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# Helicity operator in the Dirac equation

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.20.

In the Dirac equation, the spin of a particle is described using the spin operator

$\displaystyle \Sigma_{i}=\frac{1}{2}\left[\begin{array}{cc} \sigma_{i} & 0\\ 0 & \sigma_{i} \end{array}\right];\;i=x,y,z \ \ \ \ \ (1)$

(using natural units where ${\hbar=1}$) and where the Pauli matrices are

 $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle \sigma_{y}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle \sigma_{z}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (4)$

and the 0 components in 1 are ${2\times2}$ zero matrices. Spin can be oriented in any direction for particles travelling at a velocity ${v<1}$, although at the speed of light (${v=1}$), the spin is always aligned with the velocity due to length contraction effects. The relation between the directions of spin and the particle’s velocity is given by the helicity. If the 3-momentum ${\mathbf{p}}$ and spin both point in the same direction, the helicity has its maximum value (a positive quantity), while if they point in opposite directions, the helicity has its maximum negative value. If ${\mathbf{p}}$ and spin are at right angles, the helicity is zero.

These relations suggest that a helicity operator can be defined in terms of the scalar product of ${\boldsymbol{\Sigma}}$ and ${\mathbf{p}}$. The helicity operator is

 $\displaystyle \Sigma_{\mathbf{p}}$ $\displaystyle \equiv$ $\displaystyle \boldsymbol{\Sigma}\cdot\frac{\mathbf{p}}{\left|\mathbf{p}\right|}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Sigma_{1}\frac{p^{1}}{\left|\mathbf{p}\right|}+\Sigma_{2}\frac{p^{2}}{\left|\mathbf{p}\right|}+\Sigma_{3}\frac{p^{3}}{\left|\mathbf{p}\right|} \ \ \ \ \ (6)$

Since the ${\Sigma_{i}}$ are each a ${4\times4}$ matrix, the helicity ${\Sigma_{\mathbf{p}}}$ is also a ${4\times4}$ matrix, but with reference to 3-d space, it is a scalar matrix.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (7)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (8)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (9)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (10)$

For a particle moving in the ${z}$ direction, ${p^{3}>0}$, ${p^{1}=p^{2}=0}$ and we’ve seen that the states ${\left|\psi^{\left(1\right)}\right\rangle }$ and ${\left|\psi^{\left(3\right)}\right\rangle }$ are eigenstates of ${\Sigma_{3}}$ with eigenvalue ${\frac{1}{2}}$. In this case, ${p^{3}/\left|\mathbf{p}\right|=1}$ so from 6

$\displaystyle \Sigma_{\mathbf{p}}\left|\psi^{\left(1\right)}\right\rangle =\frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (11)$

Thus for a particle whose spin and velocity both point in the same direction, the maximum helicity value of ${\frac{1}{2}}$ is obtained. Similarly, if ${p^{3}<0}$, ${p^{1}=p^{2}=0}$ and ${p^{3}/\left|\mathbf{p}\right|=-1}$ so

$\displaystyle \Sigma_{\mathbf{p}}\left|\psi^{\left(1\right)}\right\rangle =-\frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (12)$

With velocity and spin pointing in opposite directions, the maximum negative value of ${-\frac{1}{2}}$ is obtained for the helicity.

Now suppose we have a particle in state ${\left|\psi^{\left(2\right)}\right\rangle }$ and that ${p^{1}\ne0}$, ${p^{2}=p^{3}=0}$. This is not a helicity eigenstate, since ${\left|\psi^{\left(2\right)}\right\rangle }$ is an eigenstate of ${\Sigma_{3}}$ with eigenvalue ${-\frac{1}{2}}$, so the spin is in the ${-z}$ direction while the velocity is in the ${x}$ direction, so the spin is not parallel to the velocity.

Mathematically, in order for ${\left|\psi^{\left(2\right)}\right\rangle }$ to be a helicity eigenstate, the spinor in ${\left|\psi^{\left(2\right)}\right\rangle }$ would have to be an eigenstate of ${\Sigma_{1}}$, which is not true, since

 $\displaystyle \Sigma_{1}\left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right]\sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}}{E+m}\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ 0\\ \frac{p^{1}}{E+m} \end{array}\right]e^{-ipx} \ \ \ \ \ (14)$

If ${p^{3}\ne0}$, ${p^{1}=p^{2}=0}$, however, we would expect ${\left|\psi^{\left(2\right)}\right\rangle }$ to be a helicity eigenstate since the spin and velocity are parallel in this case. Since ${\left|\psi^{\left(2\right)}\right\rangle }$ is an eigenstate of ${\Sigma_{3}}$ with eigenvalue ${-\frac{1}{2}}$, we have

$\displaystyle \Sigma_{\mathbf{p}}\left|\psi^{\left(1\right)}\right\rangle =\pm\frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (15)$

with the + corresponding to ${p^{3}>0}$ and the ${-}$ to ${p^{3}<0}$.

# Dirac equation: non-relativistic limit

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.19.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

In the non-relativistic limit, the relative velocity of the particle satisfies ${v\ll1}$, which means that the momentum components all satisfy ${p^{j}\ll E\approx m}$. Thus ${\sqrt{\frac{E+m}{2m}}\approx1}$ and the solutions reduce to

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (5)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (6)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array}\right]e^{ipx}\ \ \ \ \ (7)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right]e^{ipx} \ \ \ \ \ (8)$

Comparing this with the free particle solutions to the non-relativistic Schrödinger equation, we see that the ${e^{\pm ipx}=e^{\pm Et}e^{\mp\mathbf{p}\cdot\mathbf{x}}}$ factor is just what we’d get in that case. The first two components of the spinors in ${\left|\psi^{\left(1\right)}\right\rangle }$ and ${\left|\psi^{\left(2\right)}\right\rangle }$ also correspond to the eigenstates in the non-relativistic spin ${\frac{1}{2}}$ theory.

In the 4-d case, we can operate on these solutions with the 4-d spin operator ${\Sigma_{z}}$ to get

 $\displaystyle \Sigma_{z}\left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (11)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(2\right)}\right\rangle \ \ \ \ \ (14)$

Similarly, we get

 $\displaystyle \Sigma_{z}\left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(3\right)}\right\rangle \ \ \ \ \ (15)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(4\right)}\right\rangle \ \ \ \ \ (16)$

These are the same results that we get by applying the Pauli spin matrices to the 2-d spin space spinors in the non-relativistic theory.

# Dirac equation: spinors near the speed of light

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.18.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

For a particle at rest (${\mathbf{p}=0,\;E=m}$), or for a particle moving in the ${z}$ direction, all four solutions are eigenstates of the spin operator ${\Sigma_{z}}$, with eigenvalues (spins) of ${\pm\frac{1}{2}}$. If the particle is moving in the ${x}$ or ${y}$ direction, the individual spinors above aren’t eigenstates of any of the spin operators. As the speed of the particle approaches ${c}$, however, we can get some eigenstates of ${\Sigma_{x}}$ and ${\Sigma_{y}}$.

First, suppose the particle is moving in the ${x}$ direction at a speed approaching ${c}$, with any motion in the ${y}$ and ${z}$ directions much smaller by comparison. In this case, ${E\rightarrow p^{1}\rightarrow\infty}$ and the spinor components (which we’ll call ${s^{\left(n\right)}}$ for ${n=1,\ldots,4}$) of the solutions above become

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (7)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (8)$

The ${x}$ spin operator is

$\displaystyle \Sigma_{x}=\frac{1}{2}\left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right] \ \ \ \ \ (9)$

Multiplying ${\Sigma_{x}}$ into ${s^{\left(1\right)}+s^{\left(2\right)}}$, we get

 $\displaystyle \Sigma_{x}\left(s^{\left(1\right)}+s^{\left(2\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(s^{\left(1\right)}+s^{\left(2\right)}\right) \ \ \ \ \ (11)$

Similarly

 $\displaystyle \Sigma_{x}\left(s^{\left(3\right)}+s^{\left(4\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(s^{\left(3\right)}+s^{\left(4\right)}\right) \ \ \ \ \ (13)$

Thus the sums ${u_{1}+u_{2}}$ and ${v_{1}+v_{2}}$ are both eigenstates of ${\Sigma_{x}}$ with eigenvalue ${\frac{1}{2}}$.

Now suppose the particle is moving in the ${y}$ direction with ${v^{y}\rightarrow1}$ so that ${E\rightarrow p^{2}\rightarrow\infty}$. The four spinors now become

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ i \end{array}\right]\ \ \ \ \ (14)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ -i\\ 0 \end{array}\right]\ \ \ \ \ (15)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ i\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} -i\\ 0\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (17)$

The ${y}$ spin operator is

$\displaystyle \Sigma_{y}=\frac{1}{2}\left[\begin{array}{cccc} 0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{array}\right] \ \ \ \ \ (18)$

In this case

 $\displaystyle \Sigma_{y}\left(s^{\left(1\right)}+s^{\left(3\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{array}\right]\left[\begin{array}{c} 1\\ i\\ 1\\ i \end{array}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ i\\ 1\\ i \end{array}\right]\ \ \ \ \ (20)$ $\displaystyle \Sigma_{y}\left(s^{\left(2\right)}+s^{\left(4\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{array}\right]\left[\begin{array}{c} -i\\ 1\\ -i\\ 1 \end{array}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} -i\\ 1\\ -i\\ 1 \end{array}\right] \ \ \ \ \ (22)$

Thus the sums ${u_{1}+v_{2}}$ and ${v_{1}+u_{2}}$ are both eigenstates of ${\Sigma_{y}}$ with eigenvalue ${\frac{1}{2}}$. [As the ${u_{j}}$ spinors are supposed to represent particles and the ${v_{j}}$ antiparticles, I’m not sure what a mixture of the two is supposed to represent.]

# Dirac equation: spin of a moving particle

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.17.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

For a particle at rest (${\mathbf{p}=0,\;E=m}$), all four solutions are eigenstates of the spin operator ${\Sigma_{z}}$, with eigenvalues (spins) of ${\pm\frac{1}{2}}$. Here, we have a look at what happens if the particle is moving.

First, suppose the particle is moving in the ${x}$ direction, so that ${p^{1}\ne0}$ and ${p^{2}=p^{3}=0}$. In this case, the spinor components (which we’ll call ${s^{\left(n\right)}}$ for ${n=1,\ldots,4}$) of the solutions above become

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ \frac{p^{1}}{E+m} \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}}{E+m}\\ 0 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ \frac{p^{1}}{E+m}\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (7)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \frac{p^{1}}{E+m}\\ 0\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (8)$

The ${z}$ spin operator is

$\displaystyle \Sigma_{z}=\frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (9)$

Multiplying ${\Sigma_{z}}$ into the four ${s^{\left(n\right)}}$ spinors, we get

 $\displaystyle \Sigma_{z}s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 0\\ 0\\ -\frac{p^{1}}{E+m} \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle \Sigma_{z}s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 0\\ -1\\ \frac{p^{1}}{E+m}\\ 0 \end{array}\right]\ \ \ \ \ (11)$ $\displaystyle \Sigma_{z}s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 0\\ -\frac{p^{1}}{E+m}\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (12)$ $\displaystyle \Sigma_{z}s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} \frac{p^{1}}{E+m}\\ 0\\ 0\\ -1 \end{array}\right] \ \ \ \ \ (13)$

In each case, one of the non-zero components of ${s^{\left(n\right)}}$ has its sign changed, while the other non-zero component remains the same. Thus none of the ${s^{\left(n\right)}}$ spinors is an eigenstate of ${\Sigma_{z}}$.

Now suppose that ${p^{1}=p^{2}=0}$ and ${p^{3}\ne0}$. The spinors are now

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ 0 \end{array}\right]\ \ \ \ \ (14)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 0\\ -\frac{p^{3}}{E+m} \end{array}\right]\ \ \ \ \ (15)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \frac{p^{3}}{E+m}\\ 0\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (17)$

Multiplying by ${\Sigma_{z}}$ we get

 $\displaystyle \Sigma_{z}s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ 0 \end{array}\right]=\frac{1}{2}s^{\left(1\right)}\ \ \ \ \ (18)$ $\displaystyle \Sigma_{z}s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 0\\ -1\\ 0\\ \frac{p^{3}}{E+m} \end{array}\right]=-\frac{1}{2}s^{\left(2\right)}\ \ \ \ \ (19)$ $\displaystyle \Sigma_{z}s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ 0\\ 1\\ 0 \end{array}\right]=\frac{1}{2}s^{\left(3\right)}\ \ \ \ \ (20)$ $\displaystyle \Sigma_{z}s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 0\\ \frac{p^{3}}{E+m}\\ 0\\ -1 \end{array}\right]=-\frac{1}{2}s^{\left(4\right)} \ \ \ \ \ (21)$

Thus the ${s^{\left(n\right)}}$ spinors in this case are eigenstates of ${\Sigma_{z}}$.

# Dirac equation: spin of a particle at rest

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.16.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

The last time we found quantum mechanical solutions that contained column vectors, we introduced these solutions to account for spin in non-relativistic quantum mechanics. With the Dirac equation, spin finds its way into the solutions by being a consequence of the nature of the solutions, rather than by being imposed.

An experimental fact about spin is that the intrinsic spin of a particle is not affected by how fast it is moving. That is, an electron always has spin ${\frac{1}{2}}$ whether it is observed at rest or moving close to the speed of light. However, due to length contraction, the direction of an angular momentum vector for a spinning object does vary with velocity relative to the observer. This is explained more fully in Klauber’s Box 4-2, which looks at the effect of length contraction on a classical (non-quantum) rotating disk. In summary, suppose the angular momentum vector ${\mathbf{L}}$ lies in the ${x-z}$ plane at some angle ${\theta}$ to the ${x}$ axis, so that the plane of the disk, being perpendicular to ${\mathbf{L}}$, makes the same angle ${\theta}$ with the ${z}$ axis. Now suppose the disk moves along the ${x}$ axis at some speed ${v}$. As ${v}$ becomes relativistic, the angle ${\theta}$ diminishes, since the size component of the disk in the ${x}$ direction is contracted, while the component in the ${z}$ direction remains unchanged. When ${v\rightarrow c}$, the disk’s ${x}$ component tends to zero, so that ${\mathbf{L}}$ lies along the ${x}$ axis and the disk spins in the ${yz}$ plane.

Because both ${E}$ and ${\mathbf{p}}$ depend ultimately on ${v}$ in the solutions above, the Dirac solutions actually already contain this relativistic effect within the spinor components. To see this, we need the spin operators in the Dirac theory, in analogy to the Pauli matrices for non-relativistic spin ${\frac{1}{2}}$. For now, we will take these operators to be god-given. They are

$\displaystyle \Sigma_{i}=\frac{1}{2}\left[\begin{array}{cc} \sigma_{i} & 0\\ 0 & \sigma_{i} \end{array}\right];\;i=x,y,z \ \ \ \ \ (5)$

(using natural units where ${\hbar=1}$) and where the Pauli matrices are

 $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle \sigma_{y}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (7)$ $\displaystyle \sigma_{z}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (8)$

and the 0 components in 5 are ${2\times2}$ zero matrices.

As a simple example of how the ${\Sigma_{i}}$ operators work, consider the special case of a particle at rest, so that ${\mathbf{p}=0}$ and ${E=m}$. Then the four solutions at the top reduce to

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-imt}\ \ \ \ \ (9)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-imt}\ \ \ \ \ (10)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array}\right]e^{imt}\ \ \ \ \ (11)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right]e^{imt} \ \ \ \ \ (12)$

These four solutions are all eigenstates of ${\Sigma_{z}}$ as we can see by writing out ${\Sigma_{z}}$ explicitly

$\displaystyle \Sigma_{z}=\frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (13)$

Then we find

 $\displaystyle \Sigma_{z}\left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (14)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(2\right)}\right\rangle \ \ \ \ \ (15)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(3\right)}\right\rangle \ \ \ \ \ (16)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(4\right)}\right\rangle \ \ \ \ \ (17)$

Thus ${\left|\psi^{\left(1\right)}\right\rangle }$ and ${\left|\psi^{\left(3\right)}\right\rangle }$ are eigenstates of a particle with ${z}$ spin component ${\frac{1}{2}}$, and ${\left|\psi^{\left(2\right)}\right\rangle }$ and ${\left|\psi^{\left(4\right)}\right\rangle }$ represent a particle with ${z}$ spin component ${-\frac{1}{2}}$. As you might expect, the first two solutions are for particles and the last two are for antiparticles, although we haven’t actually demonstrated this yet.

# Eigenspinors of the Pauli spin matrices

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.15.

In non-relativistic quantum mechanics, the spin ${\frac{1}{2}}$ operators are given in terms of the Pauli matrices as

 $\displaystyle S_{i}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\sigma_{i}\ \ \ \ \ (1)$ $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle \sigma_{y}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle \sigma_{z}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (4)$

We’ve seen that the spinor states ${\left[\begin{array}{c} 1\\ 0 \end{array}\right]}$ and ${\left[\begin{array}{c} 0\\ 1 \end{array}\right]}$ are eigenstates of the ${S_{z}}$ operator, and that these spinors form a basis for the 2-d spinor space. We can find the eigenstates of ${S_{x}}$ and ${S_{y}}$ in the usual way from matrix algebra. For ${\sigma_{x}}$, the eigenvalues are

 $\displaystyle \left|\begin{array}{cc} -\lambda & 1\\ 1 & -\lambda \end{array}\right|$ $\displaystyle =$ $\displaystyle \lambda^{2}-1=0\ \ \ \ \ (5)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \pm1 \ \ \ \ \ (6)$

For ${\lambda=1}$, the eigenvector equation is

$\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\left[\begin{array}{c} a\\ b \end{array}\right]=\left[\begin{array}{c} a\\ b \end{array}\right] \ \ \ \ \ (7)$

which gives

$\displaystyle a=b \ \ \ \ \ (8)$

To normalize the eigenstate, we can choose ${a=b=\frac{1}{\sqrt{2}}}$, so that the state is ${\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 1 \end{array}\right]}$.

For ${\lambda=-1}$, we get

$\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\left[\begin{array}{c} a\\ b \end{array}\right]=-\left[\begin{array}{c} a\\ b \end{array}\right] \ \ \ \ \ (9)$

so the normalized eigenstate is ${\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ -1 \end{array}\right]}$.

For ${\sigma_{y}}$, we get

 $\displaystyle \left|\begin{array}{cc} -\lambda & -i\\ i & -\lambda \end{array}\right|$ $\displaystyle =$ $\displaystyle \lambda^{2}-1=0\ \ \ \ \ (10)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \pm1 \ \ \ \ \ (11)$

so the eigenvalues are the same as for ${\sigma_{x}}$ and ${\sigma_{z}}$. The eigenvector equations are

$\displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\left[\begin{array}{c} a\\ b \end{array}\right]=\pm\left[\begin{array}{c} a\\ b \end{array}\right] \ \ \ \ \ (12)$

from which we get

$\displaystyle a=\mp ib \ \ \ \ \ (13)$

Thus the two normalized eigenstates are, for ${\lambda=+1,\;-1}$ respectively:

$\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{c} i\\ -1 \end{array}\right],\;\frac{1}{\sqrt{2}}\left[\begin{array}{c} i\\ 1 \end{array}\right] \ \ \ \ \ (14)$

This can be written in terms of the ${\sigma_{z}}$ eigenstates as

$\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{c} i\\ -1 \end{array}\right]=\frac{i}{\sqrt{2}}\left[\begin{array}{c} 1\\ 0 \end{array}\right]+\frac{1}{\sqrt{2}}\left[\begin{array}{c} 0\\ -1 \end{array}\right] \ \ \ \ \ (15)$

Since the coefficients of the two ${\sigma_{z}}$ eigenstates are equal in magnitude, this means that if ${\sigma_{z}}$ is measured for a particle in an ${\sigma_{y}}$ eigenstate, it is equally likely to be spin up or spin down. The same applies to a particle in a ${\sigma_{x}}$ eigenstate.

# Dirac equation: positive probabilities and negative energies

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.14.

One problem with the Klein-Gordon equation in relativistic quantum mechanics is that the probability of finding the system in some state can be positive (for particles) or negative (for antiparticles). The Dirac equation solves this problem, as Klauber shows in his equations 4-36 through 4-38. We’ll summarize these results here.

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

Since the Dirac equation is linear, any linear combination of these solutions is also a solution, so the most general solution is

$\displaystyle \left|\psi\right\rangle =\sum_{\mathbf{p}}\sum_{r=1}^{2}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[C_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+D_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right] \ \ \ \ \ (5)$

where ${C_{r}}$ and ${D_{r}}$ are constant (independent of ${x}$) coefficients.

The probability density ${\rho}$ for the Dirac equation is

$\displaystyle \rho=\left\langle \bar{\psi}\left|\gamma^{\mu}\right|\psi\right\rangle \ \ \ \ \ (6)$

where ${\left\langle \bar{\psi}\right|}$ is an adjoint solution and despite the bracket notation, there is no integration over space involved in the definition of ${\rho}$. The total probability of the system being in some state can be found by integrating ${\rho}$ over space. This involves using the orthogonality of the solutions and inner products of the spinors (details in Klauber as mentioned above). The results for a state that consists entirely of ${C_{r}}$ terms (all ${D_{r}}$ are zero) is that

$\displaystyle \int\rho d^{3}x=\sum_{r,\mathbf{p}}\left|C_{r}\left(\mathbf{p}\right)\right|^{2}=1 \ \ \ \ \ (7)$

where we have imposed the “= 1” since the total probability must be one.

For a state consisting entirely of ${D_{r}}$ terms (all ${C_{r}}$ are zero), we get the result

$\displaystyle \int\rho d^{3}x=\sum_{r,\mathbf{p}}\left|D_{r}\left(\mathbf{p}\right)\right|^{2}=1 \ \ \ \ \ (8)$

The crucial difference between Klein-Gordon and Dirac is that with Dirac, both probabilities are positive.

The Dirac equation does, however, still give both positive and negative energies for a free particle. The Hamiltonian can be written using the operator ${i\frac{\partial}{\partial t}}$ so applying this to 1 or 2 above we get

$\displaystyle i\frac{\partial}{\partial t}\left(u_{1,2}e^{-ipx}\right)=p^{0}u_{1,2}e^{-ipx}=+E_{\mathbf{p}}u_{1,2}e^{-ipx} \ \ \ \ \ (9)$

And applying to 3 or 4 we get

$\displaystyle i\frac{\partial}{\partial t}\left(v_{1,2}e^{ipx}\right)=-p^{0}v_{1,2}e^{ipx}=-E_{\mathbf{p}}v_{1,2}e^{ipx} \ \ \ \ \ (10)$

Thus those terms in 5 with ${C_{r}}$ coefficients have positive energy and terms with ${D_{r}}$ coefficients have negative energy.

# Dirac equation: conserved probability current

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.13.

The Dirac equation is

$\displaystyle \left(i\gamma^{\mu}\partial_{\mu}-m\right)\left|\psi\right\rangle =0 \ \ \ \ \ (1)$

and the adjoint Dirac equation is

$\displaystyle i\partial_{\mu}\left\langle \bar{\psi}\right|\gamma^{\mu}+m\left\langle \bar{\psi}\right|=0 \ \ \ \ \ (2)$

where there are four solution vectors ${n=1,\ldots,4}$ and the adjoint solutions are given by

$\displaystyle \left\langle \bar{\psi}\right|=\left\langle \psi\right|\gamma^{0} \ \ \ \ \ (3)$

We’d like to find a conserved quantity analogous to that for the Klein-Gordon equation. We can do this as follows. First we multiply 1 on the left by the adjoint solutions:

$\displaystyle i\left\langle \bar{\psi}\left|\gamma^{\mu}\partial_{\mu}\right|\psi\right\rangle =m\left\langle \bar{\psi}\left|\psi\right.\right\rangle \ \ \ \ \ (4)$

Then, multiply 2 on the right by the original solutions:

$\displaystyle i\left(\partial_{\mu}\left\langle \bar{\psi}\right|\right)\gamma^{\mu}\left|\psi\right\rangle =-m\left\langle \bar{\psi}\left|\psi\right.\right\rangle \ \ \ \ \ (5)$

We’ve kept the bra in parentheses since the derivative applies only to it and not the ket portion. Adding these two equations gives

 $\displaystyle i\left\langle \bar{\psi}\left|\gamma^{\mu}\partial_{\mu}\right|\psi\right\rangle +i\left(\partial_{\mu}\left\langle \bar{\psi}\right|\right)\gamma^{\mu}\left|\psi\right\rangle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle i\partial_{\mu}\left\langle \bar{\psi}\left|\gamma^{\mu}\right|\psi\right\rangle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

where we’ve used the product rule to combine the two derivatives, so that the ${\partial_{\mu}}$ in the last line does apply to the full bracket. Note also that in the bracket in the last line, there is no integration over space, since all we’ve done is multiply the adjoint solution into the regular solution.

By the way, you might think that this result is trivial, since spacetime enters into ${\left|\psi\right\rangle }$ only in the form ${e^{\pm ipx}}$ and therefore into ${\left\langle \bar{\psi}\right|}$ in the form ${e^{\mp ipx}}$, so it would seem that the bracket ${\left\langle \bar{\psi}\left|\gamma^{\mu}\right|\psi\right\rangle }$ has no dependence on ${x}$ so obviously its derivative must be zero. However, the ${\left|\psi\right\rangle }$ here can refer to a sum of states with different momenta ${\mathbf{p}}$, so the result isn’t quite as trivial as it looks.

We can therefore define a conserved current ${j^{\mu}}$ as

$\displaystyle j^{\mu}\equiv\left\langle \bar{\psi}\left|\gamma^{\mu}\right|\psi\right\rangle \ \ \ \ \ (8)$

so that

$\displaystyle \partial_{\mu}j^{\mu}=0 \ \ \ \ \ (9)$

Again, remember that there is no integration over space in the definition of ${j^{\mu}}$.

In Klauber’s equations 4-36 and 4-37, he shows that if we consider a single particle in the state

$\displaystyle \left|\psi\right\rangle =\sum_{r,\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}C_{r}\left(\mathbf{p}\right)u_{r}e^{-ipx} \ \ \ \ \ (10)$

(remember that ${r}$ ranges over the four solutions and ${\mathbf{p}}$ over all possible discrete momenta) and integrate ${\rho\equiv j^{0}}$ over space, we get the condition

$\displaystyle \sum_{r,\mathbf{p}}\left|C_{r}\left(\mathbf{p}\right)\right|^{2}=1 \ \ \ \ \ (11)$

Thus we can interpret ${\left|C_{r}\left(\mathbf{p}\right)\right|^{2}}$ as the probability of finding the particle in state ${r}$ with momentum ${\mathbf{p}}$.

# Adjoint Dirac equation: explicit solutions

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.12.

$\displaystyle i\partial_{\mu}\left\langle \bar{\psi}^{\left(n\right)}\right|\gamma^{\mu}+m\left\langle \bar{\psi}^{\left(n\right)}\right|=0 \ \ \ \ \ (1)$

where there are four solution vectors ${n=1,\ldots,4}$ and the adjoint solutions are given by

$\displaystyle \left\langle \bar{\psi}^{\left(n\right)}\right|=\left\langle \psi^{\left(n\right)}\right|\gamma^{0} \ \ \ \ \ (2)$

 $\displaystyle \left\langle \bar{\psi}^{\left(1\right)}\right|$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} 1 & 0 & -\frac{p^{3}}{E+m} & -\frac{p^{1}-ip^{2}}{E+m}\end{array}\right]e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left\langle \bar{\psi}^{\left(2\right)}\right|$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} 0 & 1 & -\frac{p^{1}+ip^{2}}{E+m} & \frac{p^{3}}{E+m}\end{array}\right]e^{ipx}\ \ \ \ \ (4)$ $\displaystyle \left\langle \bar{\psi}^{\left(3\right)}\right|$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} \frac{p^{3}}{E+m} & \frac{p^{1}-ip^{2}}{E+m} & -1 & 0\end{array}\right]e^{-ipx}\ \ \ \ \ (5)$ $\displaystyle \left\langle \bar{\psi}^{\left(4\right)}\right|$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} \frac{p^{1}+ip^{2}}{E+m} & -\frac{p^{3}}{E+m} & 0 & -1\end{array}\right]e^{-ipx} \ \ \ \ \ (6)$

Although the derivation of the adjoint equation 1 effectively proves that these four adjoint solutions do in fact solve the adjoint equation, we can show it more directly by substituting the solutions into the equation. Because of the implied sum over ${\mu}$ we’ll need the four gamma matrices:

 $\displaystyle \gamma^{0}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right]\ \ \ \ \ (7)$ $\displaystyle \gamma^{1}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ -1 & 0 & 0 & 0 \end{array}\right]\ \ \ \ \ (8)$ $\displaystyle \gamma^{2}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 0 & -i\\ 0 & 0 & i & 0\\ 0 & i & 0 & 0\\ -i & 0 & 0 & 0 \end{array}\right]\ \ \ \ \ (9)$ $\displaystyle \gamma^{3}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cccc} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1\\ -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{array}\right] \ \ \ \ \ (10)$

The solution vectors all work in much the same way, so we’ll look at just ${\left\langle \bar{\psi}^{\left(1\right)}\right|}$ here. The derivatives ${\partial_{\mu}}$ act only on the ${e^{ipx}}$ factor, so using

$\displaystyle px=Ex_{0}-\mathbf{p}\cdot\mathbf{x} \ \ \ \ \ (11)$

we get

 $\displaystyle i\partial_{0}\left\langle \bar{\psi}^{\left(1\right)}\right|\gamma^{0}$ $\displaystyle =$ $\displaystyle -E\sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} 1 & 0 & \frac{p^{3}}{E+m} & \frac{p^{1}-ip^{2}}{E+m}\end{array}\right]e^{ipx}\ \ \ \ \ (12)$ $\displaystyle i\partial_{1}\left\langle \bar{\psi}^{\left(1\right)}\right|\gamma^{1}$ $\displaystyle =$ $\displaystyle p^{1}\sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} \frac{p^{1}-ip^{2}}{E+m} & \frac{p^{3}}{E+m} & 0 & 1\end{array}\right]e^{ipx}\ \ \ \ \ (13)$ $\displaystyle i\partial_{2}\left\langle \bar{\psi}^{\left(1\right)}\right|\gamma^{2}$ $\displaystyle =$ $\displaystyle p^{2}\sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} i\frac{p^{1}-ip^{2}}{E+m} & -i\frac{p^{3}}{E+m} & 0 & -i\end{array}\right]e^{ipx}\ \ \ \ \ (14)$ $\displaystyle i\partial_{3}\left\langle \bar{\psi}^{\left(1\right)}\right|\gamma^{3}$ $\displaystyle =$ $\displaystyle p^{3}\sqrt{\frac{E+m}{2m}}\left[\begin{array}{cccc} \frac{p^{3}}{E+m} & -\frac{p^{1}-ip^{2}}{E+m} & 1 & 0\end{array}\right]e^{ipx} \ \ \ \ \ (15)$

Adding together the first component from each of these equations, we get

 $\displaystyle \sqrt{\frac{E+m}{2m}}e^{ipx}\left[-E+\frac{\left(p^{1}\right)^{2}-ip^{1}p^{2}+ip^{1}p^{2}+\left(p^{2}\right)^{2}+\left(p^{3}\right)^{2}}{E+m}\right]$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}e^{ipx}\left(-E+\frac{\mathbf{p}^{2}}{E+m}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}e^{ipx}\left(-E+\frac{E^{2}-m^{2}}{E+m}\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sqrt{\frac{E+m}{2m}}e^{ipx}m \ \ \ \ \ (18)$

For the second component, we get

$\displaystyle \sqrt{\frac{E+m}{2m}}e^{ipx}\left[\frac{0+p^{1}p^{3}-ip^{2}p^{3}-p^{1}p^{3}+ip^{2}p^{3}}{E+m}\right]=0 \ \ \ \ \ (19)$

For the third component:

 $\displaystyle \sqrt{\frac{E+m}{2m}}e^{ipx}\left[\frac{-Ep^{3}+0+0}{E+m}+p^{3}\right]$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}e^{ipx}\left[\frac{-Ep^{3}+p^{3}\left(E+m\right)}{E+m}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle m\sqrt{\frac{E+m}{2m}}e^{ipx}\frac{p^{3}}{E+m} \ \ \ \ \ (21)$

And for the fourth component

 $\displaystyle \sqrt{\frac{E+m}{2m}}e^{ipx}\left[\frac{-E\left(p^{1}-ip^{2}\right)}{E+m}+p^{1}-ip^{2}+0\right]$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}e^{ipx}\left[\frac{\left(-E+E+m\right)\left(p^{1}-ip^{2}\right)}{E+m}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle m\sqrt{\frac{E+m}{2m}}e^{ipx}\frac{p^{1}-ip^{2}}{E+m} \ \ \ \ \ (23)$

Comparing these results with the original adjoint solution ${\left\langle \bar{\psi}^{\left(1\right)}\right|}$ we see that

$\displaystyle i\partial_{\mu}\left\langle \bar{\psi}^{\left(1\right)}\right|\gamma^{\mu}=-m\left\langle \bar{\psi}^{\left(1\right)}\right| \ \ \ \ \ (24)$

so ${\left\langle \bar{\psi}^{\left(1\right)}\right|}$ does indeed satisfy the adjoint Dirac equation. The other 3 solutions work similarly.

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.11.

The Dirac equation in condensed form is

$\displaystyle i\gamma^{\mu}\partial_{\mu}\left|\psi\right\rangle =m\left|\psi\right\rangle \ \ \ \ \ (1)$

where the gamma matrices have been defined earlier. The Hermitian conjugate of the gamma matrix ${\gamma^{\mu}}$ is given by the Hermiticity condition

$\displaystyle \gamma^{\mu\dagger}=\gamma^{0}\gamma^{\mu}\gamma^{0} \ \ \ \ \ (2)$

To get the adjoint form of the Dirac equation, we use the adjoint solutions

$\displaystyle \left\langle \bar{\psi}^{\left(n\right)}\right|\equiv\left\langle \psi^{\left(n\right)}\right|\gamma^{0} \ \ \ \ \ (3)$

The Hermitian conjugate of 1 is

$\displaystyle -i\partial_{\mu}\left\langle \psi\right|\gamma^{\mu\dagger}=m\left\langle \psi\right| \ \ \ \ \ (4)$

Remember that when we take the Hermitian conjugate of a matrix equation we must take the Hermitian conjugate of each matrix and also reverse the order of matrix multiplication, which is why the ${\gamma^{\mu\dagger}}$ term appears at the end on the LHS. The ${\partial_{\mu}}$ is a differential operator, not a matrix, so it retains its position in front of the ${\left\langle \psi\right|}$.

Using 2, we can write this as

$\displaystyle -i\partial_{\mu}\left\langle \psi\right|\gamma^{0}\gamma^{\mu}\gamma^{0}=m\left\langle \psi\right| \ \ \ \ \ (5)$

Then, by post-multiplying by ${\gamma^{0}}$ and using ${\left(\gamma^{0}\right)^{2}=I}$, the identity matrix, we get, using the definition 3

 $\displaystyle -i\partial_{\mu}\left\langle \psi\right|\gamma^{0}\gamma^{\mu}\left(\gamma^{0}\right)^{2}$ $\displaystyle =$ $\displaystyle m\left\langle \psi\right|\gamma^{0}\ \ \ \ \ (6)$ $\displaystyle -i\partial_{\mu}\left\langle \bar{\psi}\right|\gamma^{\mu}$ $\displaystyle =$ $\displaystyle m\left\langle \bar{\psi}\right|\ \ \ \ \ (7)$ $\displaystyle i\partial_{\mu}\left\langle \bar{\psi}\right|\gamma^{\mu}+m\left\langle \bar{\psi}\right|$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (8)$

Note that although the terms ${\left\langle \bar{\psi}\right|}$ in this adjoint equation are adjoint solutions, the gamma matrices ${\gamma^{\mu}}$ are the original (that is, not the Hermitian conjugate) gamma matrices.