Fall 2007 16 Qm

A system is described by the Hamiltonian $\mathcal{H}=\mathcal{H}_0 + \epsilon \mathcal{H}_1$. $\mathcal{H}_0$ has a doubly-degenerate ground state of zero energy. The corresponding eigenkets are $\left |1\right>$ and $\left |2\right>$. $\mathcal{H}_1$ has the property that $\left< 1|\mathcal{H}_1|1\right>=\left<2|\mathcal{H}_1|2\right> = 0$ and $\left<1|\mathcal{H}_1|2\right> = a$. Finally, $\epsilon$ is a small parameter.

(a) Find the two lowest energy states and the corresponding eigenkets of the Hamiltonian $\mathcal{H}$ accurate to the lowest non-vanishing order in $\epsilon$.

(b) Write down a general expression for the ket $\left |\psi(t)\right>$ in terms of the eigenkets found above.

(c) Calculate the probability of finding the system in the state $\left |1\right>$ at time t if it was in the state $\left |2\right>$ at t = 0.

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(a) The Hamiltonian $\mathcal{H}$ projected on the Hilbert space spanned by $\left |1\right>$ and $\left |2\right>$ can be written in the form of the 2 × 2 matrix

(1)
\begin{align} \mathcal{H} = \left( \begin{array}{cc}0 & a \\ a^* & 0 \end{array} \right) \end{align}

Let $a=|a|e^{i\theta/2}$. Diagonalizing the above matrix, we find that the two lowest energies are Es = |a| and Ea = -|a|. The corresponding eigenstates are

(2)
\begin{align} \left |s\right> = \frac{e^{i\theta/2}}{\sqrt{2}}\left |1\right>+\frac{e^{-i\theta/2}}{\sqrt{2}}\left |2\right>, \;\; \left |a\right>=\frac{e^{i\theta/2}}{\sqrt{2}}|1> - \frac{e^{-i\theta/2}}{\sqrt{2}}\left |2\right> \end{align}

(b) The answer is ($\hbar$ = 1):

(3)
\begin{align} \left |\psi(t)\right> = c_s e^{-i|a|t}|s>+c_a e^{i|a|t}\left |a\right>,\;\; c_s\left<s|\psi(0)\right>,\;c_a=\left<a|\psi(0)\right>. \end{align}

(c) It is easy to see that

(4)
\begin{align} c_s=\left<s|2\right>= \frac{e^{i\theta/2}}{\sqrt{2}},\; c_a=\left<a|2\right>=-\frac{e^{i\theta/2}}{\sqrt{2}} \end{align}

(5)
\begin{align} d_s\equiv \left<1|s\right>=\frac{e^{-i\theta/2}}{\sqrt{2}},\;\;d_a\equiv\left<1|a\right>=\frac{e^{-i\theta/2}}{\sqrt{2}} \end{align}

whence

(6)
\begin{align} \left<1|\phi(t)\right>=d_s e^{-i|a|t}c_s+d_a e^{i|a|t}c_a=\frac{1}{2}\left(e^{-i|a|t}-e^{i|a|t} \right) = -i\sin{(|a| t)} \end{align}

and so the probability in question is

(7)
\begin{align} |\left<1|\phi(t)\right>|^2=\sin^2{(|a| t)} \end{align}
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