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!


1 Answer 1


The general solution is $$\psi(t)=\sum_k c_k e^{-itE_k/\hbar}\psi_k$$ where the $\psi_k$ form a basis of eigenvectors with corresponding eigenvalues $E_k$, and the $c_k$ are constant.

You can match arbitrary initial conditions at $t=0$ by expanding the initial state in the eigenbasis; this will determine the valued for the $c_k$.

[Edit] To get the statistical interpretation: The expectation of the Hermitian observable $A$ at time $t$ is given in the Schroedinger picture by $$\langle A\rangle_t:=\psi(t)^*A\psi(t).$$ Here it is assumed that $\psi(t)$ has norm 1. As the squared norm is preserved by the dynamics, this gives a well-defined expectation (i.e.,, the expectation of the identity matrix is 1 at all times).

  • $\begingroup$ Ok, good that's what I imagined to do, I wasn't sure. But more importantly how do you interpret this probabilistically in matrix form? $\endgroup$
    – Freeman
    Commented Nov 6, 2012 at 20:45
  • $\begingroup$ Does your $c=0$ mean $c(t=0)$ ? $\endgroup$
    – J L
    Commented Nov 7, 2012 at 16:53
  • 1
    $\begingroup$ The $c_k$ are constant, and a $t$ was missing in the exponent of the first formula. $c=0$ meant $t=0$. I corrected both errors; sorry. $\endgroup$ Commented Nov 8, 2012 at 9:19

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