# Unitary time evolution operator and the corresponding Hamiltonian

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 $$\theta(t_0)$$.

• Are you familiar with the notion of functions of operators? There should be several posts here on SE (and I guess some Wikipedia pages etc) explaining the basics. Dec 14, 2021 at 9:15
• is $t$ also given? note that $U$ is a function of $t$, so for different times $U$ will have different eigenvalues. Therefore $\theta$ should be $\theta(t)$ somehow (or given at a fixed value of $t$)
– user275556
Dec 14, 2021 at 9:15
• @yyy; yes the value of t is also given. That is, eigenvalues of $U$ are given at one instant of time. I'll explain the exact situation: we have 2 states a and b and it is given that U evolves a into b in time t. Thus $U(t,0)$ $|a>= |b>$. We know the eigenvalues of $U(t,0)$ and I need the eigenvalues of the corresponding time independent Hamiltonian $H$. Dec 14, 2021 at 9:33

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.

• Thank you very much, I have understood and the answer has been very helpful! Dec 14, 2021 at 10:16