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# Posts Tagged ‘spinor’

## PHY456H1F: Quantum Mechanics II. Lecture 13 (Taught by Prof J.E. Sipe). Spin and spinors (cont.)

Posted by peeterjoot on October 24, 2011

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# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Multiple wavefunction spaces.

Reading: See section 26.5 in the text [1].

We identified

\begin{aligned}\psi(\mathbf{r}) = \left\langle{{ \mathbf{r}}} \vert {{\psi}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.1)

with improper basis kets

\begin{aligned}{\left\lvert {\mathbf{r}} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.2)

Now introduce many function spaces

\begin{aligned}\begin{bmatrix}\psi_1(\mathbf{r}) \\ \psi_2(\mathbf{r}) \\ \dot{v}s \\ \psi_\gamma(\mathbf{r})\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.3)

with improper (unnormalizable) basis kets

\begin{aligned}{\left\lvert {\mathbf{r} \alpha} \right\rangle}, \qquad \alpha \in 1, 2, ... \gamma\end{aligned} \hspace{\stretch{1}}(2.4)

\begin{aligned}\psi_\alpha(\mathbf{r}) = \left\langle{{ \mathbf{r}\alpha}} \vert {{\psi}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.5)

for an abstract ket ${\left\lvert {\psi} \right\rangle}$

We will try taking this Hilbert space

\begin{aligned}H = H_o \otimes H_s\end{aligned} \hspace{\stretch{1}}(2.6)

Where $H_o$ is the Hilbert space of “scalar” QM, “o” orbital and translational motion, associated with kets ${\left\lvert {\mathbf{r}} \right\rangle}$ and $H_s$ is the Hilbert space associated with the $\gamma$ components ${\left\lvert {\alpha} \right\rangle}$. This latter space we will label the “spin” or “internal physics” (class suggestion: or perhaps intrinsic). This is “unconnected” with translational motion.

We build up the basis kets for $H$ by direct products

\begin{aligned}{\left\lvert {\mathbf{r} \alpha} \right\rangle} = {\left\lvert {\mathbf{r}} \right\rangle} \otimes {\left\lvert {\alpha} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.7)

Now, for a rotated ket we seek a general angular momentum operator $\mathbf{J}$ such that

\begin{aligned}{\left\lvert {\psi'} \right\rangle} = e^{-i \theta \hat{\mathbf{n}} \cdot \mathbf{J}/\hbar} {\left\lvert {\psi} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.8)

where

\begin{aligned}\mathbf{J} = \mathbf{L} + \mathbf{S},\end{aligned} \hspace{\stretch{1}}(2.9)

where $\mathbf{L}$ acts over kets in $H_o$, “orbital angular momentum”, and $\mathbf{S}$ is the “spin angular momentum”, acting on kets in $H_s$.

Strictly speaking this would be written as direct products involving the respective identities

\begin{aligned}\mathbf{J} = \mathbf{L} \otimes I_s + I_o \otimes \mathbf{S}.\end{aligned} \hspace{\stretch{1}}(2.10)

We require

\begin{aligned}\left[{J_i},{J_j}\right] = i \hbar \sum \epsilon_{i j k} J_k\end{aligned} \hspace{\stretch{1}}(2.11)

Since $\mathbf{L}$ and $\mathbf{S}$ “act over separate Hilbert spaces”. Since these come from legacy operators

\begin{aligned}\left[{L_i},{S_j}\right] = 0\end{aligned} \hspace{\stretch{1}}(2.12)

We also know that

\begin{aligned}\left[{L_i},{L_j}\right] = i \hbar \sum \epsilon_{i j k} L_k\end{aligned} \hspace{\stretch{1}}(2.13)

so

\begin{aligned}\left[{S_i},{S_j}\right] = i \hbar \sum \epsilon_{i j k} S_k, \end{aligned} \hspace{\stretch{1}}(2.14)

as expected. We could, in principle, have more complicated operators, where this would not be true. This is a proposal of sorts. Given such a definition of operators, let’s see where we can go with it.

For matrix elements of $\mathbf{L}$ we have

\begin{aligned}{\left\langle {\mathbf{r}} \right\rvert} L_x {\left\lvert {\mathbf{r}'} \right\rangle} = -i \hbar \left( y \frac{\partial {}}{\partial {z}}-z \frac{\partial {}}{\partial {y}} \right) \delta(\mathbf{r}- \mathbf{r}')\end{aligned} \hspace{\stretch{1}}(2.15)

What are the matrix elements of ${\left\langle {\alpha} \right\rvert} S_i {\left\lvert {\alpha'} \right\rangle}$? From the commutation relationships we know

\begin{aligned}\sum_{\alpha'' = 1}^\gamma {\left\langle {\alpha} \right\rvert} S_i {\left\lvert {\alpha''} \right\rangle}{\left\langle {\alpha''} \right\rvert} S_j {\left\lvert {\alpha'} \right\rangle}-\sum_{\alpha'' = 1}^\gamma {\left\langle {\alpha} \right\rvert} S_j {\left\lvert {\alpha''} \right\rangle}{\left\langle {\alpha''} \right\rvert} S_i {\left\lvert {\alpha'} \right\rangle}=i \hbar \sum_k \epsilon_{ijk} {\left\langle {\alpha} \right\rvert} S_k {\left\lvert {\alpha''} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.16)

We see that our matrix element is tightly constrained by our choice of commutator relationships. We have $\gamma^2$ such matrix elements, and it turns out that it is possible to choose (or find) matrix elements that satisfy these constraints?

The ${\left\langle {\alpha} \right\rvert} S_i {\left\lvert {\alpha'} \right\rangle}$ matrix elements that satisfy these constraints are found by imposing the commutation relations

\begin{aligned}\left[{S_i},{S_j}\right] = i \hbar \sum \epsilon_{i j k} S_k, \end{aligned} \hspace{\stretch{1}}(2.17)

and with

\begin{aligned}S^2 = \sum_j S_j^2,\end{aligned} \hspace{\stretch{1}}(2.18)

(this is just a definition). We find

\begin{aligned}\left[{S^2},{S_i}\right] = 0\end{aligned} \hspace{\stretch{1}}(2.19)

and seeking eigenkets

\begin{aligned}S^2 {\left\lvert {s m_s} \right\rangle} &= s(s+1) \hbar^2 {\left\lvert {s m_s} \right\rangle} \\ S_z {\left\lvert {s m_s} \right\rangle} &= \hbar m_s {\left\lvert {s m_s} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.20)

Find solutions for $s = 1/2, 1, 3/2, 2, \cdots$, where $m_s \in \{-s, \cdots, s\}$. ie. $2 s + 1$ possible vectors ${\left\lvert {s m_s} \right\rangle}$ for a given $s$.

\begin{aligned}s = \frac{1}{{2}} &\implies \gamma = 2 \\ s = 1 &\implies \gamma = 3 \\ s = \frac{3}{2} &\implies \gamma = 4 \end{aligned}

We start with the algebra (mathematically the Lie algebra), and one can compute the Hilbert spaces that are consistent with these algebraic constraints.

We assume that for any type of given particle $S$ is fixed, where this has to do with the nature of the particle.

\begin{aligned}s = \frac{1}{{2}} &\qquad \text{A spinlatex 1/2particle} \\ s = 1 &\qquad \text{A spin $1$ particle} \\ s = \frac{3}{2} &\qquad \text{A spin $3/2$ particle}\end{aligned}

$S$ is fixed once we decide that we are talking about a specific type of particle.

A non-relativistic particle in this framework has two nondynamical quantities. One is the mass $m$ and we now introduce a new invariant, the spin $s$ of the particle.

This has been introduced as a kind of strategy. It is something that we are going to try, and it turns out that it does. This agrees well with experiment.

In 1939 Wigner asked, “what constraints do I get if I constrain the constraints of quantum mechanics with special relativity.” It turns out that in the non-relativistic limit, we get just this.

There’s a subtlety here, because we get into some logical trouble with the photon with a rest mass of zero ($m = 0$ is certainly allowed as a value of our invariant $m$ above). We can’t stop or slow down a photon, so orbital angular momentum is only a conceptual idea. Really, the orbital angular momentum and the spin angular momentum cannot be separated out for a photon, so talking of a spin $1$ particle really means spin as in $\mathbf{J}$, and not spin as in $\mathbf{L}$.

## Spin $1/2$ particles

Reading: See section 26.6 in the text [1].

Let’s start talking about the simplest case. This includes electrons, all leptons (integer spin particles like photons and the weakly interacting W and Z bosons), and quarks.

\begin{aligned}s &= \frac{1}{{2}} \\ m_s &= \pm \frac{1}{{2}}\end{aligned} \hspace{\stretch{1}}(2.22)

states

\begin{aligned}{\left\lvert {s m_s} \right\rangle} = {\left\lvert { \frac{1}{{2}}, \frac{1}{{2}} } \right\rangle},{\left\lvert { \frac{1}{{2}}, -\frac{1}{{2}} } \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.24)

Note there is a convention

\begin{aligned}{\left\lvert { \frac{1}{{2}} \bar{\frac{1}{{2}}} } \right\rangle} &= {\left\lvert { \frac{1}{{2}}, -\frac{1}{{2}} } \right\rangle} \\ {\left\lvert { \frac{1}{{2}} \frac{1}{{2}} } \right\rangle} &= {\left\lvert { \frac{1}{{2}} \frac{1}{{2}} } \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.25)

\begin{aligned}\begin{aligned}S^2 {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} &= \frac{1}{{2}} \left( \frac{1}{{2}} + 1 \right) \hbar^2 {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} \\ &=\frac{3}{4} \hbar^2 {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} \\ \end{aligned}\end{aligned} \hspace{\stretch{1}}(2.27)

\begin{aligned}S_z {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} = m_s \hbar {\left\lvert {\frac{1}{{2}} m_s} \right\rangle} \end{aligned} \hspace{\stretch{1}}(2.28)

For shorthand

\begin{aligned}{\left\lvert { \frac{1}{{2}} \frac{1}{{2}} } \right\rangle} &= {\left\lvert { + } \right\rangle} \\ {\left\lvert { \frac{1}{{2}} \bar{\frac{1}{{2}}} } \right\rangle} &= {\left\lvert { - } \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.29)

\begin{aligned}S^2 \rightarrow \frac{3}{4} \hbar^2 \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.31)

\begin{aligned}S_z \rightarrow \frac{\hbar}{2}\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.32)

One can easily work out from the commutation relationships that

\begin{aligned}S_x \rightarrow \frac{\hbar}{2}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.33)

\begin{aligned}S_y \rightarrow \frac{\hbar}{2}\begin{bmatrix}0 & -i \\ i & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.34)

We’ll start with adding $\mathbf{L}$ into the mix on Wednesday.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

## Dirac spinor notes.

Posted by peeterjoot on June 11, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

I was having algebraic trouble verifying orthonormality relationships for spinor solutions to the Dirac free particle equation, and initially started preparing these notes to post a question to physicsforums. However, in the process of doing so, I spotted my error. A side effect of making these notes is that I got a nice summary of some of the relationships, and it was a good starting point for some personal notes expanding on the content of these chapters.

# Context for the original question.

In Desai’s QM book [1], the non-covariant form of the free particle equation is developed as

\begin{aligned}\begin{bmatrix}E - m & - \boldsymbol{\sigma} \cdot \mathbf{p} \\ - \boldsymbol{\sigma} \cdot \mathbf{p} & E + m\end{bmatrix}u= 0,\end{aligned} \hspace{\stretch{1}}(2.1)

where each block in the matrix above is two by two. Recall that

\begin{subequations}

\begin{aligned}\sigma_1 &= \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \\ \sigma_2 &= \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \\ \sigma_3 &= \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.2a)

\end{subequations}

so

\begin{aligned}\boldsymbol{\sigma} \cdot \mathbf{p} =\begin{bmatrix}p_z & p_x - i p_y \\ p_x + i p_y & - p_z\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.3)

For spin up ${\lvert {+} \rangle}$ and spin down ${\lvert {-} \rangle}$ states, the positive energy solutions $E = {\left\lvert{E}\right\rvert} = \sqrt{\mathbf{p}^2 + m^2}$ are found to be

\begin{aligned}u^{\pm}(\mathbf{p}) =\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}{\lvert {\pm} \rangle} \\ \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {\pm} \rangle}\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.4)

and the negative energy states associated with $E = -{\left\lvert{E}\right\rvert} = -\sqrt{\mathbf{p}^2 + m^2}$ are found to be

\begin{aligned}v^{\pm}(\mathbf{p}) =\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}-\frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {\pm} \rangle} \\ {\lvert {\pm} \rangle} \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.5)

The z-axis spin up state ${\lvert {+} \rangle} = (1, 0)$ and spin down state ${\lvert {-} \rangle} = (0, 1)$ are also used to find one specific set of states for the positive energy solutions

\begin{subequations}

\begin{aligned}u^{+}(\mathbf{p}) &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}1 \\ 0 \\ \frac{p_z}{{\left\lvert{E}\right\rvert} + m} \\ \frac{p_x + i p_y}{{\left\lvert{E}\right\rvert} + m} \\ \end{bmatrix} \\ u^{-}(\mathbf{p}) &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}0 \\ 1 \\ \frac{p_x - i p_y}{{\left\lvert{E}\right\rvert} + m} \\ -\frac{p_z}{{\left\lvert{E}\right\rvert} + m} \\ \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.6a)

\end{subequations}

and negative energy solutions
\begin{subequations}

\begin{aligned}v^{+}(\mathbf{p}) &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}-\frac{p_z}{{\left\lvert{E}\right\rvert} + m} \\ -\frac{p_x + i p_y}{{\left\lvert{E}\right\rvert} + m} \\ 1 \\ 0 \\ \end{bmatrix} \\ v^{-}(\mathbf{p}) &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}-\frac{p_x - i p_y}{{\left\lvert{E}\right\rvert} + m} \\ \frac{p_z}{{\left\lvert{E}\right\rvert} + m} \\ 0 \\ 1 \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.7a)

\end{subequations}

(the book uses $u^{\pm}$ for both the negative energy states, but I’ve used $v^{\pm}$ here for the negative states for consistency with the covariant equation solutions).

Later a complete set of states $u_r(\mathbf{p}), v_r(\mathbf{p})$ are identified as solutions to the covariant Dirac equations $(\gamma \cdot p -m)u = 0$, $(\gamma \cdot p + m) v = 0$, where $p^\mu = (\mathbf{p}, {\left\lvert{E}\right\rvert})$ as follows

\begin{subequations}

\begin{aligned}u_1(\mathbf{p}) &= u^{+}(\mathbf{p}) \\ u_2(\mathbf{p}) &= u^{-}(\mathbf{p}) \\ v_1(\mathbf{p}) &= v^{+}(-\mathbf{p}) \\ v_2(\mathbf{p}) &= v^{-}(-\mathbf{p}),\end{aligned} \hspace{\stretch{1}}(2.8a)

\end{subequations}

Note very carefully the sign change above. This is important, since without that we do not have a zero inner product between all $u_r$ and $v_s$ states. Spelled out explicitly, these states for the z-axis spin up case are

\begin{subequations}

\begin{aligned}u_1(\mathbf{p}) &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}1 \\ 0 \\ \frac{p_z}{{\left\lvert{E}\right\rvert} + m} \\ \frac{p_x + i p_y}{{\left\lvert{E}\right\rvert} + m} \\ \end{bmatrix} \\ u_2(\mathbf{p}) &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}0 \\ 1 \\ \frac{p_x - i p_y}{{\left\lvert{E}\right\rvert} + m} \\ -\frac{p_z}{{\left\lvert{E}\right\rvert} + m} \\ \end{bmatrix} \\ v_1(\mathbf{p}) &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}\frac{p_z}{{\left\lvert{E}\right\rvert} + m} \\ \frac{p_x + i p_y}{{\left\lvert{E}\right\rvert} + m} \\ 1 \\ 0 \\ \end{bmatrix} \\ v_2(\mathbf{p}) &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}\frac{p_x - i p_y}{{\left\lvert{E}\right\rvert} + m} \\ -\frac{p_z}{{\left\lvert{E}\right\rvert} + m} \\ 0 \\ 1 \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.9a)

\end{subequations}

In order to construct a covariant current conservation relationship a quantity, the Dirac adjoint, was defined as

\begin{aligned}\bar{\psi} = \psi^\dagger \gamma^4,\end{aligned} \hspace{\stretch{1}}(2.10)

where

\begin{aligned}\gamma^4 = \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.11)

This Dirac adjoint can be used to form an inner product of the form

\begin{aligned}\bar{\psi}\psi\end{aligned} \hspace{\stretch{1}}(2.12)

It’s claimed in the text that we have $\bar{u_r} u_s = \delta_{rs}$, $\bar{v_r} v_s = \delta_{rs}$, and $\bar{u_r} v_s = 0$. Let’s verify all these relationships.

# Some checks.

## Verify the non-covariant solutions.

A non-relativistic approximation argument was used to determine the solutions 2.6a, but we can verify that these hold generally by substitution. For example, for the positive energy z-axis spin up state we have

\begin{aligned}&\begin{bmatrix}E - m & - \boldsymbol{\sigma} \cdot \mathbf{p} \\ - \boldsymbol{\sigma} \cdot \mathbf{p} & E + m\end{bmatrix}u^{+}(\mathbf{p}) \\ &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}E - m & 0 & -p_z & -p_x + i p_y \\ 0 & E - m & -p_x - i p_y & p_z \\ -p_z & -p_x + i p_y & E + m & 0 \\ -p_x - i p_y & p_z & 0 & E + m \end{bmatrix}\begin{bmatrix}1 \\ 0 \\ \frac{p_z}{{\left\lvert{E}\right\rvert} + m} \\ \frac{p_x + i p_y}{{\left\lvert{E}\right\rvert} + m} \\ \end{bmatrix} \\ &\sim \begin{bmatrix}E^2 - m^2 - p_x^2 - p_y^2 - p_z^2 \\ -(p_x + i p_y) p_z + p_z (p_x + i p_y) \\ - p_z( E + m ) + p_z( E + m ) \\ -(p_x + i p_y) (E + m) + (E + m)(p_x + i p_y)\end{bmatrix} \\ &= 0.\end{aligned}

Here the relationship between the free particle’s energy and momentum $E^2 - m^2 - \mathbf{p}^2 = 0$ has been used, so we have a zero as desired, and no non-relativistic approximations are required. We can show this generally too, without requiring the specifics of the z-axis spin up or down solutions. This is actually even easier. For the positive energy solutions 2.4 we have

\begin{aligned}\begin{bmatrix}E - m & - \boldsymbol{\sigma} \cdot \mathbf{p} \\ - \boldsymbol{\sigma} \cdot \mathbf{p} & E + m\end{bmatrix}u&\sim\begin{bmatrix}E - m & - \boldsymbol{\sigma} \cdot \mathbf{p} \\ - \boldsymbol{\sigma} \cdot \mathbf{p} & E + m\end{bmatrix}\begin{bmatrix}(E + m) {\lvert {\pm} \rangle} \\ (\boldsymbol{\sigma} \cdot \mathbf{p}) {\lvert {\pm} \rangle}\end{bmatrix} \\ &=\begin{bmatrix}(E^2 - m^2 - (\boldsymbol{\sigma} \cdot \mathbf{p})^2) {\lvert {\pm} \rangle} \\ 0 {\lvert {\pm} \rangle}\end{bmatrix} \\ &=\begin{bmatrix}(E^2 - m^2 - \mathbf{p}^2) {\lvert {\pm} \rangle} \\ 0 {\lvert {\pm} \rangle}\end{bmatrix} \\ &=0,\end{aligned}

where the identity $(\boldsymbol{\sigma} \cdot \mathbf{p})^2 = \mathbf{p}^2$ has been used. For the negative energy solutions 2.5 we have

\begin{aligned}\begin{bmatrix}E - m & - \boldsymbol{\sigma} \cdot \mathbf{p} \\ - \boldsymbol{\sigma} \cdot \mathbf{p} & E + m\end{bmatrix}u&\sim\begin{bmatrix}E - m & - \boldsymbol{\sigma} \cdot \mathbf{p} \\ - \boldsymbol{\sigma} \cdot \mathbf{p} & E + m\end{bmatrix}\begin{bmatrix}-(\boldsymbol{\sigma} \cdot \mathbf{p}) {\lvert {\pm} \rangle} \\ (-E + m) {\lvert {\pm} \rangle} \\ \end{bmatrix} \\ &=\begin{bmatrix}0 {\lvert {\pm} \rangle} \\ (-E^2 + m^2 + (\boldsymbol{\sigma} \cdot \mathbf{p})^2) {\lvert {\pm} \rangle} \\ \end{bmatrix} \\ &=0.\end{aligned}

## Is there something special about the z-axis orientation?

Why was the z-axis spin orientation picked? It doesn’t seem to me that there would be any reason for this. For y-axis spin, recall that our eigenstates are

\begin{aligned}{\lvert {\pm} \rangle}=\frac{1}{{\sqrt{2}}}\begin{bmatrix}1 \\ \pm i\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.13)

Our positive energy states should therefore be

\begin{aligned}u^{\pm}(\mathbf{p}) &\sim\begin{bmatrix}\begin{bmatrix}1 \\ \pm i\end{bmatrix} \\ \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} \begin{bmatrix}1 \\ \pm i\end{bmatrix} \end{bmatrix} \\ &=\begin{bmatrix}1 \\ \pm i \\ \frac{1}{{\left\lvert{E}\right\rvert} + m} \begin{bmatrix}p_z & p_x - i p_y \\ p_x + i p_y & - p_z\end{bmatrix}\begin{bmatrix}1 \\ \pm i\end{bmatrix} \end{bmatrix} \\ &\sim\begin{bmatrix}E + m \\ \pm i (E + m) \\ p_z \pm i p_x \pm p_y \\ p_x + i p_y \mp p_z \end{bmatrix}\end{aligned}

It is straightforward to verify that these are solutions. We find for example that

\begin{aligned}\begin{bmatrix}E - m & - \boldsymbol{\sigma} \cdot \mathbf{p} \\ - \boldsymbol{\sigma} \cdot \mathbf{p} & E + m\end{bmatrix}u^{+}\sim \begin{bmatrix}E^2 - m^2 - \mathbf{p}^2 \\ i (E^2 - m^2 - \mathbf{p}^2 ) \\ 0 \\ 0\end{bmatrix}= 0,\end{aligned} \hspace{\stretch{1}}(3.14)

as expected. What’s the general solution? For

\begin{aligned}\mathbf{n} = \begin{bmatrix}\sin\theta \cos\phi \\ \sin\theta \sin\phi \\ \cos\theta \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(3.15)

we find

\begin{aligned}\boldsymbol{\sigma} \cdot \mathbf{n} =\begin{bmatrix}\cos\theta & \sin\theta e^{-i\phi} \\ \sin\theta e^{i\phi} & -\cos\theta\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(3.16)

with eigenstates
\begin{subequations}

\begin{aligned}{\lvert {+} \rangle} &=\begin{bmatrix}\cos(\theta/2) e^{-i\phi/2} \\ \sin(\theta/2) e^{i\phi/2} \\ \end{bmatrix} \\ {\lvert {-} \rangle} &=\begin{bmatrix}-\sin(\theta/2) e^{-i\phi/2} \\ \cos(\theta/2) e^{i\phi/2} \\ \end{bmatrix} \end{aligned} \hspace{\stretch{1}}(3.17a)

\end{subequations}

Should we wish to consider an arbitrarily oriented spin, expressing $\mathbf{p}$ in spherical coordinates also makes sense

\begin{aligned}\mathbf{p} = {\left\lvert{\mathbf{p}}\right\rvert}\begin{bmatrix}\sin\alpha \cos\beta \\ \sin\alpha \sin\beta \\ \cos\alpha \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.18)

and we find (with $S$ and $C$ for $\sin$ and $\cos$ respectively)

\begin{subequations}

\begin{aligned}\boldsymbol{\sigma} \cdot \mathbf{p} {\lvert {+} \rangle}&={\left\lvert{\mathbf{p}}\right\rvert}\begin{bmatrix} C_\alpha C_{\theta/2} e^{-i \phi} + S_\alpha S_{\theta/2} e^{-i \beta} \\ S_\alpha C_{\theta/2} e^{i (\beta - \phi)} - C_\alpha S_{\theta/2} \end{bmatrix} \\ \boldsymbol{\sigma} \cdot \mathbf{p} {\lvert {-} \rangle}&={\left\lvert{\mathbf{p}}\right\rvert}\begin{bmatrix}- C_\alpha S_{\theta/2} e^{-i \phi} + S_\alpha C_{\theta/2} e^{-i \beta} \\ - S_\alpha S_{\theta/2} e^{i (\beta - \phi)} - C_\alpha C_{\theta/2} \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.19a)

\end{subequations}

Substitution back into 2.4, and 2.5 is then easy. Expressing these with the angles expressed as sums and differences is strongly suggested. With $\Delta = (\beta - \phi)/2$, and $\delta = (\beta + \phi)/2$ this gives

\begin{subequations}

\begin{aligned}\boldsymbol{\sigma} \cdot \mathbf{p} {\lvert {+} \rangle}&={\left\lvert{\mathbf{p}}\right\rvert}\begin{bmatrix}e^{-i\delta}\left(C_{\alpha - \theta/2} C_\Delta + i C_{\alpha + \theta/2} S_\Delta \right) \\ e^{i \Delta}\left(S_{\alpha - \theta/2} C_\Delta + i S_{\alpha + \theta/2} S_\Delta \right) \\ \end{bmatrix} \\ \boldsymbol{\sigma} \cdot \mathbf{p} {\lvert {-} \rangle}&={\left\lvert{\mathbf{p}}\right\rvert}\begin{bmatrix}e^{-i\delta}\left(S_{\alpha - \theta/2} C_\Delta - i S_{\alpha + \theta/2} S_\Delta \right) \\ e^{i \Delta}\left(-C_{\alpha - \theta/2} C_\Delta + i C_{\alpha + \theta/2} S_\Delta \right) \\ \end{bmatrix} \end{aligned} \hspace{\stretch{1}}(3.20a)

\end{subequations}

This is probably about as tidy as things can be made for the general case.

## Expanding the current equation.

With

\begin{aligned}\mathbf{j} = \psi^\dagger \boldsymbol{\alpha} \psi = \begin{bmatrix}u_1^\dagger & u_2^\dagger\end{bmatrix}\begin{bmatrix}0 & \boldsymbol{\sigma} \\ \boldsymbol{\sigma} & 0 \end{bmatrix}\begin{bmatrix}u_1 \\ u_2\end{bmatrix}= u_1^\dagger \boldsymbol{\sigma} u_2 + u_2^\dagger \boldsymbol{\sigma} u_1\end{aligned} \hspace{\stretch{1}}(3.21)

We can expand the current for a general spin up or spin down state ${\lvert {r} \rangle}$ with respect to either the positive energy or negative energy solutions.

Those (normalized) solutions are respectively

\begin{aligned}\psi_{+} &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}{\lvert {r} \rangle} \\ \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert}+ m} {\lvert {r} \rangle} \\ \end{bmatrix} \\ \psi_{-} &=\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}-\frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert}+ m} {\lvert {r} \rangle} \\ {\lvert {r} \rangle} \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.22)

For the $i$th component of the positive energy solution current we have

\begin{aligned}\psi_{+}^\dagger \boldsymbol{\alpha} \psi_{+}&=\frac{{\left\lvert{E}\right\rvert} + m}{2m}\begin{bmatrix}{\langle {r} \rvert} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m}\end{bmatrix}\begin{bmatrix}\sigma_i \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {r} \rangle} \\ \sigma_i {\lvert {r} \rangle}\end{bmatrix} \\ &=\frac{1}{2m}{\langle {r} \rvert} \left(\sigma_i (\boldsymbol{\sigma} \cdot \mathbf{p})+(\boldsymbol{\sigma} \cdot \mathbf{p}) \sigma_i \right) {\lvert {r} \rangle}\end{aligned}

Similarly for a negative energy solution we have

\begin{aligned}\psi_{-}^\dagger \boldsymbol{\alpha} \psi_{-}&=\frac{{\left\lvert{E}\right\rvert} + m}{2m}\begin{bmatrix}-{\langle {r} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} \rvert}\end{bmatrix}\begin{bmatrix}\sigma_i {\lvert {r} \rangle} \\ -\sigma_i \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {r} \rangle}\end{bmatrix} \\ &=\frac{1}{2m}{\langle {r} \rvert} \left(-\sigma_i (\boldsymbol{\sigma} \cdot \mathbf{p})-(\boldsymbol{\sigma} \cdot \mathbf{p}) \sigma_i \right) {\lvert {r} \rangle}\end{aligned}

We can expand the inner term of both easily

\begin{aligned}\sigma_i (\boldsymbol{\sigma} \cdot \mathbf{p}) + (\boldsymbol{\sigma} \cdot \mathbf{p}) \sigma_i =2 \sigma_i^2 p_i + \sum_{i \ne j} ({\sigma_i \sigma_j + \sigma_j \sigma_i}) p^j\end{aligned} \hspace{\stretch{1}}(3.24)

so that we have for the positive and negative energy solutions currents of

\begin{aligned}j_i &= {\langle {r} \rvert} \frac{p_i}{m} {\lvert {r} \rangle} \\ j_i &= -{\langle {r} \rvert} \frac{p_i}{m} {\lvert {r} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.25)

This finds the velocity dependence noted in section 33.4, but does not require taking any specific spin orientation, nor any specific momentum direction.

## Unpacking the covariant equation.

Pre-multiplication of the covariant Dirac equation by $\gamma^4$ should provide a space-time split of the Dirac equation. Let’s verify this

\begin{aligned}\gamma^4 (\gamma \cdot p - m)&=\gamma^4 (\gamma_\mu p^\mu - m) \\ &=E\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} + \gamma^4 \gamma_a p^a - m \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix},\end{aligned}

but

\begin{aligned}\gamma^4 \gamma_a =\begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix}0 & -\sigma_a \\ \sigma_a & 0 \end{bmatrix}=\begin{bmatrix}0 & -\sigma_a \\ -\sigma_a & 0 \end{bmatrix}\end{aligned}

\begin{aligned}\gamma^4 (\gamma \cdot p - m)=E\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix}0 & \sigma_a \\ \sigma_a & 0 \end{bmatrix}p^a - m \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} =\begin{bmatrix}E - m & - \sigma \cdot \mathbf{p} \\ - \sigma \cdot \mathbf{p} & E + m\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.27)

This recovers 2.1 as expected.

## Two by two form for the covariant equations.

If we put the covariant Dirac equations in two by two matrix form we get

\begin{aligned}0&= (\gamma \cdot p - m ) u \\ &= \left({\left\lvert{E}\right\rvert} \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}+ \begin{bmatrix}0 & - \sigma_a \\ \sigma_a & 0\end{bmatrix}p^a- m\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\right) u \\ &=\begin{bmatrix}{\left\lvert{E}\right\rvert} - m & - \boldsymbol{\sigma} \cdot \mathbf{p} \\ \boldsymbol{\sigma} \cdot \mathbf{p} & -{\left\lvert{E}\right\rvert} - m\end{bmatrix} u\end{aligned}

and

\begin{aligned}0 &= (\gamma \cdot p + m ) v \\ &= \left({\left\lvert{E}\right\rvert} \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}+ \begin{bmatrix}0 & - \sigma_a \\ \sigma_a & 0\end{bmatrix}p^a+ m\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\right) v \\ &=\begin{bmatrix}{\left\lvert{E}\right\rvert} + m & - \boldsymbol{\sigma} \cdot \mathbf{p} \\ \boldsymbol{\sigma} \cdot \mathbf{p} & -{\left\lvert{E}\right\rvert} + m\end{bmatrix} v\end{aligned}

This form makes it easy to verify that our solutions are

\begin{aligned}u_r(\mathbf{p}) =\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}{\lvert {r} \rangle} \\ \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {r} \rangle}\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(3.28)

and

\begin{aligned}v_r(\mathbf{p}) =\sqrt{\frac{{\left\lvert{E}\right\rvert} + m}{2m}}\begin{bmatrix}\frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {r} \rangle} \\ {\lvert {r} \rangle} \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.29)

It’s curious to consider these part of a basis for a single equation. I suppose that all together they are actually eigenstates of the equation

\begin{aligned}(\gamma \cdot p + m) (\gamma \cdot p - m) u = ((\gamma \cdot p)^2 - m^2) u = 0,\end{aligned} \hspace{\stretch{1}}(3.30)

or

\begin{aligned}(\gamma \cdot p - m) (\gamma \cdot p + m) v = ((\gamma \cdot p)^2 - m^2) v = 0,\end{aligned} \hspace{\stretch{1}}(3.31)

which have the form of the Klein-Gordan equation.

## Orthonormality.

Orthonormality for the $u$ vectors is easy to show, and we can do so without requiring any specific spin orientation

\begin{aligned}\bar{u}_r u_s &= \frac{{\left\lvert{E}\right\rvert} + m}{2m}\begin{bmatrix}{\langle {r} \rvert} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} \end{bmatrix}\gamma^4\begin{bmatrix}{\lvert {s} \rangle} \\ \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {s} \rangle}\end{bmatrix} \\ &=\frac{{\left\lvert{E}\right\rvert} + m}{2m}\begin{bmatrix}{\langle {r} \rvert} &-{\langle {r} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} \end{bmatrix}\begin{bmatrix}{\lvert {s} \rangle} \\ \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {s} \rangle}\end{bmatrix} \\ &=\frac{1}{2m({\left\lvert{E}\right\rvert} + m)}\left\langle{{r}} \vert {{s}}\right\rangle \left( E^2 + m^2 + 2 {\left\lvert{E}\right\rvert} m - \mathbf{p}^2 \right) \\ &=\left\langle{{r}} \vert {{s}}\right\rangle.\end{aligned}

It’s also easy for $v$ vectors

\begin{aligned}\bar{v}_r v_s &= \frac{{\left\lvert{E}\right\rvert} + m}{2m}\begin{bmatrix}{\langle {r} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} \rvert} \end{bmatrix}\gamma^4\begin{bmatrix}\frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {s} \rangle} \\ {\lvert {s} \rangle} \end{bmatrix} \\ &=\frac{{\left\lvert{E}\right\rvert} + m}{2m}\begin{bmatrix}{\langle {r} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} &-{\langle {r} \rvert} \end{bmatrix}\begin{bmatrix}\frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {s} \rangle} \\ {\lvert {s} \rangle} \end{bmatrix} \\ &=-\frac{1}{2m({\left\lvert{E}\right\rvert} + m)}\left\langle{{r}} \vert {{s}}\right\rangle \left( E^2 + m^2 + 2 {\left\lvert{E}\right\rvert} m - \mathbf{p}^2 \right) \\ &=-\left\langle{{r}} \vert {{s}}\right\rangle.\end{aligned}

For the cross terms we have

\begin{aligned}\bar{u}_r v_s &= \frac{{\left\lvert{E}\right\rvert} + m}{2m}\begin{bmatrix}{\langle {r} \rvert} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} \end{bmatrix}\gamma^4\begin{bmatrix}\frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {s} \rangle} \\ {\lvert {s} \rangle} \end{bmatrix} \\ &= \frac{{\left\lvert{E}\right\rvert} + m}{2m}\begin{bmatrix}{\langle {r} \rvert} &-{\langle {r} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} \end{bmatrix}\begin{bmatrix}\frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {s} \rangle} \\ {\lvert {s} \rangle} \end{bmatrix} \\ &=\frac{1}{2m}{\langle {r} \rvert} ( \boldsymbol{\sigma} \cdot \mathbf{p} - \boldsymbol{\sigma} \cdot \mathbf{p} ) {\lvert {s} \rangle} \\ &= 0\end{aligned}

## Resolution of identity.

It’s claimed that an identity representation is

\begin{aligned}\mathbf{1} = \sum_r u_r \bar{u}_r - v_r \bar{v}_r\end{aligned} \hspace{\stretch{1}}(3.32)

This makes some sense, but we can see systematically why we have this negative sign. Suppose that we have a basis ${\lvert {a_i} \rangle}$ for which we have $\left\langle{{a_i}} \vert {{a_j}}\right\rangle = \pm \delta_{ij}$ (rather than the strict orthonormality condition $\left\langle{{a_i}} \vert {{a_j}}\right\rangle = \delta_{ij}$). Consider the calculation of the Fourier coefficients of a state

\begin{aligned}{\lvert {a} \rangle} = \alpha_i {\lvert {a_i} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.33)

We have

\begin{aligned}\left\langle{{a_j}} \vert {{a}}\right\rangle = \alpha_i \left\langle{{a_j}} \vert {{a_i}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(3.34)

For $i \ne j$ $\left\langle{{a_i}} \vert {{a_j}}\right\rangle = 0$, so that the coefficient is

\begin{aligned}\alpha_j =\frac{\left\langle{{a_j}} \vert {{a}}\right\rangle}{\left\langle{{a_j}} \vert {{a_j}}\right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.35)

The coordinate representation of this state vector with respect to this basis is thus

\begin{aligned}{\lvert {a} \rangle} = \sum_i \left( \frac{\left\langle{{a_i}} \vert {{a}}\right\rangle}{\left\langle{{a_i}} \vert {{a_i}}\right\rangle} \right){\lvert {a_i} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.36)

Shuffling things around, employing the somewhat abusive seeming Dirac ket-bra operator notation, we find the general identity operation takes the form

\begin{aligned}{\lvert {a} \rangle} = \left( \frac{{\lvert {a_i} \rangle} {\langle {a_i} \rvert} }{\left\langle{{a_i}} \vert {{a_i}}\right\rangle} \right) {\lvert {a} \rangle},\end{aligned} \hspace{\stretch{1}}(3.37)

so that the identity itself has the form

\begin{aligned}\mathbf{1} = \frac{{\lvert {a_i} \rangle} {\langle {a_i} \rvert} }{\left\langle{{a_i}} \vert {{a_i}}\right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.38)

This is the sum of all the ket-bras for which the braket is one, minus the sum of all the ket-bras for which the braket is negative, showing that the form of the claimed identity is justified.

We can also verify this directly by computation, and find

\begin{aligned}\sum_r u_r \bar{u}_r &=\frac{{\left\lvert{E}\right\rvert} + m}{2m}\sum_r \begin{bmatrix}{\lvert {r} \rangle} \\ \frac{\boldsymbol{\sigma} \cdot \mathbf{p} {\lvert {r} \rangle}}{{\left\lvert{E}\right\rvert} + m}\end{bmatrix}\begin{bmatrix}{\langle {r} \rvert} &-\frac{{\langle {r} \rvert} \boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m}\end{bmatrix} \\ &=\frac{{\left\lvert{E}\right\rvert} + m}{2m}\sum_r \begin{bmatrix}{\lvert {r} \rangle}{\langle {r} \rvert} & -{\lvert {r} \rangle}{\langle {r} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} \\ \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {r} \rangle}{\langle {r} \rvert} &-\frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} {\lvert {r} \rangle}{\langle {r} \rvert} \frac{\boldsymbol{\sigma} \cdot \mathbf{p}}{{\left\lvert{E}\right\rvert} + m} \\ \end{bmatrix}\end{aligned}

We can pull the summation into the matrices and note that $\sum_r {\lvert {r} \rangle}{\langle {r} \rvert} = \mathbf{1}$ (the two by two identity), so that we are left with

\begin{aligned}\sum_r u_r \bar{u}_r =\frac{1}{{2m}}\begin{bmatrix}{\left\lvert{E}\right\rvert} + m & -\boldsymbol{\sigma} \cdot \mathbf{p} \\ \boldsymbol{\sigma} \cdot \mathbf{p} &-\frac{\mathbf{p}^2}{{\left\lvert{E}\right\rvert} + m} \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.39)

Similarly, we find

\begin{aligned}-\sum_r v_r \bar{v}_r =\frac{1}{{2m}}\begin{bmatrix}-\frac{\mathbf{p}^2}{{\left\lvert{E}\right\rvert} + m} & \boldsymbol{\sigma} \cdot \mathbf{p} \\ -\boldsymbol{\sigma} \cdot \mathbf{p} \right\rvert} + m \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(3.40)

summing the two (noting that $E^2 - \mathbf{p}^2 - m^2 = 0$) we get the block identity matrix as desired.

We’ve also just calculated the projection operators. Let’s verify that expanding the covariant form in the text produces the same result

\begin{aligned}\frac{1}{{2m}}(m \pm \gamma \cdot p) &=\frac{1}{{2m}}(m \pm \gamma^4 {\left\lvert{E}\right\rvert} \pm \gamma_a p^a ) \\ &=\frac{1}{{2m}}\left(m\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\pm {\left\lvert{E}\right\rvert}\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\pm p^a\begin{bmatrix}0 & -\sigma_a \\ \sigma_a & 0 \end{bmatrix}\right) \\ &=\frac{1}{{2m}}\begin{bmatrix}m \pm {\left\lvert{E}\right\rvert} & \mp \boldsymbol{\sigma} \cdot \mathbf{p} \\ \pm \boldsymbol{\sigma} \cdot \mathbf{p} & m \mp {\left\lvert{E}\right\rvert} \end{bmatrix}\end{aligned}

Now compare to 3.39, and 3.40, which we rewrite using $-\mathbf{p}^2/(m + {\left\lvert{E}\right\rvert}) = m - {\left\lvert{E}\right\rvert}$ as

\begin{aligned}\sum_r u_r \bar{u}_r &=\frac{1}{{2m}}\begin{bmatrix}{\left\lvert{E}\right\rvert} + m & -\boldsymbol{\sigma} \cdot \mathbf{p} \\ \boldsymbol{\sigma} \cdot \mathbf{p} &m - {\left\lvert{E}\right\rvert}\end{bmatrix} \\ -\sum_r v_r \bar{v}_r &=\frac{1}{{2m}}\begin{bmatrix}m - {\left\lvert{E}\right\rvert} & \boldsymbol{\sigma} \cdot \mathbf{p} \\ -\boldsymbol{\sigma} \cdot \mathbf{p} \right\rvert} + m \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.41)

## Lorentz transformation of Dirac equation.

Equation (35.107) in the text is missing the positional notation to show the placement of the indexes, and should be

\begin{aligned}\left[{\Sigma},{\gamma^\nu}\right] = e_\mu^{.\nu} \gamma^\mu,\end{aligned} \hspace{\stretch{1}}(3.45)

where the solution is

\begin{aligned}\Sigma = \frac{1}{{4}} \gamma^\alpha \gamma^\beta e_{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(3.45)

This does have the form I’d expect, a bivector, but we can show explicitly that this is the solution without too much trouble. Consider the commutator

\begin{aligned}\left[{ \gamma^\alpha \gamma^\beta e_{\alpha \beta} },{\gamma^\nu}\right]&=e_{\alpha \beta} \left[{ \gamma^\alpha \gamma^\beta },{\gamma^\nu}\right] \\ &=e_{\alpha \beta} \left( \gamma^\alpha \gamma^\beta \gamma^\nu-\gamma^\nu \gamma^\alpha \gamma^\beta \right) \\ &=e_{\alpha \beta} \left( \left( {\gamma^\alpha \cdot \gamma^\beta }+\gamma^\alpha \wedge \gamma^\beta \right) \gamma^\nu-\gamma^\nu \left( {\gamma^\alpha \cdot \gamma^\beta }+\gamma^\alpha \wedge \gamma^\beta \right) \right) \\ &=e_{\alpha \beta} \left(\left(\gamma^\alpha \wedge \gamma^\beta \right)\gamma^\nu-\gamma^\nu \left(\gamma^\alpha \wedge \gamma^\beta \right)\right)\\ &=e_{\alpha \beta} \left(\left(\gamma^\alpha \wedge \gamma^\beta \right) \wedge \gamma^\nu-\gamma^\nu \wedge \left(\gamma^\alpha \wedge \gamma^\beta \right)\right)+e_{\alpha \beta} \left(\left(\gamma^\alpha \wedge \gamma^\beta \right) \cdot \gamma^\nu-\gamma^\nu \cdot \left(\gamma^\alpha \wedge \gamma^\beta \right)\right)\\ &=2 e_{\alpha \beta} \left(\gamma^\alpha \wedge \gamma^\beta \right) \cdot \gamma^\nu\\ &=2 e^{\alpha \beta} \left(\gamma_\alpha \wedge \gamma_\beta \right) \cdot \gamma^\nu\\ &=2 e^{\alpha \beta} \left(\gamma_\alpha \delta_\beta^{.\nu}-\gamma_\beta \delta_\alpha^{.\nu}\right)\\ &=4 e^{\alpha \nu} \gamma_\alpha \\ &=4 e_\alpha^{.\nu} \gamma^\alpha \\ \end{aligned}

Would this be any easier to prove without utilizing the dot and wedge product identities? I used a few of them, starting with

\begin{aligned}a \cdot b &= \frac{1}{{2}} (a b + b a) = \frac{1}{{2}} \left\{{a},{b}\right\} \\ a \wedge b &= \frac{1}{{2}} (a b - b a) = \frac{1}{{2}} \left[{a},{b}\right] \\ a b &= a \cdot b + a \wedge b = \frac{1}{{2}} ( \left\{{a},{b}\right\} + \left[{a},{b}\right] )\end{aligned} \hspace{\stretch{1}}(3.45)

In matrix notation we would have to show that the anticommutator $\left\{{\gamma^\alpha},{\gamma^\beta}\right\}$ commutes with any $\gamma^\nu$ to make the first cancellation. We can do so by noting

\begin{aligned}\left[{\gamma^\alpha \gamma^\beta + \gamma^\beta \gamma^\alpha},{\gamma^\nu}\right] &= \left[{ 2 g^{\alpha \beta} \mathbf{1}},{\gamma^\nu}\right] \\ &= 2 g^{\alpha \beta} \left[{\mathbf{1}},{\gamma^\nu}\right] \\ &= 0\end{aligned}

That’s enough to get us on the path to how to prove this in matrix form

\begin{aligned}\left[{ \gamma^\alpha \gamma^\beta e_{\alpha \beta} },{\gamma^\nu}\right]&=e_{\alpha \beta} \left[{ \gamma^\alpha \gamma^\beta },{\gamma^\nu}\right] \\ &=e_{\alpha \beta} \left( \gamma^\alpha \gamma^\beta \gamma^\nu-\gamma^\nu \gamma^\alpha \gamma^\beta \right) \\ &=\frac{1}{{2}} e_{\alpha \beta} \left( \left( \left\{{\gamma^\alpha},{\gamma^\beta}\right\}+\left[{\gamma^\alpha },{ \gamma^\beta }\right]\right) \gamma^\nu-\gamma^\nu \left( \left\{{\gamma^\alpha },{\gamma^\beta }\right\}+\left[{\gamma^\alpha },{\gamma^\beta }\right]\right) \right) \\ &=\frac{1}{{2}} e_{\alpha \beta} \left( \left[{\gamma^\alpha },{ \gamma^\beta }\right] \gamma^\nu-\gamma^\nu \left[{\gamma^\alpha },{ \gamma^\beta }\right] \right) \\ &=\frac{1}{{2}} e_{\alpha \beta} \left[{\left[{\gamma^\alpha },{ \gamma^\beta }\right] },{\gamma^\nu}\right] \\ &=\frac{1}{{2}} e_{\alpha \beta} \left[{\left[{\gamma^\alpha },{ \gamma^\beta }\right] },{\gamma^\nu}\right] \\ &=\frac{1}{{2}} e_{\alpha \beta} \left[{\gamma^\alpha \gamma^\beta -\gamma^\beta \gamma^\alpha },{\gamma^\nu}\right] \\ &=e_{\alpha \beta} \left[{\gamma^\alpha \gamma^\beta },{\gamma^\nu}\right] \\ &=e_{\alpha \beta} \left(\gamma^\alpha \gamma^\beta \gamma^\nu-\gamma^\nu \gamma^\alpha \gamma^\beta \right) \\ &=e_{\alpha \beta} \left(\gamma^\alpha ( 2 g^{\beta \nu} - \gamma^\nu \gamma^\beta )-\gamma^\nu \gamma^\alpha \gamma^\beta \right) \\ &=e_{\alpha \beta} \left(2 \gamma^\alpha g^{\beta \nu} - \gamma^\alpha \gamma^\nu \gamma^\beta -\gamma^\nu \gamma^\alpha \gamma^\beta \right) \\ &=2 e_{\alpha \beta} \left(\gamma^\alpha g^{\beta \nu} - g^{\alpha \nu} \gamma^\beta \right) \\ &=2 e_{\alpha \beta} \gamma^\alpha g^{\beta \nu} + 2 e_{\beta \alpha} g^{\alpha \nu} \gamma^\beta \\ &=2 e_{\alpha}^{. \nu} \gamma^\alpha + 2 e_{\beta}^{.\nu} \gamma^\beta \\ &=4 e_{\alpha}^{. \nu} \gamma^\alpha \end{aligned}

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

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