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

October 20, 2015 phy1520 No comments , ,

### Disclaimer

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

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti.

### Dirac scattering off a potential step

For the non-relativistic case we have

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

What happens for a relativistic 1D particle?

Referring to fig. 1.

fig. 1. Potential step

the region I Hamiltonian is

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

for which the solution is

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

where
\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}

To consider the $$k_1 < 0$$ case, note that

\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}

so after flipping the signs on all the $$k_1$$ terms we find for the reflected wave

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

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

\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}.

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

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

The region I energy is

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

We must have
\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 },

so

\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}

The $$r_1$$ and $$r_2$$ branches are sketched in fig. 2.

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

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.

Fig 4. Low potential energy

fig. 5. High enough potential energy for no propagation

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.

Recall that given a wave function

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

the density and currents are respectively

\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}

Matching boundary value conditions requires

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

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

We have to match

For the relativistic case

\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},

the second two integrals are wiped out, so

\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}} }.

so we must match

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

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.

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

Try

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

This is justified by

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

so

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

At $$x = 0$$ the exponentials vanish, so equating the waves at that point means

\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}.

Solving this yields

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

This yields

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

As $$V_0 \rightarrow \infty$$ this simplifies to

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

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

## PHY1520H Graduate Quantum Mechanics. Lecture 9: Dirac equation (cont.). 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.

### Where we left off

\label{eqn:qmLecture9:20}
-i \Hbar \PD{t}{}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
=
\begin{bmatrix}
-i \Hbar c \PD{x}{} & m c^2 \\
m c^2 & i \Hbar c \PD{x}{} \\
\end{bmatrix}.

With a potential this would be

\label{eqn:qmLecture9:40}
-i \Hbar \PD{t}{}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
=
\begin{bmatrix}
-i \Hbar c \PD{x}{} + V(x) & m c^2 \\
m c^2 & i \Hbar c \PD{x}{} + V(x) \\
\end{bmatrix}.

This means that the potential is raising the energy eigenvalue of the system.

### Free Particle

Assuming a form

\label{eqn:qmLecture9:60}
\begin{bmatrix}
\psi_1(x,t) \\
\psi_2(x,t)
\end{bmatrix}
=
e^{i k x}
\begin{bmatrix}
f_1(t) \\
f_2(t) \\
\end{bmatrix},

and plugging back into the Dirac equation we have

\label{eqn:qmLecture9:80}
-i \Hbar \PD{t}{}
\begin{bmatrix}
f_1 \\
f_2
\end{bmatrix}
=
\begin{bmatrix}
k \Hbar c & m c^2 \\
m c^2 & – \Hbar k c \\
\end{bmatrix}
\begin{bmatrix}
f_1 \\
f_2
\end{bmatrix}.

We can use a diagonalizing rotation

\label{eqn:qmLecture9:100}
\begin{bmatrix}
f_1 \\
f_2
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta_k & -\sin\theta_k \\
\sin\theta_k & \cos\theta_k \\
\end{bmatrix}
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}.

Plugging this in reduces the system to the form

\label{eqn:qmLecture9:140}
-i \Hbar \PD{t}{}
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}
=
\begin{bmatrix}
E_k & 0 \\
0 & -E_k
\end{bmatrix}
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}.

Where the rotation angle is found to be given by

\label{eqn:qmLecture9:160}
\begin{aligned}
\sin(2 \theta_k) &= \frac{m c^2}{\sqrt{(\Hbar k c)^2 + m^2 c^4}} \\
\cos(2 \theta_k) &= \frac{\Hbar k c}{\sqrt{(\Hbar k c)^2 + m^2 c^4}} \\
E_k &= \sqrt{(\Hbar k c)^2 + m^2 c^4}
\end{aligned}

See fig. 1 for a sketch of energy vs momentum. The asymptotes are the limiting cases when $$m c^2 \rightarrow 0$$. The $$+$$ branch is what we usually associate with particles. What about the other energy states. For Fermions Dirac argued that the lower energy states could be thought of as “filled up”, using the Pauli principle to leave only the positive energy states available. This was called the “Dirac Sea”. This isn’t a good solution, and won’t work for example for Bosons.

fig. 1. Dirac equation solution space

Another way to rationalize this is to employ ideas from solid state theory. For example consider a semiconductor with a valence and conduction band as sketched in fig. 2.

fig. 2. Solid state valence and conduction band transition

A photon can excite an electron from the valence band to the conduction band, leaving all the valence band states filled except for one (a hole). For an electron we can use almost the same picture, as sketched in fig. 3.

fig. 3. Pair creation

A photon with energy $$E_k – (-E_k)$$ can create a positron-electron pair from the vacuum, where the energy of the electron and positron pair is $$E_k$$.

At high enough energies, we can see this pair creation occur.

### Zitterbewegung

If a particle is created at a non-eigenstate such as one on the asymptotes, then oscillations between the positive and negative branches are possible as sketched in fig. 4.

fig. 4. Zitterbewegung oscillation

Only “vertical” oscillations between the positive and negative locations on these branches is possible since those are the points that match the particle momentum. Examining this will be the aim of one of the problem set problems.

### Probability and current density

If we define a probability density

\label{eqn:qmLecture9:180}
\rho(x, t) = \Abs{\psi_1}^2 + \Abs{\psi_2}^2,

does this satisfy a probability conservation relation

\label{eqn:qmLecture9:200}
\PD{t}{\rho} + \PD{x}{j} = 0,

where $$j$$ is the probability current. Plugging in the density, we have

\label{eqn:qmLecture9:220}
\PD{t}{\rho}
=
\PD{t}{\psi_1^\conj} \psi_1
+
\psi_1^\conj \PD{t}{\psi_1}
+
\PD{t}{\psi_2^\conj} \psi_2
+
\psi_2^\conj \PD{t}{\psi_2}.

It turns out that the probability current has the form

\label{eqn:qmLecture9:240}
j(x,t) = c \lr{ \psi_1^\conj \psi_1 + \psi_2^\conj \psi_2 }.

Here the speed of light $$c$$ is the slope of the line in the plots above. We can think of this current density as right movers minus the left movers. Any state that is given can be thought of as a combination of right moving and left moving states, neither of which are eigenstates of the free particle Hamiltonian.

### Potential step

The next logical thing to think about, as in non-relativistic quantum mechanics, is to think about what occurs when the particle hits a potential step, as in fig. 5.

fig. 5. Reflection off a potential barrier

The approach is the same. We write down the wave functions for the $$V = 0$$ region (I), and the higher potential region (II).

The eigenstates are found on the solid lines above the asymptotes on the right hand movers side as sketched in fig. 6. The right and left moving designations are based on the phase velocity $$\PDi{k}{E}$$ (approaching $$\pm c$$ on the top-right and top-left quadrants respectively).

fig. 6. Right movers and left movers

For $$k > 0$$, an eigenstate for the incident wave is

\label{eqn:qmLecture9:261}
\Bpsi_{\textrm{inc}}(x) =
\begin{bmatrix}
\cos\theta_k \\
\sin\theta_k
\end{bmatrix}
e^{i k x},

For the reflected wave function, we pick a function on the left moving side of the positive energy branch.

\label{eqn:qmLecture9:260}
\Bpsi_{\textrm{ref}}(x) =
\begin{bmatrix}
? \\
?
\end{bmatrix}
e^{-i k x},

We’ll go through this in more detail next time.

## Question: Calculate the right going diagonalization

Prove (7).

To determine the relations for $$\theta_k$$ we have to solve

\label{eqn:qmLecture9:280}
\begin{bmatrix}
E_k & 0 \\
0 & -E_k
\end{bmatrix}
= R^{-1} H R.

Working with $$\Hbar = c = 1$$ temporarily, and $$C = \cos\theta_k, S = \sin\theta_k$$, that is

\label{eqn:qmLecture9:300}
\begin{aligned}
\begin{bmatrix}
E_k & 0 \\
0 & -E_k
\end{bmatrix}
&=
\begin{bmatrix}
C & S \\
-S & C
\end{bmatrix}
\begin{bmatrix}
k & m \\
m & -k
\end{bmatrix}
\begin{bmatrix}
C & -S \\
S & C
\end{bmatrix} \\
&=
\begin{bmatrix}
C & S \\
-S & C
\end{bmatrix}
\begin{bmatrix}
k C + m S & -k S + m C \\
m C – k S & -m S – k C
\end{bmatrix} \\
&=
\begin{bmatrix}
k C^2 + m S C + m C S – k S^2 & -k S C + m C^2 -m S^2 – k C S \\
-k C S – m S^2 + m C^2 – k S C & k S^2 – m C S -m S C – k C^2
\end{bmatrix} \\
&=
\begin{bmatrix}
k \cos(2 \theta_k) + m \sin(2 \theta_k) & m \cos(2 \theta_k) – k \sin(2 \theta_k) \\
m \cos(2 \theta_k) – k \sin(2 \theta_k) & -k \cos(2 \theta_k) – m \sin(2 \theta_k) \\
\end{bmatrix}.
\end{aligned}

This gives

\label{eqn:qmLecture9:320}
\begin{aligned}
E_k
\begin{bmatrix}
1 \\
0
\end{bmatrix}
&=
\begin{bmatrix}
k \cos(2 \theta_k) + m \sin(2 \theta_k) \\
m \cos(2 \theta_k) – k \sin(2 \theta_k) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
k & m \\
m & -k
\end{bmatrix}
\begin{bmatrix}
\cos(2 \theta_k) \\
\sin(2 \theta_k) \\
\end{bmatrix}.
\end{aligned}

Adding back in the $$\Hbar$$’s and $$c$$’s this is

\label{eqn:qmLecture9:340}
\begin{aligned}
\begin{bmatrix}
\cos(2 \theta_k) \\
\sin(2 \theta_k) \\
\end{bmatrix}
&=
\frac{E_k}{-(\Hbar k c)^2 -(m c^2)^2}
\begin{bmatrix}
– \Hbar k c & – m c^2 \\
– m c^2 & \Hbar k c
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\inv{E_k}
\begin{bmatrix}
\Hbar k c \\
m c^2
\end{bmatrix}.
\end{aligned}

## Question: Verify the Dirac current relationship.

Prove \ref{eqn:qmLecture9:240}.

The components of the Schrodinger equation are

\label{eqn:qmLecture9:360}
\begin{aligned}
-i \Hbar \PD{t}{\psi_1} &= -i \Hbar c \PD{x}{\psi_1} + m c^2 \psi_2 \\
-i \Hbar \PD{t}{\psi_2} &= m c^2 \psi_1 + i \Hbar c \PD{x}{\psi_2},
\end{aligned}

The conjugates of these are
\label{eqn:qmLecture9:380}
\begin{aligned}
i \Hbar \PD{t}{\psi_1^\conj} &= i \Hbar c \PD{x}{\psi_1^\conj} + m c^2 \psi_2^\conj \\
i \Hbar \PD{t}{\psi_2^\conj} &= m c^2 \psi_1^\conj – i \Hbar c \PD{x}{\psi_2^\conj}.
\end{aligned}

This gives
\label{eqn:qmLecture9:400}
\begin{aligned}
i \Hbar \PD{t}{\rho}
&=
\lr{ i \Hbar c \PD{x}{\psi_1^\conj} + m c^2 \psi_2^\conj } \psi_1 \\
&+ \psi_1^\conj \lr{ i \Hbar c \PD{x}{\psi_1} – m c^2 \psi_2 } \\
&+ \lr{ m c^2 \psi_1^\conj – i \Hbar c \PD{x}{\psi_2^\conj} } \psi_2 \\
&+ \psi_2^\conj \lr{ -m c^2 \psi_1 – i \Hbar c \PD{x}{\psi_2} }.
\end{aligned}

All the non-derivative terms cancel leaving

\label{eqn:qmLecture9:420}
\inv{c} \PD{t}{\rho}
=
\PD{x}{\psi_1^\conj} \psi_1
+ \psi_1^\conj \PD{x}{\psi_1}
– \PD{x}{\psi_2^\conj} \psi_2
– \psi_2^\conj \PD{x}{\psi_2}
=
\PD{x}{}
\lr{
\psi_1^\conj \psi_1 –
\psi_2^\conj \psi_2
}.