## Alternate Dirac equation representation

November 27, 2015 phy1520 No comments , ,

Given an alternate representation of the Dirac equation

\label{eqn:diracAlternate:20}
H =
\begin{bmatrix}
m c^2 + V_0 & c \hat{p} \\
c \hat{p} & – m c^2 + V_0
\end{bmatrix},

calculate the constant momentum solutions, the Heisenberg velocity operator $$\hat{v}$$, and find the form of the probability density current.

### Plane wave solutions

The action of the Hamiltonian on

\label{eqn:diracAlternate:40}
\psi =
e^{i k x – i E t/\Hbar}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}

is
\label{eqn:diracAlternate:60}
\begin{aligned}
H \psi
&=
\begin{bmatrix}
m c^2 + V_0 & c (-i \Hbar) i k \\
c (-i \Hbar) i k & – m c^2 + V_0
\end{bmatrix}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
e^{i k x – i E t/\Hbar} \\
&=
\begin{bmatrix}
m c^2 + V_0 & c \Hbar k \\
c \Hbar k & – m c^2 + V_0
\end{bmatrix}
\psi.
\end{aligned}

Writing

\label{eqn:diracAlternate:80}
H_k
=
\begin{bmatrix}
m c^2 + V_0 & c \Hbar k \\
c \Hbar k & – m c^2 + V_0
\end{bmatrix}

the characteristic equation is

\label{eqn:diracAlternate:100}
0
=
(m c^2 + V_0 – \lambda)
(-m c^2 + V_0 – \lambda)
– (c \Hbar k)^2
=
\lr{ (\lambda – V_0)^2 – (m c^2)^2 } – (c \Hbar k)^2,

so

\label{eqn:diracAlternate:120}
\lambda = V_0 \pm \epsilon,

where
\label{eqn:diracAlternate:140}
\epsilon^2 = (m c^2)^2 + (c \Hbar k)^2.

We’ve got

\label{eqn:diracAlternate:160}
\begin{aligned}
H – ( V_0 + \epsilon )
&=
\begin{bmatrix}
m c^2 – \epsilon & c \Hbar k \\
c \Hbar k & – m c^2 – \epsilon
\end{bmatrix} \\
H – ( V_0 – \epsilon )
&=
\begin{bmatrix}
m c^2 + \epsilon & c \Hbar k \\
c \Hbar k & – m c^2 + \epsilon
\end{bmatrix},
\end{aligned}

so the eigenkets are

\label{eqn:diracAlternate:180}
\begin{aligned}
\ket{V_0+\epsilon}
&\propto
\begin{bmatrix}
-c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&\propto
\begin{bmatrix}
-c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix}.
\end{aligned}

Up to an arbitrary phase for each, these are

\label{eqn:diracAlternate:200}
\begin{aligned}
\ket{V_0 + \epsilon}
&=
\inv{\sqrt{ 2 \epsilon ( \epsilon – m c^2) }}
\begin{bmatrix}
c \Hbar k \\
\epsilon -m c^2
\end{bmatrix} \\
\ket{V_0 – \epsilon}
&=
\inv{\sqrt{ 2 \epsilon ( \epsilon + m c^2) }}
\begin{bmatrix}
-c \Hbar k \\
\epsilon + m c^2
\end{bmatrix} \\
\end{aligned}

We can now write

\label{eqn:diracAlternate:220}
H_k =
E
\begin{bmatrix}
V_0 + \epsilon & 0 \\
0 & V_0 – \epsilon
\end{bmatrix}
E^{-1},

where
\label{eqn:diracAlternate:240}
\begin{aligned}
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
\sqrt{ \epsilon – m c^2 } & \sqrt{ \epsilon + m c^2 }
\end{bmatrix}, \qquad k > 0 \\
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
-\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
-\sqrt{ \epsilon – m c^2 } & \sqrt{ \epsilon + m c^2 }
\end{bmatrix}, \qquad k < 0. \end{aligned} Here the signs have been adjusted to ensure the transformation matrix has a unit determinant. Observe that there's redundancy in this matrix since $$\ifrac{c \Hbar \Abs{k}}{ \sqrt{ \epsilon - m c^2 } } = \sqrt{ \epsilon + m c^2 }$$, and $$\ifrac{c \Hbar \Abs{k}}{ \sqrt{ \epsilon + m c^2 } } = \sqrt{ \epsilon - m c^2 }$$, which allows the transformation matrix to be written in the form of a rotation matrix \label{eqn:diracAlternate:260} \begin{aligned} E &= \inv{\sqrt{2 \epsilon} } \begin{bmatrix} \frac{c \Hbar k}{ \sqrt{ \epsilon - m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\ \frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } & \frac{c \Hbar k}{ \sqrt{ \epsilon - m c^2 } } \end{bmatrix}, \qquad k > 0 \\
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
-\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } }
\end{bmatrix}, \qquad k < 0 \\ \end{aligned} With \label{eqn:diracAlternate:280} \begin{aligned} \cos\theta &= \frac{c \Hbar \Abs{k}}{ \sqrt{ 2 \epsilon( \epsilon - m c^2) } } = \frac{\sqrt{ \epsilon + m c^2} }{ \sqrt{ 2 \epsilon}}\\ \sin\theta &= \frac{c \Hbar k}{ \sqrt{ 2 \epsilon( \epsilon + m c^2) } } = \frac{\textrm{sgn}(k) \sqrt{ \epsilon - m c^2}}{ \sqrt{ 2 \epsilon } }, \end{aligned} the transformation matrix (and eigenkets) is $$\label{eqn:diracAlternate:300} \boxed{ E = \begin{bmatrix} \ket{V_0 + \epsilon} & \ket{V_0 - \epsilon} \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}. }$$ Observe that \ref{eqn:diracAlternate:280} can be simplified by using double angle formulas \label{eqn:diracAlternate:320} \begin{aligned} \cos(2 \theta) &= \frac{\lr{ \epsilon + m c^2} }{ 2 \epsilon } - \frac{\lr{ \epsilon - m c^2}}{ 2 \epsilon } \\ &= \frac{1}{ 2 \epsilon } \lr{ \epsilon + m c^2 - \epsilon + m c^2 } \\ &= \frac{m c^2 }{ \epsilon }, \end{aligned} and $$\label{eqn:diracAlternate:340} \sin(2\theta) = 2 \frac{1}{2 \epsilon} \textrm{sgn}(k ) \sqrt{ \epsilon^2 - (m c^2)^2 } = \frac{\Hbar k c}{\epsilon}.$$ This allows all the $$\theta$$ dependence on $$\Hbar k c$$ and $$m c^2$$ to be expressed as a ratio of momenta $$\label{eqn:diracAlternate:360} \boxed{ \tan(2\theta) = \frac{\Hbar k}{m c}. }$$

### Hyperbolic solutions

For a wave function of the form

\label{eqn:diracAlternate:380}
\psi =
e^{k x – i E t/\Hbar}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix},

some of the work above can be recycled if we substitute $$k \rightarrow -i k$$, which yields unnormalized eigenfunctions

\label{eqn:diracAlternate:400}
\begin{aligned}
\ket{V_0+\epsilon}
&\propto
\begin{bmatrix}
i c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&\propto
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix},
\end{aligned}

where

\label{eqn:diracAlternate:420}
\epsilon^2 = (m c^2)^2 – (c \Hbar k)^2.

The squared magnitude of these wavefunctions are

\label{eqn:diracAlternate:440}
\begin{aligned}
(c \Hbar k)^2 + (m c^2 \mp \epsilon)^2
&=
(c \Hbar k)^2 + (m c^2)^2 + \epsilon^2 \mp 2 m c^2 \epsilon \\
&=
(c \Hbar k)^2 + (m c^2)^2 + (m c^2)^2 \mp (c \Hbar k)^2 – 2 m c^2 \epsilon \\
&= 2 (m c^2)^2 \mp 2 m c^2 \epsilon \\
&= 2 m c^2 ( m c^2 \mp \epsilon ),
\end{aligned}

so, up to a constant phase for each, the normalized kets are

\label{eqn:diracAlternate:460}
\begin{aligned}
\ket{V_0+\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 – \epsilon ) }}
\begin{bmatrix}
i c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 + \epsilon ) }}
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix},
\end{aligned}

After the $$k \rightarrow -i k$$ substitution, $$H_k$$ is not Hermitian, so these kets aren’t expected to be orthonormal, which is readily verified

\label{eqn:diracAlternate:480}
\begin{aligned}
\braket{V_0+\epsilon}{V_0-\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 – \epsilon ) }}
\inv{\sqrt{ 2 m c^2 ( m c^2 + \epsilon ) }}
\begin{bmatrix}
-i c \Hbar k &
m c^2 – \epsilon
\end{bmatrix}
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix} \\
&=
\frac{ 2 ( c \Hbar k )^2 }{2 m c^2 \sqrt{(\Hbar k c)^2} } \\
&=
\textrm{sgn}(k)
\frac{
\Hbar k }{m c } .
\end{aligned}

### Heisenberg velocity operator

\label{eqn:diracAlternate:500}
\begin{aligned}
\hat{v}
&= \inv{i \Hbar} \antisymmetric{ \hat{x} }{ H} \\
&= \inv{i \Hbar} \antisymmetric{ \hat{x} }{ m c^2 \sigma_z + V_0 + c \hat{p} \sigma_x } \\
&= \frac{c \sigma_x}{i \Hbar} \antisymmetric{ \hat{x} }{ \hat{p} } \\
&= c \sigma_x.
\end{aligned}

### Probability current

Acting against a completely general wavefunction the Hamiltonian action $$H \psi$$ is

\label{eqn:diracAlternate:520}
\begin{aligned}
i \Hbar \PD{t}{\psi}
&= m c^2 \sigma_z \psi + V_0 \psi + c \hat{p} \sigma_x \psi \\
&= m c^2 \sigma_z \psi + V_0 \psi -i \Hbar c \sigma_x \PD{x}{\psi}.
\end{aligned}

Conversely, the conjugate $$(H \psi)^\dagger$$ is

\label{eqn:diracAlternate:540}
-i \Hbar \PD{t}{\psi^\dagger}
= m c^2 \psi^\dagger \sigma_z + V_0 \psi^\dagger +i \Hbar c \PD{x}{\psi^\dagger} \sigma_x.

These give

\label{eqn:diracAlternate:560}
\begin{aligned}
i \Hbar \psi^\dagger \PD{t}{\psi}
&=
m c^2 \psi^\dagger \sigma_z \psi + V_0 \psi^\dagger \psi -i \Hbar c \psi^\dagger \sigma_x \PD{x}{\psi} \\
-i \Hbar \PD{t}{\psi^\dagger} \psi
&= m c^2 \psi^\dagger \sigma_z \psi + V_0 \psi^\dagger \psi +i \Hbar c \PD{x}{\psi^\dagger} \sigma_x \psi.
\end{aligned}

Taking differences
\label{eqn:diracAlternate:580}
\psi^\dagger \PD{t}{\psi} + \PD{t}{\psi^\dagger} \psi
=
– c \psi^\dagger \sigma_x \PD{x}{\psi} – c \PD{x}{\psi^\dagger} \sigma_x \psi,

or

\label{eqn:diracAlternate:600}
0
=
\PD{t}{}
\lr{
\psi^\dagger \psi
}
+
\PD{x}{}
\lr{
c \psi^\dagger \sigma_x \psi
}.

The probability current still has the usual form $$\rho = \psi^\dagger \psi = \psi_1^\conj \psi_1 + \psi_2^\conj \psi_2$$, but the probability current with this representation of the Dirac Hamiltonian is

\label{eqn:diracAlternate:620}
\begin{aligned}
j
&= c \psi^\dagger \sigma_x \psi \\
&= c
\begin{bmatrix}
\psi_1^\conj &
\psi_2^\conj
\end{bmatrix}
\begin{bmatrix}
\psi_2 \\
\psi_1
\end{bmatrix} \\
&= c \lr{ \psi_1^\conj \psi_2 + \psi_2^\conj \psi_1 }.
\end{aligned}

## PHY1520H Graduate Quantum Mechanics. Lecture 19: Variational method. Taught by Prof. Arun Paramekanti

November 27, 2015 phy1520 No comments , ,

### 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{{5}} [1] content.

### Variational method

Today we want to use the variational degree of freedom to try to solve some problems that we don’t have analytic solutions for.

### Anharmonic oscillator

\label{eqn:qmLecture19:20}
V(x) = \inv{2} m \omega^2 x^2 + \lambda x^4, \qquad \lambda \ge 0.

With the potential growing faster than the harmonic oscillator, which had a ground state solution

\label{eqn:qmLecture19:40}
\psi(x) = \inv{\pi^{1/4}} \inv{a_0^{1/2} } e^{- x^2/2 a_0^2},

where
\label{eqn:qmLecture19:60}
a_0 = \sqrt{\frac{\Hbar}{m \omega}}.

Let’s try allowing $$a_0 \rightarrow a$$, to be a variational degree of freedom

\label{eqn:qmLecture19:80}
\psi_a(x) = \inv{\pi^{1/4}} \inv{a^{1/2} } e^{- x^2/2 a^2},

\label{eqn:qmLecture19:100}
\bra{\psi_a} H \ket{\psi_a}
=
\bra{\psi_a} \frac{p^2}{2m} + \inv{2} m \omega^2 x^2 + \lambda x^4 \ket{\psi_a}

We can find
\label{eqn:qmLecture19:120}
\expectation{x^2} = \inv{2} a^2

\label{eqn:qmLecture19:140}
\expectation{x^4} = \frac{3}{4} a^4

Define

\label{eqn:qmLecture19:160}
\tilde{\omega} = \frac{\Hbar}{m a^2},

so that

\label{eqn:qmLecture19:180}
\overline{{E}}_a
=
\bra{\psi_a} \lr{ \frac{p^2}{2m} + \inv{2} m \tilde{\omega}^2 x^2 }
+ \lr{
\inv{2} m \lr{ \omega^2 – \tilde{\omega}^2 } x^2
+
\lambda x^4 }
\ket{\psi_a}
=
\inv{2} \Hbar \tilde{\omega} + \inv{2} m \lr{ \omega^2 – \tilde{\omega}^2 } \inv{2} a^2 + \frac{3}{4} \lambda a^4.

Write this as
\label{eqn:qmLecture19:200}
\overline{{E}}_{\tilde{\omega}}
=
\inv{2} \Hbar \tilde{\omega} + \inv{4} \frac{\Hbar}{\tilde{\omega}} \lr{ \omega^2 – \tilde{\omega}^2 } + \frac{3}{4} \lambda \frac{\Hbar^2}{m^2 \tilde{\omega}^2 }.

This might look something like fig. 1.

fig. 1: Energy after perturbation.

Demand that

\label{eqn:qmLecture19:220}
0
= \PD{\tilde{\omega}}{ \overline{{E}}_{\tilde{\omega}}}
=
\frac{\Hbar}{2} – \frac{\Hbar}{4} \frac{\omega^2}{\tilde{\omega}^2}
– \frac{\Hbar}{4}
+ \frac{3}{4} (-2) \frac{\lambda \Hbar^2}{m^2 \tilde{\omega}^3}
=
\frac{\Hbar}{4}
\lr{
1 – \frac{\omega^2}{\tilde{\omega}^2}
– 6 \frac{\lambda \Hbar}{m^2 \tilde{\omega}^3}
}

or
\label{eqn:qmLecture19:260}
\tilde{\omega}^3 – \omega^2 \tilde{\omega} – \frac{6 \lambda \Hbar}{m^2} = 0.

for $$\lambda a_0^4 \ll \Hbar \omega$$, we have something like $$\tilde{\omega} = \omega + \epsilon$$. Expanding \ref{eqn:qmLecture19:260} to first order in $$\epsilon$$, this gives

\label{eqn:qmLecture19:280}
\omega^3 + 3 \omega^2 \epsilon – \omega^2 \lr{ \omega + \epsilon } – \frac{6 \lambda \Hbar}{m^2} = 0,

so that

\label{eqn:qmLecture19:300}
2 \omega^2 \epsilon = \frac{6 \lambda \Hbar}{m^2},

and

\label{eqn:qmLecture19:320}
\Hbar \epsilon = \frac{ 3 \lambda \Hbar^2}{m^2 \omega^2 } = 3 \lambda a_0^4.

Plugging into

\label{eqn:qmLecture19:340}
\overline{{E}}_{\omega + \epsilon}
=
\inv{2} \Hbar \lr{ \omega + \epsilon }
+ \inv{4} \frac{\Hbar}{\omega} \lr{ -2 \omega \epsilon + \epsilon^2 } + \frac{3}{4} \lambda \frac{\Hbar^2}{m^2 \omega^2 }
\approx
\inv{2} \Hbar \lr{ \omega + \epsilon }
– \inv{2} \Hbar \epsilon
+ \frac{3}{4} \lambda \frac{\Hbar^2}{m^2 \omega^2 }
=
\inv{2} \Hbar \omega + \frac{3}{4} \lambda a_0^4.

With \ref{eqn:qmLecture19:320}, that is

\label{eqn:qmLecture19:540}
\overline{{E}}_{\tilde{\omega} = \omega + \epsilon} \approx \inv{2} \Hbar \lr{ \omega + \frac{\epsilon}{2} }.

The energy levels are shifted slightly for each shift in the Hamiltonian frequency.

What do we have in the extreme anharmonic limit, where $$\lambda a_0^4 \gg \Hbar \omega$$. Now we get

\label{eqn:qmLecture19:360}
\tilde{\omega}^\conj = \lr{ \frac{ 6 \Hbar \lambda }{m^2} }^{1/3},

and
\label{eqn:qmLecture19:380}
\overline{{E}}_{\tilde{\omega}^\conj} = \frac{\Hbar^{4/3} \lambda^{1/3}}{m^{2/3}} \frac{3}{8} 6^{1/3}.

(this last result is pulled from a web treatment somewhere of the anharmonic oscillator). Note that the first factor in this energy, with $$\Hbar^4 \lambda/m^2$$ traveling together could have been worked out on dimensional grounds.

This variational method tends to work quite well in these limits. For a system where $$m = \omega = \Hbar = 1$$, for this problem, we have

tab. 1: Comparing numeric and variational solutions.

### Example: (sketch) double well potential

fig. 2: Double well potential.

\label{eqn:qmLecture19:400}
V(x) = \frac{m \omega^2}{8 a^2} \lr{ x – a }^2\lr{ x + a}^2.

Note that this potential, and the Hamiltonian, both commute with parity.

We are interested in the regime where $$a_0^2 = \frac{\Hbar}{m \omega} \ll a^2$$.

Near $$x = \pm a$$, this will be approximately

\label{eqn:qmLecture19:420}
V(x) = \inv{2} m \omega^2 \lr{ x \pm a }^2.

Guessing a wave function that is an eigenstate of parity

\label{eqn:qmLecture19:440}
\Psi_{\pm} = g_{\pm} \lr{ \phi_{\textrm{R}}(x) \pm \phi_{\textrm{L}}(x) }.

perhaps looking like the even and odd functions sketched in fig. 3, and fig. 4.

fig. 3. Even double well function

fig. 4. Odd double well function

Using harmonic oscillator functions

\label{eqn:qmLecture19:460}
\begin{aligned}
\phi_{\textrm{L}}(x) &= \Psi_{{\textrm{H}}.{\textrm{O}}.}(x + a) \\
\phi_{\textrm{R}}(x) &= \Psi_{{\textrm{H}}.{\textrm{O}}.}(x – a)
\end{aligned}

After doing a lot of integral (i.e. in the problem set), we will see a splitting of the variational energy levels as sketched in fig. 5.

fig. 5. Splitting for double well potential.

This sort of level splitting was what was used in the very first mazers.

### Perturbation theory (outline)

Given

\label{eqn:qmLecture19:480}
H = H_0 + \lambda V,

where $$\lambda V$$ is “small”. We want to figure out the eigenvalues and eigenstates of this Hamiltonian

\label{eqn:qmLecture19:500}
H \ket{n} = E_n \ket{n}.

We don’t know what these are, but do know that

\label{eqn:qmLecture19:520}
H_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}}.

We are hoping that the level transitions have adiabatic transitions between the original and perturbed levels as sketched in fig. 6.

and not crossed level transitions as sketched in fig. 7.

fig. 7. Crossed level transitions.

If we have level crossings (which can in general occur), as opposed to adiabatic transitions, then we have no hope of using perturbation theory.

# References

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

## PHY1520H Graduate Quantum Mechanics. Lecture 18: Approximation methods. Taught by Prof. Arun Paramekanti

November 26, 2015 phy1520 No comments , ,

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough, especially since I didn’t attend this class myself, and am doing a walkthrough of notes provided by Nishant.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 5 content.

### Approximation methods

Suppose we have a perturbed Hamiltonian

\label{eqn:qmLecture18:20}
H = H_0 + \lambda V,

where $$\lambda = 0$$ represents a solvable (perhaps known) system, and $$\lambda = 1$$ is the case of interest. There are two approaches of interest

1. Direct solution of $$H$$ with $$\lambda = 1$$.
2. Take $$\lambda$$ small, and do a series expansion. This is perturbation theory.

### Variational methods

Given

\label{eqn:qmLecture18:40}
H \ket{\phi_n} = E_n \ket{\phi_n},

where we don’t know $$\ket{\phi_n}$$, we can compute the expectation with respect to an arbitrary state $$\ket{\psi}$$

\label{eqn:qmLecture18:60}
\bra{\psi} H \ket{\psi}
=
\bra{\psi} H \lr{ \sum_n \ket{\phi_n} \bra{\phi_n} } \ket{\psi}
=
\sum_n E_n \braket{\psi}{\phi_n} \braket{\phi_n}{\psi}
=
\sum_n E_n \Abs{\braket{\psi}{\phi_n}}^2.

Define

\label{eqn:qmLecture18:80}
\overline{{E}}
= \frac{\bra{\psi} H \ket{\psi}}{\braket{\psi}{\psi}}.

Assuming that it is possible to express the state in the Hamiltonian energy basis

\label{eqn:qmLecture18:100}
\ket{\psi}
=
\sum_n a_n \ket{\phi_n},

this average energy is
\label{eqn:qmLecture18:120}
\overline{{E}}
= \frac{ \sum_{m,n}\bra{\phi_m} a_m^\conj H a_n \ket{\phi_n}}{ \sum_n \Abs{a_n}^2 }
= \frac{ \sum_{n} \Abs{a_n}^2 E_n }{ \sum_n \Abs{a_n}^2 }.
= \sum_{n}
\frac{\Abs{a_n}^2 }{ \sum_n \Abs{a_n}^2 }
E_n
= \sum_n \frac{P_n}{\sum_m P_m} E_n,

where $$P_m = \Abs{a_m}^2$$, which has the structure of a probability coefficient once divided by $$\sum_m P_m$$, as sketched in fig. 1.

fig. 1. A decreasing probability distribution

This average energy is a probability weighted average of the individual energy basis states. One of those energies is the ground state energy $$E_1$$, so we necessarily have

\label{eqn:qmLecture18:140}
\boxed{
\overline{{E}} \ge E_1.
}

### Example: particle in a $$[0,L]$$ box.

For the infinite potential box sketched in fig. 2.

fig. 2. Infinite potential [0,L] box.

The exact solutions for such a system are found to be

\label{eqn:qmLecture18:220}
\psi(x) = \sqrt{\frac{2}{L}} \sin\lr{ \frac{n \pi}{L} x },

where the energies are

\label{eqn:qmLecture18:240}
E = \frac{\Hbar^2}{2m} \frac{n^2 \pi^2}{L^2}.

The function $$\psi’ = x (L-x)$$ also satisfies the boundary value constraints? How close in energy is that function to the ground state?

\label{eqn:qmLecture18:260}
\overline{{E}}
=
-\frac{\Hbar^2}{2m} \frac{\int_0^L dx x (L-x) \frac{d^2}{dx^2} \lr{ x (L-x) }}{
\int_0^L dx x^2 (L-x)^2
}
=
\frac{\Hbar^2}{2m} \frac{\frac{2 L^3}{6}}{
\frac{L^5}{30}
}
=
\frac{\Hbar^2}{2m} \frac{10}{L^2}.

This average energy is quite close to the ground state energy

\label{eqn:qmLecture18:280}
\frac{\overline{{E}} }{E_1} = \frac{10}{\pi^2} = 1.014.

### Example II: particle in a $$[-L/2,L/2]$$ box.

fig. 3. Infinite potential [-L/2,L/2] box.

Shifting the boundaries, as sketched in fig. 3 doesn’t change the energy levels. For this potential let’s try a shifted trial function

\label{eqn:qmLecture18:300}
\psi(x) = \lr{ x – \frac{L}{2} } \lr{ x + \frac{L}{2} } = x^2 – \frac{L^2}{4},

without worrying about the form of the exact solution. This produces the same result as above

\label{eqn:qmLecture18:270}
\overline{{E}}
=
-\frac{\Hbar^2}{2m} \frac{\int_0^L dx \lr{ x^2 – \frac{L^2}{4} } \frac{d^2}{dx^2} \lr{ x^2 – \frac{L^2}{4} }}{
\int_0^L dx \lr{x^2 – \frac{L^2}{4} }^2
}
=
-\frac{\Hbar^2}{2m} \frac{- 2 L^3/6}{
\frac{L^5}{30}
}
=
\frac{\Hbar^2}{2m} \frac{10}{L^2}.

### Summary (Nishant)

The above example is that of a particle in a box. The actual wave function is a sin as shown. But we can
come up with a guess wave function that meets the boundary conditions and ask how accurate it is
compared to the actual one.

Basically we are assuming a wave function form and then seeing how it differs from the exact form.
We cannot do this if we have nothing to compare it against. But, we note that the variance of the
number operator in the systems eigenstate is zero. So we can still calculate the variance and try to
minimize it. This is one way of coming up with an approximate wave function. This does not necessarily
give the ground state wave function though. For this we need to minimize the energy itself.

# References

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

## PHY1520H Graduate Quantum Mechanics. Lecture 17: Clebsch-Gordan. Taught by Prof. Arun Paramekanti

### 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.

### Clebsch-Gordan

How can we related total momentum states to individual momentum states in the $$1 \otimes 2$$ space?

\label{eqn:qmLecture16:720}
\ket{l_1, l_2, l, m } = \sum_{m_1, m_2} C_{l_1 l_2 m_1 m_2}^{l_1 l_2 l m} \ket{l_1 m_1 ; l_2 m_2 }

The values $$C_{l_1 l_2 m_1 m_2}^{l_1 l_2 l m}$$ are called the Clebsch-Gordan coefficients.

### Example: spin one and spin one half

With individual momentum states $$\ket{l_1 m_1 ; l_2 m_2 }$$

\label{eqn:qmLecture16:740}
\begin{aligned}
l_1 &= 1 \\
m_1 &= \pm 1, 0 \\
l_2 &= \inv{2} \\
m_2 &= \pm \inv{2}
\end{aligned}

The total angular momentum numbers are

\label{eqn:qmLecture16:760}
l_{\textrm{tot}} \in [l_1 – l_2, l_1 + l_2] = [\ifrac{1}{2}, \ifrac{3}{2}]

The possible states $$\ket{ l_{\textrm{tot}}, m_{\textrm{tot}} }$$ are

\label{eqn:qmLecture16:780}
\ket{\inv{2}, \inv{2} }, \ket{\inv{2}, -\inv{2} },

and
\label{eqn:qmLecture16:800}
\ket{\frac{3}{2}, \frac{3}{2} }, \ket{\frac{3}{2}, \frac{1}{2} },
\ket{\frac{3}{2}, -\frac{1}{2} }
\ket{\frac{3}{2}, -\frac{3}{2} }.

The Clebsch-Gordan procedure is the search for an orthogonal angular momentum basis, built up from the individual momentum bases. For the total momentum basis we want the basis states to satisfy the ladder operators, but also have them satisfy the consistuient ladder operators for the individual particle angular momenta. This procedure is sketched in fig. 1.

fig. 1. Spin one,one-half Clebsch-Gordan procedure

Demonstrating by example, let the highest total momentum state be proportional to the highest product of individual momentum states

\label{eqn:qmLecture16:820}
\ket{ \frac{3}{2} \frac{3}{2} } = \ket{ 1 1 } \otimes \ket{ \frac{1}{2} \frac{1}{2} }.

A lowered state can be constructed in two different ways, one using the total angular momentum lowering operator

\label{eqn:qmLecture16:840}
\begin{aligned}
\ket{ \frac{3}{2} \frac{1}{2} }
&=
\hat{L}_{-}^{\textrm{tot}} \ket{ \frac{3}{2} \frac{3}{2} } \\
&= \Hbar \sqrt{\lr{\frac{3}{2} + \frac{3}{2}}\lr{\frac{3}{2} – \frac{3}{2} + 1}} \ket{ \frac{3}{2} \frac{1}{2} } \\
&= \Hbar \sqrt{3} \ket{ \frac{3}{2} \frac{1}{2} }.
\end{aligned}

On the other hand, the lowering operator can also be expressed as $$\hat{L}_{-}^{\textrm{tot}} = \hat{L}_{-}^{(1)} \otimes 1 + 1 \otimes \hat{L}_{-}^{(2)}$$. Operating with that gives

\label{eqn:qmLecture16:860}
\begin{aligned}
\ket{ \frac{3}{2} \frac{1}{2} }
&=
\hat{L}_{-}^{\textrm{tot}} \ket{ 1 1 } \otimes \ket{ \frac{1}{2} \frac{1}{2} }
\\
&=
\hat{L}_{-}^{(1)} \ket{ 1 1 } \otimes \ket{ \frac{1}{2} \frac{1}{2} }
+
\ket{ 1 1 } \otimes \hat{L}_{-}^2 \ket{ \frac{1}{2} \frac{1}{2} } \\
&=
\Hbar \sqrt{\lr{1 + 1}\lr{1 – 1 + 1}} \ket{ 1 0 } \otimes \ket{ \frac{1}{2} \frac{1}{2} }
+
\Hbar \sqrt{\lr{\frac{1}{2} + \frac{1}{2}}\lr{\frac{1}{2} – \frac{1}{2} + 1}}
\ket{ 1 1 } \otimes \ket{ \frac{1}{2} -\frac{1}{2} } \\
&=
\Hbar \sqrt{2} \ket{ 1 0 } \otimes \ket{ \frac{1}{2} \frac{1}{2} }
+
\Hbar \ket{ 1 1 } \otimes \ket{ \frac{1}{2} -\frac{1}{2} }.
\end{aligned}

Equating both sides and dispensing with the direct product notation, this is

\label{eqn:qmLecture16:880}
\sqrt{3} \ket{ \frac{3}{2} \frac{1}{2} }
=
\sqrt{2} \ket{ 1 0 ; \frac{1}{2} \frac{1}{2} }
+
\ket{ 1 1 ; \frac{1}{2} -\frac{1}{2} },

or

\label{eqn:qmLecture16:900}
\boxed{
\ket{ \frac{3}{2} \frac{1}{2} }
=
\sqrt{\frac{2}{3}} \ket{ 1 0 ; \frac{1}{2} \frac{1}{2} }
+
\inv{\sqrt{3}} \ket{ 1 1 ; \frac{1}{2} -\frac{1}{2} }.
}

This is clearly both a unit ket, and normal to $$\ket{ \frac{3}{2} \frac{3}{2} }$$. We can continue operating with the lowering operator for the total angular momentum to constuct all the states down to $$\ket{ \frac{3}{2} \frac{-3}{2} }$$. Working with $$\Hbar = 1$$ since we see it cancel out, the next lower state follows from

\label{eqn:qmLecture16:920}
\begin{aligned}
\hat{L}_{-}^{\textrm{tot}} \ket{ \frac{3}{2} \frac{1}{2} }
&=
\sqrt{2 \times 2} \ket{ \frac{3}{2} \frac{-1}{2} } \\
&=
2 \ket{ \frac{3}{2} \frac{-1}{2} }.
\end{aligned}

and from the individual lowering operators on the components of $$\ket{ \frac{3}{2} \frac{1}{2} }$$.

\label{eqn:qmLecture16:940}
\hat{L}_{-}^{(1)} \ket{ 1 0 ; \frac{1}{2} \frac{1}{2} }
=
\sqrt{ 1 \times 2 } \ket{ 1 -1 ; \frac{1}{2} \frac{1}{2} },

and

\label{eqn:qmLecture16:960}
\hat{L}_{-}^{(2)} \ket{ 1 0 ; \frac{1}{2} \frac{1}{2} }
=
\sqrt{ 1 \times 1 } \ket{ 1 0 ; \frac{1}{2} \frac{-1}{2} },

and

\label{eqn:qmLecture16:980}
\hat{L}_{-}^1
\ket{ 1 1 ; \frac{1}{2} -\frac{1}{2} }
=
\sqrt{ 2 \times 1 }
\ket{ 1 0 ; \frac{1}{2} -\frac{1}{2} }.

This gives

\label{eqn:qmLecture16:1000}
2 \ket{ \frac{3}{2} \frac{-1}{2} } =
\sqrt{\frac{2}{3}} \lr{
\sqrt{ 2 } \ket{ 1 -1 ; \frac{1}{2} \frac{1}{2} }
+
\ket{ 1 0 ; \frac{1}{2} \frac{-1}{2} }
}
+ \inv{\sqrt{3}}
\sqrt{ 2 }
\ket{ 1 0 ; \frac{1}{2} -\frac{1}{2} },

or

\label{eqn:qmLecture16:1001}
\boxed{
\ket{ \frac{3}{2} \frac{-1}{2} } =
\inv{\sqrt{ 3 }} \ket{ 1 -1 ; \frac{1}{2} \frac{1}{2} }
+
\sqrt{\frac{2}{3}}
\ket{ 1 0 ; \frac{1}{2} \frac{-1}{2} }.
}

There’s one more possible state with total angular momentum $$\frac{3}{2}$$. This time

\label{eqn:qmLecture16:1060}
\begin{aligned}
\hat{L}_{-}^{\textrm{tot}}
\ket{ \frac{3}{2} \frac{-1}{2} }
&=
\sqrt{ 1 \times 3 }
\ket{ \frac{3}{2} \frac{-3}{2} } \\
&=
\inv{\sqrt{ 3 }} \hat{L}_{-}^{(2)} \ket{ 1 -1 ; \frac{1}{2} \frac{1}{2} }
+
\sqrt{\frac{2}{3}}
\hat{L}_{-}^{(1)} \ket{ 1 0 ; \frac{1}{2} \frac{-1}{2} } \\
&=
\inv{\sqrt{ 3 }} \sqrt{1 \times 1 } \ket{ 1 -1 ; \frac{1}{2} \frac{-1}{2} }
+
\sqrt{\frac{2}{3}}
\sqrt{ 1 \times 2 } \ket{ 1 -1 ; \frac{1}{2} \frac{-1}{2} },
\end{aligned}

or
\label{eqn:qmLecture16:1080}
\boxed{
\ket{ \frac{3}{2} \frac{-3}{2} }
=
\ket{ 1 -1 ; \frac{1}{2} \frac{-1}{2} }.
}

The $$\ket{ \frac{1}{2} \frac{1}{2} }$$ state is constructed as normal to $$\ket{ \frac{3}{2} \frac{1}{2} }$$, or

\label{eqn:qmLecture17:1120}
\boxed{
\ket{ \frac{1}{2} \frac{1}{2} } =
\sqrt{\frac{1}{3}} \ket{ 1 0 ; \frac{1}{2} \frac{1}{2} }

\sqrt{ \frac{2}{3} } \ket{ 1 1 ; \frac{1}{2} -\frac{1}{2} },
}

and $$\ket{ \frac{1}{2} -\frac{1}{2} }$$ by lowering that. With

\label{eqn:qmLecture17:1160}
\hat{L}_{-}^{\textrm{tot}} \ket{ \frac{1}{2} \frac{1}{2} } = \sqrt{ 1 \times 1 } \ket{ \frac{1}{2} -\frac{1}{2} },

we have

\label{eqn:qmLecture17:1180}
\ket{ \frac{1}{2} -\frac{1}{2} } =
\sqrt{\frac{1}{3}} \lr{
\sqrt{ 1 \times 2 } \ket{ 1 -1 ; \frac{1}{2} \frac{1}{2} }
+ \ket{ 1 0 ; \frac{1}{2} -\frac{1}{2} }
}
-\sqrt{ \frac{2}{3} } \sqrt{ 2 \times 1 } \ket{ 1 0 ; \frac{1}{2} -\frac{1}{2} },

or
\label{eqn:qmLecture17:1200}
\boxed{
\ket{ \frac{1}{2} -\frac{1}{2} } =
\sqrt{ \frac{2}{3} } \ket{ 1 -1 ; \frac{1}{2} \frac{1}{2} }
– \inv{\sqrt{3}} \ket{ 1 0 ; \frac{1}{2} -\frac{1}{2} }.
}

Observe that further lowering this produces zero

\label{eqn:qmLecture17:1240}
\hat{L}_{-}^{\textrm{tot}} \ket{ \frac{1}{2} -\frac{1}{2} }
=
\sqrt{ \frac{2}{3} } \sqrt{ 1 \times 1 } \ket{ 1 -1 ; \frac{1}{2} -\frac{1}{2} }
– \inv{\sqrt{3}} \sqrt{ 1 \times 2 } \ket{ 1 -1 ; \frac{1}{2} -\frac{1}{2} }.
= 0.

All the basis elements have been determined, and are summarized in table 1.

### Example. Spin two, spin one.

With $$j_1 = 2$$ and $$j_2 = 1$$, we have $$j \in 1,2,3$$, and can proceed the same way as sketched in fig. 2.

fig. 2. Spin two,one Clebsch-Gordan procedure

# References

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

## Dirac delta function potential

November 19, 2015 phy1520 No comments , ,

Note: there’s an error below (and in the associated PDF).  10 points to anybody that finds it (I’ve fixed it in my working version of phy1520.pdf)

### Q:Dirac delta function potential

Problem 2.24/2.25 [1] introduces a Dirac delta function potential

\label{eqn:diracPotential:20}
H = \frac{p^2}{2m} – V_0 \delta(x),

which vanishes after $$t = 0$$. Solve for the bound state for $$t < 0$$ and then the time evolution after that.

### A:

The first part of this problem was assigned back in phy356, where we solved this for a rectangular potential that had the limiting form of a delta function potential. However, this problem can be solved directly by considering the $$\Abs{x} > 0$$ and $$x = 0$$ regions separately.

For $$\Abs{x} > 0$$ Schrodinger’s equation takes the form

\label{eqn:diracPotential:40}
E \psi = -\frac{\Hbar^2}{2m} \frac{d^2 \psi}{dx^2}.

With

\label{eqn:diracPotential:60}
\kappa =
\frac{\sqrt{-2 m E}}{\Hbar},

this has solutions

\label{eqn:diracPotential:80}
\psi = e^{\pm \kappa x}.

For $$x > 0$$ we must have
\label{eqn:diracPotential:100}
\psi = a e^{-\kappa x},

and for $$x < 0$$
\label{eqn:diracPotential:120}
\psi = b e^{\kappa x}.

requiring that $$\psi$$ is continuous at $$x = 0$$ means $$a = b$$, or

\label{eqn:diracPotential:140}
\psi = \psi(0) e^{-\kappa \Abs{x}}.

For the $$x = 0$$ region, consider an interval $$[-\epsilon, \epsilon]$$ region around the origin. We must have

\label{eqn:diracPotential:160}
E \int_{-\epsilon}^\epsilon \psi(x) dx = \frac{-\Hbar^2}{2m} \int_{-\epsilon}^\epsilon \frac{d^2 \psi}{dx^2} dx – V_0 \int_{-\epsilon}^\epsilon \delta(x) \psi(x) dx.

The RHS is zero

\label{eqn:diracPotential:180}
E \int_{-\epsilon}^\epsilon \psi(x) dx
=
E \frac{ e^{-\kappa (\epsilon)} – 1}{-\kappa}
-E \frac{ 1 – e^{\kappa (-\epsilon)}}{\kappa}
\rightarrow
0.

That leaves
\label{eqn:diracPotential:200}
\begin{aligned}
V_0 \int_{-\epsilon}^\epsilon \delta(x) \psi(x) dx
&=
\frac{-\Hbar^2}{2m} \int_{-\epsilon}^\epsilon \frac{d^2 \psi}{dx^2} dx \\
&=
\frac{-\Hbar^2}{2m} \evalrange{\frac{d \psi}{dx}}{-\epsilon}{\epsilon} \\
&=
\frac{-\Hbar^2}{2m}
\psi(0)
\lr
{
-\kappa e^{-\kappa (\epsilon)}

\kappa e^{\kappa (-\epsilon)}
}.
\end{aligned}

In the $$\epsilon \rightarrow 0$$ limit this gives

\label{eqn:diracPotential:220}
V_0 = \frac{\Hbar^2 \kappa}{m}.

Equating relations for $$\kappa$$ we have

\label{eqn:diracPotential:240}
\kappa = \frac{m V_0}{\Hbar^2} = \frac{\sqrt{-2 m E}}{\Hbar},

or

\label{eqn:diracPotential:260}
E = -\inv{2 m} \lr{ \frac{m V_0}{\Hbar} }^2,

with

\label{eqn:diracPotential:280}
\psi(x, t < 0) = C \exp\lr{ -i E t/\hbar – \kappa \Abs{x}}.

The normalization requires

\label{eqn:diracPotential:300}
1
= 2 \Abs{C}^2 \int_0^\infty e^{- 2 \kappa x} dx
= 2 \Abs{C}^2 \evalrange{\frac{e^{- 2 \kappa x}}{-2 \kappa}}{0}{\infty}
= \frac{\Abs{C}^2}{\kappa},

so
\label{eqn:diracPotential:320}
\boxed{
\psi(x, t < 0) = \inv{\sqrt{\kappa}} \exp\lr{ -i E t/\hbar – \kappa \Abs{x}}. } There is only one bound state for such a potential. After turning off the potential, any plane wave $$\label{eqn:diracPotential:360} \psi(x, t) = e^{i k x – i E(k) t/\Hbar},$$ where $$\label{eqn:diracPotential:380} k = \frac{\sqrt{2 m E}}{\Hbar},$$ is a solution. In particular, at $$t = 0$$, the wave packet $$\label{eqn:diracPotential:400} \psi(x,0) = \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x} A(k) dk,$$ is a solution. To solve for $$A(k)$$, we require $$\label{eqn:diracPotential:420} \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x} A(k) dk = \inv{\sqrt{\kappa}} e^{ – \kappa \Abs{x} },$$ or $$\label{eqn:diracPotential:440} \boxed{ A(k) = \inv{\sqrt{2\pi \kappa}} \int_{-\infty}^\infty e^{-i k x} e^{ – m V_0 \Abs{x}/\Hbar^2 } dx. }$$ The initial time state established by the delta function potential evolves as \label{eqn:diracPotential:480} \boxed{ \psi(x, t > 0) = \inv{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i k x – i \Hbar k^2 t/2m} A(k) dk.
}

# References

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

## PHY1520H Graduate Quantum Mechanics. Lecture 16: Addition of angular momenta. Taught by Prof. Arun Paramekanti

### 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.

• For orbital angular momentum

\label{eqn:qmLecture16:20}
\begin{aligned}
\hat{\BL}_1 &= \hat{\Br}_1 \cross \hat{\Bp}_1 \\
\hat{\BL}_1 &= \hat{\Br}_1 \cross \hat{\Bp}_1,
\end{aligned}

We can show that it is true that

\label{eqn:qmLecture16:40}
\antisymmetric{L_{1i} + L_{2i}}{L_{1j} + L_{2j}} =
i \Hbar \epsilon_{i j k} \lr{ L_{1k} + L_{2k} },

because the angular momentum of the independent particles commute. Given this is it fair to consider that the sum

\label{eqn:qmLecture16:60}
\hat{\BL}_1 + \hat{\BL}_2

is also angular momentum.

• Given $$\ket{l_1, m_1}$$ and $$\ket{l_2, m_2}$$, if a measurement is made of $$\hat{\BL}_1 + \hat{\BL}_2$$, what do we get?

Specifically, what do we get for

\label{eqn:qmLecture16:80}
\lr{\hat{\BL}_1 + \hat{\BL}_2}^2,

and for
\label{eqn:qmLecture16:100}
\lr{\hat{L}_{1z} + \hat{L}_{2z}}.

For the latter, we get

\label{eqn:qmLecture16:120}
\lr{\hat{L}_{1z} + \hat{L}_{2z}}\ket{ l_1, m_1 ; l_2, m_2 }
=
\lr{ \Hbar m_1 + \Hbar m_2 } \ket{ l_1, m_1 ; l_2, m_2 }

Given
\label{eqn:qmLecture16:140}
\hat{L}_{1z} + \hat{L}_{2z} = \hat{L}_z^{\textrm{tot}},

we find
\label{eqn:qmLecture16:160}
\begin{aligned}
\antisymmetric{\hat{L}_z^{\textrm{tot}}}{\hat{\BL}_1^2} &= 0 \\
\antisymmetric{\hat{L}_z^{\textrm{tot}}}{\hat{\BL}_2^2} &= 0 \\
\antisymmetric{\hat{L}_z^{\textrm{tot}}}{\hat{\BL}_{1z}} &= 0 \\
\antisymmetric{\hat{L}_z^{\textrm{tot}}}{\hat{\BL}_{1z}} &= 0.
\end{aligned}

We also find

\label{eqn:qmLecture16:180}
\begin{aligned}
\antisymmetric{(\hat{\BL}_1 + \hat{\BL}_2)^2}{\BL_1^2}
&=
\antisymmetric{\hat{\BL}_1^2 + \hat{\BL}_2^2 + 2 \hat{\BL}_1 \cdot
\hat{\BL}_2}{\BL_1^2} \\
&=
0,
\end{aligned}

but for
\label{eqn:qmLecture16:200}
\begin{aligned}
\antisymmetric{(\hat{\BL}_1 + \hat{\BL}_2)^2}{\hat{L}_{1z}}
&=
\antisymmetric{\hat{\BL}_1^2 + \hat{\BL}_2^2 + 2 \hat{\BL}_1 \cdot
\hat{\BL}_2}{\hat{L}_{1z}} \\
&=
2 \antisymmetric{\hat{\BL}_1 \cdot \hat{\BL}_2}{\hat{L}_{1z}} \\
&\ne 0.
\end{aligned}

Classically if we have measured $$\hat{\BL}_{1}$$ and $$\hat{\BL}_{2}$$ then we know the total angular momentum as sketched in fig. 1.

fig. 1. Classical addition of angular momenta.

In QM where we don’t know all the components of the angular momenum simultaneously, things get fuzzier. For example, if the $$\hat{L}_{1z}$$ and $$\hat{L}_{2z}$$ components have been measured, we have the angular momentum defined within a conical region as sketched in fig. 2.

fig. 2. Addition of angular momenta given measured L_z

Suppose we know $$\hat{L}_z^{\textrm{tot}}$$ precisely, but have impricise information about $$\lr{\hat{\BL}^{\textrm{tot}}}^2$$. Can we determine bounds for this? Let $$\ket{\psi} = \ket{ l_1, m_2 ; l_2, m_2 }$$, so

\label{eqn:qmLecture16:220}
\begin{aligned}
\bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi}
&=
\bra{\psi} \hat{\BL}_1^2 \ket{\psi}
+ \bra{\psi} \hat{\BL}_2^2 \ket{\psi}
+ 2 \bra{\psi} \hat{\BL}_1 \cdot \hat{\BL}_2 \ket{\psi} \\
&=
l_1 \lr{ l_1 + 1} \Hbar^2
+ l_2 \lr{ l_2 + 1} \Hbar^2
+ 2
\bra{\psi} \hat{\BL}_1 \cdot \hat{\BL}_2 \ket{\psi}.
\end{aligned}

Using the Cauchy-Schwartz inequality

\label{eqn:qmLecture16:240}
\Abs{\braket{\phi}{\psi}}^2 \le
\Abs{\braket{\phi}{\phi}}
\Abs{\braket{\psi}{\psi}},

which is the equivalent of the classical relationship
\label{eqn:qmLecture16:260}
\lr{ \BA \cdot \BB }^2 \le \BA^2 \BB^2.

Applying this to the last term, we have

\label{eqn:qmLecture16:280}
\begin{aligned}
\lr{ \bra{\psi} \hat{\BL}_1 \cdot \hat{\BL}_2 \ket{\psi} }^2
&\le
\bra{ \psi} \hat{\BL}_1 \cdot \hat{\BL}_1 \ket{\psi}
\bra{ \psi} \hat{\BL}_2 \cdot \hat{\BL}_2 \ket{\psi} \\
&=
\Hbar^4
l_1 \lr{ l_1 + 1 }
l_2 \lr{ l_2 + 2 }.
\end{aligned}

Thus for the max we have

\label{eqn:qmLecture16:300}
\bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi}
\le
\Hbar^2 l_1 \lr{ l_1 + 1 }
+\Hbar^2 l_2 \lr{ l_2 + 1 }
+ 2 \Hbar^2 \sqrt{ l_1 \lr{ l_1 + 1 } l_2 \lr{ l_2 + 2 } }

and for the min
\label{eqn:qmLecture16:360}
\bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi}
\ge
\Hbar^2 l_1 \lr{ l_1 + 1 }
+\Hbar^2 l_2 \lr{ l_2 + 1 }
– 2 \Hbar^2 \sqrt{ l_1 \lr{ l_1 + 1 } l_2 \lr{ l_2 + 2 } }.

To try to pretty up these estimate, starting with the max, note that if we replace a portion of the RHS with something bigger, we are left with a strict less than relationship.

That is

\label{eqn:qmLecture16:320}
\begin{aligned}
l_1 \lr{ l_1 + 1 } &< \lr{ l_1 + \inv{2} }^2 \\ l_2 \lr{ l_2 + 1 } &< \lr{ l_2 + \inv{2} }^2 \end{aligned} That is \label{eqn:qmLecture16:340} \begin{aligned} \bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi} &< \Hbar^2 \lr{ l_1 \lr{ l_1 + 1 } + l_2 \lr{ l_2 + 1 } + 2 \lr{ l_1 + \inv{2} } \lr{ l_2 + \inv{2} } } \\ &= \Hbar^2 \lr{ l_1^2 + l_2^2 + l_1 + l_2 + 2 l_1 l_2 + l_1 + l_2 + \inv{2} } \\ &= \Hbar^2 \lr{ \lr{ l_1 + l_2 + \inv{2} } \lr{ l_1 + l_2 + \frac{3}{2} } - \inv{4} } \end{aligned} or $$\label{eqn:qmLecture16:380} l_{\textrm{tot}} \lr{ l_{\textrm{tot}} + 1 } < \lr{ l_1 + l_2 + \inv{2} } \lr{ l_1 + l_2 + \frac{3}{2} } ,$$ which, gives $$\label{eqn:qmLecture16:400} l_{\textrm{tot}} < l_1 + l_2 + \inv{2}.$$ Finally, given a quantization requirement, that is $$\label{eqn:t:1} \boxed{ l_{\textrm{tot}} \le l_1 + l_2. }$$ Similarly, for the min, we find \label{eqn:qmLecture16:440} \begin{aligned} \bra{\psi} \lr{ \hat{\BL}_1 + \hat{\BL}_2 }^2 \ket{\psi} &>
\Hbar^2
\lr{
l_1 \lr{ l_1 + 1 }
+ l_2 \lr{ l_2 + 1 }
– 2 \lr{ l_1 + \inv{2} } \lr{ l_2 + \inv{2} }
} \\
&=
\Hbar^2
\lr{
l_1^2 + l_2^2 %+ \cancel{l_1 + l_2}
– 2 l_1 l_2
– \inv{2}
}
\end{aligned}

This will be finished Thursday, but we should get

\label{eqn:t:2}
\boxed{
\Abs{l_1 – l_2} \le l_{\textrm{tot}} \le l_1 + l_2.
}

# References

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

## Two spin time evolution

November 14, 2015 phy1520 No comments , , , ,

## Motivation

Our midterm posed a (low mark “quick question”) that I didn’t complete (or at least not properly). This shouldn’t have been a difficult question, but I spend way too much time on it, costing me time that I needed for other questions.

It turns out that there isn’t anything fancy required for this question, just perseverance and careful work.

## Guts

The question asked for the time evolution of a two particle state

\label{eqn:twoSpinHamiltonian:20}
\psi = \inv{\sqrt{2}} \lr{ \ket{\uparrow \downarrow} – \ket{\downarrow \uparrow} }

under the action of the Hamiltonian

\label{eqn:twoSpinHamiltonian:40}
H = – B S_{z,1} + 2 B S_{x,2} = \frac{\Hbar B}{2}\lr{ -\sigma_{z,1} + 2 \sigma_{x,2} } .

We have to know the action of the Hamiltonian on all the states

\label{eqn:twoSpinHamiltonian:60}
\begin{aligned}
H \ket{\uparrow \uparrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \uparrow} + 2 \ket{\uparrow \downarrow} } \\
H \ket{\uparrow \downarrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \downarrow} + 2 \ket{\uparrow \uparrow} } \\
H \ket{\downarrow \uparrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \uparrow} + 2 \ket{\downarrow \downarrow} } \\
H \ket{\downarrow \downarrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \downarrow} + 2 \ket{\downarrow \uparrow} } \\
\end{aligned}

With respect to the basis $$\setlr{ \ket{\uparrow \uparrow}, \ket{\uparrow \downarrow}, \ket{\downarrow \uparrow}, \ket{\downarrow \downarrow} }$$, the matrix of the Hamiltonian is

\label{eqn:twoSpinHamiltonian:80}
H =
\frac{ \Hbar B }{2}
\begin{bmatrix}
-1 & 2 & 0 & 0 \\
2 & -1 & 0 & 0 \\
0 & 0 & 1 & 2 \\
0 & 0 & 2 & 1 \\
\end{bmatrix}

Utilizing the block diagonal form (and ignoring the $$\Hbar B/2$$ factor for now), the characteristic equation is

\label{eqn:twoSpinHamiltonian:100}
0
=
\begin{vmatrix}
-1 -\lambda & 2 \\
2 & -1 – \lambda
\end{vmatrix}
\begin{vmatrix}
1 -\lambda & 2 \\
2 & 1 – \lambda
\end{vmatrix}
=
\lr{(1 + \lambda)^2 – 4}
\lr{(1 – \lambda)^2 – 4}.

This has solutions

\label{eqn:twoSpinHamiltonian:120}
1 \pm \lambda = \pm 2,

or, with the $$\Hbar B/2$$ factors put back in

\label{eqn:twoSpinHamiltonian:140}
\lambda = \pm \Hbar B/2 , \pm 3 \Hbar B/2.

I was thinking that we needed to compute the time evolution operator

\label{eqn:twoSpinHamiltonian:160}
U = e^{-i H t/\Hbar},

but we actually only need the eigenvectors, and the inverse relations. We can find the eigenvectors by inspection in each case from

\label{eqn:twoSpinHamiltonian:180}
\begin{aligned}
H – (1) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
-2 & 2 & 0 & 0 \\
2 & -2 & 0 & 0 \\
0 & 0 & 0 & 2 \\
0 & 0 & 2 & 0 \\
\end{bmatrix} \\
H – (-1) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
0 & 2 & 0 & 0 \\
2 & 0 & 0 & 0 \\
0 & 0 & 2 & 2 \\
0 & 0 & 2 & 2 \\
\end{bmatrix} \\
H – (3) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
-4 & 2 & 0 & 0 \\
2 & -4 & 0 & 0 \\
0 & 0 &-2 & 2 \\
0 & 0 & 2 &-2 \\
\end{bmatrix} \\
H – (-3) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
2 & 2 & 0 & 0 \\
2 & 2 & 0 & 0 \\
0 & 0 & 4 & 2 \\
0 & 0 & 2 & 1 \\
\end{bmatrix}.
\end{aligned}

The eigenkets are

\label{eqn:twoSpinHamiltonian:280}
\begin{aligned}
\ket{1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1 \\
0 \\
0 \\
\end{bmatrix} \\
\ket{-1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
0 \\
1 \\
-1 \\
\end{bmatrix} \\
\ket{3} &=
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
0 \\
1 \\
1 \\
\end{bmatrix} \\
\ket{-3} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
-1 \\
0 \\
0 \\
\end{bmatrix},
\end{aligned}

or

\label{eqn:twoSpinHamiltonian:300}
\begin{aligned}
\sqrt{2} \ket{1} &= \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} \\
\sqrt{2} \ket{-1} &= \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} \\
\sqrt{2} \ket{3} &= \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} \\
\sqrt{2} \ket{-3} &= \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow}.
\end{aligned}

We can invert these

\label{eqn:twoSpinHamiltonian:220}
\begin{aligned}
\ket{\uparrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{1} + \ket{-3} } \\
\ket{\uparrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{1} – \ket{-3} } \\
\ket{\downarrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{3} + \ket{-1} } \\
\ket{\downarrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{3} – \ket{-1} } \\
\end{aligned}

The original state of interest can now be expressed in terms of the eigenkets

\label{eqn:twoSpinHamiltonian:240}
\psi
=
\inv{2} \lr{
\ket{1} – \ket{-3} –
\ket{3} – \ket{-1}
}

The time evolution of this ket is

\label{eqn:twoSpinHamiltonian:260}
\begin{aligned}
\psi(t)
&=
\inv{2}
\lr{
e^{-i B t/2} \ket{1}
– e^{3 i B t/2} \ket{-3}
– e^{-3 i B t/2} \ket{3}
– e^{i B t/2} \ket{-1}
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
e^{-i B t/2} \lr{ \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} }
– e^{3 i B t/2} \lr{ \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow} }
– e^{-3 i B t/2} \lr{ \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} }
– e^{i B t/2} \lr{ \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} }
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
\lr{ e^{-i B t/2} – e^{3 i B t/2} } \ket{\uparrow \uparrow}
+ \lr{ e^{-i B t/2} + e^{3 i B t/2} } \ket{\uparrow \downarrow}
– \lr{ e^{-3 i B t/2} + e^{i B t/2} } \ket{\downarrow \uparrow}
+ \lr{ e^{i B t/2} – e^{-3 i B t/2} } \ket{\downarrow \downarrow}
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
e^{i B t/2} \lr{ e^{-2 i B t/2} – e^{2 i B t/2} } \ket{\uparrow \uparrow}
+ e^{i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\uparrow \downarrow}
– e^{- i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\downarrow \uparrow}
+ e^{- i B t/2} \lr{ e^{2 i B t/2} – e^{-2 i B t/2} } \ket{\downarrow \downarrow}
} \\
&=
\inv{\sqrt{2}}
\lr{
i \sin( B t )
\lr{
e^{- i B t/2} \ket{\downarrow \downarrow} – e^{i B t/2} \ket{\uparrow \uparrow}
}
+ \cos( B t ) \lr{
e^{i B t/2} \ket{\uparrow \downarrow}
– e^{- i B t/2} \ket{\downarrow \uparrow}
}
}
\end{aligned}

Note that this returns to the original state when $$t = \frac{2 \pi n}{B}, n \in \mathbb{Z}$$. I think I’ve got it right this time (although I got a slightly different answer on paper before typing it up.)

This doesn’t exactly seem like a quick answer question, at least to me. Is there some easier way to do it?

## PHY1520H Graduate Quantum Mechanics. Lecture 15: angular momentum rotation representation, and angular momentum addition. Taught by Prof. Arun Paramekanti

### 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 3. content from [1].

### Angular momentum (wrap up.)

We found

\label{eqn:qmLecture15:20}
\begin{aligned}
\hat{\BL^2} \ket{j, m} &= j(j+1) \Hbar^2 \ket{j,m} \\
\hat{L}_z \ket{j, m} &= \Hbar m \ket{j,m} \\
\hat{L}_{\pm} \ket{j, m } &= \Hbar \sqrt{(j \mp m)(j \pm m + 1)} \ket{j, m \pm 1 }
\end{aligned}

or Schwinger

\label{eqn:qmLecture15:40}
\begin{aligned}
\hat{L}_z &= \inv{2} \lr{ \hat{n}_1 – \hat{n}_2 } \Hbar \\
\hat{L}_{+} &= a_1^\dagger a_2 \Hbar \\
\hat{L}_{-} &= a_1 a_2^\dagger \Hbar \\
j &= \inv{2} \lr{ \hat{n}_1 + \hat{n}_2 },
\end{aligned}

where each of the $$a_1, a_2$$ operators obey

\label{eqn:qmLecture15:60}
\begin{aligned}
\antisymmetric{a_1}{a_1^\dagger} &= 1 \\
\antisymmetric{a_2}{a_2^\dagger} &= 1
\end{aligned}

and any pair of different index $$a$$ operators commute, as in

\label{eqn:qmLecture15:80}
\antisymmetric{a_1}{a_2^\dagger} = 0.

### Representations

It’s possible to compute matrix representations of the rotation operators

\label{eqn:qmLecture15:100}
\hat{R}_\ncap(\phi) = e^{i \hat{\BL} \cdot \ncap \phi/\Hbar}.

With respect to a ket it’s possible to find

\label{eqn:qmLecture15:120}
e^{i \hat{\BL} \cdot \ncap \phi/\Hbar} \ket{j, m}
=
\sum_{m’} d^j_{m m’}(\ncap, \phi) \ket{ j, m’ }.

This has a block diagonal form that’s sketched in fig. 1.

fig. 1. Block diagonal form for angular momentum matrix representation.

We can view $$d^j_{m m’}(\ncap, \phi)$$ as a matrix, representing the rotation. The problem of determining these matrices can be reduced to that of determining the matrix for $$\hat{\BL}$$, because once we have that we can exponentiate that.

### Example: spin 1/2

From the eigenvalue relationships, with basis states

\label{eqn:qmLecture15:160}
\begin{aligned}
\ket{\uparrow} &=
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
\ket{\downarrow} &=
\begin{bmatrix}
0 \\
1 \\
\end{bmatrix}
\end{aligned}

we find

\label{eqn:qmLecture15:180}
\begin{aligned}
\hat{L}_z &= \frac{\Hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \\
\hat{L}_{+} &= \frac{\Hbar}{2}
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} \\
\hat{L}_{-} &= \frac{\Hbar}{2}
\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}.
\end{aligned}

Rearranging we find the Pauli matrices

\label{eqn:qmLecture15:200}
\hat{L}_k = \inv{2} \Hbar \sigma_i.

Noting that $$\lr{ \Bsigma \cdot \ncap }^2 = 1$$, and $$\lr{\Bsigma \cdot \ncap }^3 = \Bsigma \cdot \ncap$$, the rotation matrix is

\label{eqn:qmLecture15:220}
e^{ i \Bsigma \cdot \ncap \phi/2 } \ket{\inv{2}, m} = \lr{ \cos( \phi/2 ) + i \Bsigma \cdot \ncap \sin(\phi/2) } \ket{\inv{2}, m}.

The steps are

1. Enumerate the states.
\label{eqn:qmLecture15:140}
j_1 = \inv{2} \leftrightarrow\, \mbox{2 states (dimension of irrep = 2)}

2. Construct the $$\hat{\BL}$$ matrices.
3. Construct $$d_{m m’}^j(\ncap, \phi)$$.

### Angular momentum operator in space representation

For $$L = 1$$ it turns out that the rotation matrices turn out to be the 3D rotation matrices. In the space representation

\label{eqn:qmLecture15:240}
\BL = \Br \cross \Bp,

the coordinates of the operator are

\label{eqn:qmLecture15:260}
\hat{L}_k = i \epsilon_{k m n} r_m \lr{ -i \Hbar \PD{r_n}{} }

We see that scaling $$\Br \rightarrow \alpha \Br$$ does not change this operator, allowing for an angular representation $$\hat{L}_k(\theta, \phi)$$ that have the form

\label{eqn:qmLecture15:280}
\begin{aligned}
\hat{L}_z &= -i \Hbar \PD{\phi}{} \\
\hat{L}_{\pm} &= \Hbar \lr{ \pm \PD{\theta}{} + i \cot \theta \PD{\phi}{} }.
\end{aligned}

Here $$\theta$$ and $$\phi$$ are the polar and azimuthal angles respectively as illustrated in fig. 2.

fig. 2. Spherical coordinate convention.

The equivalent wave function representation of the problem is

\label{eqn:qmLecture15:300}
\begin{aligned}
\hat{\BL} Y_{lm}(\theta, \phi) &= \Hbar^2 l (l + 1) Y_{lm}(\theta, \phi) \\
\hat{L}_z Y_{lm}(\theta, \phi) &= \Hbar m Y_{lm}(\theta, \phi) \\
\end{aligned}

One can find these functions

\label{eqn:qmLecture15:320}
Y_{lm}(\theta, \phi) = P_{l, m}(\cos \theta) e^{i m \phi},

where $$P_{l, m}(\cos \theta)$$ are called the associated Legendre polynomials. This can be applied whenever we have

\label{eqn:qmLecture15:340}
\antisymmetric{H}{\hat{L}_k} = 0.

where all the eigenfunctions will have the form

\label{eqn:qmLecture15:360}
\Psi(r, \theta, \phi) = R(r) Y_{lm}(\theta, \phi).

Since $$\hat{\BL}$$ is a vector we expect to be able to add angular momentum in some way similar to the addition of classical vectors as illustrated in fig. 3.

When we have a potential that depends only on the difference in position $$V(\Br_1 – \Br_2)$$ then we know from classical problems it is effective to work in center of mass coordinates

\label{eqn:qmLecture15:380}
\begin{aligned}
\hat{\BR}_{\textrm{cm}} &= \frac{\hat{\Br}_1 + \hat{\Br}_2}{2} \\
\hat{\BP}_{\textrm{cm}} &= \hat{\Bp}_1 + \hat{\Bp}_2
\end{aligned}

where

\label{eqn:qmLecture15:400}
\antisymmetric{\hat{R}_i}{\hat{P}_j} = i \Hbar \delta_{ij}.

Given

\label{eqn:qmLecture15:420}
\hat{\BL}_1 + \hat{\BL}_2 = \hat{\BL}_{\textrm{tot}},

do we have
\label{eqn:qmLecture15:440}
\antisymmetric{
\hat{L}_{\textrm{tot}, i}
}{
\hat{L}_{\textrm{tot}, j}
}
= i \Hbar \epsilon_{i j k} \hat{L}_{\textrm{tot}, k} ?

That is

\label{eqn:qmLecture15:460}
\antisymmetric{\hat{L}_{1,i} + \hat{L}_{1,j}}{\hat{L}_{2,i} + \hat{L}_{2,j}} = i \Hbar \epsilon_{i j k} \lr{ \hat{L}_{1,k} + \hat{L}_{1,k} }

FIXME: Right at the end of the lecture, there was a mention of something about whether or not $$\hat{\BL}_1^2$$ and $$\hat{L}_{1,z}$$ were sharply defined, but I missed it. Ask about this if not covered in the next lecture.

# References

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

## 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

### 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.