Quantum Mechanics Term 1 Problems Class 1

Dr A. Donos

(Michaelmas 2021)

Let $\mathcal{H}$ have dimension 2. Suppose that $\{\ket{1},\ket{2}\}$ form an orthonormal basis for $\mathcal{H}$ . $\hat{S}$ is a Hermitian operator s.t.

	$\displaystyle\hat{S}\ket{1}$	$\displaystyle=\ket{1}-i\ket{2}$
	$\displaystyle\hat{S}\ket{2}$	$\displaystyle=\ket{2}+x\ket{1},\ \ x\in\mathbb{C}$

a) Find the matrix form of $\hat{S}$
b) Fix the constant $x$
c) Find the eigenvalues $S_{i}$ of $\hat{S}$ and the normalised eigenvectors $\ket{S}_{i}$ in the $\{\ket{1},\ket{2}\}$ basis.
d) Take $\ket{\psi}=c(\ket{1}+\ket{2})$ , fix the constant $c$ and find the probability of observing the measurements of $\hat{S}$ .

Answers:
a) We know $S_{ij}=\braket{i}{\hat{S}}{j}$

	$\displaystyle S_{11}$	$\displaystyle=\braket{1}{\hat{S}}{1}=\braket{1}{1}-i\braket{1}{2}=1$
	$\displaystyle S_{12}$	$\displaystyle=\braket{1}{\hat{S}}{2}=\braket{1}{2}+x\braket{1}{1}=x$
	$\displaystyle S_{21}$	$\displaystyle=\braket{2}{\hat{S}}{1}=\braket{2}{1}-i\braket{2}{2}=-i$
	$\displaystyle S_{22}$	$\displaystyle=\braket{2}{\hat{S}}{2}=\braket{2}{2}+x\braket{2}{1}=1$
	$\displaystyle\Rightarrow S$	$\displaystyle=\left(\begin{array}[]{c c}1&x\\ -i&1\end{array}\right)$

b) Since the operator is Hermitian so is its matrix representation i.e $S_{ij}=S_{ji}^{*}$ . Hence $x=i$ .

c) Now that we have the matrix form of $\hat{S}$ , its just a matter of procedure to find the eigenvalues and eigenvectors.

	$\displaystyle\left(\begin{array}[]{c c}1-\lambda&i\\ -i&1-\lambda\end{array}\right)\left(\begin{array}[]{c}a\\ b\end{array}\right)$	$\displaystyle=0$
	$\displaystyle(1-\lambda)^{2}-(-i)i$	$\displaystyle=0$
	$\displaystyle\Rightarrow\lambda(\lambda-2)$	$\displaystyle=0$
	$\displaystyle\Rightarrow\lambda$	$\displaystyle=0\ or\ 2$

In the case $\lambda=0$ ,

	$\displaystyle\left(\begin{array}[]{c c}1&i\\ -i&1\end{array}\right)\left(\begin{array}[]{c}a\\ b\end{array}\right)$	$\displaystyle=0$
	$\displaystyle\Rightarrow a+ib$	$\displaystyle=0$
	$\displaystyle-ia+b$	$\displaystyle=0$
	$\displaystyle\Rightarrow a$	$\displaystyle=-ix,b=x,\ \ x\in\mathbb{C}$

In the case $\lambda=2$ ,

	$\displaystyle\left(\begin{array}[]{c c}-1&i\\ -i&-2\end{array}\right)\left(\begin{array}[]{c}a\\ b\end{array}\right)$	$\displaystyle=0$
	$\displaystyle\Rightarrow-a+ib$	$\displaystyle=0$
	$\displaystyle-ia-b$	$\displaystyle=0$
	$\displaystyle\Rightarrow a$	$\displaystyle=y,\ b=-iy,\ \ y\in\mathbb{C}$

Hence the normalised eigenvectors are

	$\displaystyle\ket{S_{0}}$	$\displaystyle=\frac{1}{\sqrt{2}}(\ket{1}+i\ket{2})$
	$\displaystyle\ket{S_{2}}$	$\displaystyle=\frac{1}{\sqrt{2}}(i\ket{1}+\ket{2})$

As a sanity check, we can evaluate the inner product between the two to verify if they are orthogonal.

\displaystyle\braket{S_{0}}{S_{2}}=\frac{1}{2}(i\braket{S_{0}}{1}+\braket{S_{0% }}{2})=\frac{1}{2}(i-i)=0

d) We first fix the constant by demanding the state has unit norm.

	$\displaystyle\braket{\psi}{\psi}$	$\displaystyle=2\|c\|^{2}$
	$\displaystyle\Rightarrow\|c\|$	$\displaystyle=\frac{1}{\sqrt{2}}$

To find the probabilities we use the projection operator and the fact that

	$\displaystyle\hat{P}_{S_{i}}$	$\displaystyle=\ket{S_{i}}\bra{S_{i}}$
	$\displaystyle P(S_{i})$	$\displaystyle=\braket{\psi}{\hat{P}_{S_{i}}}{\psi}$
	$\displaystyle\Rightarrow P(S_{0})$	$\displaystyle=\braket{\psi}{S_{0}}\braket{S_{0}}{\psi}$
		$\displaystyle=\|\braket{\psi}{S_{0}}\|^{2}$
	$\displaystyle\braket{\psi}{S_{0}}$	$\displaystyle=\frac{1}{2}(\braket{1}{S_{0}}+\braket{2}{S_{0}})$
		$\displaystyle=\frac{1}{2}(1+i)$
	$\displaystyle\Rightarrow P(S_{0})$	$\displaystyle=\frac{1}{2}$
	$\displaystyle\braket{\psi}{S_{2}}$	$\displaystyle=\frac{1}{2}(\braket{1}{S_{2}}+\braket{2}{S_{2}})$
		$\displaystyle=\frac{1}{2}(i+1)$
	$\displaystyle\Rightarrow P(S_{2})$	$\displaystyle=\frac{1}{2}$

As expected the two probabilities sum to 1 since the only possible measurements of $\hat{S}$ are its eigenvalues which are 0 and 2 in this case.