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

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

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

## Position operator in momentum space representation

November 8, 2015 phy1520 No comments , ,

A derivation of the position space representation of the momentum operator $$-i \Hbar \partial_x$$ is made in [1], starting with the position-momentum commutator. Here I’ll repeat that argument for the momentum space representation of the position operator.

What we want to do is expand the matrix element of the commutator. First using the definition of the commutator

\label{eqn:positionOperatorInMomentumSpace:20}
\bra{p’} X P – P X \ket{p”}
=
i \Hbar \braket{p’}{p”}
=
i \Hbar \delta{p’ – p”},

and then by inserting an identity operation in a momentum space basis

\label{eqn:positionOperatorInMomentumSpace:40}
\begin{aligned}
\bra{p’} X P – P X \ket{p”}
&=
\int dp
\bra{p’} X \ket{p}\bra{p} P \ket{p”}
-\int dp
\bra{p’} P \ket{p}\bra{p} X \ket{p”} \\
&=
\int dp
\bra{p’} X \ket{p} p \delta(p – p”)
-\int dp
p \delta(p’ – p)
\bra{p} X \ket{p”} \\
&=
\bra{p’} X \ket{p”} p”

p’ \bra{p’} X \ket{p”}.
\end{aligned}

So we have

\label{eqn:positionOperatorInMomentumSpace:60}
\bra{p’} X \ket{p”} p”

p’ \bra{p’} X \ket{p”}
=
i \Hbar \delta{p’ – p”}.

Because the RHS is zero whenever $$p’ \ne p”$$, the matrix element $$\bra{p’} X \ket{p”}$$ must also include a delta function. Let

\label{eqn:positionOperatorInMomentumSpace:80}
\bra{p’} X \ket{p”} = \delta(p’ – p”) X(p”).

Because \ref{eqn:positionOperatorInMomentumSpace:60} is an operator equation that really only takes on meaning when applied to a wave function and integrated, we do that

\label{eqn:positionOperatorInMomentumSpace:100}
\int dp” \delta(p’ – p”) X(p”) p” \psi(p”)

\int dp” p’ \delta(p’ – p”) X(p”) \psi(p”)
=
\int dp” i \Hbar \delta{p’ – p”} \psi(p”),

or
\label{eqn:positionOperatorInMomentumSpace:120}
i \Hbar \psi(p’)
=
X(p’) p’ \psi(p’)

p’
X(p’) \psi(p’).

Provided $$X(p’)$$ operates on everything to its right, this equation is solved by setting

\label{eqn:positionOperatorInMomentumSpace:140}
\boxed{
X(p’) = i \Hbar \PD{p’}{}.
}

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

## Plane wave ground state expectation for SHO

Problem [1] 2.18 is, for a 1D SHO, show that

\label{eqn:exponentialExpectationGroundState:20}
\bra{0} e^{i k x} \ket{0} = \exp\lr{ -k^2 \bra{0} x^2 \ket{0}/2 }.

Despite the simple appearance of this problem, I found this quite involved to show. To do so, start with a series expansion of the expectation

\label{eqn:exponentialExpectationGroundState:40}
\bra{0} e^{i k x} \ket{0}
=
\sum_{m=0}^\infty \frac{(i k)^m}{m!} \bra{0} x^m \ket{0}.

Let

\label{eqn:exponentialExpectationGroundState:60}
X = \lr{ a + a^\dagger },

so that

\label{eqn:exponentialExpectationGroundState:80}
x
= \sqrt{\frac{\Hbar}{2 \omega m}} X
= \frac{x_0}{\sqrt{2}} X.

Consider the first few values of $$\bra{0} X^n \ket{0}$$

\label{eqn:exponentialExpectationGroundState:100}
\begin{aligned}
\bra{0} X \ket{0}
&=
\bra{0} \lr{ a + a^\dagger } \ket{0} \\
&=
\braket{0}{1} \\
&=
0,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:120}
\begin{aligned}
\bra{0} X^2 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^2 \ket{0} \\
&=
\braket{1}{1} \\
&=
1,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:140}
\begin{aligned}
\bra{0} X^3 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^3 \ket{0} \\
&=
\bra{1} \lr{ \sqrt{2} \ket{2} + \ket{0} } \\
&=
0.
\end{aligned}

Whenever the power $$n$$ in $$X^n$$ is even, the braket can be split into a bra that has only contributions from odd eigenstates and a ket with even eigenstates. We conclude that $$\bra{0} X^n \ket{0} = 0$$ when $$n$$ is odd.

Noting that $$\bra{0} x^2 \ket{0} = \ifrac{x_0^2}{2}$$, this leaves

\label{eqn:exponentialExpectationGroundState:160}
\begin{aligned}
\bra{0} e^{i k x} \ket{0}
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \bra{0} x^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \lr{ \frac{x_0^2}{2} }^m \bra{0} X^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{1}{(2 m)!} \lr{ -k^2 \bra{0} x^2 \ket{0} }^m \bra{0} X^{2m} \ket{0}.
\end{aligned}

This problem is now reduced to showing that

\label{eqn:exponentialExpectationGroundState:180}
\frac{1}{(2 m)!} \bra{0} X^{2m} \ket{0} = \inv{m! 2^m},

or

\label{eqn:exponentialExpectationGroundState:200}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&= \frac{(2m)!}{m! 2^m} \\
&= \frac{ (2m)(2m-1)(2m-2) \cdots (2)(1) }{2^m m!} \\
&= \frac{ 2^m (m)(2m-1)(m-1)(2m-3)(m-2) \cdots (2)(3)(1)(1) }{2^m m!} \\
&= (2m-1)!!,
\end{aligned}

where $$n!! = n(n-2)(n-4)\cdots$$.

It looks like $$\bra{0} X^{2m} \ket{0}$$ can be expanded by inserting an identity operator and proceeding recursively, like

\label{eqn:exponentialExpectationGroundState:220}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&=
\bra{0} X^2 \lr{ \sum_{n=0}^\infty \ket{n}\bra{n} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^2 \lr{ \ket{0}\bra{0} + \ket{2}\bra{2} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^{2m-2} \ket{0} + \bra{0} X^2 \ket{2} \bra{2} X^{2m-2} \ket{0}.
\end{aligned}

This has made use of the observation that $$\bra{0} X^2 \ket{n} = 0$$ for all $$n \ne 0,2$$. The remaining term includes the factor

\label{eqn:exponentialExpectationGroundState:240}
\begin{aligned}
\bra{0} X^2 \ket{2}
&=
\bra{0} \lr{a + a^\dagger}^2 \ket{2} \\
&=
\lr{ \bra{0} + \sqrt{2} \bra{2} } \ket{2} \\
&=
\sqrt{2},
\end{aligned}

Since $$\sqrt{2} \ket{2} = \lr{a^\dagger}^2 \ket{0}$$, the expectation of interest can be written

\label{eqn:exponentialExpectationGroundState:260}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + \bra{0} a^2 X^{2m-2} \ket{0}.

How do we expand the second term. Let’s look at how $$a$$ and $$X$$ commute

\label{eqn:exponentialExpectationGroundState:280}
\begin{aligned}
a X
&=
\antisymmetric{a}{X} + X a \\
&=
\antisymmetric{a}{a + a^\dagger} + X a \\
&=
\antisymmetric{a}{a^\dagger} + X a \\
&=
1 + X a,
\end{aligned}

\label{eqn:exponentialExpectationGroundState:300}
\begin{aligned}
a^2 X
&=
a \lr{ a X } \\
&=
a \lr{ 1 + X a } \\
&=
a + a X a \\
&=
a + \lr{ 1 + X a } a \\
&=
2 a + X a^2.
\end{aligned}

Proceeding to expand $$a^2 X^n$$ we find
\label{eqn:exponentialExpectationGroundState:320}
\begin{aligned}
a^2 X^3 &= 6 X + 6 X^2 a + X^3 a^2 \\
a^2 X^4 &= 12 X^2 + 8 X^3 a + X^4 a^2 \\
a^2 X^5 &= 20 X^3 + 10 X^4 a + X^5 a^2 \\
a^2 X^6 &= 30 X^4 + 12 X^5 a + X^6 a^2.
\end{aligned}

It appears that we have
\label{eqn:exponentialExpectationGroundState:340}
\antisymmetric{a^2 X^n}{X^n a^2} = \beta_n X^{n-2} + 2 n X^{n-1} a,

where

\label{eqn:exponentialExpectationGroundState:360}
\beta_n = \beta_{n-1} + 2 (n-1),

and $$\beta_2 = 2$$. Some goofing around shows that $$\beta_n = n(n-1)$$, so the induction hypothesis is

\label{eqn:exponentialExpectationGroundState:380}
\antisymmetric{a^2 X^n}{X^n a^2} = n(n-1) X^{n-2} + 2 n X^{n-1} a.

Let’s check the induction
\label{eqn:exponentialExpectationGroundState:400}
\begin{aligned}
a^2 X^{n+1}
&=
a^2 X^{n} X \\
&=
\lr{ n(n-1) X^{n-2} + 2 n X^{n-1} a + X^n a^2 } X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} a X + X^n a^2 X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} \lr{ 1 + X a } + X^n \lr{ 2 a + X a^2 } \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} + 2 n X^{n} a
+ 2 X^n a
+ X^{n+1} a^2 \\
&=
X^{n+1} a^2 + (2 + 2 n) X^{n} a + \lr{ 2 n + n(n-1) } X^{n-1} \\
&=
X^{n+1} a^2 + 2(n + 1) X^{n} a + (n+1) n X^{n-1},
\end{aligned}

which concludes the induction, giving

\label{eqn:exponentialExpectationGroundState:420}
\bra{ 0 } a^2 X^{n} \ket{0 } = n(n-1) \bra{0} X^{n-2} \ket{0},

and

\label{eqn:exponentialExpectationGroundState:440}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + (2m-2)(2m-3) \bra{0} X^{2m-4} \ket{0}.

Let

\label{eqn:exponentialExpectationGroundState:460}
\sigma_{n} = \bra{0} X^n \ket{0},

so that the recurrence relation, for $$2n \ge 4$$ is

\label{eqn:exponentialExpectationGroundState:480}
\sigma_{2n} = \sigma_{2n -2} + (2n-2)(2n-3) \sigma_{2n -4}

We want to show that this simplifies to

\label{eqn:exponentialExpectationGroundState:500}
\sigma_{2n} = (2n-1)!!

The first values are

\label{eqn:exponentialExpectationGroundState:540}
\sigma_0 = \bra{0} X^0 \ket{0} = 1

\label{eqn:exponentialExpectationGroundState:560}
\sigma_2 = \bra{0} X^2 \ket{0} = 1

which gives us the right result for the first term in the induction

\label{eqn:exponentialExpectationGroundState:580}
\begin{aligned}
\sigma_4
&= \sigma_2 + 2 \times 1 \times \sigma_0 \\
&= 1 + 2 \\
&= 3!!
\end{aligned}

For the general induction term, consider

\label{eqn:exponentialExpectationGroundState:600}
\begin{aligned}
\sigma_{2n + 2}
&= \sigma_{2n} + 2 n (2n – 1) \sigma_{2n -2} \\
&= (2n-1)!! + 2n ( 2n – 1) (2n -3)!! \\
&= (2n + 1) (2n -1)!! \\
&= (2n + 1)!!,
\end{aligned}

which completes the final induction. That was also the last thing required to complete the proof, so we are done!

# References

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

## PHY1520H Graduate Quantum Mechanics. Lecture 5: time evolution of coherent states, and charged particles in a magnetic field. 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 \textchapref{{1}} [1] content.

### Coherent states (cont.)

A coherent state for the SHO $$H = \lr{ N + \inv{2} } \Hbar \omega$$ was given by

\label{eqn:qmLecture5:20}
a \ket{z} = z \ket{z},

where we showed that

\label{eqn:qmLecture5:40}
\ket{z} = c_0 e^{ z a^\dagger } \ket{0}.

In the Heisenberg picture we found

\label{eqn:qmLecture5:60}
\begin{aligned}
a_{\textrm{H}}(t) &= e^{i H t/\Hbar} a e^{-i H t/\Hbar} = a e^{-i\omega t} \\
a_{\textrm{H}}^\dagger(t) &= e^{i H t/\Hbar} a^\dagger e^{-i H t/\Hbar} = a^\dagger e^{i\omega t}.
\end{aligned}

Recall that the position and momentum representation of the ladder operators was

\label{eqn:qmLecture5:80}
\begin{aligned}
a &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} + i \hat{p} \sqrt{\inv{m \Hbar \omega}} } \\
a^\dagger &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} – i \hat{p} \sqrt{\inv{m \Hbar \omega}} },
\end{aligned}

or equivalently
\label{eqn:qmLecture5:100}
\begin{aligned}
\hat{x} &= \lr{ a + a^\dagger } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
\hat{p} &= i \lr{ a^\dagger – a } \sqrt{\frac{m \Hbar \omega}{2}}.
\end{aligned}

Given this we can compute expectation value of position operator

\label{eqn:qmLecture5:120}
\begin{aligned}
\bra{z} \hat{x} \ket{z}
&=
\sqrt{\frac{\Hbar}{ 2 m \omega}}
\bra{z}
\lr{ a + a^\dagger }
\ket{z} \\
&=
\lr{ z + z^\conj } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
&=
2 \textrm{Re} z \sqrt{\frac{\Hbar}{ 2 m \omega}} .
\end{aligned}

Similarly

\label{eqn:qmLecture5:140}
\begin{aligned}
\bra{z} \hat{p} \ket{z}
&=
i \sqrt{\frac{m \Hbar \omega}{2}}
\bra{z}
\lr{ a^\dagger – a }
\ket{z} \\
&=
\sqrt{\frac{m \Hbar \omega}{2}}
2 \textrm{Im} z.
\end{aligned}

How about the expectation of the Heisenberg position operator? That is

\label{eqn:qmLecture5:160}
\begin{aligned}
\bra{z} \hat{x}_{\textrm{H}}(t) \ket{z}
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \bra{z} \lr{ a + a^\dagger } \ket{z} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ z e^{-i \omega t} + z^\conj e^{i \omega t}} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \lr{z + z^\conj} \cos( \omega t ) -i \lr{ z – z^\conj } \sin( \omega t) } \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \expectation{x(0)} \sqrt{ \frac{2 m \omega}{\Hbar}} \cos( \omega t ) -i \expectation{p(0)} i \sqrt{\frac{2 m \omega}{\Hbar} } \sin( \omega t) } \\
&=
\expectation{x(0)} \cos( \omega t ) + \frac{\expectation{p(0)}}{m \omega} \sin( \omega t) .
\end{aligned}

We find that the average of the Heisenberg position operator evolves in time in exactly the same fashion as position in the classical Harmonic oscillator. This phase space like trajectory is sketched in fig. 1.

fig. 1. phase space like trajectory

In the text it is shown that we have the same structure for the Heisenberg operator itself, before taking expectations

\label{eqn:qmLecture5:220}
\hat{x}_{\textrm{H}}(t)
=
{x(0)} \cos( \omega t ) + \frac{{p(0)}}{m \omega} \sin( \omega t).

Where the coherent states become useful is that we will see that the second moments of position and momentum are not time dependent with respect to the coherent states. Such states remain localized.

### Uncertainty

First note that using the commutator relationship we have

\label{eqn:qmLecture5:180}
\begin{aligned}
\bra{z} a a^\dagger \ket{z}
&=
\bra{z} \lr{ \antisymmetric{a}{a^\dagger} + a^\dagger a } \ket{z} \\
&=
\bra{z} \lr{ 1 + a^\dagger a } \ket{z}.
\end{aligned}

For the second moment we have

\label{eqn:qmLecture5:200}
\begin{aligned}
\bra{z} \hat{x}^2 \ket{z}
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{a + a^\dagger } \lr{a + a^\dagger } \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + a a^\dagger + a^\dagger a
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + 2 a^\dagger a + 1
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z^2 + {(z^\conj)}^2 + 2 z^\conj z + 1} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z + z^\conj }^2
+
\frac{\Hbar}{ 2 m \omega}.
\end{aligned}

We find

\label{eqn:qmLecture5:240}
\sigma_x^2 = \frac{\Hbar}{ 2 m \omega},

and

\label{eqn:qmLecture5:260}
\sigma_p^2 = \frac{m \Hbar \omega}{2}

so

\label{eqn:qmLecture5:280}
\sigma_x^2 \sigma_p^2 = \frac{\Hbar^2}{4},

or

\label{eqn:qmLecture5:300}
\sigma_x \sigma_p = \frac{\Hbar}{2}.

This is the minimum uncertainty.

### Quantum Field theory

In Quantum Field theory the ideas of isolated oscillators is used to model particle creation. The lowest energy state (a no particle, vacuum state) is given the lowest energy level, with each additional quantum level modeling a new particle creation state as sketched in fig. 2.

fig. 2. QFT energy levels

We have to imagine many oscillators, each with a distinct vacuum energy $$\sim \Bk^2$$ . The Harmonic oscillator can be used to model the creation of particles with $$\Hbar \omega$$ energy differences from that “vacuum energy”.

### Charged particle in a magnetic field

In the classical case ( with SI units or $$c = 1$$ ) we have

\label{eqn:qmLecture5:320}
\BF = q \BE + q \Bv \cross \BB.

Alternately, we can look at the Hamiltonian view of the system, written in terms of potentials

\label{eqn:qmLecture5:340}

\label{eqn:qmLecture5:360}
\BE = – \spacegrad \phi – \PD{t}{\BA}.

Note that the curl form for the magnetic field implies one of the required Maxwell’s equations $$\spacegrad \cdot \BB = 0$$.

Ignoring time dependence of the potentials, the Hamiltonian can be expressed as

\label{eqn:qmLecture5:380}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.

In this Hamiltonian the vector $$\Bp$$ is called the canonical momentum, the momentum conjugate to position in phase space.

It is left as an exercise to show that the Lorentz force equation results from application of the Hamiltonian equations of motion, and that the velocity is given by $$\Bv = (\Bp – q \BA)/m$$.

For quantum mechanics, we use the same Hamiltonian, but promote our position, momentum and potentials to operators.

\label{eqn:qmLecture5:400}
\hat{H} = \inv{2 m} \lr{ \hat{\Bp} – q \hat{\BA}(\Br, t) }^2 + q \hat{\phi}(\Br, t).

### Gauge invariance

Can we say anything about this before looking at the question of a particle in a magnetic field?

Recall that the we can make a gauge transformation of the form

\label{eqn:qmLecture5:420a}
\label{eqn:qmLecture5:420}
\BA \rightarrow \BA + \spacegrad \chi

\label{eqn:qmLecture5:440}
\phi \rightarrow \phi – \PD{t}{\chi}

Does this notion of gauge invariance also carry over to the Quantum Hamiltonian. After gauge transformation we have

\label{eqn:qmLecture5:460}
\hat{H}’
= \inv{2 m} \lr{ \hat{\Bp} – q \BA – q \spacegrad \chi }^2 + q \lr{ \phi – \PD{t}{\chi} }

Now we are in a mess, since this function $$\chi$$ can make the Hamiltonian horribly complicated. We don’t see how gauge invariance can easily be applied to the quantum problem. Next time we will introduce a transformation that resolves some of this mess.

## Question: Lorentz force from classical electrodynamic Hamiltonian

Given the classical Hamiltonian

\label{eqn:qmLecture5:381}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.

apply the Hamiltonian equations of motion

\label{eqn:qmLecture5:480}
\begin{aligned}
\ddt{\Bp} &= – \PD{\Bq}{H} \\
\ddt{\Bq} &= \PD{\Bp}{H},
\end{aligned}

to show that this is the Hamiltonian that describes the Lorentz force equation, and to find the velocity in terms of the canonical momentum and vector potential.

The particle velocity follows easily

\label{eqn:qmLecture5:500}
\begin{aligned}
\Bv
&= \ddt{\Br} \\
&= \PD{\Bp}{H} \\
&= \inv{m} \lr{ \Bp – a \BA }.
\end{aligned}

For the Lorentz force we can proceed in the coordinate representation

\label{eqn:qmLecture5:520}
\begin{aligned}
\ddt{p_k}
&= – \PD{x_k}{H} \\
&= – \frac{2}{2m} \lr{ p_m – q A_m } \PD{x_k}{}\lr{ p_m – q A_m } – q \PD{x_k}{\phi} \\
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
\end{aligned}

We also have

\label{eqn:qmLecture5:540}
\begin{aligned}
\ddt{p_k}
&=
\ddt{} \lr{m x_k + q A_k } \\
&=
m \frac{d^2 x_k}{dt^2} + q \PD{x_m}{A_k} \frac{d x_m}{dt} + q \PD{t}{A_k}.
\end{aligned}

Putting these together we’ve got

\label{eqn:qmLecture5:560}
\begin{aligned}
m \frac{d^2 x_k}{dt^2}
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
– q \PD{x_m}{A_k} \frac{d x_m}{dt} – q \PD{t}{A_k} \\
&=
q v_m \lr{ \PD{x_k}{A_m} – \PD{x_m}{A_k} } + q E_k \\
&=
q v_m \epsilon_{k m s} B_s + q E_k,
\end{aligned}

or

\label{eqn:qmLecture5:580}
\begin{aligned}
m \frac{d^2 \Bx}{dt^2}
&=
q \Be_k v_m \epsilon_{k m s} B_s + q E_k \\
&= q \Bv \cross \BB + q \BE.
\end{aligned}

## Question: Show gauge invariance of the magnetic and electric fields

After the gauge transformation of \ref{eqn:qmLecture5:420} show that the electric and magnetic fields are unaltered.

For the magnetic field the transformed field is

\label{eqn:qmLecture5:600}
\begin{aligned}
\BB’
&= \BB.
\end{aligned}

\label{eqn:qmLecture5:620}
\begin{aligned}
\BE’
&=
– \PD{t}{\BA’} – \spacegrad \phi’ \\
&=
– \PD{t}{}\lr{\BA + \spacegrad \chi} – \spacegrad \lr{ \phi – \PD{t}{\chi}} \\
&=
– \PD{t}{\BA} – \spacegrad \phi \\
&=
\BE.
\end{aligned}

# References

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

## Question:

Given a gauge transformation of the free particle Hamiltonian to

\label{eqn:gaugeTx:20}
H = \inv{2 m} \BPi \cdot \BPi + e \phi,

where

\label{eqn:gaugeTx:40}
\BPi = \Bp – \frac{e}{c} \BA,

calculate $$m d\Bx/dt$$, $$\antisymmetric{\Pi_i}{\Pi_j}$$, and $$m d^2\Bx/dt^2$$, where $$\Bx$$ is the Heisenberg picture position operator, and the fields are functions only of position $$\phi = \phi(\Bx), \BA = \BA(\Bx)$$.

The final results for these calculations are found in [1], but seem worth deriving to exercise our commutator muscles.

### Heisenberg picture velocity operator

The first order of business is the Heisenberg picture velocity operator, but first note

\label{eqn:gaugeTx:60}
\begin{aligned}
\BPi \cdot \BPi
&= \lr{ \Bp – \frac{e}{c} \BA} \cdot \lr{ \Bp – \frac{e}{c} \BA} \\
&= \Bp^2 – \frac{e}{c} \lr{ \BA \cdot \Bp + \Bp \cdot \BA } + \frac{e^2}{c^2} \BA^2.
\end{aligned}

The time evolution of the Heisenberg picture position operator is therefore

\label{eqn:gaugeTx:80}
\begin{aligned}
\ddt{\Bx}
&= \inv{i\Hbar} \antisymmetric{\Bx}{H} \\
&= \inv{i\Hbar 2 m} \antisymmetric{\Bx}{\BPi^2} \\
&= \inv{i\Hbar 2 m} \antisymmetric{\Bx}{\Bp^2 – \frac{e}{c} \lr{ \BA \cdot \Bp
+ \Bp \cdot \BA } + \frac{e^2}{c^2} \BA^2 } \\
&= \inv{i\Hbar 2 m}
\lr{
\antisymmetric{\Bx}{\Bp^2}
– \frac{e}{c} \antisymmetric{\Bx}{ \BA \cdot \Bp + \Bp \cdot \BA }
}
.
\end{aligned}

For the $$\Bp^2$$ commutator we have

\label{eqn:gaugeTx:100}
\antisymmetric{x_r}{\Bp^2}
=
i \Hbar \PD{p_r}{\Bp^2}
=
2 i \Hbar p_r,

or
\label{eqn:gaugeTx:120}
\antisymmetric{\Bx}{\Bp^2}
=
2 i \Hbar \Bp.

Computing the remaining commutator, we’ve got

\label{eqn:gaugeTx:140}
\begin{aligned}
\antisymmetric{x_r}{\Bp \cdot \BA + \BA \cdot \Bp}
&= x_r p_s A_s – p_s A_s x_r \\
&\quad+ x_r A_s p_s – A_s p_s x_r \\
&= \lr{ \antisymmetric{x_r}{p_s} + p_s x_r } A_s – p_s A_s x_r \\
&\quad+ x_r A_s p_s – A_s \lr{ \antisymmetric{p_s}{x_r} + x_r p_s } \\
&= \antisymmetric{x_r}{p_s} A_s + {p_s A_s x_r – p_s A_s x_r} \\
&\quad+ {x_r A_s p_s – x_r A_s p_s} + A_s \antisymmetric{x_r}{p_s} \\
&= 2 i \Hbar \delta_{r s} A_s \\
&= 2 i \Hbar A_r,
\end{aligned}

so

\label{eqn:gaugeTx:160}
\antisymmetric{\Bx}{\Bp \cdot \BA + \BA \cdot \Bp} = 2 i \Hbar \BA.

Assembling these results gives

\label{eqn:gaugeTx:180}
\boxed{
\ddt{\Bx} = \inv{m} \lr{ \Bp – \frac{e}{c} \BA } = \inv{m} \BPi,
}

as asserted in the text.

### Kinetic Momentum commutators

\label{eqn:gaugeTx:200}
\begin{aligned}
\antisymmetric{\Pi_r}{\Pi_s}
&=
\antisymmetric{p_r – e A_r/c}{p_s – e A_s/c} \\
&=
{\antisymmetric{p_r}{p_s}}
– \frac{e}{c} \lr{ \antisymmetric{p_r}{A_s} + \antisymmetric{A_r}{p_s}}
+ \frac{e^2}{c^2} {\antisymmetric{A_r}{A_s}} \\
&=
– \frac{e}{c} \lr{ (-i\Hbar) \PD{x_r}{A_s} + (i\Hbar) \PD{x_s}{A_r} } \\
&=
– \frac{i e \Hbar}{c} \lr{ -\PD{x_r}{A_s} + \PD{x_s}{A_r} } \\
&=
– \frac{i e \Hbar}{c} \epsilon_{t s r} B_t,
\end{aligned}

or
\label{eqn:gaugeTx:220}
\boxed{
\antisymmetric{\Pi_r}{\Pi_s}
=
\frac{i e \Hbar}{c} \epsilon_{r s t} B_t.
}

### Quantum Lorentz force

For the force equation we have

\label{eqn:gaugeTx:240}
\begin{aligned}
m \frac{d^2 \Bx}{dt^2}
&= \ddt{\BPi} \\
&= \inv{i \Hbar} \antisymmetric{\BPi}{H} \\
&= \inv{i \Hbar 2 m } \antisymmetric{\BPi}{\BPi^2}
+ \inv{i \Hbar } \antisymmetric{\BPi}{e \phi}.
\end{aligned}

For the $$\phi$$ commutator consider one component

\label{eqn:gaugeTx:260}
\begin{aligned}
\antisymmetric{\Pi_r}{e \phi}
&=
e \antisymmetric{p_r – \frac{e}{c} A_r}{\phi} \\
&=
e \antisymmetric{p_r}{\phi} \\
&=
e (-i\Hbar) \PD{x_r}{\phi},
\end{aligned}

or
\label{eqn:gaugeTx:280}
\inv{i \Hbar} \antisymmetric{\BPi}{e \phi}
=
=
e \BE.

For the $$\BPi^2$$ commutator I initially did this the hard way (it took four notebook pages, plus two for a false start.) Realizing that I didn’t use \ref{eqn:gaugeTx:220} for that expansion was the clue to doing this more expediently.

Considering a single component

\label{eqn:gaugeTx:300}
\begin{aligned}
\antisymmetric{\Pi_r}{\BPi^2}
&=
\antisymmetric{\Pi_r}{\Pi_s \Pi_s} \\
&=
\Pi_r \Pi_s \Pi_s – \Pi_s \Pi_s \Pi_r \\
&=
\lr{ \antisymmetric{\Pi_r}{\Pi_s} + {\Pi_s \Pi_r} }
\Pi_s
– \Pi_s
\lr{ \antisymmetric{\Pi_s}{\Pi_r} + {\Pi_r \Pi_s} } \\
&= i \Hbar \frac{e}{c} \epsilon_{r s t}
\lr{ B_t \Pi_s + \Pi_s B_t },
\end{aligned}

or

\label{eqn:gaugeTx:320}
\begin{aligned}
\inv{ i \Hbar 2 m} \antisymmetric{\BPi}{\BPi^2}
&= \frac{e}{2 m c } \epsilon_{r s t} \Be_r
\lr{ B_t \Pi_s + \Pi_s B_t } \\
&= \frac{e}{ 2 m c }
\lr{
\BPi \cross \BB
– \BB \cross \BPi
}.
\end{aligned}

Putting all the pieces together we’ve got the quantum equivalent of the Lorentz force equation

\label{eqn:gaugeTx:340}
\boxed{
m \frac{d^2 \Bx}{dt^2} = e \BE + \frac{e}{2 c} \lr{
\frac{d\Bx}{dt} \cross \BB
– \BB \cross \frac{d\Bx}{dt}
}.
}

While this looks equivalent to the classical result, all the vectors here are Heisenberg picture operators dependent on position.

# References

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

## Quantum SHO ladder operators as a diagonal change of basis for the Heisenberg EOMs

Many authors pull the definitions of the raising and lowering (or ladder) operators out of their butt with no attempt at motivation. This is pointed out nicely in [1] by Eli along with one justification based on factoring the Hamiltonian.

In [2] is a small exception to the usual presentation. In that text, these operators are defined as usual with no motivation. However, after the utility of these operators has been shown, the raising and lowering operators show up in a context that does provide that missing motivation as a side effect.
It doesn’t look like the author was trying to provide a motivation, but it can be interpreted that way.

When seeking the time evolution of Heisenberg-picture position and momentum operators, we will see that those solutions can be trivially expressed using the raising and lowering operators. No special tools nor black magic is required to find the structure of these operators. Unfortunately, we must first switch to both the Heisenberg picture representation of the position and momentum operators, and also employ the Heisenberg equations of motion. Neither of these last two fit into standard narrative of most introductory quantum mechanics treatments. We will also see that these raising and lowering “operators” could also be introduced in classical mechanics, provided we were attempting to solve the SHO system using the Hamiltonian equations of motion.

I’ll outline this route to finding the structure of the ladder operators below. Because these are encountered trying to solve the time evolution problem, I’ll first show a simpler way to solve that problem. Because that simpler method depends a bit on lucky observation and is somewhat unstructured, I’ll then outline a more structured procedure that leads to the ladder operators directly, also providing the solution to the time evolution problem as a side effect.

The starting point is the Heisenberg equations of motion. For a time independent Hamiltonian $$H$$, and a Heisenberg operator $$A^{(H)}$$, those equations are

\label{eqn:harmonicOscDiagonalize:20}
\ddt{A^{(H)}} = \inv{i \Hbar} \antisymmetric{A^{(H)}}{H}.

Here the Heisenberg operator $$A^{(H)}$$ is related to the Schrodinger operator $$A^{(S)}$$ by

\label{eqn:harmonicOscDiagonalize:60}
A^{(H)} = U^\dagger A^{(S)} U,

where $$U$$ is the time evolution operator. For this discussion, we need only know that $$U$$ commutes with $$H$$, and do not need to know the specific structure of that operator. In particular, the Heisenberg equations of motion take the form

\label{eqn:harmonicOscDiagonalize:80}
\begin{aligned}
\ddt{A^{(H)}}
&= \inv{i \Hbar}
\antisymmetric{A^{(H)}}{H} \\
&= \inv{i \Hbar}
\antisymmetric{U^\dagger A^{(S)} U}{H} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} U H
– H U^\dagger A^{(S)} U
} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} H U
– U^\dagger H A^{(S)} U
} \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{A^{(S)}}{H} U.
\end{aligned}

The Hamiltonian for the harmonic oscillator, with Schrodinger-picture position and momentum operators $$x, p$$ is

\label{eqn:harmonicOscDiagonalize:40}
H = \frac{p^2}{2m} + \inv{2} m \omega^2 x^2,

so the equations of motions are

\label{eqn:harmonicOscDiagonalize:100}
\begin{aligned}
\ddt{x^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{\frac{p^2}{2m}} U \\
&= \inv{2 m i \Hbar} U^\dagger \lr{ i \Hbar \PD{p}{p^2} } U \\
&= \inv{m } U^\dagger p U \\
&= \inv{m } p^{(H)},
\end{aligned}

and
\label{eqn:harmonicOscDiagonalize:120}
\begin{aligned}
\ddt{p^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{\inv{2} m \omega^2 x^2 } U \\
&= \frac{m \omega^2}{2 i \Hbar} U^\dagger \lr{ -i \Hbar \PD{x}{x^2} } U \\
&= -m \omega^2 U^\dagger x U \\
&= -m \omega^2 x^{(H)}.
\end{aligned}

In the Heisenberg picture the equations of motion are precisely those of classical Hamiltonian mechanics, except that we are dealing with operators instead of scalars

\label{eqn:harmonicOscDiagonalize:140}
\begin{aligned}
\ddt{p^{(H)}} &= -m \omega^2 x^{(H)} \\
\ddt{x^{(H)}} &= \inv{m } p^{(H)}.
\end{aligned}

In the text the ladder operators are used to simplify the solution of these coupled equations, since they can decouple them. That’s not really required since we can solve them directly in matrix form with little work

\label{eqn:harmonicOscDiagonalize:160}
\ddt{}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix}
=
\begin{bmatrix}
0 & -m \omega^2 \\
\inv{m} & 0
\end{bmatrix}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix},

or, with length scaled variables

\label{eqn:harmonicOscDiagonalize:180}
\begin{aligned}
\ddt{}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}
&=
\begin{bmatrix}
0 & -\omega \\
\omega & 0
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\sigma_y
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}.
\end{aligned}

Writing $$y = \begin{bmatrix} \frac{p^{(H)}}{m \omega} \\ x^{(H)} \end{bmatrix}$$, the solution can then be written immediately as

\label{eqn:harmonicOscDiagonalize:200}
\begin{aligned}
y(t)
&=
\exp\lr{ -i \omega \sigma_y t } y(0) \\
&=
\lr{ \cos \lr{ \omega t } I – i \sigma_y \sin\lr{ \omega t } } y(0) \\
&=
\begin{bmatrix}
\cos\lr{ \omega t } & \sin\lr{ \omega t } \\
-\sin\lr{ \omega t } & \cos\lr{ \omega t }
\end{bmatrix}
y(0),
\end{aligned}

or

\label{eqn:harmonicOscDiagonalize:220}
\begin{aligned}
\frac{p^{(H)}(t)}{m \omega} &= \cos\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \sin\lr{ \omega t } x^{(H)}(0) \\
x^{(H)}(t) &= -\sin\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \cos\lr{ \omega t } x^{(H)}(0).
\end{aligned}

This solution depends on being lucky enough to recognize that the matrix has a Pauli matrix as a factor (which squares to unity, and allows the exponential to be evaluated easily.)

If we hadn’t been that observant, then the first tool we’d have used instead would have been to diagonalize the matrix. For such diagonalization, it’s natural to work in completely dimensionless variables. Such a non-dimensionalisation can be had by defining

\label{eqn:harmonicOscDiagonalize:240}
x_0 = \sqrt{\frac{\Hbar}{m \omega}},

and dividing the working (operator) variables through by those values. Let $$z = \inv{x_0} y$$, and $$\tau = \omega t$$ so that the equations of motion are

\label{eqn:harmonicOscDiagonalize:260}
\frac{dz}{d\tau}
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
z.

This matrix can be diagonalized as

\label{eqn:harmonicOscDiagonalize:280}
A
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
=
V
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
V^{-1},

where

\label{eqn:harmonicOscDiagonalize:300}
V =
\inv{\sqrt{2}}
\begin{bmatrix}
i & -i \\
1 & 1
\end{bmatrix}.

The equations of motion can now be written

\label{eqn:harmonicOscDiagonalize:320}
\frac{d}{d\tau} \lr{ V^{-1} z } =
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
\lr{ V^{-1} z }.

This final change of variables $$V^{-1} z$$ decouples the system as desired. Expanding that gives

\label{eqn:harmonicOscDiagonalize:340}
\begin{aligned}
V^{-1} z
&=
\inv{\sqrt{2}}
\begin{bmatrix}
-i & 1 \\
i & 1
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i \frac{p^{(H)}}{m \omega} + x^{(H)} \\
i \frac{p^{(H)}}{m \omega} + x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
a^\dagger \\
a
\end{bmatrix},
\end{aligned}

where
\label{eqn:harmonicOscDiagonalize:n}
\begin{aligned}
a^\dagger &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ -i \frac{p^{(H)}}{m \omega} + x^{(H)} } \\
a &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ i \frac{p^{(H)}}{m \omega} + x^{(H)} }.
\end{aligned}

Lo and behold, we have the standard form of the raising and lowering operators, and can write the system equations as

\label{eqn:harmonicOscDiagonalize:360}
\begin{aligned}
\ddt{a^\dagger} &= i \omega a^\dagger \\
\ddt{a} &= -i \omega a.
\end{aligned}

It is actually a bit fluky that this matched exactly, since we could have chosen eigenvectors that differ by constant phase factors, like

\label{eqn:harmonicOscDiagonalize:380}
V = \inv{\sqrt{2}}
\begin{bmatrix}
i e^{i\phi} & -i e^{i \psi} \\
1 e^{i\phi} & e^{i \psi}
\end{bmatrix},

so

\label{eqn:harmonicOscDiagonalize:341}
\begin{aligned}
V^{-1} z
&=
\frac{e^{-i(\phi + \psi)}}{\sqrt{2}}
\begin{bmatrix}
-i e^{i\psi} & e^{i \psi} \\
i e^{i\phi} & e^{i \phi}
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i e^{i\phi} \frac{p^{(H)}}{m \omega} + e^{i\phi} x^{(H)} \\
i e^{i\psi} \frac{p^{(H)}}{m \omega} + e^{i\psi} x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{i\phi} a^\dagger \\
e^{i\psi} a
\end{bmatrix}.
\end{aligned}

To make the resulting pairs of operators Hermitian conjugates, we’d want to constrain those constant phase factors by setting $$\phi = -\psi$$. If we were only interested in solving the time evolution problem no such additional constraints are required.

The raising and lowering operators are seen to naturally occur when seeking the solution of the Heisenberg equations of motion. This is found using the standard technique of non-dimensionalisation and then seeking a change of basis that diagonalizes the system matrix. Because the Heisenberg equations of motion are identical to the classical Hamiltonian equations of motion in this case, what we call the raising and lowering operators in quantum mechanics could also be utilized in the classical simple harmonic oscillator problem. However, in a classical context we wouldn’t have a justification to call this more than a change of basis.

# References

[1] Eli Lansey. The Quantum Harmonic Oscillator Ladder Operators, 2009. URL http://behindtheguesses.blogspot.ca/2009/03/quantum-harmonic-oscillator-ladder.html. [Online; accessed 18-August-2015].

[2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics, chapter {Time Development of the Oscillator}. Pearson Higher Ed, 2014.

## Question: Heisenberg picture position commutator ([1] pr. 2.5)

Evaluate

\label{eqn:positionCommutator:20}
\antisymmetric{x(t)}{x(0)},

for a Heisenberg picture operator $$x(t)$$ for a free particle.

The free particle Hamiltonian is

\label{eqn:positionCommutator:40}
H = \frac{p^2}{2m},

so the time evolution operator is

\label{eqn:positionCommutator:60}
U(t) = e^{-i p^2 t/(2 m \Hbar)}.

The Heisenberg picture position operator is

\label{eqn:positionCommutator:80}
\begin{aligned}
x^\textrm{H}
&= U^\dagger x U \\
&= e^{i p^2 t/(2 m \Hbar)} x e^{-i p^2 t/(2 m \Hbar)} \\
&= \sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i p^2 t}{2 m \Hbar} }^k
x
e^{-i p^2 t/(2 m \Hbar)} \\
&= \sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k p^{2k} x
e^{-i p^2 t/(2 m \Hbar)} \\
&=
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ \antisymmetric{p^{2k}}{x} + x p^{2k} }
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \antisymmetric{p^{2k}}{x}
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ -i \Hbar \PD{p}{p^{2k}} }
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ -i \Hbar 2 k p^{2 k -1} }
e^{-i p^2 t/(2 m \Hbar)} \\
&= x +
-2 i \Hbar p \frac{i t}{2 m \Hbar} \sum_{k = 1}^\infty \inv{(k-1)!} \lr{ \frac{i t}{2 m \Hbar} }^{k-1} p^{2(k – 1)}
e^{-i p^2 t/(2 m \Hbar)} \\
&= x + t \frac{p}{m}.
\end{aligned}

This has the structure of a classical free particle $$x(t) = x + v t$$, but in this case $$x,p$$ are operators.

The evolved position commutator is
\label{eqn:positionCommutator:100}
\begin{aligned}
\antisymmetric{x(t)}{x(0)}
&=
\antisymmetric{x + t p/m}{x} \\
&=
\frac{t}{m} \antisymmetric{p}{x} \\
&=
-i \Hbar \frac{t}{m}.
\end{aligned}

Compare this to the classical Poisson bracket
\label{eqn:positionCommutator:120}
\antisymmetric{x(t)}{x(0)}_{\textrm{classical}}
=
\PD{x}{}\lr{x + p t/m} \PD{p}{x} – \PD{p}{}\lr{x + p t/m} \PD{x}{x}
=
– \frac{t}{m}.

This has the expected relation $$\antisymmetric{x(t)}{x(0)} = i \Hbar \antisymmetric{x(t)}{x(0)}_{\textrm{classical}}$$.

# References

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

## (a)

Evaluate the classical Poisson bracket

\label{eqn:translation:420}
\antisymmetric{x}{F(p)}_{\textrm{classical}}

## (b)

Evaluate the commutator

\label{eqn:translation:440}
\antisymmetric{x}{e^{i p a/\Hbar}}

## (c)

Using the result in \ref{problem:translation:28:b}, prove that
\label{eqn:translation:460}
e^{i p a/\Hbar} \ket{x’},

is an eigenstate of the coordinate operator $$x$$.

## (a)

\label{eqn:translation:480}
\begin{aligned}
\antisymmetric{x}{F(p)}_{\textrm{classical}}
&=
\PD{x}{x} \PD{p}{F(p)} – \PD{p}{x} \PD{x}{F(p)} \\
&=
\PD{p}{F(p)}.
\end{aligned}

## (b)

Having worked backwards through these problems, the answer for this one dimensional problem can be obtained from \ref{eqn:translation:140} and is

\label{eqn:translation:500}
\antisymmetric{x}{e^{i p a/\Hbar}} = a e^{i p a/\Hbar}.

## (c)

\label{eqn:translation:520}
\begin{aligned}
x e^{i p a/\Hbar} \ket{x’}
&=
\lr{
\antisymmetric{x}{e^{i p a/\Hbar}}
e^{i p a/\Hbar} x
+
}
\ket{x’} \\
&=
\lr{ a e^{i p a/\Hbar} + e^{i p a/\Hbar} x ‘ } \ket{x’} \\
&= \lr{ a + x’ } \ket{x’}.
\end{aligned}

This demonstrates that $$e^{i p a/\Hbar} \ket{x’}$$ is an eigenstate of $$x$$ with eigenvalue $$a + x’$$.

## (a)

For power series $$F, G$$, verify

\label{eqn:translation:180}
\antisymmetric{x_k}{G(\Bp)} = i \Hbar \PD{p_k}{G}, \qquad
\antisymmetric{p_k}{F(\Bx)} = -i \Hbar \PD{x_k}{F}.

## (b)

Evaluate $$\antisymmetric{x^2}{p^2}$$, and compare to the classical Poisson bracket $$\antisymmetric{x^2}{p^2}_{\textrm{classical}}$$.

## (a)

Let

\label{eqn:translation:200}
\begin{aligned}
G(\Bp) &= \sum_{k l m} a_{k l m} p_1^k p_2^l p_3^m \\
F(\Bx) &= \sum_{k l m} b_{k l m} x_1^k x_2^l x_3^m.
\end{aligned}

It is simpler to work with a specific $$x_k$$, say $$x_k = y$$. The validity of the general result will still be clear doing so. Expanding the commutator gives

\label{eqn:translation:220}
\begin{aligned}
\antisymmetric{y}{G(\Bp)}
&=
\sum_{k l m} a_{k l m} \antisymmetric{y}{p_1^k p_2^l p_3^m } \\
&=
\sum_{k l m} a_{k l m} \lr{
y p_1^k p_2^l p_3^m – p_1^k p_2^l p_3^m y
} \\
&=
\sum_{k l m} a_{k l m} \lr{
p_1^k y p_2^l p_3^m – p_1^k y p_2^l p_3^m
} \\
&=
\sum_{k l m} a_{k l m}
p_1^k
\antisymmetric{y}{p_2^l}
p_3^m.
\end{aligned}

From \ref{eqn:translation:100}, we have $$\antisymmetric{y}{p_2^l} = l i \Hbar p_2^{l-1}$$, so

\label{eqn:translation:240}
\begin{aligned}
\antisymmetric{y}{G(\Bp)}
&=
\sum_{k l m} a_{k l m}
p_1^k
\antisymmetric{y}{p_2^l}
\lr{ l
i \Hbar p_2^{l-1}
}
p_3^m \\
&=
i \Hbar \PD{y}{G(\Bp)}.
\end{aligned}

It is straightforward to show that
$$\antisymmetric{p}{x^l} = -l i \Hbar x^{l-1}$$, allowing for a similar computation of the momentum commutator

\label{eqn:translation:260}
\begin{aligned}
\antisymmetric{p_y}{F(\Bx)}
&=
\sum_{k l m} b_{k l m} \antisymmetric{p_y}{x_1^k x_2^l x_3^m } \\
&=
\sum_{k l m} b_{k l m} \lr{
p_y x_1^k x_2^l x_3^m – x_1^k x_2^l x_3^m p_y
} \\
&=
\sum_{k l m} b_{k l m} \lr{
x_1^k p_y x_2^l x_3^m – x_1^k p_y x_2^l x_3^m
} \\
&=
\sum_{k l m} b_{k l m}
x_1^k
\antisymmetric{p_y}{x_2^l}
x_3^m \\
&=
\sum_{k l m} b_{k l m}
x_1^k
\lr{ -l i \Hbar x_2^{l-1}}
x_3^m \\
&=
-i \Hbar \PD{p_y}{F(\Bx)}.
\end{aligned}

## (b)

It isn’t clear to me how the results above can be used directly to compute $$\antisymmetric{x^2}{p^2}$$. However, when the first term of such a commutator is a monomomial, it can be expanded in terms of an $$x$$ commutator

\label{eqn:translation:280}
\begin{aligned}
\antisymmetric{x^2}{G(\Bp)}
&= x^2 G – G x^2 \\
&= x \lr{ x G } – G x^2 \\
&= x \lr{ \antisymmetric{ x }{ G } + G x } – G x^2 \\
&= x \antisymmetric{ x }{ G } + \lr{ x G } x – G x^2 \\
&= x \antisymmetric{ x }{ G } + \lr{ \antisymmetric{ x }{ G } + G x } x – G x^2 \\
&= x \antisymmetric{ x }{ G } + \antisymmetric{ x }{ G } x.
\end{aligned}

Similarily,

\label{eqn:translation:300}
\antisymmetric{x^3}{G(\Bp)} = x^2 \antisymmetric{ x }{ G } + x \antisymmetric{ x }{ G } x + \antisymmetric{ x }{ G } x^2.

An induction hypothesis can be formed

\label{eqn:translation:320}
\antisymmetric{x^k}{G(\Bp)} = \sum_{j = 0}^{k-1} x^{k-1-j} \antisymmetric{ x }{ G } x^j,

and demonstrated

\label{eqn:translation:340}
\begin{aligned}
\antisymmetric{x^{k+1}}{G(\Bp)}
&=
x^{k+1} G – G x^{k+1} \\
&=
x \lr{ x^{k} G } – G x^{k+1} \\
&=
x \lr{ \antisymmetric{x^{k}}{G} + G x^k } – G x^{k+1} \\
&=
x \antisymmetric{x^{k}}{G} + \lr{ x G } x^k – G x^{k+1} \\
&=
x \antisymmetric{x^{k}}{G} + \lr{ \antisymmetric{x}{G} + G x } x^k – G x^{k+1} \\
&=
x \antisymmetric{x^{k}}{G} + \antisymmetric{x}{G} x^k \\
&=
x \sum_{j = 0}^{k-1} x^{k-1-j} \antisymmetric{ x }{ G } x^j + \antisymmetric{x}{G} x^k \\
&=
\sum_{j = 0}^{k-1} x^{(k+1)-1-j} \antisymmetric{ x }{ G } x^j + \antisymmetric{x}{G} x^k \\
&=
\sum_{j = 0}^{k} x^{(k+1)-1-j} \antisymmetric{ x }{ G } x^j.
\end{aligned}

That was a bit overkill for this problem, but may be useful later. Application of this to the problem gives

\label{eqn:translation:360}
\begin{aligned}
\antisymmetric{x^2}{p^2}
&=
x \antisymmetric{x}{p^2}
+ \antisymmetric{x}{p^2} x \\
&=
x i \Hbar \PD{x}{p^2}
+ i \Hbar \PD{x}{p^2} x \\
&=
x 2 i \Hbar p
+ 2 i \Hbar p x \\
&= i \Hbar \lr{ 2 x p + 2 p x }.
\end{aligned}

The classical commutator is
\label{eqn:translation:380}
\begin{aligned}
\antisymmetric{x^2}{p^2}_{\textrm{classical}}
&=
\PD{x}{x^2} \PD{p}{p^2} – \PD{p}{x^2} \PD{x}{p^2} \\
&=
2 x 2 p \\
&= 2 x p + 2 p x.
\end{aligned}

This demonstrates the expected relation between the classical and quantum commutators

\label{eqn:translation:400}
\antisymmetric{x^2}{p^2} = i \Hbar \antisymmetric{x^2}{p^2}_{\textrm{classical}}.

## Question: Translation operator and position expectation. ([1] pr. 1.30)

The translation operator for a finite spatial displacement is given by

\label{eqn:translation:20}
J(\Bl) = \exp\lr{ -i \Bp \cdot \Bl/\Hbar },

where $$\Bp$$ is the momentum operator.

## (a)

Evaluate

\label{eqn:translation:40}
\antisymmetric{x_i}{J(\Bl)}.

## (b)

Demonstrate how the expectation value $$\expectation{\Bx}$$ changes under translation.

## (a)

For clarity, let’s set $$x_i = y$$. The general result will be clear despite doing so.

\label{eqn:translation:60}
\antisymmetric{y}{J(\Bl)}
=
\sum_{k= 0} \inv{k!} \lr{\frac{-i}{\Hbar}}
\antisymmetric{y}{
\lr{ \Bp \cdot \Bl }^k
}.

The commutator expands as

\label{eqn:translation:80}
\begin{aligned}
\antisymmetric{y}{
\lr{ \Bp \cdot \Bl }^k
}
+ \lr{ \Bp \cdot \Bl }^k y
&=
y \lr{ \Bp \cdot \Bl }^k \\
&=
y \lr{ p_x l_x + p_y l_y + p_z l_z } \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ p_x l_x y + y p_y l_y + p_z l_z y } \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ p_x l_x y + l_y \lr{ p_y y + i \Hbar } + p_z l_z y } \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ \Bp \cdot \Bl } y \lr{ \Bp \cdot \Bl }^{k-1}
+ i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1} \\
&= \cdots \\
&=
\lr{ \Bp \cdot \Bl }^{k-1} y \lr{ \Bp \cdot \Bl }^{k-(k-1)}
+ (k-1) i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
\lr{ \Bp \cdot \Bl }^{k} y
+ k i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1}.
\end{aligned}

In the above expansion, the commutation of $$y$$ with $$p_x, p_z$$ has been used. This gives, for $$k \ne 0$$,

\label{eqn:translation:100}
\antisymmetric{y}{
\lr{ \Bp \cdot \Bl }^k
}
=
k i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1}.

Note that this also holds for the $$k = 0$$ case, since $$y$$ commutes with the identity operator. Plugging back into the $$J$$ commutator, we have

\label{eqn:translation:120}
\begin{aligned}
\antisymmetric{y}{J(\Bl)}
&=
\sum_{k = 1} \inv{k!} \lr{\frac{-i}{\Hbar}}
k i \Hbar l_y \lr{ \Bp \cdot \Bl }^{k-1} \\
&=
l_y \sum_{k = 1} \inv{(k-1)!} \lr{\frac{-i}{\Hbar}}
\lr{ \Bp \cdot \Bl }^{k-1} \\
&=
l_y J(\Bl).
\end{aligned}

The same pattern clearly applies with the other $$x_i$$ values, providing the desired relation.

\label{eqn:translation:140}
\antisymmetric{\Bx}{J(\Bl)} = \sum_{m = 1}^3 \Be_m l_m J(\Bl) = \Bl J(\Bl).

## (b)

Suppose that the translated state is defined as $$\ket{\alpha_{\Bl}} = J(\Bl) \ket{\alpha}$$. The expectation value with respect to this state is

\label{eqn:translation:160}
\begin{aligned}
\expectation{\Bx’}
&=
\bra{\alpha_{\Bl}} \Bx \ket{\alpha_{\Bl}} \\
&=
\bra{\alpha} J^\dagger(\Bl) \Bx J(\Bl) \ket{\alpha} \\
&=
\bra{\alpha} J^\dagger(\Bl) \lr{ \Bx J(\Bl) } \ket{\alpha} \\
&=
\bra{\alpha} J^\dagger(\Bl) \lr{ J(\Bl) \Bx + \Bl J(\Bl) } \ket{\alpha} \\
&=
\bra{\alpha} J^\dagger J \Bx + \Bl J^\dagger J \ket{\alpha} \\
&=
\bra{\alpha} \Bx \ket{\alpha} + \Bl \braket{\alpha}{\alpha} \\
&=
\expectation{\Bx} + \Bl.
\end{aligned}

# References

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

## Update to old phy356 (Quantum Mechanics I) notes.

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

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

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

The official description of this course was:

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

This document contains a few things

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