## Commutators of angular momentum and a central force Hamiltonian

September 30, 2015 phy1520 No comments , , , ,

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

\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}

The $$L_x, L_z$$ commutator is

\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}

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

\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}

and for $$L_z, L_y$$

\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}

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

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

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

\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}

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

\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}

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.

## PHY1520H Graduate Quantum Mechanics. Lecture 4: Quantum Harmonic oscillator and coherent states. Taught by Prof. Arun Paramekanti

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough. This lecture reviewed a lot of quantum harmonic oscillator theory, and wouldn’t make sense without having seen raising and lowering operators (ladder operators), number operators, and the like.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 2 content.

### Classical Harmonic Oscillator

Recall the classical Harmonic oscillator equations in their Hamiltonian form

\label{eqn:qmLecture4:40}
\ddt{x} = \frac{p}{m}

\label{eqn:qmLecture4:60}
\ddt{p} = -k x.

With

\label{eqn:qmLecture4:140}
\begin{aligned}
x(t = 0) &= x_0 \\
p(t = 0) &= p_0 \\
k &= m \omega^2,
\end{aligned}

the solutions are ellipses in phase space

\label{eqn:qmLecture4:100}
x(t) = x_0 \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t)

\label{eqn:qmLecture4:120}
p(t) = p_0 \cos(\omega t) – m \omega x_0 \sin(\omega t).

After a suitable scaling of the variables, these elliptical orbits can be transformed into circular trajectories.

### Quantum Harmonic Oscillator

\label{eqn:qmLecture4:160}
\hat{H} = \frac{\hat{p}^2}{2 m} + \inv{2} k \hat{x}^2

Set

\label{eqn:qmLecture4:200}
\hat{X} = \sqrt{\frac{m \omega}{\Hbar}} \hat{x}

\label{eqn:qmLecture4:220}
\hat{P} = \sqrt{\inv{m \omega \Hbar}} \hat{p}

The commutators after this change of variables goes from

\label{eqn:qmLecture4:240}
\antisymmetric{ \hat{x}}{\hat{p}} = i \Hbar,

to
\label{eqn:qmLecture4:260}
\antisymmetric{ \hat{X}}{\hat{P}} = i.

The Hamiltonian takes the form

\label{eqn:qmLecture4:280}
\begin{aligned}
\hat{H}
&= \frac{\Hbar \omega}{2} \lr{ \hat{X}^2 + \hat{P}^2 } \\
&= \Hbar \omega \lr{ \lr{ \frac{\hat{X} -i \hat{P}}{\sqrt{2}} } \lr{ \frac{\hat{X} +i \hat{P}}{\sqrt{2}}} + \inv{2} }.
\end{aligned}

Define ladder operators (raising and lowering operators respectively)

\label{eqn:qmLecture4:320}
\hat{a}^\dagger = \frac{\hat{X} -i \hat{P}}{\sqrt{2}}

\label{eqn:qmLecture4:340}
\hat{a} = \frac{\hat{X} +i \hat{P}}{\sqrt{2}}

so

\label{eqn:qmLecture4:360}
\hat{H} = \Hbar \omega \lr{ \hat{a}^\dagger \hat{a} + \inv{2} }.

We can show

\label{eqn:qmLecture4:380}
\antisymmetric{\hat{a}}{\hat{a}^\dagger} = 1,

and

\label{eqn:qmLecture4:400}
N \ket{n} \equiv \hat{a}^\dagger a = n \ket{n},

where $$n \ge 0$$ is an integer. Recall that

\label{eqn:qmLecture4:420}
\hat{a} \ket{0} = 0,

and

\label{eqn:qmLecture4:440}
\bra{X} X + i P \ket{0} = 0.

With

\label{eqn:qmLecture4:460}
\braket{x}{0} = \Psi_0(x),

we can show

\label{eqn:qmLecture4:480}
\inv{\sqrt{2}} \lr{ X + \PD{X}{} } \Psi_0(X) = 0.

Also recall that

\label{eqn:qmLecture4:520}
\hat{a} \ket{n} = \sqrt{n} \ket{n-1}

\label{eqn:qmLecture4:540}
\hat{a}^\dagger \ket{n} = \sqrt{n + 1} \ket{n+1}

### Coherent states

Coherent states for the quantum harmonic oscillator are the eigenkets for the creation and annihilation operators

\label{eqn:qmLecture4:580}
\hat{a} \ket{z} = z \ket{z}

\label{eqn:qmLecture4:600}
\hat{a}^\dagger \ket{\tilde{z}} = \tilde{z} \ket{\tilde{z}} ,

where

\label{eqn:qmLecture4:620}
\ket{z} = \sum_{n = 0}^\infty c_n \ket{n},

and $$z$$ is allowed to be a complex number.

Looking for such a state, we compute

\label{eqn:qmLecture4:640}
\begin{aligned}
\hat{a} \ket{z}
&= \sum_{n=1}^\infty c_n \hat{a} \ket{n} \\
&= \sum_{n=1}^\infty c_n \sqrt{n} \ket{n-1}
\end{aligned}

compare this to

\label{eqn:qmLecture4:660}
\begin{aligned}
z \ket{z}
&=
z \sum_{n=0}^\infty c_n \ket{n} \\
&=
\sum_{n=1}^\infty c_n \sqrt{n} \ket{n-1} \\
&=
\sum_{n=0}^\infty c_{n+1} \sqrt{n+1} \ket{n},
\end{aligned}

so

\label{eqn:qmLecture4:680}
c_{n+1} \sqrt{n+1} = z c_n

This gives

\label{eqn:qmLecture4:700}
c_{n+1} = \frac{z c_n}{\sqrt{n+1}}

\label{eqn:qmLecture4:720}
\begin{aligned}
c_1 &= c_0 z \\
c_2 &= \frac{z c_1}{\sqrt{2}} = \frac{z^2 c_0}{\sqrt{2}} \\
\vdots &
\end{aligned}

or

\label{eqn:qmLecture4:740}
c_n = \frac{z^n}{\sqrt{n!}}.

So the desired state is

\label{eqn:qmLecture4:760}
\ket{z} = c_0 \sum_{n=0}^\infty \frac{z^n}{\sqrt{n!}} \ket{n}.

Also recall that

\label{eqn:qmLecture4:780}
\ket{n} = \frac{\lr{ \hat{a}^\dagger }^n}{\sqrt{n!}} \ket{0},

which gives

\label{eqn:qmLecture4:800}
\begin{aligned}
\ket{z}
&= c_0 \sum_{n=0}^\infty \frac{\lr{z \hat{a}^\dagger}^n }{n!} \ket{0} \\
&= c_0 e^{z \hat{a}^\dagger} \ket{0}.
\end{aligned}

The normalization is

\label{eqn:qmLecture4:820}
c_0 = e^{-\Abs{z}^2/2}.

While we have $$\braket{n_1}{n_2} = \delta_{n_1, n_2}$$, these $$\ket{z}$$ states are not orthonormal. Figuring out that this overlap

\label{eqn:qmLecture4:840}
\braket{z_1}{z_2} \ne 0,

will be left for homework.

### Dynamics

We don’t know much about these coherent states. For example does a coherent state at time zero evolve to a coherent state?

\label{eqn:qmLecture4:860}
\ket{z} \stackrel{?}{\rightarrow} \ket{z(t)}

It turns out that these questions are best tackled in the Heisenberg picture, considering

\label{eqn:qmLecture4:880}
e^{-i \hat{H} t/\Hbar } \ket{z}.

For example, what is the average of the position operator

\label{eqn:qmLecture4:900}
\bra{z} e^{i \hat{H} t/\Hbar } \hat{x} e^{-i \hat{H} t/\Hbar } \ket{z}
=
\sum_{n, n’ = 0}^\infty
\bra{n} c_n^\conj e^{i E_n t/\Hbar}
\lr{ a + a^\dagger} \sqrt{ \frac{\Hbar}{m \omega} }
c_{n’} e^{i E_{n’} t/\Hbar}
\ket{n}.

This is very messy to attempt. Instead if we know how the operator evolves we can calculate

\label{eqn:qmLecture4:920}
\bra{z} \hat{x}_{\textrm{H}}(t) \ket{z},

that is

\label{eqn:qmLecture4:940}
\expectation{\hat{x}}(t) = \bra{z} \hat{x}_{\textrm{H}}(t) \ket{z},

and for momentum

\label{eqn:qmLecture4:960}
\expectation{\hat{p}}(t) = \bra{z} \hat{p}_{\textrm{H}}(t) \ket{z}.

The question to ask is what are the expansions of

\label{eqn:qmLecture4:1000}
\hat{a}_{\textrm{H}}(t) = e^{i \hat{H} t/\Hbar} \hat{a} e^{-i \hat{H} t/\Hbar}.

\label{eqn:qmLecture4:1020}
\hat{a}^\dagger_{\textrm{H}}(t) = e^{i \hat{H} t/\Hbar} \hat{a}^\dagger e^{-i \hat{H} t/\Hbar}.

The question to ask is how do these operators ask on the basis states

\label{eqn:qmLecture4:1040}
\begin{aligned}
\hat{a}_{\textrm{H}}(t) \ket{n}
&= e^{i \hat{H} t/\Hbar} \hat{a} e^{-i \hat{H} t/\Hbar} \ket{n} \\
&= e^{i \hat{H} t/\Hbar} \hat{a} e^{-i t \omega (n + 1/2)} \ket{n} \\
&=
e^{-i t \omega (n + 1/2)}
e^{i \hat{H} t/\Hbar}
\sqrt{n} \ket{n-1} \\
&=
\sqrt{n}
e^{-i t \omega (n + 1/2)}
e^{i t \omega (n – 1/2)}
\ket{n-1} \\
&=
\sqrt{n} e^{-i \omega t} \ket{n-1} \\
&=
e^{-i \omega t} \ket{n}.
\end{aligned}

So we have found

\label{eqn:qmLecture4:1060}
\begin{aligned}
\hat{a}_{\textrm{H}}(t) &= a e^{-i\omega t} \\
\hat{a}^\dagger_{\textrm{H}}(t) &= a^\dagger e^{i\omega t}
\end{aligned}

# References

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

## Can anticommuting operators have a simulaneous eigenket?

September 28, 2015 phy1520 No comments , ,

## Question: Can anticommuting operators have a simulaneous eigenket? ([1] pr. 1.16)

Two Hermitian operators anticommute

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:20}
\symmetric{A}{B} = A B + B A = 0.

Is it possible to have a simultaneous eigenket of $$A$$ and $$B$$? Prove or illustrate your assertion.

Suppose that such a simultaneous non-zero eigenket $$\ket{\alpha}$$ exists, then

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:40}
A \ket{\alpha} = a \ket{\alpha},

and

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:60}
B \ket{\alpha} = b \ket{\alpha}

This gives

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:80}
\lr{ A B + B A } \ket{\alpha}
=
\lr{A b + B a} \ket{\alpha}
= 2 a b \ket{\alpha}.

If this is zero, one of the operators must have a zero eigenvalue. Knowing that we can construct an example of such operators. In matrix form, let

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:120}
A =
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & a \\
\end{bmatrix}

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:140}
B =
\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & b \\
\end{bmatrix}.

These are both Hermitian, and anticommute provided at least one of $$a, b$$ is zero. These have a common eigenket

\label{eqn:anticommutingOperatorWithSimulaneousEigenket:160}
\ket{\alpha} =
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.

A zero eigenvalue of one of the commuting operators may not be a sufficient condition for such anticommutation.

# References

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

## Grade 11 physics handout: “The Big Five … in Physics”?

September 26, 2015 math and physics play No comments ,

## Motivation

Check out fig. 1, a handout given to my daughter has a handout from her grade 11 physics class, titled “The Big Five”, covering some dynamics equations.

fig. 1. The Big Five… in Physics

I found this handout disorienting. Part of that disorientation is because of the weird African animal theme which I couldn’t see the rationale for. Aurora showed me that the second equation can be outlined with an elephant, apparently justifying the animal theme. The equations themselves are not in a form that I would have expected, and have a lot of redundancy built in. The assumptions required for these equations to be valid are also not stated. Those equations are

\label{eqn:theBigFivePhysics:40}
v_2 = v_1 + a \Delta t

\label{eqn:theBigFivePhysics:60}
\Delta d = \inv{2} \lr{ v_1 + v_2 } \Delta t

\label{eqn:theBigFivePhysics:80}
\Delta d = v_1 \Delta t + \inv{2} a \lr{\Delta t}^2

\label{eqn:theBigFivePhysics:100}
\Delta d = v_2 \Delta t – \inv{2} a \lr{\Delta t}^2

\label{eqn:theBigFivePhysics:120}
v_2^2 = v_1^2 + 2 a \Delta d.

## Reverse engineering “the big five”.

### Difference of velocity

The first equation \ref{eqn:theBigFivePhysics:40} is just a discrete version of the definition of scalar acceleration

\label{eqn:theBigFivePhysics:140}
a = \frac{dv}{dt}.

The approximation of that is

\label{eqn:theBigFivePhysics:160}
a = \frac{\Delta v}{\Delta t},

or

\label{eqn:theBigFivePhysics:180}
\Delta v = v_2 – v_1 = a \Delta t.

### Constant acceleration

Next in the list, it’s clear that the equation(s) for $$\Delta d$$ is really based on an assumption of constant acceleration. In fact, all four of the next equations are nothing more than variations of

\label{eqn:theBigFivePhysics:200}
v = a t.

To arrive at fig. 2.

fig. 2. Displacement as area under the curve.

Should we wish to integrate, it’s the second simplest integral we could possibly do

\label{eqn:theBigFivePhysics:220}
\begin{aligned}
\Delta x
&= \int_{t_1}^{t_2} v(t) dt \\
&= \int_{t_1}^{t_2} a t dt \\
&= \inv{2} a \lr{ t_2^2 – t_1^2 } \\
&= \inv{2} a \Delta t \lr{ t_2 + t_1 }.
\end{aligned}

This looks a little different than what’s on the formula cheet, but since (for constant acceleration) we have

\label{eqn:theBigFivePhysics:240}
t = \frac{v}{a},

this can be written as

\label{eqn:theBigFivePhysics:260}
\begin{aligned}
\Delta x
&= \inv{2} a \Delta t \lr{ \frac{v_2}{a} + \frac{v_1}{a} } \\
&= \inv{2} \Delta t \lr{ v_1 + v_2 },
\end{aligned}

as found on the formula sheet (except for them using $$\Delta d$$ for the difference in position .)

Each of the next equations follow from straight algebra

\label{eqn:theBigFivePhysics:280}
\begin{aligned}
\Delta x – v_1 \Delta t
&= \inv{2} \Delta t \lr{ v_1 + v_2 } – v_1 \Delta t \\
&= \inv{2} \Delta t \lr{ -v_1 + v_2 } \\
&= \inv{2} \lr{\Delta t}^2 \frac{\Delta v}{\Delta t} \\
&= \inv{2} a \lr{\Delta t}^2,
\end{aligned}

and

\label{eqn:theBigFivePhysics:300}
\begin{aligned}
\Delta x – v_2 \Delta t
&= \inv{2} \Delta t \lr{ v_1 + v_2 } – v_2 \Delta t \\
&= \inv{2} \Delta t \lr{ v_1 – v_2 } \\
&= -\inv{2} \lr{\Delta t}^2 \frac{\Delta v}{\Delta t} \\
&= -\inv{2} a \lr{\Delta t}^2,
\end{aligned}

and finally

\label{eqn:theBigFivePhysics:320}
\begin{aligned}
\Delta x
&= \inv{2} a \lr{ t_2^2 – t_1^2 } \\
&= \inv{2 a } \lr{ v_2^2 – v_1^2 }.
\end{aligned}

### A better set of equations.

If I had to write these “big five” equation, I’d be more inclined to write them as

\label{eqn:theBigFivePhysics:360}
a = \frac{\Delta v}{\Delta t}

\label{eqn:theBigFivePhysics:380}
v = a t = \frac{\Delta x}{\Delta t}

\label{eqn:theBigFivePhysics:400}
\Delta x = \int_{t_1}^{t_2} v dt = \inv{2} a \lr{ t_2^2 – t_1^2 }

\label{eqn:theBigFivePhysics:420}
t_1 = \frac{v_1}{a}

\label{eqn:theBigFivePhysics:440}
t_2 = \frac{v_2}{a}.

Anything more than that is just algebra. The last two could be omitted since they really follow from \ref{eqn:theBigFivePhysics:380}. For high school where calculus isn’t known, I’d swap out \ref{eqn:theBigFivePhysics:400} for \ref{eqn:theBigFivePhysics:60} which can be derived graphically by understanding that the distance is the area under the velocity curve.

I’d also leave out all mentions of big African animals, which is, just plain weird!

## Lagrangian for magnetic portion of Lorentz force

In [1] it is claimed in an Aharonov-Bohm discussion that a Lagrangian modification to include electromagnetism is

\label{eqn:magneticLorentzForceLagrangian:20}
\LL \rightarrow \LL + \frac{e}{c} \Bv \cdot \BA.

That can’t be the full Lagrangian since there is no $$\phi$$ term, so what exactly do we get?

If you have somehow, like I did, forgot the exact form of the Euler-Lagrange equations (i.e. where do the dots go), then the derivation of those equations can come to your rescue. The starting point is the action

\label{eqn:magneticLorentzForceLagrangian:40}
S = \int \LL(x, \xdot, t) dt,

where the end points of the integral are fixed, and we assume we have no variation at the end points. The variational calculation is

\label{eqn:magneticLorentzForceLagrangian:60}
\begin{aligned}
\delta S
&= \int \delta \LL(x, \xdot, t) dt \\
&= \int \lr{ \PD{x}{\LL} \delta x + \PD{\xdot}{\LL} \delta \xdot } dt \\
&= \int \lr{ \PD{x}{\LL} \delta x + \PD{\xdot}{\LL} \delta \ddt{x} } dt \\
&= \int \lr{ \PD{x}{\LL} – \ddt{}\lr{\PD{\xdot}{\LL}} } \delta x dt
+ \delta x \PD{\xdot}{\LL}.
\end{aligned}

The boundary term is killed after evaluation at the end points where the variation is zero. For the result to hold for all variations $$\delta x$$, we must have

\label{eqn:magneticLorentzForceLagrangian:80}
\boxed{
\PD{x}{\LL} = \ddt{}\lr{\PD{\xdot}{\LL}}.
}

Now lets apply this to the Lagrangian at hand. For the position derivative we have

\label{eqn:magneticLorentzForceLagrangian:100}
\PD{x_i}{\LL}
=
\frac{e}{c} v_j \PD{x_i}{A_j}.

For the canonical momentum term, assuming $$\BA = \BA(\Bx)$$ we have

\label{eqn:magneticLorentzForceLagrangian:120}
\begin{aligned}
\ddt{} \PD{\xdot_i}{\LL}
&=
\ddt{}
\lr{ m \xdot_i
+
\frac{e}{c} A_i
} \\
&=
m \ddot{x}_i
+
\frac{e}{c}
\ddt{A_i} \\
&=
m \ddot{x}_i
+
\frac{e}{c}
\PD{x_j}{A_i} \frac{dx_j}{dt}.
\end{aligned}

Assembling the results, we’ve got

\label{eqn:magneticLorentzForceLagrangian:140}
\begin{aligned}
0
&=
\ddt{} \PD{\xdot_i}{\LL}

\PD{x_i}{\LL} \\
&=
m \ddot{x}_i
+
\frac{e}{c}
\PD{x_j}{A_i} \frac{dx_j}{dt}

\frac{e}{c} v_j \PD{x_i}{A_j},
\end{aligned}

or
\label{eqn:magneticLorentzForceLagrangian:160}
\begin{aligned}
m \ddot{x}_i
&=
\frac{e}{c} v_j \PD{x_i}{A_j}

\frac{e}{c}
\PD{x_j}{A_i} v_j \\
&=
\frac{e}{c} v_j
\lr{
\PD{x_i}{A_j}

\PD{x_j}{A_i}
} \\
&=
\frac{e}{c} v_j B_k \epsilon_{i j k}.
\end{aligned}

In vector form that is

\label{eqn:magneticLorentzForceLagrangian:180}
m \ddot{\Bx}
=
\frac{e}{c} \Bv \cross \BB.

So, we get the magnetic term of the Lorentz force. Also note that this shows the Lagrangian (and the end result), was not in SI units. The $$1/c$$ term would have to be dropped for SI.

# References

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

## PHY1520H Graduate Quantum Mechanics. Lecture 3: Density matrix (cont.). Taught by Prof. Arun Paramekanti

September 24, 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 [1] chap. 3 content.

### Density matrix (cont.)

An example of a partitioned system with four total states (two spin 1/2 particles) is sketched in fig. 1.

fig. 1. Two spins

An example of a partitioned system with eight total states (three spin 1/2 particles) is sketched in fig. 2.

fig. 2. Three spins

The density matrix

\label{eqn:qmLecture3:20}
\hat{\rho} = \ket{\Psi}\bra{\Psi}

is clearly an operator as can be seen by applying it to a state

\label{eqn:qmLecture3:40}
\hat{\rho} \ket{\phi} = \ket{\Psi} \lr{ \braket{ \Psi }{\phi} }.

The quantity in braces is just a complex number.

After expanding the pure state $$\ket{\Psi}$$ in terms of basis states for each of the two partitions

\label{eqn:qmLecture3:60}
\ket{\Psi}
= \sum_{m,n} C_{m, n} \ket{m}_{\textrm{L}} \ket{n}_{\textrm{R}},

With $$\textrm{L}$$ and $$\textrm{R}$$ implied for $$\ket{m}, \ket{n}$$ indexed states respectively, this can be written

\label{eqn:qmLecture3:460}
\ket{\Psi}
= \sum_{m,n} C_{m, n} \ket{m} \ket{n}.

The density operator is

\label{eqn:qmLecture3:80}
\hat{\rho} =
\sum_{m,n}
C_{m, n}
C_{m’, n’}^\conj
\ket{m} \ket{n}
\sum_{m’,n’}
\bra{m’} \bra{n’}.

Suppose we trace over the right partition of the state space, defining such a trace as the reduced density operator $$\hat{\rho}_{\textrm{red}}$$

\label{eqn:qmLecture3:100}
\begin{aligned}
\hat{\rho}_{\textrm{red}}
&\equiv
\textrm{Tr}_{\textrm{R}}(\hat{\rho}) \\
&= \sum_{\tilde{n}} \bra{\tilde{n}} \hat{\rho} \ket{ \tilde{n}} \\
&= \sum_{\tilde{n}}
\bra{\tilde{n} }
\lr{
\sum_{m,n}
C_{m, n}
\ket{m} \ket{n}
}
\lr{
\sum_{m’,n’}
C_{m’, n’}^\conj
\bra{m’} \bra{n’}
}
\ket{ \tilde{n} } \\
&=
\sum_{\tilde{n}}
\sum_{m,n}
\sum_{m’,n’}
C_{m, n}
C_{m’, n’}^\conj
\ket{m} \delta_{\tilde{n} n}
\bra{m’ }
\delta_{ \tilde{n} n’ } \\
&=
\sum_{\tilde{n}, m, m’}
C_{m, \tilde{n}}
C_{m’, \tilde{n}}^\conj
\ket{m} \bra{m’ }
\end{aligned}

Computing the matrix element of $$\hat{\rho}_{\textrm{red}}$$, we have

\label{eqn:qmLecture3:120}
\begin{aligned}
\bra{\tilde{m}} \hat{\rho}_{\textrm{red}} \ket{\tilde{m}}
&=
\sum_{m, m’, \tilde{n}} C_{m, \tilde{n}} C_{m’, \tilde{n}}^\conj \braket{ \tilde{m}}{m} \braket{m’}{\tilde{m}} \\
&=
\sum_{\tilde{n}} \Abs{C_{\tilde{m}, \tilde{n}} }^2.
\end{aligned}

This is the probability that the left partition is in state $$\tilde{m}$$.

### Average of an observable

Suppose we have two spin half particles. For such a system the total magnetization is

\label{eqn:qmLecture3:140}
S_{\textrm{Total}} =
S_1^z
+
S_1^z,

as sketched in fig. 3.

fig. 3. Magnetic moments from two spins.

The average of some observable is

\label{eqn:qmLecture3:160}
\expectation{\hatA}
= \sum_{m, n, m’, n’} C_{m, n}^\conj C_{m’, n’}
\bra{m}\bra{n} \hatA \ket{n’} \ket{m’}.

Consider the trace of the density operator observable product

\label{eqn:qmLecture3:180}
\textrm{Tr}( \hat{\rho} \hatA )
= \sum_{m, n} \braket{m n}{\Psi} \bra{\Psi} \hatA \ket{m, n}.

Let

\label{eqn:qmLecture3:200}
\ket{\Psi} = \sum_{m, n} C_{m n} \ket{m, n},

so that

\label{eqn:qmLecture3:220}
\begin{aligned}
\textrm{Tr}( \hat{\rho} \hatA )
&= \sum_{m, n, m’, n’, m”, n”} C_{m’, n’} C_{m”, n”}^\conj
\braket{m n}{m’, n’} \bra{m”, n”} \hatA \ket{m, n} \\
&= \sum_{m, n, m”, n”} C_{m, n} C_{m”, n”}^\conj
\bra{m”, n”} \hatA \ket{m, n}.
\end{aligned}

This is just

\label{eqn:qmLecture3:240}
\boxed{
\bra{\Psi} \hatA \ket{\Psi} = \textrm{Tr}( \hat{\rho} \hatA ).
}

### Left observables

Consider

\label{eqn:qmLecture3:260}
\begin{aligned}
\bra{\Psi} \hatA_{\textrm{L}} \ket{\Psi}
&= \textrm{Tr}(\hat{\rho} \hatA_{\textrm{L}}) \\
&=
\textrm{Tr}_{\textrm{L}}
\textrm{Tr}_{\textrm{R}}
(\hat{\rho} \hatA_{\textrm{L}}) \\
&=
\textrm{Tr}_{\textrm{L}}
\lr{
\lr{
\textrm{Tr}_{\textrm{R}} \hat{\rho}
}
\hatA_{\textrm{L}})
} \\
&=
\textrm{Tr}_{\textrm{L}}
\lr{
\hat{\rho}_{\textrm{red}}
\hatA_{\textrm{L}})
}.
\end{aligned}

We see

\label{eqn:qmLecture3:280}
\bra{\Psi} \hatA_{\textrm{L}} \ket{\Psi}
=
\textrm{Tr}_{\textrm{L}} \lr{ \hat{\rho}_{\textrm{red}, \textrm{L}} \hatA_{\textrm{L}} }.

We find that we don’t need to know the state of the complete system to answer questions about portions of the system, but instead just need $$\hat{\rho}$$, a “probability operator” that provides all the required information about the partitioning of the system.

### Pure states vs. mixed states

For pure states we can assign a state vector and talk about reduced scenarios. For mixed states we must work with reduced density matrix.

## Example: Two particle spin half pure states

Consider

\label{eqn:qmLecture3:300}
\ket{\psi_1} = \inv{\sqrt{2}} \lr{ \ket{ \uparrow \downarrow } – \ket{ \downarrow \uparrow } }

\label{eqn:qmLecture3:320}
\ket{\psi_2} = \inv{\sqrt{2}} \lr{ \ket{ \uparrow \downarrow } + \ket{ \uparrow \uparrow } }.

For the first pure state the density operator is
\label{eqn:qmLecture3:360}
\hat{\rho} = \inv{2}
\lr{ \ket{ \uparrow \downarrow } – \ket{ \downarrow \uparrow } }
\lr{ \bra{ \uparrow \downarrow } – \bra{ \downarrow \uparrow } }

What are the reduced density matrices?

\label{eqn:qmLecture3:340}
\begin{aligned}
\hat{\rho}_{\textrm{L}}
&= \textrm{Tr}_{\textrm{R}} \lr{ \hat{\rho} } \\
&=
\inv{2} (-1)(-1) \ket{\downarrow}\bra{\downarrow}
+\inv{2} (+1)(+1) \ket{\uparrow}\bra{\uparrow},
\end{aligned}

so the matrix representation of this reduced density operator is

\label{eqn:qmLecture3:380}
\hat{\rho}_{\textrm{L}}
=
\inv{2}
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}.

For the second pure state the density operator is
\label{eqn:qmLecture3:400}
\hat{\rho} = \inv{2}
\lr{ \ket{ \uparrow \downarrow } + \ket{ \uparrow \uparrow } }
\lr{ \bra{ \uparrow \downarrow } + \bra{ \uparrow \uparrow } }.

This has a reduced density matrice

\label{eqn:qmLecture3:420}
\begin{aligned}
\hat{\rho}_{\textrm{L}}
&= \textrm{Tr}_{\textrm{R}} \lr{ \hat{\rho} } \\
&=
\inv{2} \ket{\uparrow}\bra{\uparrow}
+\inv{2} \ket{\uparrow}\bra{\uparrow} \\
&=
\ket{\uparrow}\bra{\uparrow} .
\end{aligned}

This has a matrix representation

\label{eqn:qmLecture3:440}
\hat{\rho}_{\textrm{L}}
=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}.

In this second example, we have more information about the left partition. That will be seen as a zero entanglement entropy in the problem set. In contrast we have less information about the first state, and will find a non-zero positive entanglement entropy in that case.

# References

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

## Constant magnetic solenoid field

September 24, 2015 phy1520 No comments , , , , ,

In [2] the following vector potential

\label{eqn:solenoidConstantField:20}
\BA = \frac{B \rho_a^2}{2 \rho} \phicap,

is introduced in a discussion on the Aharonov-Bohm effect, for configurations where the interior field of a solenoid is either a constant $$\BB$$ or zero.

I wasn’t able to make sense of this since the field I was calculating was zero for all $$\rho \ne 0$$

\label{eqn:solenoidConstantField:40}
\begin{aligned}
\BB
&= \lr{ \rhocap \partial_\rho + \zcap \partial_z + \frac{\phicap}{\rho}
\partial_\phi } \cross \frac{B \rho_a^2}{2 \rho} \phicap \\
&= \lr{ \rhocap \partial_\rho + \frac{\phicap}{\rho} \partial_\phi } \cross
\frac{B \rho_a^2}{2 \rho} \phicap \\
&=
\frac{B \rho_a^2}{2}
\rhocap \cross \phicap \partial_\rho \lr{ \inv{\rho} }
+
\frac{B \rho_a^2}{2 \rho}
\frac{\phicap}{\rho} \cross \partial_\phi \phicap \\
&=
\frac{B \rho_a^2}{2 \rho^2} \lr{ -\zcap + \phicap \cross \partial_\phi \phicap}.
\end{aligned}

Note that the $$\rho$$ partial requires that $$\rho \ne 0$$. To expand the cross product in the second term let $$j = \Be_1 \Be_2$$, and expand using a Geometric Algebra representation of the unit vector

\label{eqn:solenoidConstantField:60}
\begin{aligned}
\phicap \cross \partial_\phi \phicap
&=
\Be_2 e^{j \phi} \cross \lr{ \Be_2 \Be_1 \Be_2 e^{j \phi} } \\
&=
– \Be_1 \Be_2 \Be_3
\Be_2 e^{j \phi} (-\Be_1) e^{j \phi}
} \\
&=
\Be_1 \Be_2 \Be_3 \Be_2 \Be_1 \\
&= \Be_3 \\
&= \zcap.
\end{aligned}

So, provided $$\rho \ne 0$$, $$\BB = 0$$.

The errata [1] provides the clarification, showing that a $$\rho > \rho_a$$ constraint is required for this potential to produce the desired results. Continuity at $$\rho = \rho_a$$ means that in the interior (or at least on the boundary) we must have one of

\label{eqn:solenoidConstantField:80}
\BA = \frac{B \rho_a}{2} \phicap,

or

\label{eqn:solenoidConstantField:100}
\BA = \frac{B \rho}{2} \phicap.

The first doesn’t work, but the second does

\label{eqn:solenoidConstantField:120}
\begin{aligned}
\BB
&= \lr{ \rhocap \partial_\rho + \zcap \partial_z + \frac{\phicap}{\rho}
\partial_\phi } \cross \frac{B \rho}{2 } \phicap \\
&=
\frac{B }{2 } \rhocap \cross \phicap
+
\frac{B \rho}{2 }
\frac{\phicap}{\rho} \cross \partial_\phi \phicap \\
&= B \zcap.
\end{aligned}

So the vector potential that we want for a constant $$B \zcap$$ field in the interior $$\rho < \rho_a$$ of a cylindrical space, we need

\label{eqn:solenoidConstantField:140}
\BA =
\left\{
\begin{array}{l l}
\frac{B \rho_a^2}{2 \rho} \phicap & \quad \mbox{if $$\rho \ge \rho_a$$ } \\
\frac{B \rho}{2} \phicap & \quad \mbox{if $$\rho \le \rho_a$$.}
\end{array}
\right.

An example of the magnitude of potential is graphed in fig. 1.

fig. 1. Vector potential for constant field in cylindrical region.

# References

[1] Jun John Sakurai and Jim J Napolitano. \emph{Errata: Typographical Errors, Mistakes, and Comments, Modern Quantum Mechanics, 2nd Edition}, 2013. URL http://www.rpi.edu/dept/phys/Courses/PHYS6520/Spring2015/ErrataMQM.pdf.

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

## PHY1520H Graduate Quantum Mechanics. Lecture 2: Basic concepts, time evolution, and density operators. 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 1 (basic concepts),3 (density operator). content from [1].

### Basic concepts

We’ve reviewed the basic concepts that we will encounter in Quantum Mechanics.

1. Abstract state vector. $$\ket{ \psi}$$
2. Basis states. $$\ket{ x }$$
3. Observables, special Hermitian operators. We’ll only deal with linear observables.
4. Measurement.

We can either express the wave functions $$\psi(x) = \braket{x}{\psi}$$ in terms of a basis for the observable, or can express the observable in terms of the basis of the wave function (position or momentum for example).

We saw that the position space representation of a momentum operator (also an observable) was

\label{eqn:lecture2:20}
\hat{p} \rightarrow -i \Hbar \PD{x}{}.

In general we can find the matrix element representation of any operator by considering its representation in a given basis. For example, in a position basis, that would be

\label{eqn:lecture2:40}
\bra{x’} \hat{A} \ket{x} \leftrightarrow A_{x x’}

The Hermitian property of the observable means that $$A_{x x’} = A_{x’ x}^\conj$$

\label{eqn:lecture2:60}
\int dx \bra{x’} \hat{A} \ket{x} \braket{x }{\psi} = \braket{x’}{\phi}
\leftrightarrow
A_{x’ x} \psi_x = \phi_{x’}.

## Example: Measurement example

fig. 1. Polarizer apparatus

Consider a polarization apparatus as sketched in fig. 1, where the output is of the form $$I_{\textrm{out}} = I_{\textrm{in}} \cos^2 \theta$$.

A general input state can be written in terms of each of the possible polarizations

\label{eqn:lecture2:80}
\alpha \ket{ \updownarrow } + \beta \ket{ \leftrightarrow } \sim
\cos\theta \ket{ \updownarrow } + \sin\theta \ket{ \leftrightarrow }

Here $$\abs{\alpha}^2$$ is the probability that the input state is in the upwards polarization state, and $$\abs{\beta}^2$$ is the probability that the input state is in the downwards polarization state.

The measurement of the polarization results in an output state that has a specific polarization. That measurement is said to collapse the wavefunction.

When attempting a measurement, looking for a specific value, effects the state of the system, and is call a strong or projective measurement. Such a measurement is

• (i) Probabilistic.
• (ii) Requires many measurements.

This measurement process results a determination of the eigenvalue of the operator. The eigenvalue production of measurement is why we demand that operators be Hermitian.

It is also possible to try to do a weaker (perturbative) measurement, where some information is extracted from the input state without completely altering it.

### Time evolution

1. Schrodinger picture.
The time evolution process is governed by a Schrodinger equation of the following form\label{eqn:lecture2:100}
i \Hbar \PD{t}{} \ket{\Psi(t)} = \hat{H} \ket{\Psi(t)}.

This Hamiltonian could be, for example,

\label{eqn:lecture2:120}
\hat{H} = \frac{\hat{p}^2}{2m} + V(x),

Such a representation of time evolution is expressed in terms of operators $$\hat{x}, \hat{p}, \hat{H}, \cdots$$ that are independent of time.

2. Heisenberg picture.Suppose we have a state $$\ket{\Psi(t)}$$ and operate on this with an operator

\label{eqn:lecture2:140}
\hat{A} \ket{\Psi(t)}.

This will have time evolution of the form

\label{eqn:lecture2:160}
\hat{A} e^{-i \hat{H} t/\Hbar} \ket{\Psi(0)},

or in matrix element form

\label{eqn:lecture2:180}
\bra{\phi(t)} \hat{A} \ket{\Psi(t)}
=
\bra{\phi(0)}
e^{i \hat{H} t/\Hbar}
\hat{A} e^{-i \hat{H} t/\Hbar} \ket{\Psi(0)}.

We work with states that do not evolve in time $$\ket{\phi(0)}, \ket{\Psi(0)}, \cdots$$, but operators do evolve in time according to

\label{eqn:lecture2:200}
\hat{A}(t) =
e^{i \hat{H} t/\Hbar}
\hat{A} e^{-i \hat{H} t/\Hbar}.

### Density operator

We can have situations where it is impossible to determine a single state that describes the system. For example, given the gas in the room that you are sitting in, there are things that we can measure, but it is impossible to describe the state that describes all the particles and also impossible to construct a Hamiltonian that governs all the interactions of those many many particles.

We need a probabilistic description to even describe such a complex system.

Suppose we have a complex system that can be partitioned into two subsets, left and right, as sketched in fig. 2.

fig. 2. System partitioned into separate set of states

If the states in each partition can be enumerated separately, we can write the state of the system as sums over the probability amplitudes that for the combined states.

\label{eqn:lecture2:220}
\ket{\Psi}
=
\sum_{m, n} C_{m,n} \ket{m} \ket{n}

Here $$C_{m, n}$$ is the probability amplitude to find the state in the combined state $$\ket{m} \ket{n}$$.

As an example of such a system, we could investigate a two particle configuration where spin up or spin down can be separately measured for each particle.

\label{eqn:lecture2:240}
\ket{\psi} = \inv{\sqrt{2}} \lr{
\ket{\uparrow}\ket{\downarrow}
+
\ket{\downarrow}\ket{\rightarrow}
}

Considering such a system we could ask questions such as

• What is the probability that the left half is in state $$m$$? This would be\label{eqn:lecture2:260}
\sum_n \Abs{C_{m, n}}^2
• Probability that the left half is in state $$m$$, and the
probability that the right half is in state $$n$$? That is\label{eqn:lecture2:280}
\Abs{C_{m, n}}^2

We define the density operator

\label{eqn:lecture2:300}
\hat{\rho} = \ket{\Psi} \bra{\Psi}.

This is idempotent

\label{eqn:lecture2:320}
\hat{\rho}^2 =
\lr{ \ket{\Psi} \bra{\Psi} }
\lr{ \ket{\Psi} \bra{\Psi} }
=
\ket{\Psi} \bra{\Psi}

# References

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

## Entropy when density operator has zero eigenvalues

In the class notes and the text [1] the Von Neumann entropy is defined as

\label{eqn:densityMatrixEntropy:20}
S = -\textrm{Tr} \rho \ln \rho.

In one of our problems I had trouble evaluating this, having calculated a density operator matrix representation

\label{eqn:densityMatrixEntropy:40}
\rho = E \wedge E^{-1},

where

\label{eqn:densityMatrixEntropy:60}
E = \inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix},

and
\label{eqn:densityMatrixEntropy:100}
\wedge =
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}.

The usual method of evaluating a function of a matrix is to assume the function has a power series representation, and that a similarity transformation of the form $$A = E \wedge E^{-1}$$ is possible, so that

\label{eqn:densityMatrixEntropy:80}
f(A) = E f(\wedge) E^{-1},

however, when attempting to do this with the matrix of \ref{eqn:densityMatrixEntropy:40} leads to an undesirable result

\label{eqn:densityMatrixEntropy:120}
\ln \rho =
\inv{2}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}
\begin{bmatrix}
\ln 1 & 0 \\
0 & \ln 0
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.

The $$\ln 0$$ makes the evaluation of this matrix logarithm rather unpleasant. To give meaning to the entropy expression, we have to do two things, the first is treating the trace operation as a higher precedence than the logarithms that it contains. That is

\label{eqn:densityMatrixEntropy:140}
\begin{aligned}
-\textrm{Tr} ( \rho \ln \rho )
&=
-\textrm{Tr} ( E \wedge E^{-1} E \ln \wedge E^{-1} ) \\
&=
-\textrm{Tr} ( E \wedge \ln \wedge E^{-1} ) \\
&=
-\textrm{Tr} ( E^{-1} E \wedge \ln \wedge ) \\
&=
-\textrm{Tr} ( \wedge \ln \wedge ) \\
&=
– \sum_k \wedge_{kk} \ln \wedge_{kk}.
\end{aligned}

Now the matrix of the logarithm need not be evaluated, but we still need to give meaning to $$\wedge_{kk} \ln \wedge_{kk}$$ for zero diagonal entries. This can be done by considering a limiting scenerio

\label{eqn:densityMatrixEntropy:160}
\begin{aligned}
-\lim_{a \rightarrow 0} a \ln a
&=
-\lim_{x \rightarrow \infty} e^{-x} \ln e^{-x} \\
&=
\lim_{x \rightarrow \infty} x e^{-x} \\
&=
0.
\end{aligned}

The entropy can now be expressed in the unambiguous form, summing over all the non-zero eigenvalues of the density operator

\label{eqn:densityMatrixEntropy:180}
\boxed{
S = – \sum_{ \wedge_{kk} \ne 0} \wedge_{kk} \ln \wedge_{kk}.
}

# References

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

## PHY1520H Graduate Quantum Mechanics. Lecture 1: Lighting review. 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. 1 content.

### Classical mechanics

We’ll be talking about one body physics for most of this course. In classical mechanics we can figure out the particle trajectories using both of $$(\Br, \Bp$$, where

\label{eqn:qmLecture1:20}
\begin{aligned}
\ddt{\Br} &= \inv{m} \Bp \\
\end{aligned}

A two dimensional phase space as sketched in fig. 1 shows the trajectory of a point particle subject to some equations of motion

fig. 1. One dimensional classical phase space example

### Quantum mechanics

For this lecture, we’ll work with natural units, setting

\label{eqn:qmLecture1:480}
\boxed{
\Hbar = 1.
}

In QM we are no longer allowed to think of position and momentum, but have to start asking about state vectors $$\ket{\Psi}$$.

We’ll consider the state vector with respect to some basis, for example, in a position basis, we write

\label{eqn:qmLecture1:40}
\braket{ x }{\Psi } = \Psi(x),

a complex numbered “wave function”, the probability amplitude for a particle in $$\ket{\Psi}$$ to be in the vicinity of $$x$$.

We could also consider the state in a momentum basis

\label{eqn:qmLecture1:60}
\braket{ p }{\Psi } = \Psi(p),

a probability amplitude with respect to momentum $$p$$.

More precisely,

\label{eqn:qmLecture1:80}
\Abs{\Psi(x)}^2 dx \ge 0

is the probability of finding the particle in the range $$(x, x + dx )$$. To have meaning as a probability, we require

\label{eqn:qmLecture1:100}
\int_{-\infty}^\infty \Abs{\Psi(x)}^2 dx = 1.

The average position can be calculated using this probability density function. For example

\label{eqn:qmLecture1:120}
\expectation{x} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 x dx,

or
\label{eqn:qmLecture1:140}
\expectation{f(x)} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 f(x) dx.

Similarly, calculation of an average of a function of momentum can be expressed as

\label{eqn:qmLecture1:160}
\expectation{f(p)} = \int_{-\infty}^\infty \Abs{\Psi(p)}^2 f(p) dp.

### Transformation from a position to momentum basis

We have a problem, if we which to compute an average in momentum space such as $$\expectation{p}$$, when given a wavefunction $$\Psi(x)$$.

How do we convert

\label{eqn:qmLecture1:180}
\Psi(p)
\stackrel{?}{\leftrightarrow}
\Psi(x),

or equivalently
\label{eqn:qmLecture1:200}
\braket{p}{\Psi}
\stackrel{?}{\leftrightarrow}
\braket{x}{\Psi}.

Such a conversion can be performed by virtue of an the assumption that we have a complete orthonormal basis, for which we can introduce identity operations such as

\label{eqn:qmLecture1:220}
\int_{-\infty}^\infty dp \ket{p}\bra{p} = 1,

or
\label{eqn:qmLecture1:240}
\int_{-\infty}^\infty dx \ket{x}\bra{x} = 1

Some interpretations:

1. $$\ket{x_0} \leftrightarrow \text{sits at} x = x_0$$
2. $$\braket{x}{x’} \leftrightarrow \delta(x – x’)$$
3. $$\braket{p}{p’} \leftrightarrow \delta(p – p’)$$
4. $$\braket{x}{p’} = \frac{e^{i p x}}{\sqrt{V}}$$, where $$V$$ is the volume of the box containing the particle. We’ll define the appropriate normalization for an infinite box volume later.

The delta function interpretation of the braket $$\braket{p}{p’}$$ justifies the identity operator, since we recover any state in the basis when operating with it. For example, in momentum space

\label{eqn:qmLecture1:260}
\begin{aligned}
1 \ket{p}
&=
\lr{ \int_{-\infty}^\infty dp’
\ket{p’}\bra{p’} }
\ket{p} \\
&=
\int_{-\infty}^\infty dp’
\ket{p’}
\braket{p’}{p} \\
&=
\int_{-\infty}^\infty dp’
\ket{p’}
\delta(p – p’) \\
&=
\ket{p}.
\end{aligned}

This also the determination of an integral operator representation for the delta function

\label{eqn:qmLecture1:500}
\begin{aligned}
\delta(x – x’)
&=
\braket{x}{x’} \\
&=
\int dp \braket{x}{p} \braket{p}{x’} \\
&=
\inv{V} \int dp e^{i p x} e^{-i p x’},
\end{aligned}

or

\label{eqn:qmLecture1:520}
\delta(x – x’)
=
\inv{V} \int dp e^{i p (x- x’)}.

Here we used the fact that $$\braket{p}{x} = \braket{x}{p}^\conj$$.

FIXME: do we have a justification for that conjugation with what was defined here so far?

The conversion from a position basis to momentum space is now possible

\label{eqn:qmLecture1:280}
\begin{aligned}
\braket{p}{\Psi}
&= \Psi(p) \\
&= \int_{-\infty}^\infty \braket{p}{x} \braket{x}{\Psi} dx \\
&= \int_{-\infty}^\infty \frac{e^{-ip x}}{\sqrt{V}} \Psi(x) dx.
\end{aligned}

The momentum space to position space conversion can be written as

\label{eqn:qmLecture1:300}
\Psi(x)
= \int_{-\infty}^\infty \frac{e^{ip x}}{\sqrt{V}} \Psi(p) dp.

Now we can go back and figure out the an expectation

\label{eqn:qmLecture1:320}
\begin{aligned}
\expectation{p}
&=
\int \Psi^\conj(p) \Psi(p) p d p \\
&=
\int dp
\lr{
\int_{-\infty}^\infty \frac{e^{ip x}}{\sqrt{V}} \Psi^\conj(x) dx
}
\lr{
\int_{-\infty}^\infty \frac{e^{-ip x’}}{\sqrt{V}} \Psi(x’) dx’
}
p \\
&=\int dp dx dx’
\Psi^\conj(x)
\inv{V} e^{ip (x-x’)} \Psi(x’) p \\
&=
\int dp dx dx’
\Psi^\conj(x)
\inv{V} \lr{ -i\PD{x}{e^{ip (x-x’)}} }\Psi(x’) \\
&=
\int dp dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\inv{V} \int dx’ e^{ip (x-x’)} \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\int dx’ \lr{ \inv{V} \int dp e^{ip (x-x’)} } \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\int dx’ \delta(x – x’) \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\Psi(x)
\end{aligned}

Here we’ve essentially calculated the position space representation of the momentum operator, allowing identifications of the following form

\label{eqn:qmLecture1:380}
p \leftrightarrow -i \PD{x}{}

\label{eqn:qmLecture1:400}
p^2 \leftrightarrow – \PDSq{x}{}.

### Alternate starting point.

Most of the above results followed from the claim that $$\braket{x}{p} = e^{i p x}$$. Note that this position space representation of the momentum operator can also be taken as the starting point. Given that, the exponential representation of the position-momentum braket follows

\label{eqn:qmLecture1:540}
\bra{x} P \ket{p}
=
-i \Hbar \PD{x}{} \braket{x}{p},

but $$\bra{x} P \ket{p} = p \braket{x}{p}$$, providing a differential equation for $$\braket{x}{p}$$

\label{eqn:qmLecture1:560}
p \braket{x}{p} = -i \Hbar \PD{x}{} \braket{x}{p},

with solution

\label{eqn:qmLecture1:580}
i p x/\Hbar = \ln \braket{x}{p} + \text{const},

or
\label{eqn:qmLecture1:600}
\braket{x}{p} \propto e^{i p x/\Hbar}.

### Matrix interpretation

1. Ket’s $$\ket{\Psi} \leftrightarrow \text{column vector}$$
2. Bra’s $$\bra{\Psi} \leftrightarrow {(\text{row vector})}^\conj$$
3. Operators $$\leftrightarrow$$ matrices that act on vectors.

\label{eqn:qmLecture1:420}
\hat{p} \ket{\Psi} \rightarrow \ket{\Psi’}

### Time evolution

For a state subject to the equations of motion given by the Hamiltonian operator $$\hat{H}$$

\label{eqn:qmLecture1:440}
i \PD{t}{} \ket{\Psi} = \hat{H} \ket{\Psi},

the time evolution is given by
\label{eqn:qmLecture1:460}
\ket{\Psi(t)} = e^{-i \hat{H} t} \ket{\Psi(0)}.

### Incomplete information

We’ll need to introduce the concept of Density matrices. This will bring us to concepts like entanglement.

# References

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