# Inner product of particle-anti-particle spinor components

Suppose I have four-component spinors $\Psi$ and $\bar \Psi$ satisfying the Dirac equation with

$$\Psi(\vec x) = \int \frac{\textrm{d}^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 E_{\vec p}}} \sum_{s = \pm \frac{1}{2}} \left[ a^s_p u^s_p e^{i \vec p \cdot \vec x} + \tilde b^s_p v^s_p e^{-i \vec p \cdot \vec x} \right]$$

$$\bar \Psi(\vec x) = \int \frac{\textrm{d}^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 E_{\vec p}}} \sum_{s = \pm \frac{1}{2}} \left[ \tilde a^s_p \tilde u^s_p e^{-i \vec p \cdot \vec x} + b^s_p \tilde v^s_p e^{-i \vec p \cdot \vec x} \right] \gamma^0$$

with these definitions:

$$\tilde a \equiv a^\dagger \quad ; \quad \gamma^0 = \begin{pmatrix} 0 & 1_2 \\ 1_2 & 0 \end{pmatrix} \quad ; \quad u^s_p = \begin{pmatrix} \sqrt{p \cdot \sigma} \xi^s \\ \sqrt{p \cdot \bar \sigma} \xi^s \end{pmatrix} \quad ; \quad v^s_p = \begin{pmatrix} \sqrt{p \cdot \sigma} \eta^s \\ -\sqrt{p \cdot \bar \sigma} \eta^s \end{pmatrix}$$

$$\sigma = (1_2, \vec \sigma) \quad ; \quad \bar \sigma = (1_2 , - \vec \sigma) \quad ; \quad p = (p_0 , \vec p)$$

where $\vec \sigma$ is the usual vector of Pauli matrices and $\xi, \eta$ are two-component spinors.

The lecturer then goes on to choose an appropriate basis for $\xi$

$$\xi^1 = \begin{pmatrix}1 \\ 0\end{pmatrix} \quad ; \quad \xi^2 = \begin{pmatrix} 0 \\ 1\end{pmatrix}$$

and similarly for $\eta$ such that $\tilde \xi^r \xi^s = \delta^{rs}$ and $\tilde \eta^r \eta^s = \delta^{rs}$. This appears sensible and using this, we can compute various inner products of $u$ and $v$. In the examples discussed in the notes, mixed terms, that is, those containing products of $\xi^r$ and $\eta^s$ never occur (multiplied by 0, hence irrelevant) or cancel out.

In an example assignment, however, we are meant to calculate $\bar \psi(-i \gamma^i \partial_i + m) \psi$ which then leads me to terms such as

$$\sum_{s,t} 2 m \tilde a^t_p \tilde b^s_{-p} (p_i \sigma^i) \tilde \xi^t \eta^s$$

which also cancel out in the end, but pose the question in the meantime how $\tilde \xi^t \eta^s$ is actually defined/can be calculated. The identities gained from an appropriate choice of basis for $\eta$ and $\xi$ ‘feel’ wrong here, since, after all, $\xi$ comes from a particle-spinor and $\eta$ comes from an anti-particle spinor. I take it my question could be rephrased as to whether $\xi$ and $\eta$ live in the same space or belong to two different spaces (which would make their inner product an even more thrilling exercise).

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In a coordinate-free description, we get two 4-component spinors $u$ and $v$ in projections to orthogonal (and in particular different) eigenspaces of dimension 2.

But the above chooses suitable bases and then considers the coefficient vectors $\xi$ and $\eta$, which are vectors in the same space $C^2$. Their physical meaning is not determined by this abstract space but by the way they enter in the 4-component spinors.

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Thank you very much. Can I hence conclude that if I choose a basis for $\xi$ as above and the same for $\eta$, $\eta^1 = (1,0)^T$, $\eta^2 = (0,1)^T$, the expression $\tilde \xi^r \eta^s$ is well-defined and equal to $\delta^{rs}$? –  Claudius Nov 19 '12 at 18:06
@Claudius: Yes. –  Arnold Neumaier Nov 19 '12 at 18:09