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


1 Answer 1


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 $$\Psi(0)^{-1} = \Psi(0)^* = \frac{1}{3} \pmatrix{2 & 2 & 1\cr 2 & -1 & -2\cr 1 & -2 & 2\cr}$$

  • $\begingroup$ Firstly thank you for your answer, this looks very helpful. So why do we have a 3x3 matrix instead of a linear combination of the three columns as eigenvectors? However what I am incredibly confused about is how you interpret this probabilistically? $\endgroup$
    – Freeman
    Commented Nov 6, 2012 at 20:46
  • $\begingroup$ When you do $\Psi(t) v$, you are taking a linear combination of the three columns of $\Psi(t)$ with coefficients $v_1,v_2,v_3$. $\endgroup$ Commented Nov 6, 2012 at 22:10
  • $\begingroup$ The probability of the solution $\psi(t)$ being in the state corresponding to unit vector $u$ is $|u^* \psi(t)|^2$. $\endgroup$ Commented Nov 6, 2012 at 22:12

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