We know that the solution for $i\hbar \frac{\partial}{\partial t}|\psi (t) \rangle = H|\psi (t)\rangle $ where $H$ is time-independent Hamiltonian, is $|\psi(t)\rangle = e^{-iHt/\hbar}|\psi(t=0)\rangle$.

Now, suppose $|\psi\rangle = \frac{1}{\sqrt{N}}(|\psi_1\rangle + |\psi_2\rangle)$ with some proper normalization factor $N$. A simple example would be $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$, where $|+\rangle$ is $+$ eigenvector of Pauli-X operator and $|0\rangle, |1\rangle$ are $\pm$ eigenvectors of Pauli-Z operator. (I used the notation that is commonly used in quantum information area)

If $|\psi\rangle = \frac{1}{\sqrt{N}}(|\psi_1\rangle + |\psi_2\rangle)$, then does the solution of Schrodinger equation for $|\psi(t)\rangle$ is simply

$|\psi(t)\rangle = \frac{1}{\sqrt{N}} e^{-iHt/\hbar}(|\psi_1(t=0)\rangle + |\psi_2(t=0)\rangle)$?

Can we generalize this to time-dependent Hamiltonian case?


1 Answer 1


Yes, indeed. In fact, for a two-by-two Hamiltonian one could evaluate the operator $e^{-iHt}$ exactly and compare the result with the solution of the SE for the same Hamiltonian (when solving it as a system of differential equations).

A 2-by-2 Hamiltonian can be written as a sum of Pauli matrices: $$ H=\mathbf{\omega}\cdot\mathbf{\sigma}=\omega_x\sigma_x+\omega_y\sigma_y+\omega_z\sigma_z=\begin{bmatrix}\omega_z&\omega_x-i\omega_y\\\omega_x+i\omega_y&-\omega_z\end{bmatrix} $$ One can than evaluate the exponential operator $$ e^{-iHt}=e^{-i\mathbf{\omega}\cdot\mathbf{\sigma}t}= \sum_{n=0}^{+\infty}\frac{(-i\mathbf{\omega}\cdot\mathbf{\sigma}t)^n}{n!} $$ using the properties of Pauli matrices $$ \sigma_x\sigma_y=i\sigma_z, \sigma_y\sigma_z=i\sigma_x, \sigma_z\sigma_x=i\sigma_y, $$ reducing it to a 2-by-2 matrix.

Alternatively the same result can be obtained by transforming the Hamiltonian to the diagonal basis, exponentiating and then transforming back.

The Schrödinger equation is in this case: $$i\partial_t|\psi(t)\rangle = H|\psi(t)\rangle,\\ |\psi(t)\rangle=a(t)|+\rangle + b(t)|-\rangle\\ \Rightarrow \begin{bmatrix}\partial_t a(t)\\ \partial_t b(t)\end{bmatrix}= \begin{bmatrix}\omega_z&\omega_x-i\omega_y\\\omega_x+i\omega_y&-\omega_z\end{bmatrix} \begin{bmatrix} a(t)\\ b(t)\end{bmatrix} $$


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