$ and $\theta(t_0)$.

Answer 1

I will try to give a more general answer: given an operator $A$, how can we interpret a function of the operator $f(A)$?

There are two possible options:

i) if the function $f(x)$ admits to a power-series expansion about $x=0$, we can use this to define the $f(A) = \sum \frac{f^{(n)}(0)}{n!} A^n$ and we know how to take integer powers of operators - we just apply the operator $n$ times.

ii) if the operator $A$ can be diagonalized and has eigenstates $|a\rangle$ with eigenvalues $a$ where the eigenstates form a basis, then we can define $f(A)$ via its operation on the eigenstates $f(A)|a\rangle = f(a) |a\rangle$. Then, for any state $|\psi\rangle$ we can know how $f(A)$ acts upon it by expanding $|\psi\rangle$ in the eigenbasis $|\psi\rangle = \sum_a c_a |a\rangle$ and then acting with $f(A)$ on each of the states in the expansion.

If both $A$ has a basis of eigenvectors and $f(x)$ has a power series, it is easy to see that both definitions are consistent with one another. If none of the things apply, then the function may be ill-defined (for example - $\ln(a^{\dagger})$, where $a^{\dagger}$ is the Harmonic oscillator raising operator, is not well defined).

Now for the case at hand -- you have the function $U = \exp(-i H t/\hbar)$ of the Hamiltonian. As both definitions can be applied here, we can choose to focus on the second. It is clear the the eigenvectors of $U$ are the eigenvectors of $H$, and if $H$ has an eigenvalue $E$ when acting on a state $H|\psi_E\rangle = E |\psi_E\rangle$, then $U|\psi_E\rangle = \exp(-i H t/\hbar) |\psi_E\rangle = \exp(-iEt/\hbar)|\psi_E\rangle$, which gives you the answer.