Spinor solutions with alternate $$\gamma^0$$ representation.

January 2, 2019 phy2403 No comments , ,

This follows an interesting derivation of the $$u, v$$ spinors [2], adding some details.

In class (QFT I) and [3] we used a non-diagonal $$\gamma^0$$ representation
\label{eqn:spinorSolutions:20}
\gamma^0 =
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix},

whereas in [2] a diagonal representation is used
\label{eqn:spinorSolutions:40}
\gamma^0 =
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}.

This representation makes it particularly simple to determine the form of the $$u, v$$ spinors. We seek solutions of the Dirac equation
\label{eqn:spinorSolutions:60}
\begin{aligned}
0 &= \lr{ i \gamma^\mu \partial_\mu – m } u(p) e^{-i p \cdot x} \\
0 &= \lr{ i \gamma^\mu \partial_\mu – m } v(p) e^{i p \cdot x},
\end{aligned}

or
\label{eqn:spinorSolutions:80}
\begin{aligned}
0 &= \lr{ \gamma^\mu p_\mu – m } u(p) e^{-i p \cdot x} \\
0 &= -\lr{ \gamma^\mu p_\mu + m } v(p) e^{i p \cdot x}.
\end{aligned}

In the rest frame where $$\gamma^\mu p_\mu = E \gamma^0$$, where $$E = m = \omega_\Bp$$, these take the particularly simple form
\label{eqn:spinorSolutions:100}
\begin{aligned}
0 &= \lr{ \gamma^0 – 1 } u(E, \Bzero) \\
0 &= \lr{ \gamma^0 + 1 } v(E, \Bzero).
\end{aligned}

This is a nice relation, as we can determine a portion of the structure of the rest frame $$u, v$$ that is independent of the Dirac matrix representation
\label{eqn:spinorSolutions:120}
\begin{aligned}
u(E, \Bzero) &= (\gamma^0 + 1) \psi \\
v(E, \Bzero) &= (\gamma^0 – 1) \psi
\end{aligned}

Similarly, and more generally, we have
\label{eqn:spinorSolutions:140}
\begin{aligned}
u(p) &= (\gamma^\mu p_\mu + m) \psi \\
v(p) &= (\gamma^\mu p_\mu – m) \psi
\end{aligned}

also independent of the representation of $$\gamma^\mu$$. Looking forward to non-matrix representations of the Dirac equation ([1]) note that we have not yet imposed a spinorial structure on the solution
\label{eqn:spinorSolutions:260}
\psi
=
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix},

where $$\phi, \chi$$ are two component matrices.

The particular choice of the diagonal representation \ref{eqn:spinorSolutions:40} for $$\gamma^0$$ makes it simple to determine additional structure for $$u, v$$. Consider the rest frame first, where
\label{eqn:spinorSolutions:160}
\begin{aligned}
\gamma^0 – 1 &=
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}

\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 \\
0 & 2
\end{bmatrix} \\
\gamma^0 + 1 &=
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
=
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix},
\end{aligned}

so we have
\label{eqn:spinorSolutions:280}
\begin{aligned}
u(E, \Bzero) &=
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix} \\
v(E, \Bzero) &=
\begin{bmatrix}
0 & 0 \\
0 & 2
\end{bmatrix}
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix}
\end{aligned}

Therefore a basis for the spinors $$u$$ (in the rest frame), is
\label{eqn:spinorSolutions:180}
u(E, \Bzero) \in \setlr{
\begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix}
},

and a basis for the rest frame spinors $$v$$ is
\label{eqn:spinorSolutions:200}
v(E, \Bzero) \in \setlr{
\begin{bmatrix}
0 \\
0 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix}
}.

Using the two spinor bases $$\zeta^a, \eta^a$$ notation from class, we can write these
\label{eqn:spinorSolutions:220}
\begin{aligned}
u^a(E, \Bzero) &=
\begin{bmatrix}
\zeta^a \\
0
\end{bmatrix},
v^a(E, \Bzero) &=
\begin{bmatrix}
0 \\
\eta^a \\
\end{bmatrix}.
\end{aligned}

For the non-rest frame solutions, [2] opts not to boost, as in [3], but to use the geometry of $$\gamma^\mu p_\mu \pm m$$. With their diagonal representation of $$\gamma^0$$ those are
\label{eqn:spinorSolutions:240}
\begin{aligned}
\gamma^\mu p_\mu – m
&=
p_0
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
p_k
\begin{bmatrix}
0 & \sigma^k \\
– \sigma^k & 0
\end{bmatrix}

m
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
E – m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E – m
\end{bmatrix} \\
\gamma^\mu p_\mu + m
&=
p_0
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
p_k
\begin{bmatrix}
0 & \sigma^k \\
– \sigma^k & 0
\end{bmatrix}
+
m
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
E + m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E + m
\end{bmatrix} \\
\end{aligned}

Let’s assume that the arbitrary momentum solutions \ref{eqn:spinorSolutions:140} are each proportional to the rest frame solutions
\label{eqn:spinorSolutions:300}
\begin{aligned}
u^a(p) &= (\gamma^\mu p_\mu + m) u^a(E, \Bzero) \\
v^a(p) &= (\gamma^\mu p_\mu – m) u^a(E, \Bzero).
\end{aligned}

Plugging in \ref{eqn:spinorSolutions:240} gives
\label{eqn:spinorSolutions:320}
\begin{aligned}
u^a(p) &=
\begin{bmatrix}
(E + m) \zeta^a \\
(\Bsigma \cdot \Bp ) \zeta^a
\end{bmatrix} \\
v^a(p) &=
\begin{bmatrix}
(\Bsigma \cdot \Bp) \eta^a \\
(E + m) \eta^a
\end{bmatrix},
\end{aligned}

where an overall sign on $$v^a(p)$$ has been dropped. Let’s check the assumption that the rest frame and general solutions are so simply related
\label{eqn:spinorSolutions:340}
\begin{aligned}
\lr{ \gamma^\mu p_\mu – m } u^a(p)
&=
\begin{bmatrix}
E – m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E – m
\end{bmatrix}
\begin{bmatrix}
(E + m) \zeta^a \\
(\Bsigma \cdot \Bp ) \zeta^a
\end{bmatrix} \\
&=
\begin{bmatrix}
(E^2 – m^2 – \Bp^2) \zeta^a \\
0
\end{bmatrix} \\
&= 0,
\end{aligned}

and
\label{eqn:spinorSolutions:360}
\begin{aligned}
\lr{ \gamma^\mu p_\mu + m } v^a(p)
&=
\begin{bmatrix}
E + m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E + m
\end{bmatrix}
\begin{bmatrix}
(\Bsigma \cdot \Bp ) \eta^a \\
(E + m) \eta^a \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 \\
\Bp^2 + m^2 – E^2
\end{bmatrix} \\
&= 0.
\end{aligned}

Everything works out nicely. The form of the solution for this representation of $$\gamma^0$$ is much simpler than the Chiral solution that we found in class. We end up with an explicit split of energy and spatial momentum components in the spinor solutions, instead of factors involving $$p \cdot \sigma$$ and $$p \cdot \overline{\sigma}$$, which are arguably nicer from a Lorentz invariance point of view.

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Claude Itzykson and Jean-Bernard Zuber. Quantum field theory. McGraw-Hill, 1980.

[3] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

PHY2403H Quantum Field Theory. Lecture 22: Dirac sea, charges, angular momentum, spin, U(1) symmetries, electrons and positrons. Taught by Prof. Erich Poppitz

This post is a synopsis of the material from the second last lecture of QFT I. I missed that class, but worked from notes kindly provided by Emily Tyhurst, and Stefan Divic, filling in enough details that it made sense to me.

[Click here for an unabrided PDF of my full notes on this day’s lecture material.]

Topics covered include

• The Hamiltonian action on single particle states showed that the Hamiltonian was an energy eigenoperator
\label{eqn:qftLecture22:140}
H \ket{\Bp, r}
=
\omega_\Bp \ket{\Bp, r}.
• The conserved Noether current and charge for spatial translations, the momentum operator, was found to be
\label{eqn:momentumDirac:260}
\BP =
\int d^3 x

which could be written in creation and anhillation operator form as
\label{eqn:momentumDirac:261}
\BP = \sum_{s = 1}^2
\int \frac{d^3 q}{(2\pi)^3} \Bp \lr{
a_\Bp^{s\dagger}
a_\Bp^{s}
+
b_\Bp^{s\dagger}
b_\Bp^{s}
}.

Single particle states were found to be the eigenvectors of this operator, with momentum eigenvalues
\label{eqn:momentumDirac:262}
\BP a_\Bq^{s\dagger} \ket{0} = \Bq (a_\Bq^{s\dagger} \ket{0}).
• The conserved Noether current and charge for a rotation was found. That charge is
\label{eqn:qftLecture22:920}
\BJ = \int d^3 x \Psi^\dagger(x) \lr{ \underbrace{\Bx \cross (-i \spacegrad)}_{\text{orbital angular momentum}} + \inv{2} \underbrace{\mathbf{1} \otimes \Bsigma}_{\text{spin angular momentum}} } \Psi,

where
\label{eqn:qftLecture22:260}
\mathbf{1} \otimes \Bsigma =
\begin{bmatrix}
\Bsigma & 0 \\
0 & \Bsigma
\end{bmatrix},

which has distinct orbital and spin angular momentum components. Unlike NRQM, we see both types of angular momentum as components of a single operator. It is argued in [3] that for a particle at rest the single particle state is an eigenvector of this operator, with eigenvalues $$\pm 1/2$$ — the Fermion spin eigenvalues!
• We examined two $$U(1)$$ global symmetries. The Noether charge for the “vector” $$U(1)$$ symmetry is
\label{eqn:qftLecture22:380}
Q
=
\int \frac{d^3 q}{(2\pi)^3} \sum_{s = 1}^2
\lr{
a_\Bp^{s \dagger} a_\Bp^s

b_\Bp^{s \dagger}
b_\Bp^s
},

This charge operator characterizes the $$a, b$$ operators. $$a$$ particles have charge $$+1$$, and $$b$$ particles have charge $$-1$$, or vice-versa depending on convention. We call $$a$$ the operator for the electron, and $$b$$ the operator for the positron.
• CPT (Charge-Parity-TimeReversal) symmetries were also mentioned, but not covered in class. We were pointed to [2], [3], [4] to start studying that topic.

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Dr. Michael Luke. Quantum Field Theory., 2011. URL https://www.physics.utoronto.ca/~luke/PHY2403F/References_files/lecturenotes.pdf. [Online; accessed 05-Dec-2018].

[3] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

[4] Dr. David Tong. Quantum Field Theory. URL http://www.damtp.cam.ac.uk/user/tong/qft.html.

Explicit form of the square root of p . sigma.

December 10, 2018 phy2403 No comments , , , ,

With the help of Mathematica, a fairly compact form was found for the root of $$p \cdot \sigma$$
\label{eqn:DiracUVmatricesExplicit:121}
\sqrt{ p \cdot \sigma }
=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 } & – p^1 + i p^2 \\
– p^1 – i p^2 & \omega_\Bp+ p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 }
\end{bmatrix}.

A bit of examination shows that we can do much better. The leading scalar term can be simplified by squaring it
\label{eqn:squarerootpsigma:140}
\begin{aligned}
\lr{ \sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} } }^2
&=
\omega_\Bp- \Norm{\Bp} + \omega_\Bp+ \Norm{\Bp} + 2 \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
&=
2 \omega_\Bp + 2 m,
\end{aligned}

where the on-shell value of the energy $$\omega_\Bp^2 = m^2 + \Bp^2$$ has been inserted. Using that again in the matrix, we have
\label{eqn:squarerootpsigma:160}
\begin{aligned}
\sqrt{ p \cdot \sigma }
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\begin{bmatrix}
\omega_\Bp- p^3 + m & – p^1 + i p^2 \\
– p^1 – i p^2 & \omega_\Bp+ p^3 + m
\end{bmatrix} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0
-p^1 \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
-p^2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
-p^3 \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}
} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0
-p^1 \sigma^1
-p^2 \sigma^2
-p^3 \sigma^3
} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0 – \Bsigma \cdot \Bp
}.
\end{aligned}

We’ve now found a nice algebraic form for these matrix roots
\label{eqn:squarerootpsigma:180}
\boxed{
\begin{aligned}
\sqrt{p \cdot \sigma} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \sigma } \\
\sqrt{p \cdot \overline{\sigma}} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \overline{\sigma}}.
\end{aligned}}

As a check, let’s square one of these explicitly
\label{eqn:squarerootpsigma:101}
\begin{aligned}
\lr{ \sqrt{p \cdot \sigma} }^2
&= \inv{2 \omega_\Bp + 2 m }
\lr{ m^2 + (p \cdot \sigma)^2 + 2 m (p \cdot \sigma) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ m^2 + (\omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + \Bp^2) + 2 m (p \cdot \sigma) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ 2 \omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + 2 m (\omega_\Bp – \Bsigma \cdot \Bp) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ 2 \omega_\Bp \lr{ \omega_\Bp + m } – (2 \omega_\Bp + 2 m) \Bsigma \cdot \Bp } \\
&=
\omega_\Bp – \Bsigma \cdot \Bp \\
&=
p \cdot \sigma,
\end{aligned}

which validates the result.

Momentum operator for the Dirac field?

In the borrowed notes I have for last Monday’s lecture (which I missed) I see the momentum operator defined by
\label{eqn:momentumDirac:20}
\BP = \sum_{s = 1}^2
\int \frac{d^3 q}{(2\pi)^3} \Bp \lr{
a_\Bp^{s\dagger}
a_\Bp^{s}
+
b_\Bp^{s\dagger}
b_\Bp^{s}
}.

There’s a “use Noether’s theorem” comment associated with this. For the scalar field, using Noether’s theorem, we identified the conserved charge of a spacetime translation as the momentum operator
\label{eqn:momentumDirac:40}
P^i = \int d^3 x T^{0i} = – \int d^3 x \pi(x) \spacegrad \phi(x),

and if we plugged in the creation and anhillation operator representation of $$\pi, \phi$$, out comes
\label{eqn:momentumDirac:60}
\BP =
\inv{2} \int \frac{d^3 q}{(2\pi)^3} \Bp \lr{ a_\Bp^\dagger a_\Bp + a_\Bp a_\Bp^\dagger},

(plus $$e^{\pm 2 i \omega_\Bp t}$$ terms that we can argue away.)

It wasn’t clear to me how this worked with the Dirac field, but it turns out that this does follow systematically as expected. For a spacetime translation
\label{eqn:momentumDirac:80}
x^\mu \rightarrow x^\mu + a^\mu,

we find
\label{eqn:momentumDirac:100}
\delta \Psi = -a^\mu \partial_\mu \Psi,

so for the Dirac Lagrangian, we have
\label{eqn:momentumDirac:120}
\begin{aligned}
\delta \LL
&= \delta \lr{ \overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \Psi } \\
&=
(\delta \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \delta \Psi \\
&=
(-a^\sigma \partial_\sigma \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } (-a^\sigma \partial_\sigma \Psi ) \\
&=
-a^\sigma \partial_\sigma \LL \\
&=
\partial_\sigma (-a^\sigma \LL),
\end{aligned}

i.e. $$J^\mu = -a^\mu \LL$$.
To plugging this into the Noether current calculating machine, we have
\label{eqn:momentumDirac:160}
\begin{aligned}
\PD{(\partial_\mu \Psi)}{\LL}
&=
\PD{(\partial_\mu \Psi)}{} \lr{ \overline{\Psi} i \gamma^\sigma \partial_\sigma \Psi – m \overline{\Psi} \Psi } \\
&=
\overline{\Psi} i \gamma^\mu,
\end{aligned}

and
\label{eqn:momentumDirac:180}
\PD{(\partial_\mu \overline{\Psi})}{\LL} = 0,

so
\label{eqn:momentumDirac:140}
\begin{aligned}
j^\mu
&=
(\delta \overline{\Psi}) \PD{(\partial_\mu \overline{\Psi})}{\LL}
+
\PD{(\partial_\mu \Psi)}{\LL} (\delta \Psi)
– a^\mu \LL \\
&=
\overline{\Psi} i \gamma^\mu (-a^\sigma \partial_\sigma \Psi)
– a^\sigma {\delta^{\mu}}_{\sigma} \LL \\
&=
– a^\sigma
\lr{
\overline{\Psi} i \gamma^\mu \partial_\sigma \Psi
+ {\delta^{\mu}}_{\sigma} \LL
} \\
&=
-a_\nu
\lr{
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL
}.
\end{aligned}

We can now define an energy-momentum tensor
\label{eqn:momentumDirac:200}
T^{\mu\nu}
=
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL.

A couple things are of notable in this tensor. One is that it is not symmetric, and there’s doesn’t appear to be any hope
of making it so. For example, the space+time components are way different
\label{eqn:momentumDirac:220}
\begin{aligned}
T^{0k} &= \overline{\Psi} i \gamma^0 \partial^k \Psi \\
T^{k0} &= \overline{\Psi} i \gamma^k \partial^0 \Psi,
\end{aligned}

so if we want a momentum like creature, we have to use $$T^{0k}$$, not $$T^{k0}$$. The charge associated with that current is
\label{eqn:momentumDirac:240}
\begin{aligned}
Q^k
&=
\int d^3 x
\overline{\Psi} i \gamma^0 \partial^k \Psi \\
&=
\int d^3 x
\Psi^\dagger (-i \partial_k) \Psi,
\end{aligned}

or translating from component to vector form
\label{eqn:momentumDirac:260}
\BP =
\int d^3 x

which is the how the momentum operator is first stated in [2]. Here the vector notation doesn’t have any specific representation, but it is interesting to observe how this is directly related to the massless Dirac Lagrangian

\label{eqn:momentumDirac:280}
\begin{aligned}
\LL(m = 0)
&=
\overline{\Psi} i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i (\partial_0 + \gamma_0 \gamma^k \partial_k) \Psi \\
&=
\Psi^\dagger i (\partial_0 – \gamma_0 \gamma_k \partial_k ) \Psi,
\end{aligned}

but since $$\gamma_0 \gamma_k$$ is a $$4 \times 4$$ representation of the Pauli matrix $$\sigma_k$$ Lagrangian itself breaks down into
\label{eqn:momentumDirac:300}
\LL(m = 0)
=
\Psi^\dagger i \partial_0 \Psi
+
\Bsigma \cdot \lr{ \Psi^\dagger (-i\spacegrad) \Psi },

components, and lo and behold, out pops the momentum operator density! There is ambiguity as to what order of products $$\gamma_0 \gamma_k$$, or $$\gamma_k \gamma_0$$ to pick to represent the Pauli basis ([1] uses $$\gamma_k \gamma_0$$), but we also have sign ambiguity in assembling a Noether charge from the conserved current, so I don’t think that matters. Some part of this should be expected this since the Dirac equation in momentum space is just $$\gamma \cdot p – m = 0$$, so there is an intimate connection with the operator portion and momentum.

The last detail to fill in is going from \ref{eqn:momentumDirac:260} to \ref{eqn:momentumDirac:20} using the $$a, b$$ representation of the field. That’s an algebraically messy looking job that I don’t feel like trying at the moment.

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

Alternate Dirac equation representation

November 27, 2015 phy1520 No comments , ,

Given an alternate representation of the Dirac equation

\label{eqn:diracAlternate:20}
H =
\begin{bmatrix}
m c^2 + V_0 & c \hat{p} \\
c \hat{p} & – m c^2 + V_0
\end{bmatrix},

calculate the constant momentum solutions, the Heisenberg velocity operator $$\hat{v}$$, and find the form of the probability density current.

Plane wave solutions

The action of the Hamiltonian on

\label{eqn:diracAlternate:40}
\psi =
e^{i k x – i E t/\Hbar}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}

is
\label{eqn:diracAlternate:60}
\begin{aligned}
H \psi
&=
\begin{bmatrix}
m c^2 + V_0 & c (-i \Hbar) i k \\
c (-i \Hbar) i k & – m c^2 + V_0
\end{bmatrix}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
e^{i k x – i E t/\Hbar} \\
&=
\begin{bmatrix}
m c^2 + V_0 & c \Hbar k \\
c \Hbar k & – m c^2 + V_0
\end{bmatrix}
\psi.
\end{aligned}

Writing

\label{eqn:diracAlternate:80}
H_k
=
\begin{bmatrix}
m c^2 + V_0 & c \Hbar k \\
c \Hbar k & – m c^2 + V_0
\end{bmatrix}

the characteristic equation is

\label{eqn:diracAlternate:100}
0
=
(m c^2 + V_0 – \lambda)
(-m c^2 + V_0 – \lambda)
– (c \Hbar k)^2
=
\lr{ (\lambda – V_0)^2 – (m c^2)^2 } – (c \Hbar k)^2,

so

\label{eqn:diracAlternate:120}
\lambda = V_0 \pm \epsilon,

where
\label{eqn:diracAlternate:140}
\epsilon^2 = (m c^2)^2 + (c \Hbar k)^2.

We’ve got

\label{eqn:diracAlternate:160}
\begin{aligned}
H – ( V_0 + \epsilon )
&=
\begin{bmatrix}
m c^2 – \epsilon & c \Hbar k \\
c \Hbar k & – m c^2 – \epsilon
\end{bmatrix} \\
H – ( V_0 – \epsilon )
&=
\begin{bmatrix}
m c^2 + \epsilon & c \Hbar k \\
c \Hbar k & – m c^2 + \epsilon
\end{bmatrix},
\end{aligned}

so the eigenkets are

\label{eqn:diracAlternate:180}
\begin{aligned}
\ket{V_0+\epsilon}
&\propto
\begin{bmatrix}
-c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&\propto
\begin{bmatrix}
-c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix}.
\end{aligned}

Up to an arbitrary phase for each, these are

\label{eqn:diracAlternate:200}
\begin{aligned}
\ket{V_0 + \epsilon}
&=
\inv{\sqrt{ 2 \epsilon ( \epsilon – m c^2) }}
\begin{bmatrix}
c \Hbar k \\
\epsilon -m c^2
\end{bmatrix} \\
\ket{V_0 – \epsilon}
&=
\inv{\sqrt{ 2 \epsilon ( \epsilon + m c^2) }}
\begin{bmatrix}
-c \Hbar k \\
\epsilon + m c^2
\end{bmatrix} \\
\end{aligned}

We can now write

\label{eqn:diracAlternate:220}
H_k =
E
\begin{bmatrix}
V_0 + \epsilon & 0 \\
0 & V_0 – \epsilon
\end{bmatrix}
E^{-1},

where
\label{eqn:diracAlternate:240}
\begin{aligned}
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
\sqrt{ \epsilon – m c^2 } & \sqrt{ \epsilon + m c^2 }
\end{bmatrix}, \qquad k > 0 \\
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
-\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
-\sqrt{ \epsilon – m c^2 } & \sqrt{ \epsilon + m c^2 }
\end{bmatrix}, \qquad k < 0. \end{aligned} Here the signs have been adjusted to ensure the transformation matrix has a unit determinant. Observe that there's redundancy in this matrix since $$\ifrac{c \Hbar \Abs{k}}{ \sqrt{ \epsilon - m c^2 } } = \sqrt{ \epsilon + m c^2 }$$, and $$\ifrac{c \Hbar \Abs{k}}{ \sqrt{ \epsilon + m c^2 } } = \sqrt{ \epsilon - m c^2 }$$, which allows the transformation matrix to be written in the form of a rotation matrix \label{eqn:diracAlternate:260} \begin{aligned} E &= \inv{\sqrt{2 \epsilon} } \begin{bmatrix} \frac{c \Hbar k}{ \sqrt{ \epsilon - m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\ \frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } & \frac{c \Hbar k}{ \sqrt{ \epsilon - m c^2 } } \end{bmatrix}, \qquad k > 0 \\
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
-\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } }
\end{bmatrix}, \qquad k < 0 \\ \end{aligned} With \label{eqn:diracAlternate:280} \begin{aligned} \cos\theta &= \frac{c \Hbar \Abs{k}}{ \sqrt{ 2 \epsilon( \epsilon - m c^2) } } = \frac{\sqrt{ \epsilon + m c^2} }{ \sqrt{ 2 \epsilon}}\\ \sin\theta &= \frac{c \Hbar k}{ \sqrt{ 2 \epsilon( \epsilon + m c^2) } } = \frac{\textrm{sgn}(k) \sqrt{ \epsilon - m c^2}}{ \sqrt{ 2 \epsilon } }, \end{aligned} the transformation matrix (and eigenkets) is $$\label{eqn:diracAlternate:300} \boxed{ E = \begin{bmatrix} \ket{V_0 + \epsilon} & \ket{V_0 - \epsilon} \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}. }$$ Observe that \ref{eqn:diracAlternate:280} can be simplified by using double angle formulas \label{eqn:diracAlternate:320} \begin{aligned} \cos(2 \theta) &= \frac{\lr{ \epsilon + m c^2} }{ 2 \epsilon } - \frac{\lr{ \epsilon - m c^2}}{ 2 \epsilon } \\ &= \frac{1}{ 2 \epsilon } \lr{ \epsilon + m c^2 - \epsilon + m c^2 } \\ &= \frac{m c^2 }{ \epsilon }, \end{aligned} and $$\label{eqn:diracAlternate:340} \sin(2\theta) = 2 \frac{1}{2 \epsilon} \textrm{sgn}(k ) \sqrt{ \epsilon^2 - (m c^2)^2 } = \frac{\Hbar k c}{\epsilon}.$$ This allows all the $$\theta$$ dependence on $$\Hbar k c$$ and $$m c^2$$ to be expressed as a ratio of momenta $$\label{eqn:diracAlternate:360} \boxed{ \tan(2\theta) = \frac{\Hbar k}{m c}. }$$

Hyperbolic solutions

For a wave function of the form

\label{eqn:diracAlternate:380}
\psi =
e^{k x – i E t/\Hbar}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix},

some of the work above can be recycled if we substitute $$k \rightarrow -i k$$, which yields unnormalized eigenfunctions

\label{eqn:diracAlternate:400}
\begin{aligned}
\ket{V_0+\epsilon}
&\propto
\begin{bmatrix}
i c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&\propto
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix},
\end{aligned}

where

\label{eqn:diracAlternate:420}
\epsilon^2 = (m c^2)^2 – (c \Hbar k)^2.

The squared magnitude of these wavefunctions are

\label{eqn:diracAlternate:440}
\begin{aligned}
(c \Hbar k)^2 + (m c^2 \mp \epsilon)^2
&=
(c \Hbar k)^2 + (m c^2)^2 + \epsilon^2 \mp 2 m c^2 \epsilon \\
&=
(c \Hbar k)^2 + (m c^2)^2 + (m c^2)^2 \mp (c \Hbar k)^2 – 2 m c^2 \epsilon \\
&= 2 (m c^2)^2 \mp 2 m c^2 \epsilon \\
&= 2 m c^2 ( m c^2 \mp \epsilon ),
\end{aligned}

so, up to a constant phase for each, the normalized kets are

\label{eqn:diracAlternate:460}
\begin{aligned}
\ket{V_0+\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 – \epsilon ) }}
\begin{bmatrix}
i c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 + \epsilon ) }}
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix},
\end{aligned}

After the $$k \rightarrow -i k$$ substitution, $$H_k$$ is not Hermitian, so these kets aren’t expected to be orthonormal, which is readily verified

\label{eqn:diracAlternate:480}
\begin{aligned}
\braket{V_0+\epsilon}{V_0-\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 – \epsilon ) }}
\inv{\sqrt{ 2 m c^2 ( m c^2 + \epsilon ) }}
\begin{bmatrix}
-i c \Hbar k &
m c^2 – \epsilon
\end{bmatrix}
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix} \\
&=
\frac{ 2 ( c \Hbar k )^2 }{2 m c^2 \sqrt{(\Hbar k c)^2} } \\
&=
\textrm{sgn}(k)
\frac{
\Hbar k }{m c } .
\end{aligned}

Heisenberg velocity operator

\label{eqn:diracAlternate:500}
\begin{aligned}
\hat{v}
&= \inv{i \Hbar} \antisymmetric{ \hat{x} }{ H} \\
&= \inv{i \Hbar} \antisymmetric{ \hat{x} }{ m c^2 \sigma_z + V_0 + c \hat{p} \sigma_x } \\
&= \frac{c \sigma_x}{i \Hbar} \antisymmetric{ \hat{x} }{ \hat{p} } \\
&= c \sigma_x.
\end{aligned}

Probability current

Acting against a completely general wavefunction the Hamiltonian action $$H \psi$$ is

\label{eqn:diracAlternate:520}
\begin{aligned}
i \Hbar \PD{t}{\psi}
&= m c^2 \sigma_z \psi + V_0 \psi + c \hat{p} \sigma_x \psi \\
&= m c^2 \sigma_z \psi + V_0 \psi -i \Hbar c \sigma_x \PD{x}{\psi}.
\end{aligned}

Conversely, the conjugate $$(H \psi)^\dagger$$ is

\label{eqn:diracAlternate:540}
-i \Hbar \PD{t}{\psi^\dagger}
= m c^2 \psi^\dagger \sigma_z + V_0 \psi^\dagger +i \Hbar c \PD{x}{\psi^\dagger} \sigma_x.

These give

\label{eqn:diracAlternate:560}
\begin{aligned}
i \Hbar \psi^\dagger \PD{t}{\psi}
&=
m c^2 \psi^\dagger \sigma_z \psi + V_0 \psi^\dagger \psi -i \Hbar c \psi^\dagger \sigma_x \PD{x}{\psi} \\
-i \Hbar \PD{t}{\psi^\dagger} \psi
&= m c^2 \psi^\dagger \sigma_z \psi + V_0 \psi^\dagger \psi +i \Hbar c \PD{x}{\psi^\dagger} \sigma_x \psi.
\end{aligned}

Taking differences
\label{eqn:diracAlternate:580}
\psi^\dagger \PD{t}{\psi} + \PD{t}{\psi^\dagger} \psi
=
– c \psi^\dagger \sigma_x \PD{x}{\psi} – c \PD{x}{\psi^\dagger} \sigma_x \psi,

or

\label{eqn:diracAlternate:600}
0
=
\PD{t}{}
\lr{
\psi^\dagger \psi
}
+
\PD{x}{}
\lr{
c \psi^\dagger \sigma_x \psi
}.

The probability current still has the usual form $$\rho = \psi^\dagger \psi = \psi_1^\conj \psi_1 + \psi_2^\conj \psi_2$$, but the probability current with this representation of the Dirac Hamiltonian is

\label{eqn:diracAlternate:620}
\begin{aligned}
j
&= c \psi^\dagger \sigma_x \psi \\
&= c
\begin{bmatrix}
\psi_1^\conj &
\psi_2^\conj
\end{bmatrix}
\begin{bmatrix}
\psi_2 \\
\psi_1
\end{bmatrix} \\
&= c \lr{ \psi_1^\conj \psi_2 + \psi_2^\conj \psi_1 }.
\end{aligned}

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.

Second update of aggregate notes for phy1520, Graduate Quantum Mechanics

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.

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
}.

PHY1520H Graduate Quantum Mechanics. Lecture 8: Dirac equation in 1D. 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.

Schrodinger Derivation

Recall that a “derivation” of the Schrodinger equation can be associated with the following equivalences

\label{eqn:qmLecture8:300}
E \leftrightarrow \Hbar \omega \leftrightarrow i \Hbar \PD{t}{}

\label{eqn:qmLecture8:320}
p \leftrightarrow \Hbar k \leftrightarrow -i \Hbar \PD{t}{}

so that the classical energy relationship

\label{eqn:qmLecture8:20}
E = \frac{p^2}{2m}

takes the form

\label{eqn:qmLecture8:40}
i \Hbar \PD{t}{} = -\frac{\Hbar^2}{2m}.

How do we do this in a relativistic context where the energy momentum relationship is

\label{eqn:qmLecture8:60}
E = \sqrt{ p^2 c^2 + m^2 c^4 } \approx m c^2 + \frac{p^2}{2m} + \cdots

where $$m$$ is the rest mass and $$c$$ is the speed of light.

Attempt I

\label{eqn:qmLecture8:80}
E = m c^2 + \frac{p^2}{2m} + (…) p^4 + (…) p^6 + \cdots

First order in time, but infinite order in space $$\partial/\partial x$$. Useless.

Attempt II

\label{eqn:qmLecture8:100}
E^2 = p^2 c^2 + m^2 c^4.

This gives

\label{eqn:qmLecture8:120}
-\Hbar^2 \PDSq{t}{\psi} = – \Hbar^2 c^2 \PDSq{x}{\psi} + m^2 c^4 \psi.

This is the Klein-Gordon equation, which is second order in time.

Attempt III

Suppose that we have the matrix

\label{eqn:qmLecture8:140}
\begin{bmatrix}
p c & m c^2 \\
m c^2 & – p c
\end{bmatrix},

or

\label{eqn:qmLecture8:160}
\begin{bmatrix}
m c^2 & i p c \\
-i p c & – m c^2
\end{bmatrix},

These both happen to have eigenvalues $$\lambda_{\pm} = \pm \sqrt{p^2 c^2}$$. For those familiar with the Dirac matrices, this amounts to a choice for different representations of the gamma matrices.

Working with \ref{eqn:qmLecture8:140}, which has some nicer features than other possible representations, we seek a state

\label{eqn:qmLecture8:180}
\Bpsi =
\begin{bmatrix}
\psi_1(x, t) \\
\psi_2(x, t) \\
\end{bmatrix},

where we aim to write down an equation for this composite state.

\label{eqn:qmLecture8:200}
i \Hbar \PD{t}{\Bpsi} = \BH \Bpsi

Assuming the matrix is the Hamiltonian, multiplying that with the composite state gives

\label{eqn:qmLecture8:220}
\begin{aligned}
\begin{bmatrix}
i \Hbar \PD{t}{\psi_1} \\
i \Hbar \PD{t}{\psi_1}
\end{bmatrix}
&=
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix}
\begin{bmatrix}
\psi_1(x, t) \\
\psi_2(x, t) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\hat{p} c \psi_1 + m c^2 \psi_2 \\
m c^2 \psi_1 – \hat{p} c \psi_2
\end{bmatrix}.
\end{aligned}

What happens when we square this?

\label{eqn:qmLecture8:240}
\begin{aligned}
\lr{ i \Hbar \PD{t}{} }^2 \Bpsi
&= \BH \BH \Bpsi \\
&=
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix}
\begin{bmatrix}
\hat{p} c & m c^2 \\
m c^2 & – \hat{p} c
\end{bmatrix}
\Bpsi \\
&=
\begin{bmatrix}
\hat{p}^2 c^2 + m^2 c^4 & 0 \\
0 & \hat{p}^2 c^2 + m^2 c^4 \\
\end{bmatrix}.
\end{aligned}

That is
\label{eqn:qmLecture8:260}
– \Hbar^2 \PDSq{t}{} \Bpsi
=
\lr{ \hat{p}^2 c^2 + m^2 c^4 } \mathbf{1} \Bpsi,

or more exactly

\label{eqn:qmLecture8:280}
– \Hbar^2 \PDSq{t}{} \psi_{1,2}
=
\lr{ \hat{p}^2 c^2 + m^2 c^4 } \psi_{1,2}.

This recovers the Klein Gordon equation for each of the wave functions $$\psi_1, \psi_2$$.

Instead of squaring the operators, lets try to solve the first order equation. To do so we’ll want to diagonalize $$\BH$$.

Before doing that, let’s write out the Hamiltonian in an alternate but useful form

\label{eqn:qmLecture8:340}
\BH
=
\hat{p} c
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
m c^2
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
= \hat{p} c \hat{\sigma}_z + m c^2 \hat{\sigma}_x.

We have two types of operators in the mix here. We have matrix operators that act on the wave function matrices, as well as derivative operators that act on the components of those matrices.

We have

\label{eqn:qmLecture8:360}
\hat{\sigma}_z
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}
=
\begin{bmatrix}
\psi_1 \\
-\psi_2 \\
\end{bmatrix},

and

\label{eqn:qmLecture8:380}
\hat{\sigma}_x
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}
=
\begin{bmatrix}
\psi_2 \\
\psi_1 \\
\end{bmatrix}.

Because the derivative actions of $$\hat{p}$$ and the matrix operators are independent, we see that these operators commute. For example

\label{eqn:qmLecture8:400}
\hat{\sigma}_z \hat{p}
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}
=
\hat{\sigma}_z
\begin{bmatrix}
-i \Hbar \PD{x}{\psi_1} \\
-i \Hbar \PD{x}{\psi_2} \\
\end{bmatrix}
=
\begin{bmatrix}
-i \Hbar \PD{x}{\psi_1} \\
i \Hbar \PD{x}{\psi_2} \\
\end{bmatrix}
=
\hat{p}
\hat{\sigma}_z
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\end{bmatrix}.

Diagonalizing it

Suppose the wave function matrix has the structure

\label{eqn:qmLecture8:420}
\Bpsi =
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}
e^{i k x}.

We’ll plug this into the Schrodinger equation and see what we get.