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

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.

• Can you expand on the difficulty you are having? The right hand side of the first equation is often taken as the definition of the matrix exponential in the second, so exactly what definitions you are using and exactly what approach you are taking matter for us to be able to help. Jun 8, 2021 at 14:31
• The definition of the matrix exponentiation I used is simply $e^M = \sum_i^\infty \frac{M^i}{i!}$. The answer below was just fine. So simple! I should have found it myself :-(
– Jean
Jun 11, 2021 at 2:25

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}