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Q: [1] pr 4.11
(a) Given a time reversal invariant Hamiltonian, show that for any energy eigenket
\begin{equation}\label{eqn:totallyAsymmetricPotential:20}
\expectation{\BL} = 0.
\end{equation}
(b) If the wave function of such a state is expanded as
\begin{equation}\label{eqn:totallyAsymmetricPotential:40}
\sum_{l,m} F_{l m} Y_{l m}(\theta, \phi),
\end{equation}
what are the phase restrictions on \( F_{lm} \)?
A: part (a)
For a time reversal invariant Hamiltonian \( H \) we have
\begin{equation}\label{eqn:totallyAsymmetricPotential:60}
H \Theta = \Theta H.
\end{equation}
If \( \ket{\psi} \) is an energy eigenstate with eigenvalue \( E \), we have
\begin{equation}\label{eqn:totallyAsymmetricPotential:80}
\begin{aligned}
H \Theta \ket{\psi}
&= \Theta H \ket{\psi} \\
&= \lambda \Theta \ket{\psi},
\end{aligned}
\end{equation}
so \( \Theta \ket{\psi} \) is also an eigenvalue of \( H \), so can only differ from \( \ket{\psi} \) by a phase factor. That is
\begin{equation}\label{eqn:totallyAsymmetricPotential:100}
\begin{aligned}
\ket{\psi’}
&=
\Theta \ket{\psi} \\
&= e^{i\delta} \ket{\psi}.
\end{aligned}
\end{equation}
Now consider the expectation of \( \BL \) with respect to a time reversed state
\begin{equation}\label{eqn:totallyAsymmetricPotential:120}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\bra{ \psi} \Theta^{-1} \BL \Theta \ket{\psi} \\
&=
\bra{ \psi} (-\BL) \ket{\psi},
\end{aligned}
\end{equation}
however, we also have
\begin{equation}\label{eqn:totallyAsymmetricPotential:140}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\lr{ \bra{ \psi} e^{-i\delta} } \BL \lr{ e^{i\delta} \ket{\psi} } \\
&=
\bra{\psi} \BL \ket{\psi},
\end{aligned}
\end{equation}
so we have \( \bra{\psi} \BL \ket{\psi} = -\bra{\psi} \BL \ket{\psi} \) which is only possible if \( \expectation{\BL} = \bra{\psi} \BL \ket{\psi} = 0\).
A: part (b)
Consider the expansion of the wave function of a time reversed energy eigenstate
\begin{equation}\label{eqn:totallyAsymmetricPotential:160}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} e^{i\delta} \ket{\psi} \\
&=
e^{i\delta} \braket{\Bx}{\psi},
\end{aligned}
\end{equation}
and then consider the same state expanded in the position basis
\begin{equation}\label{eqn:totallyAsymmetricPotential:180}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \ket{\Bx’}\bra{\Bx’} } \ket{\psi} \\
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} } \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \Theta \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \ket{\Bx’} \\
&=
\int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \braket{\Bx}{\Bx’} \\
&=
\int d^3 \Bx’ \braket{\psi}{\Bx’} \delta(\Bx- \Bx’) \\
&=
\braket{\psi}{\Bx}.
\end{aligned}
\end{equation}
This demonstrates a relationship between the wave function and its complex conjugate
\begin{equation}\label{eqn:totallyAsymmetricPotential:200}
\braket{\Bx}{\psi} = e^{-i\delta} \braket{\psi}{\Bx}.
\end{equation}
Now expand the wave function in the spherical harmonic basis
\begin{equation}\label{eqn:totallyAsymmetricPotential:220}
\begin{aligned}
\braket{\Bx}{\psi}
&=
\int d\Omega \braket{\Bx}{\ncap}\braket{\ncap}{\psi} \\
&=
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) \\
&=
e^{-i\delta}
\lr{
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) }^\conj \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj Y_{lm}^\conj(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj (-1)^m Y_{l,-m}(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{l,-m}(r)}^\conj (-1)^m Y_{l,m}(\theta, \phi),
\end{aligned}
\end{equation}
so the \( F_{lm} \) functions are constrained by
\begin{equation}\label{eqn:totallyAsymmetricPotential:240}
F_{lm}(r) = e^{-i\delta} \lr{ F_{l,-m}(r)}^\conj (-1)^m.
\end{equation}
References
[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.