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### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering chap. 4 content from [1].

### Parity (review)

\begin{equation}\label{eqn:qmLecture12:20}

\hat{\Pi} \hat{x} \hat{\Pi} = – \hat{x}

\end{equation}

\begin{equation}\label{eqn:qmLecture12:40}

\hat{\Pi} \hat{p} \hat{\Pi} = – \hat{p}

\end{equation}

These are polar vectors, in contrast to an axial vector such as \( \BL = \Br \cross \Bp \).

\begin{equation}\label{eqn:qmLecture12:60}

\hat{\Pi}^2 = 1

\end{equation}

\begin{equation}\label{eqn:qmLecture12:80}

\Psi(x) \rightarrow \Psi(-x)

\end{equation}

If \( \antisymmetric{\hat{\Pi}}{\hat{H}} = 0 \) then all the eigenstates are either

- even: \( \hat{\Pi} \) eigenvalue is \( + 1 \).
- odd: \( \hat{\Pi} \) eigenvalue is \( – 1 \).

We are done with discrete symmetry operators for now.

### Translations

Define a (continuous) translation operator

\begin{equation}\label{eqn:qmLecture12:100}

\hat{T}_\epsilon \ket{x} = \ket{x + \epsilon}

\end{equation}

The action of this operator is sketched in fig. 1.

This is a unitary operator

\begin{equation}\label{eqn:qmLecture12:120}

\hat{T}_{-\epsilon} = \hat{T}_{\epsilon}^\dagger = \hat{T}_{\epsilon}^{-1}

\end{equation}

In a position basis, the action of this operator is

\begin{equation}\label{eqn:qmLecture12:140}

\bra{x} \hat{T}_{\epsilon} \ket{\psi} = \braket{x-\epsilon}{\psi} = \psi(x – \epsilon)

\end{equation}

\begin{equation}\label{eqn:qmLecture12:160}

\Psi(x – \epsilon) \approx \Psi(x) – \epsilon \PD{x}{\Psi}

\end{equation}

\begin{equation}\label{eqn:qmLecture12:180}

\bra{x} \hat{T}_{\epsilon} \ket{\Psi}

= \braket{x}{\Psi} – \frac{\epsilon}{\Hbar} \bra{ x} i \hat{p} \ket{\Psi}

\end{equation}

\begin{equation}\label{eqn:qmLecture12:200}

\hat{T}_{\epsilon} \approx \lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{p} }

\end{equation}

A non-infinitesimal translation can be composed of many small translations, as sketched in fig. 2.

For \( \epsilon \rightarrow 0, N \rightarrow \infty, N \epsilon = a \), the total translation operator is

\begin{equation}\label{eqn:qmLecture12:220}

\begin{aligned}

\hat{T}_{a}

&= \hat{T}_{\epsilon}^N \\

&= \lim_{\epsilon \rightarrow 0, N \rightarrow \infty, N \epsilon = a }

\lr{ 1 – \frac{\epsilon}{\Hbar} \hat{p} }^N \\

&= e^{-i a \hat{p}/\Hbar}

\end{aligned}

\end{equation}

The momentum \( \hat{p} \) is called a “Generator” generator of translations. If a Hamiltonian \( H \) is translationally invariant, then

\begin{equation}\label{eqn:qmLecture12:240}

\antisymmetric{\hat{T}_{a}}{H} = 0, \qquad \forall a.

\end{equation}

This means that momentum will be a good quantum number

\begin{equation}\label{eqn:qmLecture12:260}

\antisymmetric{\hat{p}}{H} = 0.

\end{equation}

### Rotations

Rotations form a non-Abelian group, since the order of rotations \( \hatR_1 \hatR_2 \ne \hatR_2 \hatR_1 \).

Given a rotation acting on a ket

\begin{equation}\label{eqn:qmLecture12:280}

\hatR \ket{\Br} = \ket{R \Br},

\end{equation}

observe that the action of the rotation operator on a wave function is inverted

\begin{equation}\label{eqn:qmLecture12:300}

\bra{\Br} \hatR \ket{\Psi}

=

\bra{R^{-1} \Br} \ket{\Psi}

= \Psi(R^{-1} \Br).

\end{equation}

## Example: Z axis normal rotation

Consider an infinitesimal rotation about the z-axis as sketched in fig. 3(a),(b)

\begin{equation}\label{eqn:qmLecture12:320}

\begin{aligned}

x’ &= x – \epsilon y \\

y’ &= y + \epsilon y \\

z’ &= z

\end{aligned}

\end{equation}

The rotated wave function is

\begin{equation}\label{eqn:qmLecture12:340}

\tilde{\Psi}(x,y,z)

= \Psi( x + \epsilon y, y – \epsilon x, z )

=

\Psi( x, y, z )

+

\epsilon y \underbrace{\PD{x}{\Psi}}_{i \hat{p}_x/\Hbar}

–

\epsilon x \underbrace{\PD{y}{\Psi}}_{i \hat{p}_y/\Hbar}.

\end{equation}

The state must then transform as

\begin{equation}\label{eqn:qmLecture12:360}

\ket{\tilde{\Psi}}

=

\lr{

1

+ i \frac{\epsilon}{\Hbar} \hat{y} \hat{p}_x

– i \frac{\epsilon}{\Hbar} \hat{x} \hat{p}_y

}

\ket{\Psi}.

\end{equation}

Observe that the combination \( \hat{x} \hat{p}_y – \hat{y} \hat{p}_x \) is the \( \hat{L}_z \) component of angular momentum \( \hat{\BL} = \hat{\Br} \cross \hat{\Bp} \), so the infinitesimal rotation can be written

\begin{equation}\label{eqn:qmLecture12:380}

\boxed{

\hatR_z(\epsilon) \ket{\Psi}

=

\lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{L}_z } \ket{\Psi}.

}

\end{equation}

For a finite rotation \( \epsilon \rightarrow 0, N \rightarrow \infty, \phi = \epsilon N \), the total rotation is

\begin{equation}\label{eqn:qmLecture12:420}

\hatR_z(\phi)

=

\lr{ 1 – \frac{i \epsilon}{\Hbar} \hat{L}_z }^N,

\end{equation}

or

\begin{equation}\label{eqn:qmLecture12:440}

\boxed{

\hatR_z(\phi)

=

e^{-i \frac{\phi}{\Hbar} \hat{L}_z}.

}

\end{equation}

Note that \( \antisymmetric{\hat{L}_x}{\hat{L}_y} \ne 0 \).

By construction using Euler angles or any other method, a general rotation will include contributions from components of all the angular momentum operator, and will have the structure

\begin{equation}\label{eqn:qmLecture12:480}

\boxed{

\hatR_\ncap(\phi)

=

e^{-i \frac{\phi}{\Hbar} \lr{ \hat{\BL} \cdot \ncap }}.

}

\end{equation}

### Rotationally invariant \( \hat{H} \).

Given a rotationally invariant Hamiltonian

\begin{equation}\label{eqn:qmLecture12:520}

\antisymmetric{\hat{R}_\ncap(\phi)}{\hat{H}} = 0 \qquad \forall \ncap, \phi,

\end{equation}

then every

\begin{equation}\label{eqn:qmLecture12:540}

\antisymmetric{\BL \cdot \ncap}{\hat{H}} = 0,

\end{equation}

or

\begin{equation}\label{eqn:qmLecture12:560}

\antisymmetric{L_i}{\hat{H}} = 0,

\end{equation}

Non-Abelian implies degeneracies in the spectrum.

### Time-reversal

Imagine that we have something moving along a curve at time \( t = 0 \), and ending up at the final position at time \( t = t_f \).

Imagine that we flip the direction of motion (i.e. flipping the velocity) and run time backwards so the final-time state becomes the initial state.

If the time reversal operator is designated \( \hat{\Theta} \), with operation

\begin{equation}\label{eqn:qmLecture12:580}

\hat{\Theta} \ket{\Psi} = \ket{\tilde{\Psi}},

\end{equation}

so that

\begin{equation}\label{eqn:qmLecture12:600}

\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(t)} = \ket{\Psi(0)},

\end{equation}

or

\begin{equation}\label{eqn:qmLecture12:620}

\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(0)} = \ket{\Psi(-t)}.

\end{equation}

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.