If we have a Hilbert space of $\mathbb{C}^3$ so that a wave function is a 3-component column vector
$$\psi_t=(\psi_1(t),\psi_2(t),\psi_3(t),)$$
with Hamiltonian $H$ given by
$$H=\hbar\omega
\begin{pmatrix}
1 & 2 & 0 \\
2 & 0 & 2 \\
0 & 2 & -1
\end{pmatrix}$$
With
$$\psi_t(0)=(1,0,0)^T$$
So I proceeded to find the stationary states of $H$ by finding it's eigenvectors and eigenvalues. $H$ has eigenvalues and eigenvectors:
$$3\hbar\omega,0,-3\hbar\omega$$
$$\psi_+=\frac{1}{3}(2,2,1)^T,\psi_0=\frac{1}{3}(2,-1,-2)^T,\psi_-=\frac{1}{3}(1,-2,2)^T$$
Respectively.
Could anyone explain to me how to go from this to a general time dependent solution, and compute probabilities of location? I have only ever encountered $\Psi=\Psi(x,y,z,t)$ before, so I am extremely confused by this matrix format.
I would be extremely grateful for any help! Many thanks in advance
Best Answer
If $\psi$ is an eigenvector of $H$ for eigenvalue $\lambda$, then $e^{-i\lambda t/\hbar} \psi$ is a solution of the Schrodinger equation. Place the three solutions thus obtained side-by-side as columns of a $3 \times 3$ matrix $$ \Psi(t) = \frac{1}{3} \pmatrix{ 2 e^{-3\omega it} & 2 & e^{3\omega it}\cr 2 e^{-3\omega it} & -1 & -2 e^{3\omega it}\cr e^{-3\omega it} & -2 & 2 e^{3\omega it}\cr}$$ that satisfies the matrix version of the equation: $ i \hbar \dfrac{d\Psi}{dt} = H \Psi$. The general time-dependent solution is then $\psi(t) = \Psi(t) v$ for any constant $3$-vector $v$. If you want $\psi(0) = (1,0,0)^T$ then take $v = \Psi(0)^{-1} (1,0,0)^T$. Your eigenvectors are orthonormal, so $\Psi(t)$ is unitary, and thus $$\Psi(0)^{-1} = \Psi(0)^* = \frac{1}{3} \pmatrix{2 & 2 & 1\cr 2 & -1 & -2\cr 1 & -2 & 2\cr}$$