How quantum mechanical states evolve with time

A disarmingly simple-looking differential equation which determines how a quantum mechanical state evolves with time was first elaborated by Erwin Schrödinger:

Time evolution of a quantum state $$i\hbar\frac{d}{dt}\ket{\psi(t)} = \hat H(t)\ket{\psi(t)}.$$

Guessing a solution

How to find a solution to Schrödinger's equation? $$i\hbar\frac{d}{dt}\ket{\psi(t)} = \hat H(t)\ket{\psi(t)}.$$

One way to solve any Diff Eq is to make a guess and then check to see if it works. Let's guess that the soluton to the time-dependent function can be written as a product of two pieces: a function of time alone, $T(t)$, times a time-independent piece, which we could write as $\ket(0)$. that describes other characteristics of the system. For $\ket{\psi(0)}$ let's consider $\ket +$, since we know it's an eigenstate of $\hat H$: $\hat H\ket + = E_+\ket +=\omega_0\hbar/2\ket +$. So consider: $$\ket{\psi(t)}=T(t)\ket{\psi(0)}=T(t)\ket +.$$

  1. Substitute this function into the differential equation above, and see if you can derive a simpler differential equation for $T(t)$ alone... $$\ket{\psi(t)}=T(t)\ket +=T(t)\begincv 1\\0\endcv$$ Does the Hamiltonian for our spin-1/2-in-a-$B$-field depend in any explicit way on time?
  2. What kinds of functions obey the differential equation for $T(t)$ that you came up with?

We found $\frac{d}{dt}T(t)=-\frac{i\omega_0}{2}T(t)$. The solution to this diffeq is $T(t)=e^{-\frac{i\omega_0}{2}t}$, so, $$\ket{\psi(t)}=e^{-\frac{i\omega_0}{2}t}\ket +.$$

Use the same approach to find $\ket{\psi(t)}$ if $\ket{\psi(0)}=\ket -$.

$$\ket{\psi(t)}=e^{+\frac{i\omega_0}{2}t}\ket -.$$

Time dependence of a quantum state

More generally, any eigenstate of the Hamiltonian, $\ket{E_j}$ labelled with its eigenvalue $E_j$, satisfies the eigenvalue equation: $$\hat H \ket{E_j}=E_j\ket{E_j}$$ and if a system starts at $t=0$ in an energy eigenstate $\psi(0)=\ket{E_j}$, it will evolve in time according to: $$\ket \psi(t)=e^{-iE_jt/\hbar}\ket{E_j}.\label{eigenvector}$$

Since the Hamiltonian is an observable, its basis states form an orthonormal basis, so, we can write any state as a superposition of these "Energy" basis states. So, let's say at $t=0$ the quantum state is given by: $$\ket{\psi(0)} = c_1\ket{E_1}+c_2\ket{E_2}+c_3\ket{ E_3}+...$$ We know from Eq $\ref{eigenvector}$ how each ket above evolves in time, so putting this together: $$\ket{\psi(t)} = c_1e^{-iE_1t/\hbar}\ket{E_1} +c_2e^{-iE_2t/\hbar}\ket{E_2} +c_3e^{-iE_3t/\hbar}\ket{E_3}+...$$

Time dependence of a quantum state

When $\hat H$ is time-independent,

  1. Diagonalize the Hamiltonian to find its eigenstates, $\{\ket{E_n}\}$,
  2. Express the initial state as a superposition of these energy eigenstates $$\ket{\psi(0)}= c_1\ket{E_1}+c_2\ket{E_2}+c_3\ket{ E_3}+...=\begincv c_1\\c_2\\c_3\\.\\.\endcv$$
  3. "tack on" the time-dependent term $e^{-iE_jt/\hbar}$ to each coefficient: $$\ket{\psi(t)}=\begincv c_1e^{-iE_1t/\hbar} \\c_2e^{-iE_2t/\hbar}\\c_3e^{-iE_3t/\hbar}\\.\\.\endcv$$