# 4-vector from a spinor

Currently reading Aitchison's book on SUSY, and on page 35 (section 2.2) he asks the reader to prove that $\bar{\Psi}\gamma^\mu\Psi=\psi^\dagger\sigma^\mu\psi+\chi^\dagger\bar{\sigma}^\mu\chi$ transforms as a 4-vector since the quantities $\psi^\dagger\sigma^\mu\psi$ and $\chi^\dagger\bar{\sigma}^\mu\chi$ transform separately as 4-vectors. I'm familiar with such questions in Dirac/QFT courses, where a more rigorous formalism ($\Lambda^\mu_{\,\nu}$ and $S$ matrices, for instance) is used, but I'm unsure about how to proceed using Aitchison's notation, where we have (for a general Lorentz transformation):

$$V=(1+i\boldsymbol{\epsilon}\cdot\boldsymbol{\sigma}/2-\boldsymbol{\eta}\cdot\boldsymbol{\sigma}/2)$$ $$\psi\to\psi'=V\psi\qquad\chi\to\chi'=V^{\dagger^{-1}}\chi$$

From this, and following e.g. (2.32) in the book, we have:

$$\psi^\dagger\sigma^\mu\psi\to\psi^{\dagger '}\sigma^\mu\psi'=\psi^\dagger V^\dagger\sigma^\mu V\psi$$

and it's not clear to me how this is the transformation law of a 4-vector in terms of V's.

EDIT: I believe this question to differ enough from How to prove that Weyl spinors equations are Lorentz invariant? in term of notation to deserve its own answer (which might very well comment on said notational differences).

• @AccidentalFourierTransform see edit. – Demosthene Dec 31 '16 at 12:14
• Thanks @AccidentalFourierTransform :) Obviously there should be a way of translating this notation into the one of the question you linked and still get the right answer, but I think this would be missing the author's point. So either I'm missing something, and an answer would have educational value, or the author's notation isn't very practical after all, and pointing it out would have informative value :) – Demosthene Dec 31 '16 at 12:24
• @AccidentalFourierTransform up? :) – Demosthene Jun 20 '17 at 21:35

I know this post is 4y old, but I still couldn't find the answer to this question online, and this page still shows up in the first results when googling for this problem.

the problem is nearly identical to exercise (2.3) from Maggiore's "A modern introduction to quantum field theory", which requires you to prove that, given a left-handed or right-handed spinor $$\psi_L$$, $$\psi_R$$, then $${\psi_R}^\dagger \sigma^\mu \psi_R$$ and $${\psi_L}^\dagger {\bar\sigma}^\mu \psi_L$$ transform as 4-vectors: $$$$(\Lambda_R\;{\psi_R})^\dagger \sigma^\mu (\Lambda_R\;\psi_R) = {\Lambda^\mu}_\nu \;{\psi_R}^\dagger \sigma^\nu \psi_R \\ (\Lambda_L\;{\psi_L})^\dagger \bar\sigma^\mu (\Lambda_L\;\psi_L) = {\Lambda^\mu}_\nu \;{\psi_L}^\dagger \bar\sigma^\nu \psi_L$$$$

Where $$\sigma^\mu = (1_2,\sigma^i)$$ and $$\bar\sigma^\mu = (1_2,-\sigma^i)$$. $$\Lambda_R, \Lambda_L$$ are elements of the Lorentz group in the (0,1/2) and (1/2,0) representations, while $${\Lambda^\mu}_\nu$$ is the $$\mu,\nu$$ component of the same element of the group, but in the (1/2,1/2) representation. Hopefully this will help other people that are reading Maggiore or other books that use Aitchison's notation.

A general Lorentz transformation in any representation can be written as

$$U(\Lambda) = exp\{ -i\boldsymbol{\theta}\cdot\boldsymbol{J}+i\boldsymbol{\eta}\cdot\boldsymbol{K}\}$$ where $$\boldsymbol\theta$$ and $$\boldsymbol\eta$$ are two 3-vectors of constants which identify the transformation, and $$\{J^i,K^i\}_{i=1,2,3}$$ are the generators of the Lorentz group ($$J^i$$ generate the rotations, while $$K^i$$ generate the boosts). In a representation of dimension n, they are nxn matrices.

I will demonstrate only the left-handed identity (the procedure is identical for the right-handed one). In the (1/2,0) representation, $$\boldsymbol J = \boldsymbol\sigma/2$$ and $$\boldsymbol K = i\boldsymbol\sigma/2$$, therefore $$\Lambda_L = exp\{ (-i\boldsymbol\theta - \boldsymbol\eta)\cdot\boldsymbol{\sigma}/2\}$$

Since the left-handed equation must hold for every left-handed spinor $$\psi_L$$, then we have to prove that

$${\Lambda_L}^\dagger \bar\sigma^\mu \Lambda_L = {\Lambda^\mu}_\nu \bar\sigma^\nu$$

for every element of the Lorentz group. Moreover, since every finite Lorentz transformation can by obtained by chaining infinitesimal Lorentz transformations, we just need to prove that the identity holds at the first order in $$\boldsymbol\theta$$ and $$\boldsymbol\eta$$: $$U(\Lambda) \approx 1- i\theta_jJ^j+i\eta_jK^j\\ \Lambda_L \approx 1_2- i/2 \;\theta_j\sigma^j-1/2 \;\eta_j\sigma^j$$ I then proceeded to bruteforce the problem, by expanding separately the rhs and the lhs of the equation and showing that they are equal.

Oh, by the way - I've been using the Einstein notation for sums; moreover, my latin indexes span from 1 to 3, while my greek indexes span from 0 to 3.

I separated the cases $$\mu = 0$$ and $$\mu = i$$ for convenience, so that I could substitute $$\bar\sigma^0 = 1_2$$ in the first case and $$\bar\sigma^i = -\sigma^i$$ in the second case. By neglecting the quadratic terms in $$\theta$$ and $$\eta$$, the expansion of the lhs yelds $${\Lambda_L}^\dagger \bar\sigma^0 \Lambda_L = 1_2 - \eta_j\sigma^j \stackrel{?}{=} {\Lambda^0}_\nu \bar\sigma^\nu \\ {\Lambda_L}^\dagger \bar\sigma^i \Lambda_L = -\sigma^i - \theta_j\epsilon^{ijk}\sigma^k - \eta^i \stackrel{?}{=} {\Lambda^i}_\nu \bar\sigma^\nu$$ where I used $$\sigma^i \sigma^j = 1_2 \delta^{ij} + i\epsilon^{ijk}\sigma^k$$ in the computation of the $$\mu=i$$ case.

We then need to expand the rhs side of both equations: as shown, $$U({\Lambda}) = 1- i\theta_j{J^j}+i\eta_j{K^j}\\ {\Lambda^\mu}_\nu = {\delta^\mu}_\nu- i\theta_j{(J^j)^\mu}_\nu+i\eta_j{(K^j)^\mu}_\nu$$ In order to proceed with the calculation, we need a clean formula for the $$\mu,\nu$$ components of the rotations and boosts generators $$J^i$$ and $$K^i$$ in the (1/2,1/2) (4-vector) representation. I actually couldn't find them anywhere, even though I'm still sure they can be found in tons of qft books.

Instead, the Lorentz (1/2,1/2) representation is commonly written in terms of the $$J^{\rho\sigma}$$ matrices, a set of 4x4 matrices antisymmetric in $$(\rho,\sigma)$$ with $$\mu,\nu$$ component given by $${(J^{\rho\sigma})^\mu}_\nu = i(\eta^{\rho\mu}\;{\delta^\sigma}_\nu - \eta^{\sigma\mu}\;{\delta^\rho}_\nu)$$ where $$\eta^{\mu\nu}$$ denotes the Minkowski metric (there is no risk of confusing it with the rapidity vector $$\eta^i$$ since the latter one has only one index). In this convention, $$\eta^{\mu\nu} = diag\{1,-1,-1,-1\}$$.

The $$J^{\rho\sigma}$$ matrices satisfy different commutation relations from the $$(K^i,J^i)$$ ones, but the $$(K^i,J^i)$$ algebra can be restored by defining $$J^i = 1/2\; \epsilon^{ijk}J^{jk} \\ K^i = J^{i0}$$ We can then find $${\Lambda^\mu}_\nu = {\delta^\mu}_\nu- i\theta_i(\;\;1/2\;\epsilon^{ijk}({J^{jk})^\mu}_\nu\;\;) + i\eta_i({J^{i0})^\mu}_\nu$$ and proceed by plugging the formula for the $$\mu,\nu$$ components of the $$J^{\rho\sigma}$$ matrices. The calculation yelds $${\Lambda^\mu}_\nu = {\delta^\mu}_\nu + 1/2\;\theta_i\epsilon^{ijk}(\delta^{k\mu}{\delta^j}_\nu - \delta^{j\mu}{\delta^k}_\nu) + \eta_i(\delta^{i\mu}{\delta^0}_\nu + \delta^{0\mu}{\delta^i}_\nu)$$ For the latter and the following computations, I've used that $$\eta^{0\mu} = \delta^{0\mu}$$, $$\eta^{i\mu} = -\delta^{i\mu}$$ and, of course, $$\delta^{0i} = 0$$.

The last thing to do is to calculate $${\Lambda^0}_\nu \bar\sigma^\nu$$ and $${\Lambda^i}_\nu \bar\sigma^\nu$$, and verify that they're equal to, respectively, $$1 - \eta_j\sigma^j$$ and $$-\sigma^i - \theta_j\epsilon^{ijk}\sigma^k - \eta^i$$.