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## 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.