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1D SHO linear superposition that maximizes expectation

October 7, 2015 phy1520 , , , , , ,

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Question: 1D SHO linear superposition that maximizes expectation ([1] pr. 2.17)

For a 1D SHO

(a)

Construct a linear combination of \( \ket{0}, \ket{1} \) that maximizes \( \expectation{x} \) without using wave functions.

(b)

How does this state evolve with time?

(c)

Evaluate \( \expectation{x} \) using the Schrodinger picture.

(d)

Evaluate \( \expectation{x} \) using the Heisenberg picture.

(e)

Evaluate \( \expectation{(\Delta x)^2} \).

Answer

(a)

Forming

\begin{equation}\label{eqn:shoSuperposition:20}
\ket{\psi} = \frac{\ket{0} + \sigma \ket{1}}{\sqrt{1 + \Abs{\sigma}^2}}
\end{equation}

the position expectation is

\begin{equation}\label{eqn:shoSuperposition:40}
\bra{\psi} x \ket{\psi}
=
\inv{1 + \Abs{\sigma}^2} \lr{ \bra{0} + \sigma^\conj \bra{1} } \frac{x_0}{\sqrt{2}} \lr{ a^\dagger + a } \lr{ \ket{0} + \sigma \ket{1} }.
\end{equation}

Evaluating the action of the operators on the kets, we’ve got

\begin{equation}\label{eqn:shoSuperposition:60}
\lr{ a^\dagger + a } \lr{ \ket{0} + \sigma \ket{1} }
=
\ket{1} + \sqrt{2} \sigma \ket{2} + \sigma \ket{0}.
\end{equation}

The \( \ket{2} \) term is killed by the bras, leaving

\begin{equation}\label{eqn:shoSuperposition:80}
\begin{aligned}
\expectation{x}
&=
\inv{1 + \Abs{\sigma}^2} \frac{x_0}{\sqrt{2}} \lr{ \sigma + \sigma^\conj} \\
&=
\frac{\sqrt{2} x_0 \textrm{Re} \sigma}{1 + \Abs{\sigma}^2}.
\end{aligned}
\end{equation}

Any imaginary component in \( \sigma \) will reduce the expectation, so we are constrained to picking a real value.

The derivative of

\begin{equation}\label{eqn:shoSuperposition:100}
f(\sigma) = \frac{\sigma}{1 + \sigma^2},
\end{equation}

is

\begin{equation}\label{eqn:shoSuperposition:120}
f'(\sigma) = \frac{1 – \sigma^2}{(1 + \sigma^2)^2}.
\end{equation}

That has zeros at \( \sigma = \pm 1 \). The second derivative is

\begin{equation}\label{eqn:shoSuperposition:140}
f”(\sigma) = \frac{-2 \sigma (3 – \sigma^2)}{(1 + \sigma^2)^3}.
\end{equation}

That will be negative (maximum for the extreme value) at \( \sigma = 1 \), so the linear superposition of these first two energy eigenkets that maximizes the position expectation is

\begin{equation}\label{eqn:shoSuperposition:160}
\psi = \inv{\sqrt{2}}\lr{ \ket{0} + \ket{1} }.
\end{equation}

That maximized position expectation is

\begin{equation}\label{eqn:shoSuperposition:180}
\expectation{x}
=
\frac{x_0}{\sqrt{2}}.
\end{equation}

(b)

The time evolution is given by

\begin{equation}\label{eqn:shoSuperposition:200}
\begin{aligned}
\ket{\Psi(t)}
&= e^{-iH t/\Hbar} \inv{\sqrt{2}}\lr{ \ket{0} + \ket{1} } \\
&= \inv{\sqrt{2}}\lr{ e^{-i(0+ \ifrac{1}{2})\Hbar \omega t/\Hbar} \ket{0} +
e^{-i(1+ \ifrac{1}{2})\Hbar \omega t/\Hbar} \ket{1} } \\
&= \inv{\sqrt{2}}\lr{ e^{-i \omega t/2} \ket{0} + e^{-3 i \omega t/2} \ket{1} }.
\end{aligned}
\end{equation}

(c)

The position expectation in the Schrodinger representation is

\begin{equation}\label{eqn:shoSuperposition:220}
\begin{aligned}
\expectation{x(t)}
&=
\inv{2}
\lr{ e^{i \omega t/2} \bra{0} + e^{3 i \omega t/2} \bra{1} } \frac{x_0}{\sqrt{2}} \lr{ a^\dagger + a }
\lr{ e^{-i \omega t/2} \ket{0} + e^{-3 i \omega t/2} \ket{1} } \\
&=
\frac{x_0}{2\sqrt{2}}
\lr{ e^{i \omega t/2} \bra{0} + e^{3 i \omega t/2} \bra{1} }
\lr{ e^{-i \omega t/2} \ket{1} + e^{-3 i \omega t/2} \sqrt{2} \ket{2} + e^{-3 i \omega t/2} \ket{0} } \\
&=
\frac{x_0}{\sqrt{2}} \cos(\omega t).
\end{aligned}
\end{equation}

(d)

\begin{equation}\label{eqn:shoSuperposition:240}
\begin{aligned}
\expectation{x(t)}
&=
\inv{2}
\lr{ \bra{0} + \bra{1} } \frac{x_0}{\sqrt{2}}
\lr{ a^\dagger e^{i\omega t} + a e^{-i \omega t} }
\lr{ \ket{0} + \ket{1} } \\
&=
\frac{x_0}{2 \sqrt{2}}
\lr{ \bra{0} + \bra{1} }
\lr{ e^{i\omega t} \ket{1} + \sqrt{2} e^{i\omega t} \ket{2} + e^{-i \omega t} \ket{0} } \\
&=
\frac{x_0}{\sqrt{2}} \cos(\omega t),
\end{aligned}
\end{equation}

matching the calculation using the Schrodinger picture.

(e)

Let’s use the Heisenberg picture for the uncertainty calculation. Using the calculation above we have

\begin{equation}\label{eqn:shoSuperposition:260}
\begin{aligned}
\expectation{x^2}
&=
\inv{2} \frac{x_0^2}{2}
\lr{ e^{-i\omega t} \bra{1} + \sqrt{2} e^{-i\omega t} \bra{2} + e^{i \omega t} \bra{0} }
\lr{ e^{i\omega t} \ket{1} + \sqrt{2} e^{i\omega t} \ket{2} + e^{-i \omega t} \ket{0} } \\
&=
\frac{x_0^2}{4} \lr{ 1 + 2 + 1} \\
&=
x_0^2.
\end{aligned}
\end{equation}

The uncertainty is
\begin{equation}\label{eqn:shoSuperposition:280}
\begin{aligned}
\expectation{(\Delta x)^2}
&=
\expectation{x^2} – \expectation{x}^2 \\
&=
x_0^2 – \frac{x_0^2}{2} \cos^2(\omega t) \\
&=
\frac{x_0^2}{2} \lr{ 2 – \cos^2(\omega t) } \\
&=
\frac{x_0^2}{2} \lr{ 1 + \sin^2(\omega t) }
\end{aligned}
\end{equation}

References

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

PHY1520H Graduate Quantum Mechanics. Lecture 6: Electromagnetic gauge transformation and Aharonov-Bohm effect. Taught by Prof. Arun Paramekanti

October 6, 2015 phy1520 , , ,

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

Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

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

Particle with \( \BE, \BB \) fields

We express our fields with vector and scalar potentials

\begin{equation}\label{eqn:qmLecture6:20}
\BE, \BB \rightarrow \BA, \phi
\end{equation}

and apply a gauge transformed Hamiltonian

\begin{equation}\label{eqn:qmLecture6:40}
H = \inv{2m} \lr{ \Bp – q \BA }^2 + q \phi.
\end{equation}

Recall that in classical mechanics we have

\begin{equation}\label{eqn:qmLecture6:60}
\Bp – q \BA = m \Bv
\end{equation}

where \( \Bp \) is not gauge invariant, but the classical momentum \( m \Bv \) is.

If given a point in phase space we must also specify the gauge that we are working with.

For the quantum case, temporarily considering a Hamiltonian without any scalar potential, but introducing a gauge transformation

\begin{equation}\label{eqn:qmLecture6:80}
\BA \rightarrow \BA + \spacegrad \chi,
\end{equation}

which takes the Hamiltonian from

\begin{equation}\label{eqn:qmLecture6:100}
H = \inv{2m} \lr{ \Bp – q \BA }^2,
\end{equation}

to
\begin{equation}\label{eqn:qmLecture6:120}
H = \inv{2m} \lr{ \Bp – q \BA -q \spacegrad \chi }^2.
\end{equation}

We care that the position and momentum operators obey

\begin{equation}\label{eqn:qmLecture6:140}
\antisymmetric{\hat{r}_i}{\hat{p}_j} = i \Hbar \delta_{i j}.
\end{equation}

We can apply a transformation that keeps \( \Br \) the same, but changes the momentum

\begin{equation}\label{eqn:qmLecture6:160}
\begin{aligned}
\hat{\Br}’ &= \hat{\Br} \\
\hat{\Bp}’ &= \hat{\Bp} – q \spacegrad \chi(\Br)
\end{aligned}
\end{equation}

This maps the Hamiltonian to

\begin{equation}\label{eqn:qmLecture6:101}
H = \inv{2m} \lr{ \Bp’ – q \BA -q \spacegrad \chi }^2,
\end{equation}

We want to check if the commutator relationships have the desired structure, that is

\begin{equation}\label{eqn:qmLecture6:180}
\begin{aligned}
\antisymmetric{r_i’}{r_j’} &= 0 \\
\antisymmetric{p_i’}{p_j’} &= 0
\end{aligned}
\end{equation}

This is confirmed in \ref{problem:qmLecture6:1}.

Another thing of interest is how are the wave functions altered by this change of variables? The wave functions must change in response to this transformation if the energies of the Hamiltonian are to remain the same.

Considering a plane wave specified by

\begin{equation}\label{eqn:qmLecture6:200}
e^{i \Bk \cdot \Br},
\end{equation}

where we alter the momentum by

\begin{equation}\label{eqn:qmLecture6:220}
\Bk \rightarrow \Bk – e \spacegrad \chi.
\end{equation}

This takes the plane wave to

\begin{equation}\label{eqn:qmLecture6:240}
e^{i \lr{ \Bk – q \spacegrad \chi } \cdot \Br}.
\end{equation}

We want to try to find a wave function for the new Hamiltonian

\begin{equation}\label{eqn:qmLecture6:260}
H’ = \inv{2m} \lr{ \Bp’ – q \BA -q \spacegrad \chi }^2,
\end{equation}

of the form

\begin{equation}\label{eqn:qmLecture6:280}
\psi'(\Br)
\stackrel{?}{=}
e^{i \theta(\Br)} \psi(\Br),
\end{equation}

where the new wave function differs from a wave function for the original Hamiltonian by only a position dependent phase factor.

Let’s look at the action of the Hamiltonian on the new wave function

\begin{equation}\label{eqn:qmLecture6:300}
H’ \psi'(\Br) .
\end{equation}

Looking at just the first action

\begin{equation}\label{eqn:qmLecture6:320}
\begin{aligned}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } e^{i \theta(\Br)} \psi(\Br)
&=
e^{i\theta}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi }
\psi(\Br)
+
\lr{
-i \Hbar i \spacegrad \theta
}
e^{i\theta}
\psi(\Br) \\
&=
e^{i\theta}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi
+ \Hbar \spacegrad \theta
}
\psi(\Br).
\end{aligned}
\end{equation}

If we choose

\begin{equation}\label{eqn:qmLecture6:340}
\theta = \frac{q \chi}{\Hbar},
\end{equation}

then we are left with

\begin{equation}\label{eqn:qmLecture6:360}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } e^{i \theta(\Br)} \psi(\Br)
=
e^{i\theta}
\lr{ -i \Hbar \spacegrad – q \BA }
\psi(\Br).
\end{equation}

Let \( \BM = -i \Hbar \spacegrad – q \BA \), and act again with \( \lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } \)

\begin{equation}\label{eqn:qmLecture6:700}
\begin{aligned}
\lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi } e^{i \theta} \BM \psi
&=
e^{i\theta}
\lr{ -i \Hbar i \spacegrad \theta – q \BA – q \spacegrad \chi } e^{i \theta} \BM \psi
+
e^{i\theta}
\lr{ -i \Hbar \spacegrad } \BM \psi \\
&=
e^{i\theta}
\lr{ -i \Hbar \spacegrad -q \BA + \spacegrad \lr{ \Hbar \theta – q \chi} } \BM \psi \\
&=
e^{i\theta} \BM^2 \psi.
\end{aligned}
\end{equation}

Restoring factors of \( m \), we’ve shown that for a choice of \( \Hbar \theta – q \chi \), we have

\begin{equation}\label{eqn:qmLecture6:400}
\inv{2m} \lr{ -i \Hbar \spacegrad – q \BA – q \spacegrad \chi }^2 e^{i \theta} \psi = e^{i\theta}
\inv{2m} \lr{ -i \Hbar \spacegrad – q \BA }^2 \psi.
\end{equation}

When \( \psi \) is an energy eigenfunction, this means

\begin{equation}\label{eqn:qmLecture6:420}
H’ e^{i\theta} \psi = e^{i \theta} H \psi = e^{i\theta} E\psi = E (e^{i\theta} \psi).
\end{equation}

We’ve found a transformation of the wave function that has the same energy eigenvalues as the corresponding wave functions for the original untransformed Hamiltonian.

In summary
\begin{equation}\label{eqn:qmLecture6:440}
\boxed{
\begin{aligned}
H’ &= \inv{2m} \lr{ \Bp – q \BA – q \spacegrad \chi}^2 \\
\psi'(\Br) &= e^{i \theta(\Br)} \psi(\Br), \qquad \text{where}\, \theta(\Br) = q \chi(\Br)/\Hbar
\end{aligned}
}
\end{equation}

Aharonov-Bohm effect

Consider a periodic motion in a fixed ring as sketched in fig. 1.

fig. 1. particle confined to a ring

fig. 1. particle confined to a ring

Here the displacement around the perimeter is \( s = R \phi \) and the Hamiltonian

\begin{equation}\label{eqn:qmLecture6:460}
H = – \frac{\Hbar^2}{2 m} \PDSq{s}{} = – \frac{\Hbar^2}{2 m R^2} \PDSq{\phi}{}.
\end{equation}

Now assume that there is a magnetic field squeezed into the point at the origin, by virtue of a flux at the origin

\begin{equation}\label{eqn:qmLecture6:480}
\BB = \Phi_0 \delta(\Br) \zcap.
\end{equation}

We know that

\begin{equation}\label{eqn:qmLecture6:500}
\oint \BA \cdot d\Bl = \Phi_0,
\end{equation}

so that

\begin{equation}\label{eqn:qmLecture6:520}
\BA = \frac{\Phi_0}{2 \pi r} \phicap.
\end{equation}

The Hamiltonian for the new configuration is

\begin{equation}\label{eqn:qmLecture6:540}
\begin{aligned}
H
&= – \lr{ -i \Hbar \spacegrad – q \frac{\Phi_0}{2 \pi r } \phicap }^2 \\
&= – \inv{2 m} \lr{ -i \Hbar \inv{R} \PD{\phi}{} – q \frac{\Phi_0}{2 \pi R } }^2.
\end{aligned}
\end{equation}

Here the replacement \( r \rightarrow R \) makes use of the fact that this problem as been posed with the particle forced to move around the ring at the fixed radius \( R \).

For this transformed Hamiltonian, what are the wave functions?

\begin{equation}\label{eqn:qmLecture6:560}
\psi(\phi)’
\stackrel{?}{=}
e^{i n \phi}.
\end{equation}

\begin{equation}\label{eqn:qmLecture6:580}
\begin{aligned}
H \psi
&= \inv{2 m}
\lr{ -i \Hbar \inv{R} (i n) – q \frac{\Phi_0}{2 \pi R } }^2 e^{i n \phi} \\
&=
\underbrace{\inv{2 m}
\lr{ \frac{\Hbar n}{R} – q \frac{\Phi_0}{2 \pi R } }^2}_{E_n} e^{i n \phi}.
\end{aligned}
\end{equation}

This is very unclassical, since the energy changes in a way that depends on the flux, because particles are seeing magnetic fields that are not present at the point of the particle.

This is sketched in fig. 2.

fig. 2. Energy variation with flux.

fig. 2. Energy variation with flux.

we see that there are multiple points that the energies hit the minimum levels

Question:

Show that after a transformation of position and momentum of the following form

\begin{equation}\label{eqn:qmLecture6:600}
\begin{aligned}
\hat{\Br}’ &= \hat{\Br} \\
\hat{\Bp}’ &= \hat{\Bp} – q \spacegrad \chi(\Br)
\end{aligned}
\end{equation}

all the commutators have the expected values.

Answer

The position commutators don’t need consideration. Of interest is the momentum-position commutators

\begin{equation}\label{eqn:qmLecture6:620}
\begin{aligned}
\antisymmetric{\hat{p}_k’}{\hat{x}_k’}
&=
\antisymmetric{\hat{p}_k – q \partial_k \chi}{\hat{x}_k} \\
&=
\antisymmetric{\hat{p}_k}{\hat{x}_k} – q \antisymmetric{\partial_k \chi}{\hat{x}_k} \\
&=
\antisymmetric{\hat{p}_k}{\hat{x}_k},
\end{aligned}
\end{equation}

and the momentum commutators

\begin{equation}\label{eqn:qmLecture6:640}
\begin{aligned}
\antisymmetric{\hat{p}_k’}{\hat{p}_j’}
&=
\antisymmetric{\hat{p}_k – q \partial_k \chi}{\hat{p}_j – q \partial_j \chi} \\
&=
\antisymmetric{\hat{p}_k}{\hat{p}_j}
– q \lr{ \antisymmetric{\partial_k \chi}{\hat{p}_j} + \antisymmetric{\hat{p}_k}{\partial_j \chi} }.
\end{aligned}
\end{equation}

That last sum of commutators is

\begin{equation}\label{eqn:qmLecture6:660}
\begin{aligned}
\antisymmetric{\partial_k \chi}{\hat{p}_j} + \antisymmetric{\hat{p}_k}{\partial_j \chi}
&=
– i \Hbar \lr{ \PD{k}{(\partial_j \chi)} – \PD{j}{(\partial_k \chi)} } \\
&= 0.
\end{aligned}
\end{equation}

We’ve shown that

\begin{equation}\label{eqn:qmLecture6:680}
\begin{aligned}
\antisymmetric{\hat{p}_k’}{\hat{x}_k’} &= \antisymmetric{\hat{p}_k}{\hat{x}_k} \\
\antisymmetric{\hat{p}_k’}{\hat{p}_j’} &= \antisymmetric{\hat{p}_k}{\hat{p}_j}.
\end{aligned}
\end{equation}

All the other commutators clearly have the desired transformation properties.

References

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

First update of aggregate notes for phy1520, Graduate Quantum Mechanics

October 2, 2015 phy1520 ,

I’ve posted a first update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. This includes lecture notes up to lecture 5, my ungraded solutions for the first problem set, and a number of worked problems from chapter 1 and 2 of Sakurai done as review preparation for the course (since I hadn’t done QM since 2011).

Most of the content was posted individually in the following locations. Bug fixes and enhancements to the original documents will only be made in the aggregate notes.

PHY1520H Graduate Quantum Mechanics. Lecture 5: time evolution of coherent states, and charged particles in a magnetic field. Taught by Prof. Arun Paramekanti

October 1, 2015 phy1520 , , , , , , , , , , , ,

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

Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering \textchapref{{1}} [1] content.

Coherent states (cont.)

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

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

where we showed that

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

In the Heisenberg picture we found

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

or equivalently
\begin{equation}\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}
\end{equation}

Given this we can compute expectation value of position operator

\begin{equation}\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}
\end{equation}

Similarly

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

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

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

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

\begin{equation}\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}
\end{equation}

For the second moment we have

\begin{equation}\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}
\end{equation}

We find

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

and

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

so

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

or

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

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

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

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

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

\begin{equation}\label{eqn:qmLecture5:340}
\BB = \spacegrad \cross \BA,
\end{equation}
\begin{equation}\label{eqn:qmLecture5:360}
\BE = – \spacegrad \phi – \PD{t}{\BA}.
\end{equation}

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

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

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.

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

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}
\begin{equation}\label{eqn:qmLecture5:420}
\BA \rightarrow \BA + \spacegrad \chi
\end{equation}
\begin{equation}\label{eqn:qmLecture5:440}
\phi \rightarrow \phi – \PD{t}{\chi}
\end{equation}

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

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

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

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

apply the Hamiltonian equations of motion

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

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.

Answer

The particle velocity follows easily

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

For the Lorentz force we can proceed in the coordinate representation

\begin{equation}\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}
\end{equation}

We also have

\begin{equation}\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}
\end{equation}

Putting these together we’ve got

\begin{equation}\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}
\end{equation}

or

\begin{equation}\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}
\end{equation}

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.

Answer

For the magnetic field the transformed field is

\begin{equation}\label{eqn:qmLecture5:600}
\begin{aligned}
\BB’
&= \spacegrad \cross \lr{ \BA + \spacegrad \chi } \\
&= \spacegrad \cross \BA + \spacegrad \cross \lr{ \spacegrad \chi } \\
&= \spacegrad \cross \BA \\
&= \BB.
\end{aligned}
\end{equation}

\begin{equation}\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}
\end{equation}

References

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

Commutators of angular momentum and a central force Hamiltonian

September 30, 2015 phy1520 , , , ,

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

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

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

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

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

The \( L_x, L_z \) commutator is

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

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

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

and for \( L_z, L_y \)

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

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

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

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

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

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

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

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

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

References

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