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 $\theta(t_0)$.
Quantum Mechanics – Unitary Time Evolution Operator and Corresponding Hamiltonian
hamiltonianhomework-and-exercisesoperatorsquantum mechanicstime evolution
Related Solutions
Yes, you are on the right track. The series you have there is called Dyson's series.
First note that the $n$'th term looks like $$ U_n = \left(-\frac{i}{\hbar}\right)^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n) $$
The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing us to write:
\begin{align} U_n = \left(-\frac{i}{\hbar}\right)^n \int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n}\ H(t_1)\cdots H(t_n) = \frac{\left(-\frac{i}{\hbar}\right)^n}{n!}\int_0^t dt_1 \cdots\int_0^t dt_{n} \mathcal{T}\left[H(t_1)\cdots H(t_n)\right] \end{align}
Two things happened: first, we "overcount" by making the upper limits equal to $t$ on all the integrals. This is compensated by the factor of $\frac{1}{n!}$. You'll need to convince yourself why this factor is needed ;)
Second, by this change of integration area we mess up the ordering of the Hamiltonians in the process. This is where the time-ordering symbol $\mathcal{T}$ comes in. Basically, this operator ensures that the Hamiltonians are always ordered in the correct way. For instance for $n=2$ it operates as
\begin{align} \mathcal{T}[H(t_1) H(t_2)] = \begin{cases} H(t_1) H(t_2) & t_2 > t_1\\ H(t_2) H(t_1) & t_2 < t_1 \end{cases} \end{align}
Putting everything together we have
$$ U(t, t') = 1 + \sum_{n=1}^\infty \frac{\left(-\frac{i}{\hbar}\right)^n}{n!} \int_{t'}^t dt_1 \cdots\int_{t'}^t dt_n \mathcal{T}[H(t_1)\cdots H(t_n)] $$ Frequently, this is denoted symbolically as
$$ U(t, t') = \mathcal{T}\exp\left(-\frac{i}{\hbar} \int_{t'}^t H(t_1) dt_1\right) $$ This notation is understood as representing the power series.
You can prove this without looking at any of the specific cases by doing a first-order perturbation of the differential equation that defines the time-evolution operator.
We start with
$$ i \hbar \frac{\partial}{\partial t} U(t, t_0) = H U(t, t_0). $$
Or, rearranging some terms,
$$ \frac{\partial}{\partial t} U(t, t_0) = - \frac{i}{\hbar} H U(t, t_0). $$
At a time $t + \delta t$, the above equation tells us that to first order in $\delta t$ we have
$$ U(t + \delta t, t_0) = U(t, t_0) - \frac{i}{\hbar} H U(t, t_0) \delta t. $$
If $U$ is unitary, then we have
$$ I = U^{\dagger} U (t + \delta t, t_0) = \left[ U^{\dagger}(t, t_0) + \frac{i}{\hbar} U^{\dagger}(t, t_0) H^{\dagger} \delta t\right] \left[ U(t, t_0) - \frac{i}{\hbar} H U(t, t_0) \delta t \right]. $$
Expanding this and keeping only terms to first order in $\delta t$ yields
$$ I = U^{\dagger} U (t, t_0) - \frac{i}{\hbar} U^{\dagger} (t, t_0) H U (t, t_0) \delta t + \frac{i}{\hbar} U^{\dagger}(t, t_0) H^{\dagger} U (t, t_0) \delta t. $$
Since we demanded that $U$ is unitary for all time, we should also have $U^{\dagger} U (t, t_0) = I.$ Substituting this and canceling $I$ from both sides of the equation yields
$$ 0 = \frac{i}{\hbar} U^{\dagger}(t, t_0) \left(H^{\dagger} - H \right) U(t, t_0) \delta t. $$
We must therefore have $H^{\dagger} = H$. So we have shown that if $U$ is unitary, then $H$ must be Hermitian.
For the other direction, we return to our first order expansion:
$$ U^{\dagger} U (t + \delta t, t_0) = U^{\dagger} U (t, t_0) - \frac{i}{\hbar} U^{\dagger} (t, t_0) H U (t, t_0) \delta t + \frac{i}{\hbar} U^{\dagger}(t, t_0) H^{\dagger} U (t, t_0) \delta t. $$
If $H$ is Hermitian, then
$$ U^{\dagger} U (t + \delta t, t_0) = U^{\dagger} U(t, t_0). $$
So $U^{\dagger} U(t, t_0)$ is constant for all $t$. Since $U(t_0, t_0)$ is the identity, we must have $U^{\dagger} U(t, t_0) = I$ for all $t$. We have thus proven that if $H$ is Hermitian, then $U$ must be unitary.
Best Answer
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.