Chapter 5, The Dirac Equation Video Solutions, Quarks and leptons: introductory course in modern particle physics

Problem 1

Prove that the $\alpha_i$ and $\beta$ are hermitian, traceless matrices of even dimensionality, with eigenvalues \pm 1 .

The lowest dimensionality matrices satisfying all these requirements are $4 \times 4$. The choice of the four matrices $(\alpha, \beta)$ is not unique. The Dirac-Pauli representation is most frequently used:
$$
\alpha=\left(\begin{array}{ll}
0 & \boldsymbol{\sigma} \\
\boldsymbol{\sigma} & 0
\end{array}\right), \quad \beta=\left(\begin{array}{rr}
I & 0 \\
0 & -I
\end{array}\right)
$$
where $I$ denotes the unit $2 \times 2$ matrix (which is frequently written as 1) and where $\boldsymbol{\sigma}$ are the Pauli matrices:
$$
\sigma_1=\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right), \quad \sigma_2=\left(\begin{array}{rr}
0 & -i \\
i & 0
\end{array}\right), \quad \sigma_3=\left(\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right) .
$$

Another possible representation, the Weyl representation, is
$$
\alpha=\left(\begin{array}{rr}
-\sigma & 0 \\
0 & \sigma
\end{array}\right), \quad \beta=\left(\begin{array}{ll}
0 & I \\
I & 0
\end{array}\right)
$$

Most of the results are independent of the choice of representation. Certainly, all the physics depends only on the properties listed in (5.3). In fact, not until we exhibit explicit solutions of the Dirac equation in Section 5.3 will we use a particular representation. Unless stated otherwise, we shall always choose the Dirac-Pauli representation, (5.4).

A four-component column vector $\psi$ which satisfies the Dirac equation (5.1) is called a Dirac spinor. We might have anticipated two independent solutions (particles and antiparticles), but instead we have four!

Maybe the surprise should not have been total. We know at least one other example where a field with more components appears when linearizing the equation. The covariant Maxwell equations $\square^2 A_\mu=0$ are second-order but can be written in a linear form $\partial_\mu F^{\mu \nu}=0$ by introducing the field strength $F_{\mu \nu}$, which has more components than $A_{\mu^*}$

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Problem 2

Operate on (5.7) with $\gamma^\nu \partial_v$ and show that each of the four components $\psi_i$ satisfies the Klein-Gordon equation
$$
\left(\square^2+m^2\right) \psi_i=0 .
$$

For a free particle, we can therefore seek four-momentum eigensolutions of Dirac's equation of the form
$$
\psi=u(\mathbf{p}) e^{-i p \cdot x},
$$
where $u$ is a four-component spinor independent of $x$. Substituting in (5.7), we have
$$
\left(\gamma^\mu p_\mu-m\right) u(\mathbf{p})=0
$$
or, using the abbreviated notation $\mathcal{A} \equiv \gamma^\mu A_\mu$ for any four-vector $A_\mu$,
$$
(p-m) u=0 \text {. }
$$

Since we are seeking energy eigenvectors, it is easier to use the original form, (5.1),
$$
H u=(\boldsymbol{\alpha} \cdot \mathbf{p}+\beta m) u=E u .
$$

There are four independent solutions of this equation, two with $E>0$ and two with $E<0$. This is particularly easy to see in the Dirac-Pauli representation of $\alpha$ and $\beta$. First, take the particle at rest, $\mathbf{p}=0$. Using (5.4), we have
$$
H u=\beta m u=\left(\begin{array}{cc}
m I & 0 \\
0 & -m I
\end{array}\right) u
$$
with eigenvalues $E=m, m,-m,-m$, and eigenvectors
$$
\left(\begin{array}{l}
1 \\
0 \\
0 \\
0
\end{array}\right), \quad\left(\begin{array}{l}
0 \\
1 \\
0 \\
0
\end{array}\right), \quad\left(\begin{array}{l}
0 \\
0 \\
1 \\
0
\end{array}\right), \quad\left(\begin{array}{l}
0 \\
0 \\
0 \\
1
\end{array}\right) .
$$

As we have just mentioned, the electron, charge $-e$, is regarded as the particle. The first two solutions therefore describe an $E>0$ electron. The $E<0$ particle solutions are to be interpreted, as before, as describing an $E>0$ antiparticle (positron) (see Section 3.6).
For $\mathbf{p} \neq 0,(5.21)$ becomes, using (5.4),
$$
H u=\left(\begin{array}{cc}
m & \boldsymbol{\sigma} \cdot \mathbf{p} \\
\boldsymbol{\sigma} \cdot \mathbf{p} & -m
\end{array}\right)\left(\begin{array}{l}
u_A \\
u_B
\end{array}\right)=E\left(\begin{array}{l}
u_A \\
u_B
\end{array}\right),
$$
where $u$ has been divided into two two-component spinors, $u_A$ and $u_B$. This EXERCISE 5.2 Operate on (5.7) with $\gamma^\nu \partial_v$ and show that each of the four components $\psi_i$ satisfies the Klein-Gordon equation
$$
\left(\square^2+m^2\right) \psi_i=0 .
$$

Since we are seeking energy eigenvectors, it is easier to use the original form, (5.1),
$$
H u=(\boldsymbol{\alpha} \cdot \mathbf{p}+\beta m) u=E u .
$$

As we have just mentioned, the electron, charge $-e$, is regarded as the particle. The first two solutions therefore describe an $E>0$ electron. The $E<0$ particle solutions are to be interpreted, as before, as describing an $E>0$ antiparticle (positron) (see Section 3.6).
For $\mathbf{p} \neq 0,(5.21)$ becomes, using (5.4),
$$
H u=\left(\begin{array}{cc}
m & \boldsymbol{\sigma} \cdot \mathbf{p} \\
\boldsymbol{\sigma} \cdot \mathbf{p} & -m
\end{array}\right)\left(\begin{array}{l}
u_A \\
u_B
\end{array}\right)=E\left(\begin{array}{l}
u_A \\
u_B
\end{array}\right),
$$
where $u$ has been divided into two two-component spinors, $u_A$ and $u_B$. This call this quantum number the helicity of the state. The possible eigenvalues $\lambda$ of the helicity operator $\frac{1}{2} \boldsymbol{\sigma} \cdot \hat{\mathbf{p}}$ are
$$
\lambda=\left\{\begin{array}{l}
+\frac{1}{2} \text { positive helicity, } \\
-\frac{1}{2} \text { negative helicity. }
\end{array}\right.
$$

We see that no other component of $\boldsymbol{\sigma}$ has eigenvalues which are good quantum numbers.

With the above choice (5.25), of the spinors $\chi^{(s)}$, it is appropriate to choose $\mathbf{p}$ along the $z$ axis, $\mathbf{p}=(0,0, p)$. Then
$$
\frac{1}{2} \mathbf{\sigma} \cdot \hat{\mathbf{p}} \chi^{(s)}=\frac{1}{2} \sigma_3 \chi^{(s)}=\lambda \chi^{(s)}
$$
with $\lambda= \pm \frac{1}{2}$ corresponding to $s=1,2$, respectively.

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02:29

Problem 3

Calculate the $\lambda=+\frac{1}{2}$ helicity eigenspinor of an electron of momentum $\mathbf{p}^{\prime}=(p \sin \theta, 0, p \cos \theta)$.

Mayukh Banik

Numerade Educator

01:06

Problem 4

Confirm the desired result that the Dirac equation describes "intrinsic" angular momentum ( $\equiv$ spin)- $\frac{1}{2}$ particles.

Narayan Hari

Numerade Educator

09:39

Problem 5

For a nonrelativistic electron of velocity $v$, use (5.24) to show that $u_A$ is larger than $u_B$ by a factor of the order $v / c$. In nonrelativistic problems, $\psi_A$ and $\psi_B$ are referred to as the "large" and "small" components of the electron wave function $\psi$.
In the nonrelativistic limit, show that the Dirac equation for an electron (charge $-e)$ in an electromagnetic field $A^\mu=\left(A^0, A\right)$ reduces to the Schrödinger-Pauli equation
$$
\left(\frac{1}{2 m}(\mathbf{P}+e \mathbf{A})^2+\frac{e}{2 m} \mathbf{\sigma} \cdot \mathbf{B}-e A^0\right) \psi_A=E_{N R} \psi_A,
$$
where the magnetic field $\mathbf{B}=\nabla \times \mathbf{A}$ and $E_{N R}=E-m$. Assume $\left|e A^0\right|$ $\& m$.

Khoobchandra Agrawal

Numerade Educator

Problem 6

In representation (5.4) and (5.8) of the $\gamma$-matrices, show that a possible choice of $C$ is
$$
C \gamma^0=i \gamma^2=\left(\begin{array}{llll}
& & & 1 \\
& -1 & &
\end{array}\right) .
$$

Try this operation out on a particular spinor, and show, for instance, that
$$
\psi_C^{(1)}=i \gamma^2\left[u^{(1)}(\mathbf{p}) e^{-i p \cdot x}\right]^*=u^{(4)}(-\mathbf{p}) e^{i p \cdot x}=v^{(1)}(\mathbf{p}) e^{i p \cdot x} .
$$

Further, show that in this representation
$$
\begin{aligned}
C^{-1} \gamma^\mu C & =\left(-\gamma^\mu\right)^T, \\
C & =-C^{-1}=-C^{\dagger}=-C^T, \\
\bar{\psi}_C & =-\psi^T C^{-1} .
\end{aligned}
$$

We have seen that the electron current is
$$
j^\mu=-e \bar{\psi} \gamma^\mu \psi \text {. }
$$

The current associated with the charge conjugate field is therefore
$$
\begin{aligned}
j_C^\mu & =-e \bar{\psi}_C \gamma^\mu \psi_C \\
& =+e \psi^T C^{-1} \gamma^\mu C \bar{\psi}^T \\
& =-e \psi^T\left(\gamma^\mu\right)^T \bar{\psi}^T \\
& =-(-) e \bar{\psi} \gamma^\mu \psi
\end{aligned}
$$
[see (5.39)]. The origin of the extra minus sign introduced in the last line is subtle but important. It is clearly necessary for a physically meaningful result, that is, if $j_C^\mu$ is to be the positron current. The minus sign is related to the connection between spin and statistics; in field theory, it occurs because of the antisymmetric nature of the fermion fields. In field theory, the charge conjugation operator $C$ changes a positive-energy electron into a positive-energy positron, and the formalism is completely $\mathrm{e}^{+} \leftrightarrow \mathrm{e}^{-}$symmetric. However, in a single-particle (electron) theory, positron states are not allowed; rather, $C$ changes a positive-energy electron state into a negative-energy electron state. As a result, we can show that we must add to our Feynman rules the requirement that we insert by hand an extra minus sign for every negative-energy electron in the final state of the process. The $C$ invariance of electromagnetic interactions then follows:
$$
j_\mu^C\left(A^\mu\right)^C=\left(-j_\mu\right)\left(-A^\mu\right)=j_\mu A^\mu .
$$

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01:11

Problem 7

Use (5.41) to show that
$$
\bar{u}^{(s)} u^{(s)}=2 m, \quad \bar{v}^{(s)} v^{(s)}=-2 m .
$$

Raj Bala

Numerade Educator

01:02

Problem 8

Show that $(\sigma \cdot p)^2=|\mathbf{p}|^2$.

Raj Bala

Numerade Educator

Problem 9

Derive the completeness relations
$$
\begin{aligned}
& \sum_{s=1,2} u^{(s)}(p) \bar{u}^{(s)}(p)=p+m, \\
& \sum_{s=1,2} v^{(s)}(p) \bar{v}^{(s)}(p)=p-m .
\end{aligned}
$$
These $4 \times 4$ matrix relations are used extensively in the evaluation of Feynman diagrams (see Chapter 6).

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01:33

Problem 10

Show that $p p=p^2$.

Adriano Chikande

Numerade Educator

Problem 11

Show that
$$
\Lambda_{+}=\frac{p+m}{2 m}, \quad \Lambda_{-}=\frac{-p+m}{2 m}
$$
project over positive and negative energy states, respectively. Recall that the projection operators must satisfy
$$
\Lambda_{ \pm}^2=\Lambda_{ \pm} \quad \text { and } \quad \Lambda_{+}+\Lambda_{-}=1
$$

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Problem 12

For a proper infinitesimal Lorentz transformation
$$
\Lambda_\mu^\nu=\delta_\mu^\nu+\varepsilon_\mu^\nu,
$$
show that an $S$ which satisfies (5.57) is
$$
S_L=1-\frac{i}{4} \sigma_{\mu v} \varepsilon^{\mu \nu} .
$$
Hence, show that
$$
\begin{aligned}
S_L^{-1} & =\gamma^0 S_L^{\dagger} \gamma^0 \\
\gamma^5 S_L & =S_L \gamma^5 .
\end{aligned}
$$

For space inversion, or the parity operation,
$$
\Lambda_\mu^\nu=\left(\begin{array}{cccc}
1 & & & \\
& -1 & & \\
& & -1 & \\
& & & -1
\end{array}\right) .
$$

Then, (5.57) becomes
$$
\begin{aligned}
& S_P^{-1} \gamma^0 S_P=\gamma^0, \\
& S_P^{-1} \gamma^k S_P=-\gamma^k \quad \text { for } k=1,2,3,
\end{aligned}
$$
which is satisfied by
$$
S_P=\gamma^0 \text {. }
$$

In the Dirac-Pauli representation of $\gamma^0,(5.51)$, the behavior of the four components of $\psi$ under parity is therefore
$$
\psi_{1,2}^{\prime}=\psi_{1,2} \quad \text { and } \quad \psi_{3,4}^{\prime}=-\psi_{3,4} \text {. }
$$

The "at rest" states, $(5.22)$, are therefore eigenstates of parity, with the positive and negative energy states (that is, the electron and the positron) having opposite intrinsic parities.

Armed with $S_I$ and $S_P$, we can now check the claimed properties of the bilinear covariants. First, we note that
$$
\begin{aligned}
\bar{\psi}^{\prime} & =\psi^{\dagger} \gamma^0=\psi^{\dagger} S^{\dagger} \gamma^0=\psi^{\dagger} \gamma^0 S^{-1} \\
& =\bar{\psi} S^{-1},
\end{aligned}
$$
where we have used (5.56) and (5.60). As an example, let us establish the character of $\bar{\psi} \gamma^\mu \psi$. Under Lorentz transformations,
$$
\bar{\psi}^{\prime} \gamma^\mu \psi^{\prime}=\bar{\psi} S_L^{-1} \gamma^\mu S_L \psi=\Lambda_\nu^\mu\left(\bar{\psi} \gamma^\nu \psi\right),
$$
using (5.64) and (5.57), while under the parity operation,
$$
\bar{\psi}^{\prime} \gamma^\mu \psi^{\prime}=\bar{\psi} S_P^{-1} \gamma^\mu S_P \psi=\left\{\begin{array}{r}
\bar{\psi} \gamma^0 \psi, \\
-\bar{\psi} \gamma^k \psi .
\end{array}\right.
$$

These transformation properties are precisely what we expect for a Lorentz four-vector.

From (5.56) and (5.64), if follows immediately that $\bar{\psi} \psi$ is a Lorentz scalar. The probability density $\rho=\psi^{\dagger} \psi$ is not a scalar, but is the time-like component of the

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Problem 13

Show that the operators
$$
P_R \equiv \frac{1}{2}\left(1+\gamma^5\right), \quad P_L \equiv \frac{1}{2}\left(1-\gamma^5\right)
$$
have the appropriate properties to be (right- and left-hand) projection operators, that is,
$$
P_i^2=P_i, \quad P_L+P_R=1, \quad P_R P_L=0 .
$$

Here, $\gamma^5$ is called the chirality operator.

For a massive fermion, we define the projections $\frac{1}{2}\left(1 \pm \gamma^5\right) u$ to be right- and left-handed components of $u$.

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Problem 14

For a massive fermion, show that handedness is not a good quantum number. That is, show that $\gamma^5$ does not commute with the Hamiltonian. However, verify that helicity is conserved but is frame dependent. In particular, show that the helicity is reversed by "overtaking" the particle concerned.

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Problem 15

Working in the Dirac-Pauli representation of $\gamma$-matrices, (5.51), show that at high energies
$$
\gamma^5 u^{(s)}=\left(\begin{array}{cc}
\boldsymbol{\sigma} \cdot \hat{\mathbf{p}} & 0 \\
0 & \boldsymbol{\sigma} \cdot \hat{\mathbf{p}}
\end{array}\right) u^{(s)},
$$
where $u^{(s)}$ is the electron spinor of (5.27). That is, show that in the extreme relativistic limit, the chirality operator $\left(\gamma^5\right)$ is equal to the helicity operator; and so, for example, $\frac{1}{2}\left(1-\gamma^5\right) u=u_L$ corresponds to an electron of negative helicity.

Of course, the fact that $\frac{1}{2}\left(1-\gamma^5\right)$ projects out negative helicity fermions at high energies does not depend on the choice of representation. We need only choose a representation if we wish to show explicit spinors. The particular advantage of the Dirac-Pauli representation is that it diagonalizes the energy in the nonrelativistic limit $\left(\gamma^0\right.$ is diagonal), whereas the Weyl representation diagonalizes the helicity in the extreme relativistic limit ( $\gamma^5$ is diagonal).

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