I Spin matrices and Field transformations

[Markus Kahn]

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

 

[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.

 

[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...

 

[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.