Question

The electron inan Hatomis in the state ψ(t)- E-E.00 e E I expressed using eigenstates of the form EJ.m) . What is the expectation value of the L- operator and the parity operator E, ,1,1) which is

Answer 1

The eigenvalue equation for $\hat{L}_z$ operator is given by
  $\hat{L}_z|E,l,m_l\rangle=m_l|E,l,m_l\rangle$
And so, for the given state
  $|\psi(t)\rangle =\frac{1}{\sqrt{3}}e^{-iE_1t/\hbar}|E_1,0,0\rangle+\sqrt{\frac{2}{3}}e^{-iE_2 t/\hbar}|E_2,1,1\rangle$
And so,
  $\hat{L}_z|\psi(t)\rangle =\frac{1}{\sqrt{3}}e^{-iE_1t/\hbar}\hat{L}_z|E_1,0,0\rangle+\sqrt{\frac{2}{3}}e^{-iE_2 t/\hbar}\hat{L}_z|E_2,1,1\rangle$
  $\Rightarrow \hat{L}_z|\psi(t)\rangle =\sqrt{\frac{2}{3}}e^{-iE_2 t/\hbar}|E_2,1,1\rangle$
And so,
  $\langle \psi(t)|\hat{L}_z|\psi(t)\rangle =\left ( \frac{1}{\sqrt{3}}e^{iE_1t/\hbar}\langle E_1,0,0|+\sqrt{\frac{2}{3}}e^{iE_2 t/\hbar}\langle E_2,1,1|\right )\sqrt{\frac{2}{3}}e^{-iE_2 t/\hbar}|E_2,1,1\rangle$
And we use the normalisation
$\langle E_1,l_1,m_{l_1}|E_2,l_2,m_{l_2}\rangle=\delta_{E_1,E_2}\delta_{l_1,l_2}\delta_{m_{l_1},m_{l_2}}$
So, we get
$\Rightarrow \langle \psi(t)|\hat{L}_z|\psi(t)\rangle =\frac{\sqrt{2}}{3}e^{i(E_1-E_2)t/\hbar}\langle E_1,0,0|E_2,1,1\rangle+\frac{2}{3}\langle E_2,1,1|E_2,1,1\rangle$
$\Rightarrow \langle \psi(t)|\hat{L}_z|\psi(t)\rangle =\frac{2}{3}$

And the eigenvalue equation for $\hat{P}$ operator is given by
  $\hat{P}|E,l,m_l\rangle=(-1)^{l}|E,l,m_l\rangle$
And so, for the given state
  $|\psi(t)\rangle =\frac{1}{\sqrt{3}}e^{-iE_1t/\hbar}|E_1,0,0\rangle+\sqrt{\frac{2}{3}}e^{-iE_2 t/\hbar}|E_2,1,1\rangle$
And so,
  $\hat{P}|\psi(t)\rangle =\frac{1}{\sqrt{3}}e^{-iE_1t/\hbar}\hat{P}|E_1,0,0\rangle+\sqrt{\frac{2}{3}}e^{-iE_2 t/\hbar}\hat{P}|E_2,1,1\rangle$
  $\Rightarrow \hat{P}|\psi(t)\rangle =\frac{1}{\sqrt{3}}e^{-iE_1t/\hbar}(-1)^{0}|E_1,0,0\rangle+\sqrt{\frac{2}{3}}e^{-iE_2 t/\hbar}(-1)^{1}|E_2,1,1\rangle$
$\Rightarrow \hat{P}|\psi(t)\rangle =\frac{1}{\sqrt{3}}e^{-iE_1t/\hbar}|E_1,0,0\rangle-\sqrt{\frac{2}{3}}e^{-iE_2 t/\hbar}|E_2,1,1\rangle$
And so,
  $\langle \psi(t)|\hat{P}|\psi(t)\rangle =\left ( \frac{1}{\sqrt{3}}e^{iE_1t/\hbar}\langle E_1,0,0|+\sqrt{\frac{2}{3}}e^{iE_2 t/\hbar}\langle E_2,1,1|\right )\left ( \frac{1}{\sqrt{3}}e^{-iE_1t/\hbar}|E_1,0,0\rangle-\sqrt{\frac{2}{3}}e^{-iE_2 t/\hbar}|E_2,1,1\rangle \right )$
And we use the normalisation
$\langle E_1,l_1,m_{l_1}|E_2,l_2,m_{l_2}\rangle=\delta_{E_1,E_2}\delta_{l_1,l_2}\delta_{m_{l_1},m_{l_2}}$
So, we get
$\Rightarrow \langle \psi(t)|\hat{P}|\psi(t)\rangle =\frac{1}{3}\langle E_1,0,0|E_1,0,0\rangle-\frac{2}{3}\langle E_2,1,1|E_2,1,1\rangle$
$\Rightarrow \langle \psi(t)|\hat{P}|\psi(t)\rangle =\frac{1}{3}-\frac{2}{3}$
  $\Rightarrow \langle \psi(t)|\hat{P}|\psi(t)\rangle = -\frac{1}{3}$​​​​