Hamiltonian

PHY1520H Graduate Quantum Mechanics. Lecture 12: Symmetry (cont.). Taught by Prof. Arun Paramekanti

November 5, 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 chap. 4 content from [1].

Parity (review)

\begin{equation}\label{eqn:qmLecture12:20}
\hat{\Pi} \hat{x} \hat{\Pi} = – \hat{x}
\end{equation}
\begin{equation}\label{eqn:qmLecture12:40}
\hat{\Pi} \hat{p} \hat{\Pi} = – \hat{p}
\end{equation}

These are polar vectors, in contrast to an axial vector such as \( \BL = \Br \cross \Bp \).

\begin{equation}\label{eqn:qmLecture12:60}
\hat{\Pi}^2 = 1
\end{equation}

\begin{equation}\label{eqn:qmLecture12:80}
\Psi(x) \rightarrow \Psi(-x)
\end{equation}

If \( \antisymmetric{\hat{\Pi}}{\hat{H}} = 0 \) then all the eigenstates are either

  • even: \( \hat{\Pi} \) eigenvalue is \( + 1 \).
  • odd: \( \hat{\Pi} \) eigenvalue is \( – 1 \).

We are done with discrete symmetry operators for now.

Translations

Define a (continuous) translation operator

\begin{equation}\label{eqn:qmLecture12:100}
\hat{T}_\epsilon \ket{x} = \ket{x + \epsilon}
\end{equation}

The action of this operator is sketched in fig. 1.

lecture12Fig1

fig. 1. Translation operation.

 

This is a unitary operator

\begin{equation}\label{eqn:qmLecture12:120}
\hat{T}_{-\epsilon} = \hat{T}_{\epsilon}^\dagger = \hat{T}_{\epsilon}^{-1}
\end{equation}

In a position basis, the action of this operator is

\begin{equation}\label{eqn:qmLecture12:140}
\bra{x} \hat{T}_{\epsilon} \ket{\psi} = \braket{x-\epsilon}{\psi} = \psi(x – \epsilon)
\end{equation}

\begin{equation}\label{eqn:qmLecture12:160}
\Psi(x – \epsilon) \approx \Psi(x) – \epsilon \PD{x}{\Psi}
\end{equation}

\begin{equation}\label{eqn:qmLecture12:180}
\bra{x} \hat{T}_{\epsilon} \ket{\Psi}
= \braket{x}{\Psi} – \frac{\epsilon}{\Hbar} \bra{ x} i \hat{p} \ket{\Psi}
\end{equation}

\begin{equation}\label{eqn:qmLecture12:200}
\hat{T}_{\epsilon} \approx \lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{p} }
\end{equation}

A non-infinitesimal translation can be composed of many small translations, as sketched in fig. 2.

fig. 2. Composition of small translations

fig. 2. Composition of small translations

For \( \epsilon \rightarrow 0, N \rightarrow \infty, N \epsilon = a \), the total translation operator is

\begin{equation}\label{eqn:qmLecture12:220}
\begin{aligned}
\hat{T}_{a}
&= \hat{T}_{\epsilon}^N \\
&= \lim_{\epsilon \rightarrow 0, N \rightarrow \infty, N \epsilon = a }
\lr{ 1 – \frac{\epsilon}{\Hbar} \hat{p} }^N \\
&= e^{-i a \hat{p}/\Hbar}
\end{aligned}
\end{equation}

The momentum \( \hat{p} \) is called a “Generator” generator of translations. If a Hamiltonian \( H \) is translationally invariant, then

\begin{equation}\label{eqn:qmLecture12:240}
\antisymmetric{\hat{T}_{a}}{H} = 0, \qquad \forall a.
\end{equation}

This means that momentum will be a good quantum number

\begin{equation}\label{eqn:qmLecture12:260}
\antisymmetric{\hat{p}}{H} = 0.
\end{equation}

Rotations

Rotations form a non-Abelian group, since the order of rotations \( \hatR_1 \hatR_2 \ne \hatR_2 \hatR_1 \).

Given a rotation acting on a ket

\begin{equation}\label{eqn:qmLecture12:280}
\hatR \ket{\Br} = \ket{R \Br},
\end{equation}

observe that the action of the rotation operator on a wave function is inverted

\begin{equation}\label{eqn:qmLecture12:300}
\bra{\Br} \hatR \ket{\Psi}
=
\bra{R^{-1} \Br} \ket{\Psi}
= \Psi(R^{-1} \Br).
\end{equation}

Example: Z axis normal rotation

Consider an infinitesimal rotation about the z-axis as sketched in fig. 3(a),(b)

lecture12Fig3

fig 3(a). Rotation about z-axis.

fig 3(b). Rotation about z-axis.

fig 3(b). Rotation about z-axis.

\begin{equation}\label{eqn:qmLecture12:320}
\begin{aligned}
x’ &= x – \epsilon y \\
y’ &= y + \epsilon y \\
z’ &= z
\end{aligned}
\end{equation}

The rotated wave function is

\begin{equation}\label{eqn:qmLecture12:340}
\tilde{\Psi}(x,y,z)
= \Psi( x + \epsilon y, y – \epsilon x, z )
=
\Psi( x, y, z )
+
\epsilon y \underbrace{\PD{x}{\Psi}}_{i \hat{p}_x/\Hbar}
–
\epsilon x \underbrace{\PD{y}{\Psi}}_{i \hat{p}_y/\Hbar}.
\end{equation}

The state must then transform as

\begin{equation}\label{eqn:qmLecture12:360}
\ket{\tilde{\Psi}}
=
\lr{
1
+ i \frac{\epsilon}{\Hbar} \hat{y} \hat{p}_x
– i \frac{\epsilon}{\Hbar} \hat{x} \hat{p}_y
}
\ket{\Psi}.
\end{equation}

Observe that the combination \( \hat{x} \hat{p}_y – \hat{y} \hat{p}_x \) is the \( \hat{L}_z \) component of angular momentum \( \hat{\BL} = \hat{\Br} \cross \hat{\Bp} \), so the infinitesimal rotation can be written

\begin{equation}\label{eqn:qmLecture12:380}
\boxed{
\hatR_z(\epsilon) \ket{\Psi}
=
\lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{L}_z } \ket{\Psi}.
}
\end{equation}

For a finite rotation \( \epsilon \rightarrow 0, N \rightarrow \infty, \phi = \epsilon N \), the total rotation is

\begin{equation}\label{eqn:qmLecture12:420}
\hatR_z(\phi)
=
\lr{ 1 – \frac{i \epsilon}{\Hbar} \hat{L}_z }^N,
\end{equation}

or
\begin{equation}\label{eqn:qmLecture12:440}
\boxed{
\hatR_z(\phi)
=
e^{-i \frac{\phi}{\Hbar} \hat{L}_z}.
}
\end{equation}

Note that \( \antisymmetric{\hat{L}_x}{\hat{L}_y} \ne 0 \).

By construction using Euler angles or any other method, a general rotation will include contributions from components of all the angular momentum operator, and will have the structure

\begin{equation}\label{eqn:qmLecture12:480}
\boxed{
\hatR_\ncap(\phi)
=
e^{-i \frac{\phi}{\Hbar} \lr{ \hat{\BL} \cdot \ncap }}.
}
\end{equation}

Rotationally invariant \( \hat{H} \).

Given a rotationally invariant Hamiltonian

\begin{equation}\label{eqn:qmLecture12:520}
\antisymmetric{\hat{R}_\ncap(\phi)}{\hat{H}} = 0 \qquad \forall \ncap, \phi,
\end{equation}

then every

\begin{equation}\label{eqn:qmLecture12:540}
\antisymmetric{\BL \cdot \ncap}{\hat{H}} = 0,
\end{equation}

or
\begin{equation}\label{eqn:qmLecture12:560}
\antisymmetric{L_i}{\hat{H}} = 0,
\end{equation}

Non-Abelian implies degeneracies in the spectrum.

Time-reversal

Imagine that we have something moving along a curve at time \( t = 0 \), and ending up at the final position at time \( t = t_f \).

fig. 4. Time reversal trajectory.

fig. 4. Time reversal trajectory.

Imagine that we flip the direction of motion (i.e. flipping the velocity) and run time backwards so the final-time state becomes the initial state.

If the time reversal operator is designated \( \hat{\Theta} \), with operation

\begin{equation}\label{eqn:qmLecture12:580}
\hat{\Theta} \ket{\Psi} = \ket{\tilde{\Psi}},
\end{equation}

so that

\begin{equation}\label{eqn:qmLecture12:600}
\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(t)} = \ket{\Psi(0)},
\end{equation}

or

\begin{equation}\label{eqn:qmLecture12:620}
\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(0)} = \ket{\Psi(-t)}.
\end{equation}

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

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.

Quantum Virial Theorem

August 31, 2015 phy1520 , , , , ,

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Question: Quantum Virial Theorem ([1] pr. 2.7)

Consider a particle with Hamiltonian

\begin{equation}\label{eqn:qmVirialTheorem:20}
H = \frac{\Bp^2}{2 m} + V(\Bx),
\end{equation}

By calculating the time evolution of \( \antisymmetric{\Bx \cdot \Bp}{H} \), identify the quantum virial theorem and show the conditions where it is satisfied.

Answer

\begin{equation}\label{eqn:qmVirialTheorem:40}
\begin{aligned}
\antisymmetric{\Bx \cdot \Bp}{H}
&=
\inv{2 m} \antisymmetric{\Bx \cdot \Bp}{\Bp^2} + \antisymmetric{\Bx \cdot \Bp}{V(\Bx)} \\
&=
\inv{2 m} \lr{ x_r p_r \Bp^2 – \Bp^2 x_r p_r}
+
\lr{ x_r p_r V(\Bx) – V(\Bx) x_r p_r } \\
&=
\inv{2 m} \antisymmetric{ x_r }{\Bp^2} p_r
+
x_r \antisymmetric{ p_r}{ V(\Bx)},
\end{aligned}
\end{equation}

Evaluating those commutators separately, gives

\begin{equation}\label{eqn:qmVirialTheorem:60}
\begin{aligned}
\antisymmetric{ x_r }{\Bp^2}
&=
\antisymmetric{ x_r }{p_r^2}\qquad \text{no sum} \\
&=
2 i \Hbar p_r,
\end{aligned}
\end{equation}

and

\begin{equation}\label{eqn:qmVirialTheorem:80}
\antisymmetric{ p_r}{ V(\Bx)}
= -i \Hbar \PD{x_r}{V(\Bx)},
\end{equation}

so
\begin{equation}\label{eqn:qmVirialTheorem:100}
\begin{aligned}
\ddt{}\lr{\Bx \cdot \Bp}
&=
\inv{i \Hbar}
\antisymmetric{\Bx \cdot \Bp}{H} \\
&=
\inv{2 m} 2 p_r p_r – x_r \PD{x_r}{V(\Bx)} \\
&=
\frac{\Bp^2}{m} – \Bx \cdot \spacegrad V(\Bx).
\end{aligned}
\end{equation}

Taking expectation values, assuming that the states are independent of time, we have

\begin{equation}\label{eqn:qmVirialTheorem:120}
\begin{aligned}
0
&= \ddt{} \expectation{ \Bx \cdot \Bp } \\
&= \expectation{\frac{\Bp^2}{m}} – \expectation{\Bx \cdot \spacegrad V(\Bx)}.
\end{aligned}
\end{equation}

Note that taking the expectation with respect to stationary states was required to reverse the order of the time derivative with the expectation operation.

The right hand side is the quantum equivalent of the virial theorem, relating the average kinetic energy to the potential

\begin{equation}\label{eqn:qmVirialTheorem:140}
2 \expectation{T} = \expectation{\Bx \cdot \spacegrad V(\Bx)}
\end{equation}

References

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

A symmetric real Hamiltonian

August 31, 2015 phy1520 , ,

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Question: A symmetric real Hamiltonian ([1] pr. 2.9)

Find the time evolution for the state \( \ket{a’} \) for a Hamiltian of the form

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:20}
H = \delta \lr{ \ket{a’}\bra{a’} + \ket{a”}\bra{a”} }
\end{equation}

Answer

This Hamiltonian has the matrix representation

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:40}
H =
\begin{bmatrix}
0 & \delta \\
\delta & 0
\end{bmatrix},
\end{equation}

which has a characteristic equation of

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:60}
\lambda^2 -\delta^2 = 0,
\end{equation}

so the energy eigenvalues are \( \pm \delta \).

The diagonal basis states are respectively

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:80}
\ket{\pm\delta} =
\inv{\sqrt{2}}
\begin{bmatrix}
\pm 1 \\
1
\end{bmatrix}.
\end{equation}

The time evolution operator is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:100}
\begin{aligned}
U
&= e^{-i H t/\Hbar} \\
&=
e^{-i \delta t/\Hbar} \ket{+\delta}\bra{+\delta}
+ e^{i \delta t/\Hbar} \ket{-\delta}\bra{-\delta} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
+ \frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
-1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 \\
1
\end{bmatrix} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}
+\frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & -1 \\
-1 & 1
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}.
\end{aligned}
\end{equation}

The desired time evolution in the original basis is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:140}
\begin{aligned}
\ket{a’, t}
&=
e^{-i H t/\Hbar}
\ket{a’, 0} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar)
\end{bmatrix} \\
&=
\cos(\delta t/\Hbar) \ket{a’,0} -i \sin(\delta t/\Hbar) \ket{a”,0}.
\end{aligned}
\end{equation}

This evolution has the same structure as left circularly polarized light.

The probability of finding the system in state \( \ket{a”} \) given an initial state of \( \ket{a’,0} \) is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:160}
P
=
\Abs{\braket{a”}{a’,t}}^2
=
\sin^2 \lr{ \delta t/\Hbar }.
\end{equation}

References

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