0
$\begingroup$

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 Answers 1

2

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

  • 0
    The probability of the solution $\psi(t)$ being in the state corresponding to unit vector $u$ is $|u^* \psi(t)|^2$.2012-11-06