## The emphasis of remembrance day is so wrong!

I’ve deposited my stepson for the day at the Unionville public school warfare celebration and indoctrination center for the day. He’s got his scouts uniform with him so he can put it on for the celebration of war, and make the veterans feel good. He was sent off with instructions to enjoy the celebration, and how much this day is about respect.

The flag is at half mast today. In reverent tones other fathers tell their sons that this is to show respect. I stand there listening, just barely able to keep from gagging.

Needless to say, I feel significantly different about remembrance day than most people I know. It seems so obvious to me that this day was designed as war propaganda, but I’m not allowed to have that opinion. My opinion is viewed as one of disrespect. I’m not allowed to view veterans as unwitting pawns in the actions of evil men. I am supposed to respect the fact that they had the misfortune to have to go off to war and kill other people for the psychopaths that run governments “in our names”.

I am especially not allowed to have an opinion that the great and holy world war II shouldn’t have been fought. I must love Hilter for thinking something like that. It’s true that I consider it a tragedy that so many civilian populations in Germany were bombed in the name of bringing down Hilter. I also consider it equally tragic that civilian populations in Britain were also bombed by Germany. Warfare should not involve civilians, but it always does. That is one of the reasons that it is so profitable.

We will never know what history would have been like if North American forces did not submit the propaganda of glorious warfare, but there are a few things that we can know. We know that Allied support was given to Stalin, killer of more of his own people than Hilter killed. We know that Churchill gifted still more victims to Stalin when all was done. We know that the United States engineered to have Japan enter the war by imposing brutal sanctions. We know that US companies like Ford and IBM supported Hitler’s war and genocide actions (respectively). As with all the current enemies of the United States, you can almost always find a time when those enemies were incubated by the same people who later turn on them as warfare fodder. We know from the admissions of Germans interviewed after the war, that full fledged war was used as the justification for the Jewish genocide. We know that psychopaths in the United States government killed hundreds of people in Japan with needless atomic bombs. Those atomic bombs were explicitly dropped on civilian populations, because they wanted test sites that had not already been ravaged by conventional carpet bombing. Japan was ready to give up when these bombs were dropped, but the atomic bombs were a great way to show power, especially to Russia, who was ready to move in and take desired resources. We know a lot about the blatant evil that did occur because “we” joined the war. Despite that one is not allowed to question the holiness of world war II.

It seems especially despicable to me that remembrance day is pushed on us and on the kids without any context of history. Don’t look at the root causes for the wars that turned your grandfathers into pawns. Don’t think about all the civilians that are were killed and displaced as they served.

Close your eyes and observe the holy moment of silence, but don’t think. Thank and respect the veterans for their service, but never look at the underlying issues. Celebrate the goodness of war.

## PHY1520H Graduate Quantum Mechanics. Lecture 14: Angular momentum (cont.). Taught by Prof. Arun Paramekanti

[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 [1] chap. 3 content.

### Review: Angular momentum

Given eigenket $$\ket{a, b}$$, where

\label{eqn:qmLecture14:20}
\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}

We were looking for

\label{eqn:qmLecture14:40}
\hat{L}_{x,y} \ket{a,b} = \sum_{b’} \mathcal{A}^{x,y}_{a; b, b’} \ket{a,b’},

by applying

\label{eqn:qmLecture14:60}
\hat{L}_{\pm} = \hat{L}_x \pm i \hat{L}_y.

We found

\label{eqn:qmLecture14:80}
\hat{L}_{\pm} \propto \ket{a, b \pm 1}.

Let

\label{eqn:qmLecture14:100}
\ket{\phi_\pm} = \hat{L}_{\pm} \ket{a, b}.

We want

\label{eqn:qmLecture14:120}
\braket{\phi_\pm}{\phi_\pm} \ge 0,

or
\label{eqn:qmLecture14:140}
\begin{aligned}
\bra{a,b} \hat{L}_{+} \hat{L}_{-} \ket{a, b} &\ge 0 \\
\bra{a,b} \hat{L}_{-} \hat{L}_{+} \ket{a, b} &\ge 0
\end{aligned}

We found

\label{eqn:qmLecture14:160}
\begin{aligned}
\hat{L}_{+} \hat{L}_{-} =
\lr{ \hat{L}_x + i \hat{L}_y } \lr{ \hat{L}_x – i \hat{L}_y }
&= \lr{ \hat{\BL}^2 – \hat{L}_z^2 } -i \antisymmetric{\hat{L}_x}{\hat{L}_y} \\
&= \lr{ \hat{\BL}^2 – \hat{L}_z^2 } -i \lr{ i \Hbar \hat{L}_z } \\
&= \lr{ \hat{\BL}^2 – \hat{L}_z^2 } + \Hbar \hat{L}_z,
\end{aligned}

so

\label{eqn:qmLecture14:180}
\bra{a,b} \hat{L}_{+} \hat{L}_{-} \ket{a, b}
=
\expectation{ \hat{\BL}^2 – \hat{L}_z^2 + \Hbar \hat{L}_z }.

Similarly
\label{eqn:qmLecture14:200}
\bra{a,b} \hat{L}_{-} \hat{L}_{+} \ket{a, b}
=
\expectation{ \hat{\BL}^2 – \hat{L}_z^2 – \Hbar \hat{L}_z }.

### Constraints

\label{eqn:qmLecture14:220}
\begin{aligned}
a – b^2 + b &\ge 0 \\
a – b^2 – b &\ge 0
\end{aligned}

If these are satisfied at the equality extreme we have

\label{eqn:qmLecture14:240}
\begin{aligned}
b_{\textrm{max}} \lr{ b_{\textrm{max}} + 1 } &= a \\
b_{\textrm{min}} \lr{ b_{\textrm{min}} – 1 } &= a.
\end{aligned}

Rearranging this to solve, we can rewrite the equality as

\label{eqn:qmLecture14:680}
\lr{ b_{\textrm{max}} + \inv{2} }^2 – \inv{4} = \lr{ b_{\textrm{min}} – \inv{2} }^2 – \inv{4},

which has solutions at

\label{eqn:qmLecture14:700}
b_{\textrm{max}} + \inv{2} = \pm \lr{ b_{\textrm{min}} – \inv{2} }.

One of the solutions is

\label{eqn:qmLecture14:260}
-b_{\textrm{min}} = b_{\textrm{max}}.

The other solution is $$b_{\textrm{max}} = b_{\textrm{min}} – 1$$, which we discard.

The final constraint is therefore

\label{eqn:qmLecture14:280}
\boxed{
– b_{\textrm{max}} \le b \le b_{\textrm{max}},
}

and

\label{eqn:qmLecture14:320}
\begin{aligned}
\hat{L}_{+} \ket{a, b_{\textrm{max}}} &= 0 \\
\hat{L}_{-} \ket{a, b_{\textrm{min}}} &= 0
\end{aligned}

If we had the sequence, which must terminate at $$b_{\textrm{min}}$$ or else it will go on forever

\label{eqn:qmLecture14:340}
\ket{a, b_{\textrm{max}}}
\overset{\hat{L}_{-}}{\rightarrow}
\ket{a, b_{\textrm{max}} – 1}
\overset{\hat{L}_{-}}{\rightarrow}
\ket{a, b_{\textrm{max}} – 2}
\cdots
\overset{\hat{L}_{-}}{\rightarrow}
\ket{a, b_{\textrm{min}}},

then we know that $$b_{\textrm{max}} – b_{\textrm{min}} \in \mathbb{Z}$$, or

\label{eqn:qmLecture14:360}
b_{\textrm{max}} – n = b_{\textrm{min}} = -b_{\textrm{max}}

or

\label{eqn:qmLecture14:380}
b_{\textrm{max}} = \frac{n}{2},

this is either an integer or a $$1/2$$ odd integer, depending on whether $$n$$ is even or odd. These are called “orbital” or “spin” respectively.

The convention is to write

\label{eqn:qmLecture14:400}
\begin{aligned}
b_{\textrm{max}} &= j \\
a &= j(j + 1).
\end{aligned}

so for $$m \in -j, -j + 1, \cdots, +j$$

\label{eqn:qmLecture14:420}
\boxed{
\begin{aligned}
\hat{\BL}^2 \ket{j, m} &= \Hbar^2 j (j + 1) \ket{j, m} \\
L_z \ket{j, m} &= \Hbar m \ket{j, m}.
\end{aligned}
}

## Schwinger’s Harmonic oscillator representation of angular momentum operators.

In [2] a powerful method for describing angular momentum with harmonic oscillators was introduced, which will be outlined here. The question is whether we can construct a set of harmonic oscillators that allows a mapping from

\label{eqn:qmLecture14:460}
\hat{L}_{+} \leftrightarrow a^{+}?

Picture two harmonic oscillators, one with states counted from one zero towards $$\infty$$ and another with states counted from a different zero towards $$-\infty$$, as pictured in fig. 1.

fig. 1. Overlapping SHO domains

Is it possible that such an overlapping set of harmonic oscillators can provide the properties of the angular momentum operators? Let’s relabel the counting so that we have two sets of positive counted SHO systems, each counted in a positive direction as sketched in fig. 2.

fig. 2. Relabeling the counting for overlapping SHO systems

It turns out that given a constraint there the number of ways to distribute particles between a pair of SHO systems, the process that can be viewed as reproducing the angular momentum action is a transfer of particles from one harmonic oscillator to the other. For $$\hat{L}_z = +j$$

\label{eqn:qmLecture14:480}
\begin{aligned}
n_1 &= n_{\textrm{max}} \\
n_2 &= 0,
\end{aligned}

and for $$\hat{L}_z = -j$$

\label{eqn:qmLecture14:500}
\begin{aligned}
n_1 &= 0 \\
n_2 &= n_{\textrm{max}}.
\end{aligned}

We can make the identifications

\label{eqn:qmLecture14:520}
\hat{L}_z = \lr{ n_1 – n_2 } \frac{\Hbar}{2},

and
\label{eqn:qmLecture14:540}
j = \inv{2} n_{\textrm{max}},

or

\label{eqn:qmLecture14:560}
n_1 + n_2 = \text{fixed} = n_{\textrm{max}}

Changes that keep $$n_1 + n_2$$ fixed are those that change $$n_1$$, $$n_2$$ by $$+1$$ or $$-1$$ respectively, as sketched in fig. 3.

fig. 3. Number conservation constraint.

Can we make an identification that takes

\label{eqn:qmLecture14:580}
\ket{n_1, n_2} \overset{\hat{L}_{-}}{\rightarrow} \ket{n_1 – 1, n_2 + 1}?

What operator in the SHO problem has this effect? Let’s try

\boxedEquation{eqn:qmLecture14:620}{
\begin{aligned}
\hat{L}_{-} &= \Hbar a_2^\dagger a_1 \\
\hat{L}_{+} &= \Hbar a_1^\dagger a_2 \\
\hat{L}_z &= \frac{\Hbar}{2} \lr{ n_1 – n_2 }
\end{aligned}
}

Is this correct? Do we need to make any scalar adjustments? We want

\label{eqn:qmLecture14:640}
\antisymmetric{\hat{L}_z}{\hat{L}_{\pm}} = \pm \Hbar \hat{L}_{\pm}.

First check this with the $$\hat{L}_{+}$$ commutator

\label{eqn:qmLecture14:660}
\begin{aligned}
\antisymmetric{\hat{L}_z}{\hat{L}_{+}}
&=
\inv{2} \Hbar^2 \antisymmetric{ n_1 – n_2}{a_1^\dagger a_2 } \\
&=
\inv{2} \Hbar^2 \antisymmetric{ a_1^\dagger a_1 – a_2^\dagger a_2 }{a_1^\dagger a_2 } \\
&=
\inv{2} \Hbar^2
\lr{
\antisymmetric{ a_1^\dagger a_1 }{a_1^\dagger a_2 }
-\antisymmetric{ a_2^\dagger a_2 }{a_1^\dagger a_2 }
} \\
&=
\inv{2} \Hbar^2
\lr{
a_2 \antisymmetric{ a_1^\dagger a_1 }{a_1^\dagger }
-a_1^\dagger \antisymmetric{ a_2^\dagger a_2 }{a_2 }
}.
\end{aligned}

But

\label{eqn:qmLecture14:720}
\begin{aligned}
\antisymmetric{ a^\dagger a }{a^\dagger }
&=
a^\dagger a
a^\dagger

a^\dagger
a^\dagger a \\
&=
a^\dagger \lr{ 1 +
a^\dagger a}

a^\dagger
a^\dagger a \\
&=
a^\dagger,
\end{aligned}

and
\label{eqn:qmLecture14:740}
\begin{aligned}
\antisymmetric{ a^\dagger a }{a}
&=
a^\dagger a a
-a a^\dagger a \\
&=
a^\dagger a a
-\lr{ 1 + a^\dagger a } a \\
&=
-a,
\end{aligned}

so
\label{eqn:qmLecture14:760}
\antisymmetric{\hat{L}_z}{\hat{L}_{+}} = \Hbar^2 a_2 a_1^\dagger = \Hbar \hat{L}_{+},

as desired. Similarly

\label{eqn:qmLecture14:780}
\begin{aligned}
\antisymmetric{\hat{L}_z}{\hat{L}_{-}}
&=
\inv{2} \Hbar^2 \antisymmetric{ n_1 – n_2}{a_2^\dagger a_1 } \\
&=
\inv{2} \Hbar^2 \antisymmetric{ a_1^\dagger a_1 – a_2^\dagger a_2 }{a_2^\dagger a_1 } \\
&=
\inv{2} \Hbar^2 \lr{
a_2^\dagger \antisymmetric{ a_1^\dagger a_1 }{a_1 }
– a_1 \antisymmetric{ a_2^\dagger a_2 }{a_2^\dagger }
} \\
&=
\inv{2} \Hbar^2 \lr{
a_2^\dagger (-a_1)
– a_1 a_2^\dagger
} \\
&=
– \Hbar^2 a_2^\dagger a_1 \\
&=
– \Hbar \hat{L}_{-}.
\end{aligned}

With

\label{eqn:qmLecture14:800}
\begin{aligned}
j &= \frac{n_1 + n_2}{2} \\
m &= \frac{n_1 – n_2}{2} \\
\end{aligned}

We can make the identification

\label{eqn:qmLecture14:820}
\ket{n_1, n_2} = \ket{ j+ m , j – m}.

### Another way

With

\label{eqn:qmLecture14:840}
\hat{L}_{+} \ket{j, m} = d_{j,m}^{+} \ket{j, m+1}

or

\label{eqn:qmLecture14:860}
\Hbar a_1^\dagger a_2 \ket{j + m, j-m} = d_{j,m}^{+} \ket{ j + m + 1, j- m-1},

we can seek this factor $$d_{j,m}^{+}$$ by operating with $$\hat{L}_{+}$$

\label{eqn:qmLecture14:880}
\begin{aligned}
\hat{L}_{+} \ket{j, m}
&=
\Hbar a_1^\dagger a_2 \ket{n_1, n_2} \\
&=
\Hbar a_1^\dagger a_2 \ket{j+m,j-m} \\
&=
\Hbar \sqrt{ n + 1 } \sqrt{n_2} \ket{j+m +1,j-m-1} \\
&=
\Hbar \sqrt{ \lr{ j+ m + 1}\lr{ j – m } } \ket{j+m +1,j-m-1}
\end{aligned}

That gives
\label{eqn:qmLecture14:900}
\begin{aligned}
d_{j,m}^{+} &= \Hbar \sqrt{\lr{ j – m } \lr{ j+ m + 1} } \\
d_{j,m}^{-} &= \Hbar \sqrt{\lr{ j + m } \lr{ j- m + 1} }.
\end{aligned}

This equivalence can be used to model spin interaction in crystals as harmonic oscillators. This equivalence of lattice vibrations and spin oscillations is called “spin waves”.

# References

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

[2] J Schwinger. Quantum theory of angular momentum. biedenharn l., van dam h., editors, 1955. URL http://www.ifi.unicamp.br/ cabrera/teaching/paper_schwinger.pdf.