# Spin matrices and Field transformations

• I
Let us for a moment look a field transformations of the type
$$\phi(x)\longmapsto \exp\left(\frac{1}{2}\omega_{\mu\nu}S^{\mu\nu}\right)\phi(x),$$
where ##\omega## is anti-symmetric and ##S^{\mu\nu}## satisfy the commutation relations of the Lorentz group, namely
$$\left[S_{\mu \nu}, S_{\rho \sigma}\right]=-\eta_{\mu \rho} S_{\nu \sigma}+\eta_{\mu \sigma} S_{\nu \rho}-\eta_{\nu \sigma} S_{\mu \rho}+\eta_{\nu \rho} S_{\mu \sigma}.$$
These matrices ##S^{\mu\nu}## are sometimes (in my lecture notes for example) called "Spin matrices" but I'm having a hard time associating the existence and their specific form (f.e. for a scalar field ##S^{\mu\nu}=0## and for a bispinor ##S^{\mu\nu}=\frac{1}{4}[\gamma^\mu,\gamma^\nu]##, where the ##\gamma##-matrices satisfy the Clifford algebra, etc.) with the actual spin of the particle. I've already been pointed out to [1], but I honestly don't really see how this is related to the question (since there we construct a projector that gives us a spinor pointing into a specific direction, instead of explaing what this has to do with the transformation matrices at all).

I've already taken a look into Perskin & Schroeders and Bjorken & Drell's book, but I'm having a hard time finding the passage that explains this connection... I obviously don't expect anyone to give me an incredibly long and complicated explanation if this should be necessary, I'd already be happy if someone could point me to a specific chapter/subsection of a book that discusses this.

[1] https://en.wikipedia.org/wiki/Dirac...spinor_with_a_given_spin_direction_and_charge

## Answers and Replies

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samalkhaiat
bispinor ##S^{\mu\nu}=\frac{1}{4}[\gamma^\mu,\gamma^\nu]##, where the ##\gamma##-matrices satisfy the Clifford algebra, etc.) with the actual spin of the particle.
This follows from the form invariance of the Dirac equation under Lorentz transformations. Now define and calculate $\Sigma^{k} \equiv \frac{i}{2}\epsilon^{kij}S^{ij}$. That is

$$\Sigma^{k} = \frac{i}{8} \epsilon^{kij} \big[ \gamma^{i} , \gamma^{j} \big] = \frac{1}{2} \begin{pmatrix} \sigma^{k} & 0 \\ 0 & \sigma^{k} \end{pmatrix} .$$

In non-relativistic QM the operator $s^{k} = \frac{1}{2} \sigma^{k}$ acts on 2-component wave function (spinor) which describes spin-1/2 non-relativistic particle. Similarly $$\Sigma^{k} = \begin{pmatrix} s^{k} & 0 \\ 0 & s^{k} \end{pmatrix} ,$$ acts on 4-component “wave function” (bispinor) which describes relativistic spin-1/2 particle.

In field theory, the conserved charge associated with Lorentz symmetry is given (for Dirac field) by

$$J^{\mu\nu} = \int d^{3}x \ \psi^{\dagger}(x) \left( i x^{\mu}\partial^{\nu} - i x^{\nu}\partial^{\mu} + i S^{\mu\nu}\right) \psi (x) ,$$ and (for the same reason above) is called the angular momentum tensor of the Dirac field. Indeed, if you define the 3-vector $J^{k} = \frac{1}{2}\epsilon^{kij}J^{ij}$ you fined

$$\vec{J} = \int d^{3}x \ \psi^{\dagger}(x) \left( \vec{x} \times \frac{1}{i} \vec{\nabla} \right) \psi (x) + \int d^{3}x \ \psi^{\dagger}(x) \ \vec{\Sigma} \ \psi (x) ,$$ which is nothing but the usual $\vec{J} = \vec{L} + \vec{S}$ relation.

dextercioby and Markus Kahn
First of all, thank you for answering! Your answer provided some new questions, so I hope you can address them as well if you find time...
In non-relativistic QM the operator ##s^{k} = \frac{1}{2} \sigma^{k}## acts on 2-component wave function (spinor) which describes spin-1/2 non-relativistic particle.
I only know the operator ##s^k=\frac{1}{2}\sigma^k## as the fundamental representation of ##SU(2)##, or more precisely of ##H_{1/2}## (where ##H_{i}## for ##i\in \frac{1}{2}\mathbb{N}## are the irreducible representations of ##SU(2)##). Now by definition (if I'm not mistaken) elements of ##H_{1/2}## act on
$$\left\vert\frac{1}{2},\frac{1}{2}\right\rangle =\begin{pmatrix}1\\0\end{pmatrix}\quad \text{and}\quad \left\vert\frac{1}{2},-\frac{1}{2}\right\rangle =\begin{pmatrix}0\\1\end{pmatrix},$$
which describe different sates of a spin 1/2 particle. If I understood you correctly then ##\vert1/2,1/2\rangle## and ##\vert 1/2,-1/2\rangle## would be spinors, but I know spinors only as fields that transform like
$$\psi_L'(x')=\exp\left(\frac{1}{2}\omega_{\mu\nu}(S^{\mu\nu})_L\right)\psi_L(x)\quad\quad \psi_R'(x') = \exp\left(\frac{1}{2}\omega_{\mu\nu}(S^{\mu\nu})_R\right)\psi_R(x),$$
where ##\left( S _ { k \ell } \right) _ { L } = \frac { 1 } { 2 } \varepsilon _ { j k \ell } \sigma _ { j } = \left( S _ { k \ell } \right) _ { R }## and ##\left( S _ { 0 k } \right) _ { L } = \frac { 1 } { 2 } i \sigma _ { k } = \left( S ^ { 0 k } \right) _ { R }## under a Lorentz transformation, which is basically another set of Spin matrices.

So my question in this case would be: The representation ##H_{1/2}## tells me that I'm dealing with spin 1/2 particles, but when I look at the definition of a spinor using fields I don't really understand how I can see that the ##\Sigma^k##, resulting from the set ##(S^{\mu\nu})_L ##, will have anything to do with ##H_{1/2}##, so how can I be sure that I deal with spin 1/2 particles.

PS:
I tried calculating ##\Sigma^k## for the set of ##(S^{\mu\nu})_L## and basically found ##\Sigma_L^k=s^k##, which I probably should have gotten...

dextercioby
samalkhaiat
You are not asking new questions. The vectors $|\frac{1}{2}, \pm \frac{1}{2}\rangle$, or equivalently the elementary spinors $$\chi_{+} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} , \ \ \chi_{-} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} ,$$ span the lowest-dimensional representation space (your $H_{1/2}$) of the spin group $SU(2)$. This means that any vector $|\psi \rangle \in H_{1/2}$ can be written as $$|\psi \rangle = \sum_{m = \pm \frac{1}{2}} |\frac{1}{2} , m \rangle \langle \frac{1}{2} , m |\psi \rangle ,$$ or equivalently represented by the following (2-component) spinor $$\psi = \psi_{+}\chi_{+} + \psi_{-}\chi_{-} = \begin{pmatrix} \psi_{+} \\ \psi_{-} \end{pmatrix} ,$$ where $$\psi_{\pm} = \langle \frac{1}{2} , \pm \frac{1}{2}|\psi \rangle .$$ The above is just mathematics. To describe spin-1/2 particles we must take into account both the coordinate degrees of freedom and the spin degrees of freedom. In other words, spin-1/2 particles live in the direct product space of the infinite-dimensional space $\big\{ |x \rangle \big\}$ and the 2-dimensional spin space $H_{1/2}$. So, in $\big\{|x\rangle \big\} \otimes H_{1/2}$, we have for the base ket $$|x , \frac{1}{2} , \pm \frac{1}{2} \rangle = |x \rangle \otimes |\frac{1}{2} , \pm \frac{1}{2} \rangle ,$$ and for the inner product we write $$\langle \psi |\phi \rangle = \sum_{m = \pm \frac{1}{2}} \int dx \ \langle \psi |x , \frac{1}{2}, m \rangle \langle x , \frac{1}{2} , m |\phi \rangle = \sum_{a = 1,2} \int dx \ \psi^{\ast}_{a}(x) \phi_{a}(x) .$$ Only in $\big\{|x\rangle \big\} \otimes H_{1/2}$ we can talk about the (2-component) wave function (or spinor “field”) for a particle with spin: $$\Psi (x) = \begin{pmatrix} \psi_{1}(x) \\ \psi_{2}(x) \end{pmatrix} = \begin{pmatrix} \langle x , \frac{1}{2} , + \frac{1}{2}|\psi \rangle \\ \langle x , \frac{1}{2} , - \frac{1}{2}|\psi \rangle \end{pmatrix} .$$

Now, let’s talk about your concerns regarding the spin-1/2 nature of spinors and the relation to the spin transformation matrices. I will give you two convincing methods:

1) Dirac bispinor as a quantum-mechanical wave function

In order for the Dirac equation to be form invariant under Lorentz transformation $x \to \Lambda x$, the Dirac bispinor must transform according to $$\Psi (x) \to e^{- \frac{i}{4}S^{\mu\nu}\omega_{\mu\nu}} \Psi (\Lambda^{-1}x),$$ with $S^{\mu\nu} = \frac{i}{2}[\gamma^{\mu} , \gamma^{\nu}]$. In QM, the above transformation is implemented by unitary transformation $$\Psi (x) \to \mathcal{U}(\Lambda) \Psi (x) ,$$ on the Hilbert space of the Dirac wave function $\Psi (x)$. Writing $\mathcal{U} = e^{- \frac{i}{2}M^{\mu\nu}\omega_{\mu\nu}}$, and expanding in the infinitesimal parameter $\omega$, we get $$\left( 1 - \frac{i}{2} \omega_{\mu\nu}M^{\mu\nu} \right) \Psi (x) = \left( 1 - \frac{i}{4} \omega_{\mu\nu}S^{\mu\nu} \right) \Psi ( x - \omega x ) .$$ Comparing first order terms, we get the following expression for the Hermitian operator $M^{\mu\nu}$ as the infinitesimal generator of Lorentz transformations $$M^{\mu\nu} = \frac{1}{2} S^{\mu\nu} + i ( x^{\mu}\partial^{\nu} - x^{\nu}\partial^{\mu}) . \ \ \ \ (1)$$

Recall that the representations of the Poincare group are labelled by the eigenvalues of two Casimir operators $P^{2} = P_{\mu}P^{\mu}$ and $W^{2} = W_{\mu}W^{\mu}$: $P^{\mu}$ is the energy-momentum operator (the infinitesimal generator of translations), whereas $W_{\mu}$ (the Pauli-Lubanski vector) is defined in terms of the infinitesimal generator of Lorentz transformations $M^{\mu\nu}$, as $$W_{\mu} = - \frac{1}{2} \epsilon_{\mu\nu\rho\sigma} \ M^{\nu\rho}P^{\sigma} . \ \ \ \ \ \ (2)$$ Now, if you substitute (1) and $P^{\sigma} = i \partial^{\sigma}$ in (2), you get $$W_{\mu} = \frac{i}{4} \epsilon_{\mu\nu\rho\sigma} \ S^{\nu\rho} \ \partial^{\sigma} .$$ Notice that the orbital contribution has disappeared, implying that $S^{\mu\nu}$ and, therefore, $W_{\mu}$ corresponds to spin angular momentum. Indeed, if we calculate $W^{2}\Psi (x)$ using the identity $$\epsilon_{\mu\nu\rho\sigma}\epsilon^{\mu}{}_{\bar{\nu}\bar{\rho}\bar{\sigma}} = - \begin{vmatrix}\eta_{\nu \bar{\nu}} & \eta_{\nu \bar{\rho}} & \eta_{\nu \bar{\sigma}} \\ \eta_{\rho \bar{\nu}} & \eta_{\rho \bar{\rho}} & \eta_{\rho \bar{\sigma}} \\ \eta_{\sigma \bar{\nu}} & \eta_{\sigma \bar{\rho}} & \eta_{\sigma \bar{\sigma}} \end{vmatrix} ,$$ and $P^{2}\Psi = -\partial^{2}\Psi = m^{2}\Psi$, we obtain $$W^{2}\Psi = - \frac{3}{4}m^{2}\Psi = - \frac{1}{2} \left( \frac{1}{2} + 1 \right) m^{2} \Psi .$$ Do you recognise the relation $S^{2} = s (s + 1)$? And where did $\frac{1}{2} ( \frac{1}{2} + 1)$ above come from? In general, if $m^{2}$ is the eigenvalue of $P^{2}$, then $W^{2}$ takes only values of the form $$W^{2} = - m^{2} s (s + 1) ,$$ where $s$ is integer or half integer.

2) Dirac bispinor as field operator

This method is complicated and lengthy so I will use over-simplified version of it. The full treatment can be found in section 3-3-2 of the QFT text by Itzykson & Zuber.

In my previous post, we had the expression $$\vec{J} = \int d^{3}x \ \psi^{\dagger}(x) \left( \vec{x} \times \frac{1}{i} \vec{\nabla} + \vec{\Sigma} \right) \psi (x) .$$ Now, using the canonical equal-time anti-commutation relations for Dirac operators, it is easy to show that $$\big[ \vec{J} , \psi (x) \big] = \left( \vec{x} \times \frac{1}{i} \vec{\nabla} + \vec{\Sigma} \right) \psi (x).$$ Applying this to a left-handed spinor $\chi_{a}(0)$, we get $$\big[ J^{k} , \chi_{a}(0)\big] = \frac{1}{2} ( \sigma^{k})_{a}{}^{b} \chi_{b}(0).$$ In particular, for $k = 3$ we have $$\big[ J^{3} , \chi_{1}(0) \big] = \frac{1}{2} \chi_{1}(0), \ \ \ \ \ (3)$$ $$\big[ J^{3} , \chi_{2}(0) \big] = - \frac{1}{2} \chi_{2}(0). \ \ \ \ \$$ These are eigenvalues equations: Basically $\chi_{1}(0) \sim a^{\dagger} + b$ . So, if we use $b |0 \rangle = 0$, write $a^{\dagger}|0\rangle = |a \rangle$ and use the fact that Lorentz symmetry is not broken spontaneously, i.e., $J^{3}|0 \rangle = 0$, we see from (3) that type-a particle $$J^{3}|a \rangle = \frac{1}{2} |a \rangle ,$$ carries $+ \frac{1}{2}$ unit of angular momentum along the third axis.

Markus Kahn and Spinnor