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PHY1520H Graduate Quantum Mechanics. Lecture 11: Symmetries in QM. Taught by Prof. Arun Paramekanti

October 29, 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{{4}} [1] content.

Symmetry in classical mechanics

In a classical context considering a Hamiltonian

\begin{equation}\label{eqn:qmLecture11:20}
H(q_i, p_i),
\end{equation}

a symmetry means that certain \( q_i \) don’t appear. In that case the rate of change of one of the generalized momenta is zero

\begin{equation}\label{eqn:qmLecture11:40}
\ddt{p_k} = – \PD{q_k}{H} = 0,
\end{equation}

so \( p_k \) is a constant of motion. This simplifies the problem by reducing the number of degrees of freedom. Another aspect of such a symmetry is that it \underline{relates trajectories}. For example, assuming a rotational symmetry as in fig. 1.

fig. 1. Trajectory under rotational symmetry

fig. 1. Trajectory under rotational symmetry

the trajectory of a particle after rotation is related by rotation to the trajectory of the unrotated particle.

Symmetry in quantum mechanics

Suppose that we have a symmetry operation that takes states from

\begin{equation}\label{eqn:qmLecture11:60}
\ket{\psi} \rightarrow \ket{U \psi}
\end{equation}
\begin{equation}\label{eqn:qmLecture11:80}
\ket{\phi} \rightarrow \ket{U \phi},
\end{equation}

we expect that

\begin{equation}\label{eqn:qmLecture11:100}
\Abs{\braket{ \psi}{\phi} }^2 = \Abs{\braket{ U\psi}{ U\phi} }^2.
\end{equation}

This won’t hold true for a general operator. Two cases where this does hold true is when

  • \( \braket{\psi}{\phi} = \braket{ U\psi}{ U\phi} \). Here \( U \) is unitary, and the equivalence follows from

    \begin{equation}\label{eqn:qmLecture11:120}
    \braket{ U\psi}{ U\phi} = \bra{ \psi} U^\dagger U { \phi} = \bra{ \psi} 1 { \phi} = \braket{\psi}{\phi}.
    \end{equation}

  • \( \braket{\psi}{\phi} = \braket{ U\psi}{ U\phi}^\conj \). Here \( U \) is anti-unitary.

Unitary case

If an “observable” is not changed by a unitary operation representing a symmetry we must have

\begin{equation}\label{eqn:qmLecture11:140}
\bra{\psi} \hat{A} \ket{\psi}
\rightarrow
\bra{U \psi} \hat{A} \ket{U \psi}
=
\bra{\psi} U^\dagger \hat{A} U \ket{\psi},
\end{equation}

so
\begin{equation}\label{eqn:qmLecture11:160}
U^\dagger \hat{A} U = \hat{A},
\end{equation}

or
\begin{equation}\label{eqn:qmLecture11:180}
\boxed{
\hat{A} U = U \hat{A}.
}
\end{equation}

An observable that is unchanged by a unitary symmetry commutes \( \antisymmetric{\hat{A}}{U} \) with the operator \( U \) for that transformation.

Symmetries of the Hamiltonian

Given
\begin{equation}\label{eqn:qmLecture11:200}
\antisymmetric{H}{U} = 0,
\end{equation}

\( H \) is invariant.

Given

\begin{equation}\label{eqn:qmLecture11:220}
H \ket{\phi_n} = \epsilon_n \ket{\phi_n} .
\end{equation}

\begin{equation}\label{eqn:qmLecture11:240}
\begin{aligned}
U H \ket{\phi_n}
&= H U \ket{\phi_n} \\
&= \epsilon_n U \ket{\phi_n}
\end{aligned}
\end{equation}

Such a state

\begin{equation}\label{eqn:qmLecture11:260}
\ket{\psi_n} = U \ket{\phi_n}
\end{equation}

is also an eigenstate with the \underline{same} energy.

Suppose this process is repeated, finding other states

\begin{equation}\label{eqn:qmLecture11:280}
U \ket{\psi_n} = \ket{\chi_n}
\end{equation}
\begin{equation}\label{eqn:qmLecture11:300}
U \ket{\chi_n} = \ket{\alpha_n}
\end{equation}

Because such a transformation only generates states with the initial energy, this process cannot continue forever. At some point this process will enumerate a fixed size set of states. These states can be orthonormalized.

We can say that symmetry operations are generators of a \underlineAndIndex{group}. For a set of symmetry operations we can

  • Form products that lie in a closed set

    \begin{equation}\label{eqn:qmLecture11:320}
    U_1 U_2 = U_3
    \end{equation}

  • can define an inverse
    \begin{equation}\label{eqn:qmLecture11:340}
    U \leftrightarrow U^{-1}.
    \end{equation}

  • obeys associative rules for multiplication
    \begin{equation}\label{eqn:qmLecture11:360}
    U_1 ( U_2 U_3 ) = (U_1 U_2) U_3.
    \end{equation}

  • has an identity operation.

When \( H \) has a symmetry, then degenerate eigenstates form \underlineAndIndex{irreducible} representations (which cannot be further block diagonalized).

Some simple examples

Example: Inversion.

{example:qmLecture11:1}

Given a state and a parity operation \( \hat{\Pi} \), with the transformation

\begin{equation}\label{eqn:qmLecture11:380}
\ket{\psi} \rightarrow \hat{\Pi} \ket{\psi}
\end{equation}

In one dimension, the parity operation is just inversion. In two dimensions, this is a set of flipping operations on two axes fig. 2.

fig. 2. 2D parity operation

fig. 2. 2D parity operation

The operational effects of this operator are

\begin{equation}\label{eqn:qmLecture11:400}
\begin{aligned}
\hat{x} &\rightarrow – \hat{x} \\
\hat{p} &\rightarrow – \hat{p}.
\end{aligned}
\end{equation}

Acting again with the parity operator produces the original value, so it is its own inverse, and \( \hat{\Pi}^\dagger = \hat{\Pi} = \hat{\Pi}^{-1} \). In an expectation value

\begin{equation}\label{eqn:qmLecture11:420}
\bra{ \hat{\Pi} \psi } \hat{x} \ket{ \hat{\Pi} \psi } = – \bra{\psi} \hat{x} \ket{\psi}.
\end{equation}

This means that

\begin{equation}\label{eqn:qmLecture11:440}
\hat{\Pi}^\dagger \hat{x} \hat{\Pi} = – \hat{x},
\end{equation}

or
\begin{equation}\label{eqn:qmLecture11:460}
\hat{x} \hat{\Pi} = – \hat{\Pi} \hat{x},
\end{equation}

\begin{equation}\label{eqn:qmLecture11:480}
\begin{aligned}
\hat{x} \hat{\Pi} \ket{x_0}
&= – \hat{\Pi} \hat{x} \ket{x_0} \\
&= – \hat{\Pi} x_0 \ket{x_0} \\
&= – x_0 \hat{\Pi} \ket{x_0}
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:qmLecture11:500}
\hat{\Pi} \ket{x_0} = \ket{-x_0}.
\end{equation}

Acting on a wave function

\begin{equation}\label{eqn:qmLecture11:520}
\begin{aligned}
\bra{x} \hat{\Pi} \ket{\psi}
&=
\braket{-x}{\psi} \\
&= \psi(-x).
\end{aligned}
\end{equation}

What does this mean for eigenfunctions. Eigenfunctions are supposed to form irreducible representations of the group. The group has just two elements

\begin{equation}\label{eqn:qmLecture11:540}
\setlr{ 1, \hat{\Pi} },
\end{equation}

where \( \hat{\Pi}^2 = 1 \).

Suppose we have a Hamiltonian

\begin{equation}\label{eqn:qmLecture11:560}
H = \frac{\hat{p}^2}{2m} + V(\hat{x}),
\end{equation}

where \( V(\hat{x}) \) is even ( \( \antisymmetric{V(\hat{x})}{\hat{\Pi} } = 0 \) ). The squared momentum commutes with the parity operator

\begin{equation}\label{eqn:qmLecture11:580}
\begin{aligned}
\antisymmetric{\hat{p}^2}{\hat{\Pi}}
&=
\hat{p}^2 \hat{\Pi}
– \hat{\Pi} \hat{p}^2 \\
&=
\hat{p}^2 \hat{\Pi}
– (\hat{\Pi} \hat{p}) \hat{p} \\
&=
\hat{p}^2 \hat{\Pi}
-(- \hat{p} \hat{\Pi}) \hat{p} \\
&=
\hat{p}^2 \hat{\Pi}
+ \hat{p} (-\hat{p} \hat{\Pi}) \\
&=
0.
\end{aligned}
\end{equation}

Only two functions are possible in the symmetry set \( \setlr{ \Psi(x), \hat{\Pi} \Psi(x) } \), since

\begin{equation}\label{eqn:qmLecture11:600}
\begin{aligned}
\hat{\Pi}^2 \Psi(x)
&= \hat{\Pi} \Psi(-x) \\
&= \Psi(x).
\end{aligned}
\end{equation}

This symmetry severely restricts the possible solutions, making it so there can be only one dimensional forms of this problem with solutions that are either even or odd respectively

\begin{equation}\label{eqn:qmLecture11:620}
\begin{aligned}
\phi_e(x) &= \psi(x ) + \psi(-x) \\
\phi_o(x) &= \psi(x ) – \psi(-x).
\end{aligned}
\end{equation}

References

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

Degeneracy in non-commuting observables that both commute with the Hamiltonian.

October 22, 2015 phy1520 ,

[Click here for a PDF of an older version of post with nicer formatting]. Updates will be made in my old grad quantum notes.

In problem 1.17 of [2] we are to show that non-commuting operators that both commute with the Hamiltonian, have, in general, degenerate energy eigenvalues. That is

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:320}
[A,H] = [B,H] = 0,
\end{equation}

but

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:340}
[A,B] \ne 0.
\end{equation}

Matrix example of non-commuting commutators

I thought perhaps the problem at hand would be easier if I were to construct some example matrices representing operators that did not commute, but did commuted with a Hamiltonian. I came up with

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:360}
\begin{aligned}
A &=
\begin{bmatrix}
\sigma_z & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix} \\
B &=
\begin{bmatrix}
\sigma_x & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix} \\
H &=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
\end{aligned}
\end{equation}

This system has \( \antisymmetric{A}{H} = \antisymmetric{B}{H} = 0 \), and

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:380}
\antisymmetric{A}{B}
=
\begin{bmatrix}
0 & 2 & 0 \\
-2 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
\end{equation}

There is one shared eigenvector between all of \( A, B, H \)

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:400}
\ket{3} =
\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.
\end{equation}

The other eigenvectors for \( A \) are
\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:420}
\begin{aligned}
\ket{a_1} &=
\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} \\
\ket{a_2} &=
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix},
\end{aligned}
\end{equation}

and for \( B \)
\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:440}
\begin{aligned}
\ket{b_1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix} \\
\ket{b_2} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix},
\end{aligned}
\end{equation}

This clearly has the degeneracy sought.

Looking to [1], it appears that it is possible to construct an even simpler example. Let

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:460}
\begin{aligned}
A &=
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} \\
B &=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix} \\
H &=
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}.
\end{aligned}
\end{equation}

Here \( \antisymmetric{A}{B} = -A \), and \( \antisymmetric{A}{H} = \antisymmetric{B}{H} = 0 \), but the Hamiltonian isn’t interesting at all physically.

A less boring example builds on this. Let

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:480}
\begin{aligned}
A &=
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix} \\
B &=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix} \\
H &=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}.
\end{aligned}
\end{equation}

Here \( \antisymmetric{A}{B} \ne 0 \), and \( \antisymmetric{A}{H} = \antisymmetric{B}{H} = 0 \). I don’t see a way for any exception to be constructed.

The problem

The concrete examples above give some intuition for solving the more abstract problem. Suppose that we are working in a basis that simultaneously diagonalizes operator \( A \) and the Hamiltonian \( H \). To make life easy consider the simplest case where this basis is also an eigenbasis for the second operator \( B \) for all but two of that operators eigenvectors. For such a system let’s write

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:160}
\begin{aligned}
H \ket{1} &= \epsilon_1 \ket{1} \\
H \ket{2} &= \epsilon_2 \ket{2} \\
A \ket{1} &= a_1 \ket{1} \\
A \ket{2} &= a_2 \ket{2},
\end{aligned}
\end{equation}
where \( \ket{1}\), and \( \ket{2} \) are not eigenkets of \( B \). Because \( B \) also commutes with \( H \), we must have

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:180}
H B \ket{1}
= H \sum_n \ket{n}\bra{n} B \ket{1}
= \sum_n \epsilon_n \ket{n} B_{n 1},
\end{equation}

and
\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:200}
B H \ket{1}
= B \epsilon_1 \ket{1}
= \epsilon_1 \sum_n \ket{n}\bra{n} B \ket{1}
= \epsilon_1 \sum_n \ket{n} B_{n 1}.
\end{equation}

We can now compute the action of the commutators on \( \ket{1}, \ket{2} \),
\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:220}
\antisymmetric{B}{H} \ket{1}
=
\sum_n \lr{ \epsilon_1 – \epsilon_n } \ket{n} B_{n 1}.
\end{equation}

Similarly
\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:240}
\antisymmetric{B}{H} \ket{2}
=
\sum_n \lr{ \epsilon_2 – \epsilon_n } \ket{n} B_{n 2}.
\end{equation}

However, for those kets \( \ket{m} \in \setlr{ \ket{3}, \ket{4}, \cdots } \) that are eigenkets of \( B \), with \( B \ket{m} = b_m \ket{m} \), we have

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:280}
\antisymmetric{B}{H} \ket{m}
=
B \epsilon_m \ket{m} – H b_m \ket{m}
=
b_m \epsilon_m \ket{m} – \epsilon_m b_m \ket{m}
=
0,
\end{equation}

The sums in \ref{eqn:angularMomentumAndCentralForceCommutators:220} and \ref{eqn:angularMomentumAndCentralForceCommutators:240} reduce to
\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:500}
\antisymmetric{B}{H} \ket{1}
=
\sum_{n=1}^2 \lr{ \epsilon_1 – \epsilon_n } \ket{n} B_{n 1}
=
\lr{ \epsilon_1 – \epsilon_2 } \ket{2} B_{2 1},
\end{equation}
and
\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:520}
\antisymmetric{B}{H} \ket{2}
=
\sum_{n=1}^2 \lr{ \epsilon_2 – \epsilon_n } \ket{n} B_{n 2}
=
\lr{ \epsilon_2 – \epsilon_1 } \ket{1} B_{1 2}.
\end{equation}
Since the commutator is zero, the matrix elements of the commutator must all be zero, in particular
\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:260}
\begin{aligned}
\bra{1} \antisymmetric{B}{H} \ket{1} &= \lr{ \epsilon_1 – \epsilon_2 } B_{2 1} \braket{1}{2} = 0 \\
\bra{2} \antisymmetric{B}{H} \ket{1} &= \lr{ \epsilon_1 – \epsilon_2 } B_{2 1} \braket{1}{1} \\
\bra{1} \antisymmetric{B}{H} \ket{2} &= \lr{ \epsilon_2 – \epsilon_1 } B_{1 2} \braket{1}{2} = 0 \\
\bra{2} \antisymmetric{B}{H} \ket{2} &= \lr{ \epsilon_2 – \epsilon_1 } B_{1 2} \braket{2}{2}.
\end{aligned}
\end{equation}
We must either have

  • \( B_{2 1} = B_{1 2} = 0 \), or
  • \( \epsilon_1 = \epsilon_2 \).

If the first condition were true we would have

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:300}
B \ket{1}
=
\ket{n}\bra{n} B \ket{1}
=
\ket{n} B_{n 1}
=
\ket{1} B_{1 1},
\end{equation}

and \( B \ket{2} = B_{2 2} \ket{2} \). This contradicts the requirement that \( \ket{1}, \ket{2} \) not be eigenkets of \( B \), leaving only the second option. That second option means there must be a degeneracy in the system.

References

[1] Ronald M. Aarts. Commuting Matrices, 2015. URL http://mathworld.wolfram.com/CommutingMatrices.html. [Online; accessed 22-Oct-2015].

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

PHY1520H Graduate Quantum Mechanics. Lecture 10: 1D Dirac scattering off potential step. Taught by Prof. Arun Paramekanti

October 20, 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.

Dirac scattering off a potential step

For the non-relativistic case we have

\begin{equation}\label{eqn:qmLecture10:20}
\begin{aligned}
E < V_0 &\Rightarrow T = 0, R = 1 \\ E > V_0 &\Rightarrow T > 0, R < 1.
\end{aligned}
\end{equation}

What happens for a relativistic 1D particle?

Referring to fig. 1.

fig. 1. Potential step

fig. 1. Potential step

the region I Hamiltonian is

\begin{equation}\label{eqn:qmLecture10:40}
H =
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix},
\end{equation}

for which the solution is

\begin{equation}\label{eqn:qmLecture10:60}
\Phi = e^{i k_1 x }
\begin{bmatrix}
\cos \theta_1 \\
\sin \theta_1
\end{bmatrix},
\end{equation}

where
\begin{equation}\label{eqn:qmLecture10:80}
\begin{aligned}
\cos 2 \theta_1 &= \frac{ \Hbar c k_1 }{E_{k_1}} \\
\sin 2 \theta_1 &= \frac{ m c^2 }{E_{k_1}} \\
\end{aligned}
\end{equation}

To consider the \( k_1 < 0 \) case, note that

\begin{equation}\label{eqn:qmLecture10:100}
\begin{aligned}
\cos^2 \theta_1 – \sin^2 \theta_1 &= \cos 2 \theta_1 \\
2 \sin\theta_1 \cos\theta_1 &= \sin 2 \theta_1
\end{aligned}
\end{equation}

so after flipping the signs on all the \( k_1 \) terms we find for the reflected wave

\begin{equation}\label{eqn:qmLecture10:120}
\Phi = e^{-i k_1 x}
\begin{bmatrix}
\sin\theta_1 \\
\cos\theta_1
\end{bmatrix}.
\end{equation}

FIXME: this reasoning doesn’t entirely make sense to me. Make sense of this by trying this solution as was done for the form of the incident wave solution.

The region I wave has the form

\begin{equation}\label{eqn:qmLecture10:140}
\Phi_I
=
A e^{i k_1 x}
\begin{bmatrix}
\cos\theta_1 \\
\sin\theta_1 \\
\end{bmatrix}
+
B e^{-i k_1 x}
\begin{bmatrix}
\sin\theta_1 \\
\cos\theta_1 \\
\end{bmatrix}.
\end{equation}

By the time we are done we want to have computed the reflection coefficient

\begin{equation}\label{eqn:qmLecture10:160}
R =
\frac{\Abs{B}^2}{\Abs{A}^2}.
\end{equation}

The region I energy is

\begin{equation}\label{eqn:qmLecture10:180}
E = \sqrt{ \lr{ m c^2}^2 + \lr{ \Hbar c k_1 }^2 }.
\end{equation}

We must have
\begin{equation}\label{eqn:qmLecture10:200}
E
=
\sqrt{ \lr{ m c^2}^2 + \lr{ \Hbar c k_2 }^2 } + V_0
=
\sqrt{ \lr{ m c^2}^2 + \lr{ \Hbar c k_1 }^2 },
\end{equation}

so

\begin{equation}\label{eqn:qmLecture10:220}
\begin{aligned}
\lr{ \Hbar c k_2 }^2
&=
\lr{ E – V_0 }^2 – \lr{ m c^2}^2 \\
&=
\underbrace{\lr{ E – V_0 + m c }}_{r_1}\underbrace{\lr{ E – V_0 – m c }}_{r_2}.
\end{aligned}
\end{equation}

The \( r_1 \) and \( r_2 \) branches are sketched in fig. 2.

fig. 2. Energy signs

fig. 2. Energy signs

For low energies, we have a set of potentials for which we will have propagation, despite having a potential barrier. For still higher values of the potential barrier the product \( r_1 r_2 \) will be negative, so the solutions will be decaying. Finally, for even higher energies, there will again be propagation.

The non-relativistic case is sketched in fig. 3.

fig. 3. Effects of increasing potential for non-relativistic case

fig. 3. Effects of increasing potential for non-relativistic case

For the relativistic case we must consider three different cases, sketched in fig 4, fig 5, and fig 6 respectively. For the low potential energy, a particle with positive group velocity (what we’ve called right moving) can be matched to an equal energy portion of the potential shifted parabola in region II. This is a case where we have transmission, but no antiparticle creation. There will be an energy region where the region II wave function has only a dissipative term, since there is no region of either of the region II parabolic branches available at the incident energy. When the potential is shifted still higher so that \( V_0 > E + m c^2 \), a positive group velocity in region I with a given energy can be matched to an antiparticle branch in the region II parabolic energy curve.

lecture10Fig4a

Fig 4. Low potential energy

lecture10Fig4b

fig. 5. High enough potential energy for no propagation

lecture10Fig4c

fig 6. High potential energy

 

Boundary value conditions

We want to ensure that the current across the barrier is conserved (no particles are lost), as sketched in fig. 7.

 

fig. 7. Transmitted, reflected and incident components.

fig. 7. Transmitted, reflected and incident components.

Recall that given a wave function

\begin{equation}\label{eqn:qmLecture10:240}
\Psi =
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix},
\end{equation}

the density and currents are respectively

\begin{equation}\label{eqn:qmLecture10:260}
\begin{aligned}
\rho &= \psi_1^\conj \psi_1 + \psi_2^\conj \psi_2 \\
j &= \psi_1^\conj \psi_1 – \psi_2^\conj \psi_2
\end{aligned}
\end{equation}

Matching boundary value conditions requires

  1. For both the relativistic and non-relativistic cases we must have\begin{equation}\label{eqn:qmLecture10:280}
    \Psi_{\textrm{L}} = \Psi_{\textrm{R}}, \qquad \mbox{at \( x = 0 \).}
    \end{equation}
  2. For the non-relativistic case we want
    \begin{equation}\label{eqn:qmLecture10:300}
    \int_{-\epsilon}^\epsilon -\frac{\Hbar^2}{2m} \PDSq{x}{\Psi} =
    {\int_{-\epsilon}^\epsilon \lr{ E – V(x) } \Psi(x)}.
    \end{equation}The RHS integral is zero, so

    \begin{equation}\label{eqn:qmLecture10:320}
    -\frac{\Hbar^2}{2m} \lr{ \evalbar{\PD{x}{\Psi}}{{\textrm{R}}} – \evalbar{\PD{x}{\Psi}}{{\textrm{L}}} } = 0.
    \end{equation}

    We have to match

    For the relativistic case

    \begin{equation}\label{eqn:qmLecture10:460}
    -i \Hbar \sigma_z \int_{-\epsilon}^\epsilon \PD{x}{\Psi} +
    {m c^2 \sigma_x \int_{-\epsilon}^\epsilon \psi}
    =
    {\int_{-\epsilon}^\epsilon \lr{ E – V_0 } \psi},
    \end{equation}

the second two integrals are wiped out, so

\begin{equation}\label{eqn:qmLecture10:340}
-i \Hbar c \sigma_z \lr{ \psi(\epsilon) – \psi(-\epsilon) }
=
-i \Hbar c \sigma_z \lr{ \psi_{\textrm{R}} – \psi_{\textrm{L}} }.
\end{equation}

so we must match

\begin{equation}\label{eqn:qmLecture10:360}
\sigma_z \psi_{\textrm{R}} = \sigma_z \psi_{\textrm{L}} .
\end{equation}

It appears that things are simpler, because we only have to match the wave function values at the boundary, and don’t have to match the derivatives too. However, we have a two component wave function, so there are still two tasks.

Solving the system

Let’s look for a solution for the \( E + m c^2 > V_0 \) case on the right branch, as sketched in fig. 8.

 

fig. 8. High potential region. Anti-particle transmission.

fig. 8. High potential region. Anti-particle transmission.

While the right branch in this case is left going, this might work out since that is an antiparticle. We could try both.

Try

\begin{equation}\label{eqn:qmLecture10:480}
\Psi_{II} = D e^{i k_2 x}
\begin{bmatrix}
-\sin\theta_2 \\
\cos\theta_2
\end{bmatrix}.
\end{equation}

This is justified by

\begin{equation}\label{eqn:qmLecture10:500}
+E \rightarrow
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix},
\end{equation}

so

\begin{equation}\label{eqn:qmLecture10:520}
-E \rightarrow
\begin{bmatrix}
-\sin\theta \\
\cos\theta \\
\end{bmatrix}
\end{equation}

At \( x = 0 \) the exponentials vanish, so equating the waves at that point means

\begin{equation}\label{eqn:qmLecture10:380}
\begin{bmatrix}
\cos\theta_1 \\
\sin\theta_1 \\
\end{bmatrix}
+
\frac{B}{A}
\begin{bmatrix}
\sin\theta_1 \\
\cos\theta_1 \\
\end{bmatrix}
=
\frac{D}{A}
\begin{bmatrix}
-\sin\theta_2 \\
\cos\theta_2
\end{bmatrix}.
\end{equation}

Solving this yields

\begin{equation}\label{eqn:qmLecture10:400}
\frac{B}{A} = – \frac{\cos(\theta_1 – \theta_2)}{\sin(\theta_1 + \theta_2)}.
\end{equation}

This yields

\begin{equation}\label{eqn:qmLecture10:420}
\boxed{
R = \frac{1 + \cos( 2 \theta_1 – 2 \theta_2) }{1 – \cos( 2 \theta_1 – 2 \theta_2)}.
}
\end{equation}

As \( V_0 \rightarrow \infty \) this simplifies to

\begin{equation}\label{eqn:qmLecture10:440}
R = \frac{ E – \sqrt{ E^2 – \lr{ m c^2 }^2 } }{ E + \sqrt{ E^2 – \lr{ m c^2 }^2 } }.
\end{equation}

Filling in the details for these results part of problem set 4.

Second update of aggregate notes for phy1520, Graduate Quantum Mechanics

October 20, 2015 phy1520 , , , , , , , , , , , ,

I’ve posted a second 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 9, my ungraded solutions for the second problem set, and some additional worked practise problems.

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

Plane wave ground state expectation for SHO

October 18, 2015 phy1520 , , , , , , , , , , ,

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

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

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

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

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

Let

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

so that

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

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

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

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

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

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

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

This problem is now reduced to showing that

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

or

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

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

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

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

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

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

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

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

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

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

Proceeding to expand \( a^2 X^n \) we find
\begin{equation}\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}
\end{equation}

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

where

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

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

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

Let’s check the induction
\begin{equation}\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}
\end{equation}

which concludes the induction, giving

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

and

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

Let

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

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

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

We want to show that this simplifies to

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

The first values are

\begin{equation}\label{eqn:exponentialExpectationGroundState:540}
\sigma_0 = \bra{0} X^0 \ket{0} = 1
\end{equation}
\begin{equation}\label{eqn:exponentialExpectationGroundState:560}
\sigma_2 = \bra{0} X^2 \ket{0} = 1
\end{equation}

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

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

For the general induction term, consider

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

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.