### 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)

\label{eqn:qmLecture12:20}
\hat{\Pi} \hat{x} \hat{\Pi} = – \hat{x}

\label{eqn:qmLecture12:40}
\hat{\Pi} \hat{p} \hat{\Pi} = – \hat{p}

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

\label{eqn:qmLecture12:60}
\hat{\Pi}^2 = 1

\label{eqn:qmLecture12:80}
\Psi(x) \rightarrow \Psi(-x)

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

\label{eqn:qmLecture12:100}
\hat{T}_\epsilon \ket{x} = \ket{x + \epsilon}

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

fig. 1. Translation operation.

This is a unitary operator

\label{eqn:qmLecture12:120}
\hat{T}_{-\epsilon} = \hat{T}_{\epsilon}^\dagger = \hat{T}_{\epsilon}^{-1}

In a position basis, the action of this operator is

\label{eqn:qmLecture12:140}
\bra{x} \hat{T}_{\epsilon} \ket{\psi} = \braket{x-\epsilon}{\psi} = \psi(x – \epsilon)

\label{eqn:qmLecture12:160}
\Psi(x – \epsilon) \approx \Psi(x) – \epsilon \PD{x}{\Psi}

\label{eqn:qmLecture12:180}
\bra{x} \hat{T}_{\epsilon} \ket{\Psi}
= \braket{x}{\Psi} – \frac{\epsilon}{\Hbar} \bra{ x} i \hat{p} \ket{\Psi}

\label{eqn:qmLecture12:200}
\hat{T}_{\epsilon} \approx \lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{p} }

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

fig. 2. Composition of small translations

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

\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}

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

\label{eqn:qmLecture12:240}
\antisymmetric{\hat{T}_{a}}{H} = 0, \qquad \forall a.

This means that momentum will be a good quantum number

\label{eqn:qmLecture12:260}
\antisymmetric{\hat{p}}{H} = 0.

### 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

\label{eqn:qmLecture12:280}
\hatR \ket{\Br} = \ket{R \Br},

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

\label{eqn:qmLecture12:300}
\bra{\Br} \hatR \ket{\Psi}
=
\bra{R^{-1} \Br} \ket{\Psi}
= \Psi(R^{-1} \Br).

## Example: Z axis normal rotation

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

\label{eqn:qmLecture12:320}
\begin{aligned}
x’ &= x – \epsilon y \\
y’ &= y + \epsilon y \\
z’ &= z
\end{aligned}

The rotated wave function is

\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}.

The state must then transform as

\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}.

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

\label{eqn:qmLecture12:380}
\boxed{
\hatR_z(\epsilon) \ket{\Psi}
=
\lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{L}_z } \ket{\Psi}.
}

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

\label{eqn:qmLecture12:420}
\hatR_z(\phi)
=
\lr{ 1 – \frac{i \epsilon}{\Hbar} \hat{L}_z }^N,

or
\label{eqn:qmLecture12:440}
\boxed{
\hatR_z(\phi)
=
e^{-i \frac{\phi}{\Hbar} \hat{L}_z}.
}

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

\label{eqn:qmLecture12:480}
\boxed{
\hatR_\ncap(\phi)
=
e^{-i \frac{\phi}{\Hbar} \lr{ \hat{\BL} \cdot \ncap }}.
}

### Rotationally invariant $$\hat{H}$$.

Given a rotationally invariant Hamiltonian

\label{eqn:qmLecture12:520}
\antisymmetric{\hat{R}_\ncap(\phi)}{\hat{H}} = 0 \qquad \forall \ncap, \phi,

then every

\label{eqn:qmLecture12:540}
\antisymmetric{\BL \cdot \ncap}{\hat{H}} = 0,

or
\label{eqn:qmLecture12:560}
\antisymmetric{L_i}{\hat{H}} = 0,

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 $$t = t_f$$.

fig. 4. Time reversal trajectory.

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

\label{eqn:qmLecture12:580}
\hat{\Theta} \ket{\Psi} = \ket{\tilde{\Psi}},

so that

\label{eqn:qmLecture12:600}
\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(t)} = \ket{\Psi(0)},

or

\label{eqn:qmLecture12:620}
\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(0)} = \ket{\Psi(-t)}.

# References

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