How to reconcile two different derivations of the time-independent Schrödinger equation?

Question

On one hand, using the Spectral decomposition of the Hamiltonian operator $H$, assumed to be an Hermitian operator, it is relatively simple to derive the equation $U(t) = \sum |v_j\rangle\langle v_j| e^{-i \lambda_j t / \hslash}$, where $\lambda_j$ is the eigenvalue and $|v_j\rangle$ is the corresponding eigenvector - assuming discrete decomposition and non-degenerate case.

On the other hand, using the matrix exponential and solving the linear partial differential equation, $U(t) = e^{-i H t / \hslash}$. Assuming that Hamiltonian is an Hermitian operator defined over a finite-dimensional Hilbert space $\mathcal{H}_n$, the operator is bounded and the matrix exponential converges.

I'm trying to reconcile both equations. Plugging the Spectral decomposition of $H$ in the second equation does not seem to give me the first one.

Answer 1

Since $H|v_j\rangle =\lambda_j |v_j\rangle$, you have : $$H^n|v_j\rangle = \lambda^n|v_j\rangle \qquad \text{and therefore}\qquad e^{\alpha H}|v_j\rangle = e^{\alpha\lambda_j}|v_j\rangle$$

Furthermore, the spectral theorem ensures that the $|v_j\rangle$ can be taken to be an orthonormal basis and therefore : $$1 = \sum_j |v_j\rangle \langle v_j|$$

Therefore, you have : \begin{align} U(t) &= e^{-iHt/\hbar} \\ &= \sum_j e^{-iHt/\hbar} |v_j\rangle\langle v_j| \\ &= \sum_j e^{-i\lambda_j t/\hbar}|v_j\rangle\langle v_j| \end{align}