angular momentum

Third update of aggregate notes for phy1520, Graduate Quantum Mechanics.

November 9, 2015 phy1520 , , , , , , , , , , , , , , , ,

I’ve posted a third update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 13, my solutions for the third problem set, and some additional worked practice problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

PHY1520H Graduate Quantum Mechanics. Lecture 13: Time reversal (cont.), and angular momentum. Taught by Prof. Arun Paramekanti

November 7, 2015 phy1520 , , , ,

[Click here for a PDF of this post with nicer formatting]

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 \textchapref{{4}}, \textchapref{{3}} [1] content.

Time reversal (cont.)

Given a time reversed state

\begin{equation}\label{eqn:qmLecture13:20}
\ket{\tilde{\Psi}(t)} = \Theta \ket{\Psi(0)}
\end{equation}

which can alternately be written

\begin{equation}\label{eqn:qmLecture13:40}
\Theta^{-1} \ket{\tilde{\Psi}(t)} = \ket{\Psi(-t)} = e^{i \hat{H} t/\Hbar} \ket{\Psi(0)}
\end{equation}

The left hand side can be expanded as the evolution of the state as found at time \( -t \)

\begin{equation}\label{eqn:qmLecture13:60}
\begin{aligned}
\Theta^{-1} \ket{\tilde{\Psi}(t)}
&=
\Theta^{-1} e^{-i \hat{H} t/\Hbar} \ket{\tilde{\Psi}(-t)} \\
&=
\Theta^{-1} e^{-i \hat{H} t/\Hbar} \Theta \ket{\Psi(0)}.
\end{aligned}
\end{equation}

To first order for a small time increment \( \delta t \), we have

\begin{equation}\label{eqn:qmLecture13:80}
\lr{ 1 + i \frac{\hat{H}}{\Hbar} \delta t } \ket{\Psi(0)} =
\Theta^{-1} \lr{ 1 – i \frac{\hat{H}}{\Hbar} \delta t } \Theta \ket{\Psi(0)},
\end{equation}

or

\begin{equation}\label{eqn:qmLecture13:120}
i \frac{\hat{H}}{\Hbar} \delta t \ket{\Psi(0)}
=
\Theta^{-1} (- i) \frac{\hat{H}}{\Hbar} \delta t \Theta \ket{\Psi(0)}.
\end{equation}

Since this holds for any state \( \ket{\Psi(0)} \), the time reversal operator satisfies

\begin{equation}\label{eqn:qmLecture13:140}
i \hat{H}
=
\Theta^{-1} (- i) \hat{H} \Theta.
\end{equation}

Note that the factors of \( i \) have not been canceled on purpose, since we are allowing for the time reversal operator to not necessarily commute with imaginary numbers.

There are two possible solutions

  • If \( \Theta \) is unitary where \( \Theta i = i \Theta \), then

    \begin{equation}\label{eqn:qmLecture13:160}
    \hat{H}
    =
    -\Theta^{-1} \hat{H} \Theta,
    \end{equation}

    or
    \begin{equation}\label{eqn:qmLecture13:180}
    \Theta \hat{H}
    =
    – \hat{H} \Theta.
    \end{equation}

    Consider the implications of this on energy eigenstates
    \begin{equation}\label{eqn:qmLecture13:200}
    \hat{H} \ket{\Psi_n} = E_n \ket{\Psi_n},
    \end{equation}

    \begin{equation}\label{eqn:qmLecture13:220}
    \Theta \hat{H} \ket{\Psi_n} = E_n \Theta \ket{\Psi_n},
    \end{equation}

    but

    \begin{equation}\label{eqn:qmLecture13:240}
    -\hat{H} \Theta \ket{\Psi_n} = E_n \Theta \ket{\Psi_n},
    \end{equation}

    or

    \begin{equation}\label{eqn:qmLecture13:260}
    \hat{H} \lr{ \Theta \ket{\Psi_n}} = -E_n \lr{ \Theta \ket{\Psi_n} }.
    \end{equation}

    This would mean that \( \lr{ \Theta \ket{\Psi_n}} \) is an eigenket of \( \hat{H} \), but with a negative energy eigenvalue.

  • \( \Theta \) is antiunitary, where \( \Theta i = -i \Theta \).

    This time
    \begin{equation}\label{eqn:qmLecture13:280}
    i \hat{H} = i \Theta^{-1} \hat{H} \Theta,
    \end{equation}

    so

    \begin{equation}\label{eqn:qmLecture13:300}
    \Theta \hat{H} = \hat{H} \Theta.
    \end{equation}

    Acting on an energy eigenket, we’ve got

    \begin{equation}\label{eqn:qmLecture13:1400}
    \Theta \hat{H} \ket{\Psi_n}
    =
    E_n \lr{ \Theta \ket{\Psi_n} },
    \end{equation}

    and
    \begin{equation}\label{eqn:qmLecture13:1420}
    \lr{ \hat{H} \Theta } \ket{\Psi_n}
    =
    \hat{H} \lr{ \Theta \ket{\Psi_n} },
    \end{equation}

    so \( \Theta \ket{\Psi_n} \) is an eigenstate with energy \( E_n \).

What properties do we expect from \( \Theta \)?

We expect
\begin{equation}\label{eqn:qmLecture13:320}
\begin{aligned}
\hat{x} &\rightarrow \hat{x} \\
\hat{p} &\rightarrow -\hat{p} \\
\hat{\BL} &\rightarrow -\hat{\BL}
\end{aligned}
\end{equation}

where we have a sign flip in the time dependent momentum operator (and therefore angular momentum), but not for position. If we have

\begin{equation}\label{eqn:qmLecture13:340}
\Theta^{-1} \hat{x} \Theta = \hat{x},
\end{equation}

if that’s true, then how about the momentum operator in the position basis
\begin{equation}\label{eqn:qmLecture13:360}
\begin{aligned}
\Theta^{-1} \hat{p} \Theta
&=
\Theta^{-1} \lr{ -i \Hbar \PD{x}{} } \Theta \\
&=
\Theta^{-1} \lr{ -i \Hbar } \Theta \PD{x}{} \\
&=
i \Hbar \Theta^{-1} \Theta \PD{x}{} \\
&=
-\hat{p}.
\end{aligned}
\end{equation}

How about the \( x,p \) commutator? For that we have

\begin{equation}\label{eqn:qmLecture13:380}
\begin{aligned}
\Theta^{-1} \antisymmetric{\hat{x}}{\hat{p}} \Theta
&=
\Theta^{-1} \lr{ i \Hbar } \Theta \\
&=
-i \Hbar \Theta^{-1} \Theta \\
&=
– \antisymmetric{\hat{x}}{\hat{p}}.
\end{aligned}
\end{equation}

For the the angular momentum operators

\begin{equation}\label{eqn:qmLecture13:420}
\hat{L}_i = \epsilon_{ijk} \hat{r}_j \hat{p}_k,
\end{equation}

the time reversal operator should flip the sign due to its action on \( \hat{p}_k \).

Time reversal acting on spin 1/2 (Fermions). Attempt I.

Consider two spin states \( \ket{\uparrow}, \ket{\downarrow} \). What should the action of the time reversal operator on such a state be? Let’s (incorrectly) start by supposing that the time reversal operator effects are

\begin{equation}\label{eqn:qmLecture13:440}
\begin{aligned}
\Theta \ket{\uparrow} &= \ket{\downarrow} \\
\Theta \ket{\downarrow} &= \ket{\uparrow}.
\end{aligned}
\end{equation}

Given a general state
so that if

\begin{equation}\label{eqn:qmLecture13:740}
\ket{\Psi} = a \ket{\uparrow} + b \ket{\downarrow},
\end{equation}

the action of the time reversal operator would be

\begin{equation}\label{eqn:qmLecture13:760}
\Theta \ket{\Psi} = a^\conj \ket{\downarrow} + b^\conj \ket{\uparrow}.
\end{equation}

That action is:

\begin{equation}\label{eqn:qmLecture13:460}
\begin{aligned}
a \rightarrow b^\conj \\
b \rightarrow a^\conj
\end{aligned}
\end{equation}

Let’s consider whether or not such an action a spin operator with properties

\begin{equation}\label{eqn:qmLecture13:480}
\antisymmetric{\hat{S}_i}{\hat{S}_j} = i \epsilon_{ijk} \hat{S}_k.
\end{equation}

produce the desired inversion of sign

\begin{equation}\label{eqn:qmLecture13:500}
\Theta^{-1} \hat{S}_i \Theta = – \hat{S}_i.
\end{equation}

The expectations of the spin operators (without any application of time reversal) are

\begin{equation}\label{eqn:qmLecture13:1440}
\begin{aligned}
\bra{\Psi} \hat{S}_x \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\sigma_x
\lr{ a \ket{\uparrow} + b \ket{\downarrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\lr{ a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj b + b^\conj a },
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:qmLecture13:1460}
\begin{aligned}
\bra{\Psi} \hat{S}_y \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\sigma_y
\lr{ a \ket{\uparrow} + b \ket{\downarrow} } \\
&=
\frac{i\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\lr{ a \ket{\downarrow} – b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2 i} \lr{ a^\conj b – b^\conj a },
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:qmLecture13:1480}
\begin{aligned}
\bra{\Psi} \hat{S}_z \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\uparrow} + b^\conj \bra{\downarrow} }
\sigma_z
\lr{ a \ket{\uparrow} – b \ket{\downarrow} } \\
&=
\frac{\Hbar}{2} \lr{ \Abs{a}^2 – \Abs{b}^2 }
\end{aligned}
\end{equation}

The time reversed actions are

\begin{equation}\label{eqn:qmLecture13:1560}
\begin{aligned}
\bra{\Psi} \Theta^{-1} \hat{S}_x \Theta \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\sigma_x
\lr{ a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\lr{ a \ket{\uparrow} + b \ket{\downarrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj b + b^\conj a },
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:qmLecture13:1580}
\begin{aligned}
\bra{\Psi} \Theta^{-1} \hat{S}_y \Theta \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\sigma_y
\lr{ a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{i\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\lr{ -a \ket{\uparrow} + b \ket{\downarrow} } \\
&=
\frac{\Hbar}{2 i} \lr{ -a^\conj b + b^\conj a },
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:qmLecture13:1600}
\begin{aligned}
\bra{\Psi} \Theta^{-1} \hat{S}_z \Theta \ket{\Psi}
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\sigma_z
\lr{ a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2}
\lr{ a^\conj \bra{\downarrow} + b^\conj \bra{\uparrow} }
\lr{ -a \ket{\downarrow} + b \ket{\uparrow} } \\
&=
\frac{\Hbar}{2} \lr{ -\Abs{a}^2 + \Abs{b}^2 }
\end{aligned}
\end{equation}

We see that this is not right, because the sign for the x component has not been flipped.

Spin 1/2 (Fermions). Attempt II.

Again assuming

\begin{equation}\label{eqn:qmLecture13:580}
\ket{\Psi} = a \ket{\uparrow} + b \ket{\downarrow},
\end{equation}

now try the action

\begin{equation}\label{eqn:qmLecture13:780}
\Theta \ket{\Psi} = a^\conj \ket{\downarrow} – b^\conj \ket{\uparrow}.
\end{equation}

This is the action:

\begin{equation}\label{eqn:qmLecture13:600}
\begin{aligned}
a \rightarrow -b^\conj \\
b \rightarrow a^\conj
\end{aligned}
\end{equation}

The correct action of time reversal on the basis states (up to a phase choice) is

\begin{equation}\label{eqn:qmLecture13:630}
\boxed{
\begin{aligned}
\Theta \ket{\uparrow} &= \ket{\downarrow} \\
\Theta \ket{\downarrow} &= -\ket{\uparrow} \\
\end{aligned}
}
\end{equation}

Note that acting the time reversal operator twice has the effects

\begin{equation}\label{eqn:qmLecture13:660}
\Theta^2 \ket{\uparrow} = \Theta \ket{\downarrow} = – \ket{\uparrow}
\end{equation}
\begin{equation}\label{eqn:qmLecture13:680}
\Theta^2 \ket{\downarrow} = \Theta (-\ket{\uparrow}) = – \ket{\uparrow}.
\end{equation}

We end up with the same state we started with, but with the opposite sign. This means that as an operator

\begin{equation}\label{eqn:qmLecture13:700}
\boxed{
\Theta^2 = -1.
}
\end{equation}

This is try for half integer particles (Fermions) \( S = 1/2, 3/2, 5/2, \cdots \), but for Bosons with integer spin \( S \).

\begin{equation}\label{eqn:qmLecture13:720}
\boxed{
\Theta^2 = 1.
}
\end{equation}

Kramer’s degeneracy for Spin 1/2 (Fermions)

Suppose we imagine there is state for which the action of the time reversal operator products the same state, just different in phase

\begin{equation}\label{eqn:qmLecture13:800}
\begin{aligned}
\Theta \ket{\Psi_n}
&= \ket{\tilde{\Psi}_n} \\
&= e^{i \delta} \ket{\tilde{\Psi}_n},
\end{aligned}
\end{equation}

then
\begin{equation}\label{eqn:qmLecture13:840}
\begin{aligned}
\Theta^2 \ket{\Psi_n}
&= \Theta e^{i \delta} \ket{\tilde{\Psi}_n} \\
&= e^{i \delta} e^{i \delta} \ket{\tilde{\Psi}_n},
\end{aligned}
\end{equation}

but

\begin{equation}\label{eqn:qmLecture13:860}
\begin{aligned}
\Theta e^{i \delta} \ket{\tilde{\Psi}_n}
&=
e^{-i \delta} \Theta \ket{\tilde{\Psi}_n} \\
&=
e^{-i \delta} e^{i \delta} \ket{\tilde{\Psi}_n} \\
&=
\ket{\tilde{\Psi}_n}
\ne
– \ket{\tilde{\Psi}_n}.
\end{aligned}
\end{equation}

This is a contradiction, so we must have at least a two-fold degeneracy. This is called Kramer’s degeneracy. In the homework we will show that this is not the case for integer spin particles.

Angular momentum

In classical mechanics the (orbital) angular momentum is

\begin{equation}\label{eqn:qmLecture13:880}
\BL = \Br \cross \Bp.
\end{equation}

Here “orbital” is to distinguish from spin angular momentum.

In quantum mechanics, the mapping to operators, in component form, is

\begin{equation}\label{eqn:qmLecture13:900}
\hat{L}_i = \epsilon_{ijk} \hat{r}_j \hat{p}_k.
\end{equation}

These operators do not commute
\begin{equation}\label{eqn:qmLecture13:920}
\antisymmetric{\hat{L}_i}{\hat{L}_j}
=
i \Hbar \epsilon_{ijk} \hat{L}_k.
\end{equation}

which means that we can’t simultaneously determine \( \hat{L}_i \) for all \( i \).

Aside: In quantum mechanics, we define an operator \( \hat{\BV} \) to be a vector operator if

\begin{equation}\label{eqn:qmLecture13:940}
\antisymmetric{\hat{L}_i}{\hatV_j}
=
i \Hbar \epsilon_{ijk} \hatV_k.
\end{equation}

The commutator of the squared angular momentum operator with any \( \hat{L}_i \), say \( \hat{L}_x \) is zero

\begin{equation}\label{eqn:qmLecture13:960}
\begin{aligned}
\antisymmetric{
\hat{L}_x^2 +
\hat{L}_y^2 +
\hat{L}_z^2
}
{\hat{L}_x}
&=
\hat{L}_y \hat{L}_y \hat{L}_x
– \hat{L}_x \hat{L}_y \hat{L}_y
+
\hat{L}_z \hat{L}_z \hat{L}_x
– \hat{L}_x \hat{L}_z \hat{L}_z \\
&=
\hat{L}_y \lr{ \antisymmetric{\hat{L}_y}{\hat{L}_x} + {\hat{L}_x \hat{L}_y} }
-\lr{ \antisymmetric{\hat{L}_x}{\hat{L}_y} + {\hat{L}_y \hat{L}_x} } \hat{L}_y \\
&\quad +\hat{L}_z \lr{ \antisymmetric{\hat{L}_z}{\hat{L}_x} + {\hat{L}_x \hat{L}_z} }
-\lr{ \antisymmetric{\hat{L}_x}{\hat{L}_z} + {\hat{L}_z \hat{L}_x} } \hat{L}_z \\
&=
\hat{L}_y \antisymmetric{\hat{L}_y}{\hat{L}_x}
-\antisymmetric{\hat{L}_x}{\hat{L}_y} \hat{L}_y
+\hat{L}_z \antisymmetric{\hat{L}_z}{\hat{L}_x}
-\antisymmetric{\hat{L}_x}{\hat{L}_z} \hat{L}_z \\
&=
i \Hbar \lr{
-\hat{L}_y \hat{L}_z
– \hat{L}_z \hat{L}_y
+\hat{L}_z \hat{L}_y
+ \hat{L}_y \hat{L}_z
} \\
&=
0.
\end{aligned}
\end{equation}

Suppose we have a state \( \ket{\Psi} \) with a well defined \( \hat{L}_z \) eigenvalue and well defined \( \hat{\BL^2} \) eigenvalue, written as

\begin{equation}\label{eqn:qmLecture13:1000}
\ket{\Psi} = \ket{a, b},
\end{equation}

where the label \( a \) is used for the eigenvalue of \( \hat{\BL}^2 \) and \( b \) labels the eigenvalue of \( \hat{L}_z \). Then

\begin{equation}\label{eqn:qmLecture13:1020}
\begin{aligned}
\hat{\BL}^2 \ket{a , b} &= \Hbar^2 a \ket{a ,b} \\
\hat{L}_z \ket{a , b} &= \Hbar b \ket{a ,b}.
\end{aligned}
\end{equation}

Things aren’t so nice when we act with other angular momentum operators, producing a scrambled mess

\begin{equation}\label{eqn:qmLecture13:1040}
\begin{aligned}
\hat{L}_x \ket{a , b} &= \sum_{a’, b’} \mathcal{A}^x_{a, b, a’, b’} \ket{a’, b’} \\
\hat{L}_y \ket{a , b} &= \sum_{a’, b’} \mathcal{A}^y_{a, b, a’, b’} \ket{a’, b’} \\
\end{aligned}
\end{equation}

With this representation, we have

\begin{equation}\label{eqn:qmLecture13:1060}
\hat{L}_x \hat{\BL}^2 \ket{a, b}
=
\hat{L}_x \Hbar^2 a
\sum_{a’, b’} \mathcal{A}^x_{a, b, a’, b’} \ket{a’, b’}.
\end{equation}

\begin{equation}\label{eqn:qmLecture13:1080}
\hat{\BL}^2 \hat{L}_x \ket{a, b}
=
\Hbar^2
\sum_{a’, b’} a’ \mathcal{A}^x_{a, b, a’, b’} \ket{a’, b’}.
\end{equation}

Since \( \hat{\BL}^2, \hat{L}_x \) commute, we must have

\begin{equation}\label{eqn:qmLecture13:1100}
\mathcal{A}^x_{a, b, a’, b’} = \delta_{a, a’} \mathcal{A}^x_{a’; b, b’},
\end{equation}

and similarly
\begin{equation}\label{eqn:qmLecture13:1120}
\mathcal{A}^y_{a, b, a’, b’} = \delta_{a, a’} \mathcal{A}^y_{a’; b, b’}.
\end{equation}

Simplifying things we can write the action of \( \hat{L}_x, \hat{L}_y \) on the state as

\begin{equation}\label{eqn:qmLecture13:1140}
\begin{aligned}
\hat{L}_x \ket{a , b} &= \sum_{ b’} \mathcal{A}^x_{a; b, b’} \ket{a, b’} \\
\hat{L}_y \ket{a , b} &= \sum_{ b’} \mathcal{A}^y_{a; b, b’} \ket{a, b’} \\
\end{aligned}
\end{equation}

Let’s define
\begin{equation}\label{eqn:qmLecture13:1160}
\begin{aligned}
\hat{L}_{+} &\equiv \hat{L}_x + i \hat{L}_y \\
\hat{L}_{-} &\equiv \hat{L}_x – i \hat{L}_y \\
\end{aligned}
\end{equation}

Because these are sums of \( \hat{L}_x, \hat{L}_y \) they must also commute with \( \hat{\BL}^2 \)

\begin{equation}\label{eqn:qmLecture13:1180}
\antisymmetric{\hat{\BL}^2}{\hat{L}_{\pm}} = 0.
\end{equation}

The commutators with \( \hat{L}_z \) are non-zero

\begin{equation}\label{eqn:qmLecture13:1740}
\begin{aligned}
\antisymmetric{\hat{L}_z}{\hat{L}_{\pm}}
&=
\hat{L}_z \lr{ \hat{L}_x \pm i \hat{L}_y }
– \lr{ \hat{L}_x \pm i \hat{L}_y } \hat{L}_z \\
&=
\antisymmetric{\hat{L}_z}{\hat{L}_x}
\pm i
\antisymmetric{\hat{L}_z}{\hat{L}_y} \\
&=
i \Hbar \lr{
\hat{L}_y \mp i \hat{L}_x
} \\
&=
\Hbar \lr{ i \hat{L}_y \pm \hat{L}_x } \\
&=
\pm \Hbar \lr{ \hat{L}_x \pm i \hat{L}_y } \\
&=
\pm \Hbar \hat{L}_{\pm}.
\end{aligned}
\end{equation}

Explicitly, that is

\begin{equation}\label{eqn:qmLecture13:1220}
\begin{aligned}
\hat{L}_z \hat{L}_{+} – \hat{L}_{+} \hat{L}_z &= \Hbar \hat{L}_{+} \\
\hat{L}_z \hat{L}_{-} – \hat{L}_{-} \hat{L}_z &= -\Hbar \hat{L}_{-}
\end{aligned}
\end{equation}

Now we are set to compute actions of these (assumed) raising and lowering operators on the eigenstate of \( \hat{L}_z, \hat{\BL}^2 \)

\begin{equation}\label{eqn:qmLecture13:1240}
\begin{aligned}
\hat{L}_z \hat{L}_{\pm} \ket{a, b}
&=
\Hbar \hat{L}_{\pm} \ket{a,b} \pm \hat{L}_{\pm} \hat{L}_z \ket{a,b} \\
&=
\Hbar \hat{L}_{\pm} \ket{a,b} \pm \Hbar b \hat{L}_{\pm} \ket{a,b} \\
&=
\Hbar \lr{ b \pm 1 } \hat{L}_{\pm} \ket{a, b} .
\end{aligned}
\end{equation}

There must be a proportionality of the form

\begin{equation}\label{eqn:qmLecture13:1260}
\ket{\hat{L}_{\pm}} \propto \ket{a, b \pm 1},
\end{equation}

The products of the raising and lowering operators are

\begin{equation}\label{eqn:qmLecture13:1280}
\begin{aligned}
\hat{L}_{-} \hat{L}_{+}
&=
\lr{ \hat{L}_x – i \hat{L}_y }
\lr{ \hat{L}_x + i \hat{L}_y } \\
&=
\hat{L}_x^2 + \hat{L}_y^2 + i \hat{L}_x \hat{L}_y – i \hat{L}_y \hat{L}_x \\
&=
\lr{ \hat{\BL}^2 – \hat{L}_z^2 } + i \antisymmetric{\hat{L}_x}{\hat{L}_y} \\
&=
\hat{\BL}^2 – \hat{L}_z^2 – \Hbar \hat{L}_z,
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:qmLecture13:1300}
\begin{aligned}
\hat{L}_{+} \hat{L}_{-}
&=
\lr{ \hat{L}_x + i \hat{L}_y }
\lr{ \hat{L}_x – i \hat{L}_y } \\
&=
\hat{L}_x^2 + \hat{L}_y^2 – i \hat{L}_x \hat{L}_y + i \hat{L}_y \hat{L}_x \\
&=
\lr{ \hat{\BL}^2 – \hat{L}_z^2 } – i \antisymmetric{\hat{L}_x}{\hat{L}_y} \\
&=
\hat{\BL}^2 – \hat{L}_z^2 + \Hbar \hat{L}_z,
\end{aligned}
\end{equation}

So we must have

\begin{equation}\label{eqn:qmLecture13:1320}
\begin{aligned}
0
&\le \bra{a, b} \hat{L}_{-} \hat{L}_{+} \ket{a, b} \\
&=
\bra{a, b}
\lr{ \hat{\BL}^2 – \hat{L}_z^2 – \Hbar \hat{L}_z }
\ket{a, b} \\
&=
\Hbar^2 a – \Hbar^2 b^2 – \Hbar^2 b,
\end{aligned}
\end{equation}

and

\begin{equation}\label{eqn:qmLecture13:1340}
\begin{aligned}
0
&\le \bra{a, b} \hat{L}_{+} \hat{L}_{-} \ket{a, b} \\
&=
\bra{a, b}
\lr{ \hat{\BL}^2 – \hat{L}_z^2 + \Hbar \hat{L}_z }
\ket{a, b} \\
&=
\Hbar^2 a – \Hbar^2 b^2 + \Hbar^2 b.
\end{aligned}
\end{equation}

This puts constraints on \( a, b \), roughly of the form

  1. \begin{equation}\label{eqn:qmLecture13:1360}
    a – b( b + 1) \ge 0
    \end{equation}

    With \( b_{\textrm{max}} > 0 \), \( b_{\textrm{max}} \approx \sqrt{a} \).

  2. \begin{equation}\label{eqn:qmLecture13:1380}
    a – b( b – 1) \ge 0
    \end{equation}

    With \( b_{\textrm{min}} < 0 \), \( b_{\textrm{max}} \approx -\sqrt{a} \).

Question: Angular momentum commutators

Using \( \hat{L}_i = \epsilon_{ijk} \hat{r}_j \hat{p}_k \), show that

\begin{equation}\label{eqn:qmLecture13:1620}
\antisymmetric{\hat{L}_i}{\hat{L}_j} = i \Hbar \epsilon_{ijk} \hat{L}_k
\end{equation}

Answer

Let’s start without using abstract index expressions, computing the commutator for \( \hat{L}_1, \hat{L}_2 \), which should show the basic steps required

\begin{equation}\label{eqn:qmLecture13:1640}
\begin{aligned}
\antisymmetric{\hat{L}_1}{\hat{L}_2}
&=
\antisymmetric{\hat{r}_2 \hat{p}_3 – \hat{r}_3 \hat{p}_2}{\hat{r}_3 \hat{p}_1
– \hat{r}_1 \hat{p}_3} \\
&=
\antisymmetric{\hat{r}_2 \hat{p}_3}{\hat{r}_3 \hat{p}_1}
-\antisymmetric{\hat{r}_2 \hat{p}_3}{\hat{r}_1 \hat{p}_3}
-\antisymmetric{\hat{r}_3 \hat{p}_2}{\hat{r}_3 \hat{p}_1}
+\antisymmetric{\hat{r}_3 \hat{p}_2}{\hat{r}_1 \hat{p}_3}.
\end{aligned}
\end{equation}

The first of these commutators is

\begin{equation}\label{eqn:qmLecture13:1660}
\begin{aligned}
\antisymmetric{\hat{r}_2 \hat{p}_3}{\hat{r}_3 \hat{p}_1}
&=
{\hat{r}_2 \hat{p}_3}{\hat{r}_3 \hat{p}_1}
–
{\hat{r}_3 \hat{p}_1}
{\hat{r}_2 \hat{p}_3} \\
&=
\hat{r}_2 \hat{p}_1 \antisymmetric{\hat{p}_3}{\hat{r}_3} \\
&=
-i \Hbar \hat{r}_2 \hat{p}_1.
\end{aligned}
\end{equation}

We see that any factors in the commutator don’t have like indexes (i.e. \( \hat{r}_k, \hat{p}_k \)) on both position and momentum terms, can be pulled out of the commutator. This leaves

\begin{equation}\label{eqn:qmLecture13:1680}
\begin{aligned}
\antisymmetric{\hat{L}_1}{\hat{L}_2}
&=
\hat{r}_2 \hat{p}_1 \antisymmetric{\hat{p}_3}{\hat{r}_3}
-{\antisymmetric{\hat{r}_2 \hat{p}_3}{\hat{r}_1 \hat{p}_3}}
-{\antisymmetric{\hat{r}_3 \hat{p}_2}{\hat{r}_3 \hat{p}_1}}
+\hat{r}_1 \hat{p}_2 \antisymmetric{\hat{r}_3}{\hat{p}_3} \\
&=
i \Hbar \lr{ \hat{r}_1 \hat{p}_2 – \hat{r}_2 \hat{p}_1 } \\
&=
i \Hbar \hat{L}_3.
\end{aligned}
\end{equation}

With cyclic permutation this is really enough to consider \ref{eqn:qmLecture13:1620} proven. However, can we do this in the general case with the abstract index expression? The quantity to simplify looks forbidding

\begin{equation}\label{eqn:qmLecture13:1700}
\antisymmetric{\hat{L}_i}{\hat{L}_j}
=
\epsilon_{i a b }
\epsilon_{j s t }
\antisymmetric{ \hat{r}_a \hat{p}_b }{ \hat{r}_s \hat{p}_t }
\end{equation}

Because there are no repeated indexes, this doesn’t submit to any of the normal reduction identities. Note however, since we only care about the \( i \ne j \) case, that one of the indexes \( a, b \) must be \( j \) for this quantity to be non-zero. Therefore (for \( i \ne j \))

\begin{equation}\label{eqn:qmLecture13:1720}
\begin{aligned}
\antisymmetric{\hat{L}_i}{\hat{L}_j}
&=
\epsilon_{i j b }
\epsilon_{j s t }
\antisymmetric{ \hat{r}_j \hat{p}_b }{ \hat{r}_s \hat{p}_t }
+
\epsilon_{i a j }
\epsilon_{j s t }
\antisymmetric{ \hat{r}_a \hat{p}_j }{ \hat{r}_s \hat{p}_t } \\
&=
\epsilon_{i j b }
\epsilon_{j s t }
\lr{
\antisymmetric{ \hat{r}_j \hat{p}_b }{ \hat{r}_s \hat{p}_t }
–
\antisymmetric{ \hat{r}_b \hat{p}_j }{ \hat{r}_s \hat{p}_t }
} \\
&=
-\delta^{s t}_{[i b]}
\antisymmetric{ \hat{r}_j \hat{p}_b – \hat{r}_b \hat{p}_j }{ \hat{r}_s
\hat{p}_t } \\
&=
\antisymmetric{ \hat{r}_j \hat{p}_b – \hat{r}_b \hat{p}_j }{ \hat{r}_b
\hat{p}_i – \hat{r}_i \hat{p}_b } \\
&=
\antisymmetric{ \hat{r}_j \hat{p}_b }{ \hat{r}_b \hat{p}_i }
– {\antisymmetric{ \hat{r}_j \hat{p}_b }{ \hat{r}_i \hat{p}_b }}
– {\antisymmetric{ \hat{r}_b \hat{p}_j }{ \hat{r}_b \hat{p}_i }}
+ \antisymmetric{ \hat{r}_b \hat{p}_j }{ \hat{r}_i \hat{p}_b } \\
&=
\hat{r}_j \hat{p}_i \antisymmetric{ \hat{p}_b }{ \hat{r}_b }
+ \hat{r}_i \hat{p}_j \antisymmetric{ \hat{r}_b }{ \hat{p}_b } \\
&=
i \Hbar \lr{ \hat{r}_i \hat{p}_j – \hat{r}_j \hat{p}_i } \\
&=
i \Hbar \epsilon_{i j k} \hat{r}_i \hat{p}_j .
\end{aligned}
\end{equation}

References

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

Commutators of angular momentum and a central force Hamiltonian

September 30, 2015 phy1520 , , , ,

[Click here for a PDF of this post with nicer formatting]

In problem 1.17 of [1] we are to show that non-commuting operators that both commute with the Hamiltonian, have, in general, degenerate energy eigenvalues. It suggests considering \( L_x, L_z \) and a central force Hamiltonian \( H = \Bp^2/2m + V(r) \) as examples.

Let’s just demonstrate these commutators act as expected in these cases.

With \( \BL = \Bx \cross \Bp \), we have

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:20}
\begin{aligned}
L_x &= y p_z – z p_y \\
L_y &= z p_x – x p_z \\
L_z &= x p_y – y p_x.
\end{aligned}
\end{equation}

The \( L_x, L_z \) commutator is

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:40}
\begin{aligned}
\antisymmetric{L_x}{L_z}
&=
\antisymmetric{y p_z – z p_y }{x p_y – y p_x} \\
&=
\antisymmetric{y p_z}{x p_y}
-\antisymmetric{y p_z}{y p_x}
-\antisymmetric{z p_y }{x p_y}
+\antisymmetric{z p_y }{y p_x} \\
&=
x p_z \antisymmetric{y}{p_y}
+ z p_x \antisymmetric{p_y }{y} \\
&=
i \Hbar \lr{ x p_z – z p_x } \\
&=
– i \Hbar L_y
\end{aligned}
\end{equation}

cyclicly permuting the indexes shows that no pairs of different \( \BL \) components commute. For \( L_y, L_x \) that is

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:60}
\begin{aligned}
\antisymmetric{L_y}{L_x}
&=
\antisymmetric{z p_x – x p_z }{y p_z – z p_y} \\
&=
\antisymmetric{z p_x}{y p_z}
-\antisymmetric{z p_x}{z p_y}
-\antisymmetric{x p_z }{y p_z}
+\antisymmetric{x p_z }{z p_y} \\
&=
y p_x \antisymmetric{z}{p_z}
+ x p_y \antisymmetric{p_z }{z} \\
&=
i \Hbar \lr{ y p_x – x p_y } \\
&=
– i \Hbar L_z,
\end{aligned}
\end{equation}

and for \( L_z, L_y \)

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:80}
\begin{aligned}
\antisymmetric{L_z}{L_y}
&=
\antisymmetric{x p_y – y p_x }{z p_x – x p_z} \\
&=
\antisymmetric{x p_y}{z p_x}
-\antisymmetric{x p_y}{x p_z}
-\antisymmetric{y p_x }{z p_x}
+\antisymmetric{y p_x }{x p_z} \\
&=
z p_y \antisymmetric{x}{p_x}
+ y p_z \antisymmetric{p_x }{x} \\
&=
i \Hbar \lr{ z p_y – y p_z } \\
&=
– i \Hbar L_x.
\end{aligned}
\end{equation}

If these angular momentum components are also shown to commute with themselves (which they do), the commutator relations above can be summarized as

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:100}
\antisymmetric{L_a}{L_b} = i \Hbar \epsilon_{a b c} L_c.
\end{equation}

In the example to consider, we’ll have to consider the commutators with \( \Bp^2 \) and \( V(r) \). Picking any one component of \( \BL \) is sufficent due to the symmetries of the problem. For example

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:120}
\begin{aligned}
\antisymmetric{L_x}{\Bp^2}
&=
\antisymmetric{y p_z – z p_y}{p_x^2 + p_y^2 + p_z^2} \\
&=
\antisymmetric{y p_z}{{p_x^2} + p_y^2 + {p_z^2}}
-\antisymmetric{z p_y}{{p_x^2} + {p_y^2} + p_z^2} \\
&=
p_z \antisymmetric{y}{p_y^2}
-p_y \antisymmetric{z}{p_z^2} \\
&=
p_z 2 i \Hbar p_y
2 i \Hbar p_y
-p_y 2 i \Hbar p_z \\
&=
0.
\end{aligned}
\end{equation}

How about the commutator of \( \BL \) with the potential? It is sufficient to consider one component again, for example

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:140}
\begin{aligned}
\antisymmetric{L_x}{V}
&=
\antisymmetric{y p_z – z p_y}{V} \\
&=
y \antisymmetric{p_z}{V} – z \antisymmetric{p_y}{V} \\
&=
-i \Hbar y \PD{z}{V(r)} + i \Hbar z \PD{y}{V(r)} \\
&=
-i \Hbar y \PD{r}{V}\PD{z}{r} + i \Hbar z \PD{r}{V}\PD{y}{r} \\
&=
-i \Hbar y \PD{r}{V} \frac{z}{r} + i \Hbar z \PD{r}{V}\frac{y}{r} \\
&=
0.
\end{aligned}
\end{equation}

We’ve shown that all the components of \( \BL \) commute with a central force Hamiltonian, and each different component of \( \BL \) do not commute.

The next step will be figuring out how to use this to show that there are energy degeneracies.

References

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

Update to old phy356 (Quantum Mechanics I) notes.

February 12, 2015 math and physics play , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

It’s been a long time since I took QM I. My notes from that class were pretty rough, but I’ve cleaned them up a bit.

The main value to these notes is that I worked a number of introductory Quantum Mechanics problems.

These were my personal lecture notes for the Fall 2010, University of Toronto Quantum mechanics I course (PHY356H1F), taught by Prof. Vatche Deyirmenjian.

The official description of this course was:

The general structure of wave mechanics; eigenfunctions and eigenvalues; operators; orbital angular momentum; spherical harmonics; central potential; separation of variables, hydrogen atom; Dirac notation; operator methods; harmonic oscillator and spin.

This document contains a few things

• My lecture notes.
Typos, if any, are probably mine(Peeter), and no claim nor attempt of spelling or grammar correctness will be made. The first four lectures had chosen not to take notes for since they followed the text very closely.
• Notes from reading of the text. This includes observations, notes on what seem like errors, and some solved problems. None of these problems have been graded. Note that my informal errata sheet for the text has been separated out from this document.
• Some assigned problems. I have corrected some the errors after receiving grading feedback, and where I have not done so I at least recorded some of the grading comments as a reference.
• Some worked problems associated with exam preparation.