Fall 2007 16 Qm

Fall 2007

Fall 2007 1 Mech
Fall 2007 2 Mech
Fall 2007 3 EM
Fall 2007 4 EM
Fall 2007 5 QM
Fall 2007 6 QM
Fall 2007 7 SM
Fall 2007 8 SM
Fall 2007 9 Misc
Fall 2007 10 Misc
Fall 2007 11 Mech
Fall 2007 12 Mech
Fall 2007 13 EM
Fall 2007 14 EM
Fall 2007 15 QM
Fall 2007 16 QM
Fall 2007 17 SM
Fall 2007 18 SM
Fall 2007 19 Misc
Fall 2007 20 Misc

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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Answer

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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 E_s = |a| and E_a = -|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}

In addition,

(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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