Quantum Mechanics Term 1 Problems Class 2

June 2010 Problem 9
A Quantum Mechanical system has an observable $\widehat{S}$ that can take values $\pm 1$. The Hamiltonian acts on the eigenstates $\pm 1$ as follows

$\widehat{H}|-1⟩$ $=E_0(-\cos (\theta )|-1⟩+i\sin (\theta )|1⟩)$

$\widehat{H}|1⟩$ $=E_0(\cos (\theta )|1⟩-i\sin (\theta )|-1⟩)$

$\theta ,E_0$ $\in ℝ,\theta \ne 0,m\pi ,m\in \mathbb{Z}$

a) Find the eigenvalues and eigenstates of $\widehat{H}$
b) At $t=0$, the value of $\widehat{S}$ is measured and found to be +1. The system is left undisturbed for a time $t$ and measured again. What are the probabilities at time $t$ of measuring $+1,-1$?
c) What is the expectation of measuring $\widehat{S}$ at time $t$?

Answers:
a) To find the eigenvalues and eigenstates of $\widehat{H}$ we first find the matrix representation in the $|\pm 1⟩$ basis.

$⟨1|\widehat{H}|1⟩$ $=E_0\cos (\theta )$

$⟨1|\widehat{H}|-1⟩$ $=i{E}_0\sin (\theta )$

$⟨-1|\widehat{H}|1⟩$ $=-i{E}_0\sin (\theta )$

$⟨-1|\widehat{H}|-1⟩$ $=-E_0\cos (\theta )$

$\Rightarrow H$

$det(H-EI)$ $=0$

$\Rightarrow E^2$ $=E_0^2$

$\Rightarrow E$ $=\pm E_0$

This is rather unfortunate for notational purposes as the eigenvalues of $\widehat{H}$ coincide with that of $\widehat{S}$.
For $E=E_0$,

$|E=E_0⟩$ $=a|-1⟩+b|1⟩$

$\widehat{H}|E=E_0⟩$ $=|E=E_0⟩$

$=\left(\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \end{array}\right)$

$\Rightarrow -a\cos (\theta )-ib\sin (\theta )$ $=a$

$\Rightarrow a(1+\cos (\theta ))$ $=-ib\sin (\theta )$

$\Rightarrow a({\cos }^2\left({\displaystyle \frac{\theta }{2}}\right))$ $=-i\sin \left({\displaystyle \frac{\theta }{2}}\right)\cos \left({\displaystyle \frac{\theta }{2}}\right)$

$\Rightarrow a$ $=-ib\tan \left({\displaystyle \frac{\theta }{2}}\right)$

We choose $b=\cos \left(\frac{\theta }{2}\right)$.

$\Rightarrow |E=E_0⟩$ $=\left(\begin{array}{c}\hfill -i\sin \left(\frac{\theta }{2}\right)\hfill \\ \hfill \cos \left(\frac{\theta }{2}\right)\hfill \end{array}\right)$

$=-i\sin \left({\displaystyle \frac{\theta }{2}}\right)|-1⟩+\cos \left({\displaystyle \frac{\theta }{2}}\right)|1⟩$

$\Rightarrow |E=-E_0⟩$ $=\left(\begin{array}{c}\hfill -\cos \left(\frac{\theta }{2}\right)\hfill \\ \hfill i\sin \left(\frac{\theta }{2}\right)\hfill \end{array}\right)$

$=-\cos \left({\displaystyle \frac{\theta }{2}}\right)|-1⟩+i\sin \left({\displaystyle \frac{\theta }{2}}\right)|1⟩$

We used a shortcut in finding $|E=-E_0⟩$ by exploiting the orthogonality between $|E=E_0⟩$ and $|E=-E_0⟩$.

b) Right after the measurement, the state collapses to $\psi =c⟨{\widehat{P}}_{S=1}|{\psi }_{before}⟩$ whereby $c$ is the normalization constant.

${\widehat{P}}_{S=1}$ $=|1⟩⟨1|$

$\Rightarrow {\psi }_{after}$ $=|1⟩⟨1|{\psi }_{before}⟩$

$=\lambda |1⟩,\lambda \in ℂ$

$\Rightarrow {\psi }_{after}$ $=|1⟩$

$\Rightarrow |\psi (t=0)⟩$ $=|1⟩$

We perform time evolution with this initial condition.

$|\psi (t)⟩$ $=e^{-\frac{it}{\mathrm{\hslash }}\widehat{H}}|1⟩$

$e^{-\frac{it}{\mathrm{\hslash }}\widehat{H}}$ $={\displaystyle \sum _{j=1}^2}{e}^{-\frac{it}{\mathrm{\hslash }}{E}_j}{\widehat{P}}_{E_j}$

$=e^{-\frac{i{E}_{0}t}{\mathrm{\hslash }}}|E=E_0⟩⟨E=E_0|+e^{\frac{i{E}_{0}t}{\mathrm{\hslash }}}|E=-E_0⟩⟨E=-E_0|$

$\Rightarrow |\psi (t)⟩$ $=e^{-\frac{i{E}_{0}t}{\mathrm{\hslash }}}⟨E=E_0|1⟩|E=E_0⟩+e^{\frac{i{E}_{0}t}{\mathrm{\hslash }}}⟨E=-E_0|1⟩|E=-E_0⟩$

$=\cos \left({\displaystyle \frac{\theta }{2}}\right)e^{-\frac{i{E}_{0}t}{\mathrm{\hslash }}}|E=E_0⟩-i\sin \left({\displaystyle \frac{\theta }{2}}\right)e^{\frac{i{E}_{0}t}{\mathrm{\hslash }}}|E=-E_0⟩$

$P(S=1)$ $=⟨\psi (t)|{\widehat{P}}_{S=1}|\psi (t)⟩$

$={|⟨\psi (t)|1⟩|}^2$

$⟨1|\psi (t)⟩$ $={\cos }^2\left({\displaystyle \frac{\theta }{2}}\right)e^{-\frac{i{E}_{0}t}{\mathrm{\hslash }}}+{\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)e^{\frac{i{E}_{0}t}{\mathrm{\hslash }}}$

$\Rightarrow {|⟨\psi (t)|1⟩|}^2$ $={\cos }^4\left({\displaystyle \frac{\theta }{2}}\right)+{\sin }^4\left({\displaystyle \frac{\theta }{2}}\right)+{\cos }^2\left({\displaystyle \frac{\theta }{2}}\right){\sin }^2\left({\displaystyle \frac{\theta }{2}}\right)(e^{\frac{2i{E}_{0}t}{\mathrm{\hslash }}}+e^{\frac{-2i{E}_{0}t}{\mathrm{\hslash }}})$

$={\displaystyle \frac{1}{2}}(1+{\cos }^2(\theta ))+{\displaystyle \frac{1}{4}}({\sin }^2(\theta ))(e^{\frac{2i{E}_{0}t}{\mathrm{\hslash }}}+e^{\frac{-2i{E}_{0}t}{\mathrm{\hslash }}})$

$={\displaystyle \frac{1}{2}}(1+{\cos }^2(\theta ))+{\displaystyle \frac{1}{4}}({\sin }^2(\theta ))({(e^{\frac{i{E}_{0}t}{\mathrm{\hslash }}}-e^{\frac{-i{E}_{0}t}{\mathrm{\hslash }}})}^2+2)$

$={\displaystyle \frac{1}{2}}(1+{\cos }^2(\theta ))-({\sin }^2(\theta ))\left({\left({\displaystyle \frac{e^{\frac{i{E}_{0}t}{\mathrm{\hslash }}}-e^{\frac{-i{E}_{0}t}{\mathrm{\hslash }}}}{2i}}\right)}^2+{\displaystyle \frac{1}{2}}\right)$

$={\displaystyle \frac{1}{2}}(1+{\cos }^2(\theta )+{\sin }^2(\theta ))-{\sin }^2(\theta ){\sin }^2\left({\displaystyle \frac{E_{0}t}{\mathrm{\hslash }}}\right)$

$=1-{\sin }^2(\theta ){\sin }^2\left({\displaystyle \frac{E_{0}t}{\mathrm{\hslash }}}\right)$

$\Rightarrow P(S=-1)$ $={\sin }^2(\theta ){\sin }^2\left({\displaystyle \frac{E_{0}t}{\mathrm{\hslash }}}\right)$

c) For the expectation, we could use the formula $⟨\widehat{S}⟩=⟨\psi (t)|\widehat{S}|\psi (t)⟩$. However, this involves a lot of computation. So instead we can just use the formula from probability theory.

$⟨\widehat{S}⟩$ $=(-1)P(S=-1)+P(S=1)$

$⟨\widehat{S}⟩$ $=-{\sin }^2(\theta ){\sin }^2\left({\displaystyle \frac{t}{\mathrm{\hslash }}}\right)+\left(1-{\sin }^2(\theta ){\sin }^2\left({\displaystyle \frac{t}{\mathrm{\hslash }}}\right)\right)$