The quantized Dirac field at a certain space-time point can be written (roughly) as a linear combination of creation operators acting on the Hilbert space of physical states, with coefficient that are free field harmonic solutions. These free field harmonic solutions are Dirac spinors, which are four element complex vectors (not in the sense of four-vectors...). This means that when operating with the Dirac field operator on a state in the Hilbert space, we get some linear combination of states with spinor coefficients.

Here are my question:

a) Is my description correct?

b) Is there a mathematical framework to describe such a linear space? (I am aware of linear spaces over fields of numbers, not over spinors...)

c) How can the vacuum state, for example, be considered as an entity with spinor coefficients?

d) (For this question, I'll be more specific, hoping that an answer will give me insight) It is known that the chirality operator does not commute with the free Dirac Hamiltonian. But the Hamiltonian is a one dimensional quantity (energy...) whereas the chirality operator is a 4x4 matrix. So what is the meaning of performing both on a quantum state in a certain order? (I mean, what does it mean to operate the 4x4 matrix on a Hilbert space state...)



1 Answer 1


This is at the current limit of my understanding of fermionic QFT. It's such a good question that it deserves an answer from an expert.
a) Fourier analysing the second quantized Dirac operator field gives, $\hat{\psi}^{a}(x)$, \begin{equation} \hat{\psi}^{a}(x)=\int \frac{d^{3}p}{(2\pi)^{3/2}}\exp(ix^{r}p^{r})\hat{\psi}^{a}(p) \end{equation} where the label $a=1,2,3,4$ runs over the four components of the Dirac spinor and the label $r=1,2,3$ runs over the three spatial coords and I'm working at a constant time. So, borrowing Lior's introductory remarks, "The quantized Dirac field $\hat{\psi}^{a}(x)$ at a certain spatial point can be written as a linear combination of creation operators $\hat{\psi}^{a}(p)$ acting on the Hilbert space of physical states, with coefficients that are free field harmonic solutions $\exp(ix^{r}p^{r})$." I don't think the free field harmonic solutions are Dirac spinors because the spinor index is on the creation/annihilation operators. A state $|\Psi\rangle$ will be a sum of polynomials in the creation operators applied to a vacuum state $|S\rangle$, \begin{equation} |\Psi\rangle= (\ldots +c\hat{\psi}^{a}(p_{1})\hat{\psi}^{b}(p_{2})...\hat{\psi}^{c}(p_{n})+\ldots )|S\rangle \end{equation} The field operator $\hat{\psi}^{a}(x)$ applied to the state changes the state by modifying the polynomial just by multiplication.

b) The mathematical framework is the space of polynomials in the variables $\hat{\psi}^{a}(p)$.

c) The vacuum state $|S\rangle$ is the trivial polynomial $1$.

d) The Hamiltonian for the free electron field is, \begin{equation} \hat{H}=\int d^{3}x (-i\hat{\psi}^{\dagger}\gamma^{0}\gamma^{r}\frac{\partial\hat{\psi}}{\partial x^{r}}+m\hat{\psi}^{\dagger}\gamma^{0}\psi) \end{equation} where I've suppressed the spinor indices and so I'm using matrix notation for anything to do with the spinors. The chirality operator is $\gamma^{5}$. \begin{equation} \gamma^{5}= \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right] \end{equation} where the entries are 2x2 block matrices. $\gamma^{0}$ is, \begin{equation} \gamma^{0}= \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] \end{equation} The action of the chirality operator is to change the Dirac spinors $\hat{\psi}\rightarrow \gamma^{5}\hat{\psi}$. Nothing happens to the gamma matrices. For simplicity, just look at what happens to the mass term. \begin{equation} \hat{\psi}^{\dagger}\gamma^{0}\hat{\psi}\rightarrow (\gamma^{5}\hat{\psi})^{\dagger}\gamma^{0}\gamma^{5}\hat{\psi}=\hat{\psi}^{\dagger}(\gamma^{5}\gamma^{0}\gamma^{5})\hat{\psi}=\hat{\psi}^{\dagger}(-\gamma^{0})\hat{\psi}=-\hat{\psi}^{\dagger}\gamma^{0}\hat{\psi} \end{equation} So, the mass term changes sign under the chirality transformation so the Hamiltonian is not invariant. Although the Hamiltonian looks like a "scalar" because we've summed over all the spinor indices, transforming each component of the spinor gives a result which is minus the thing we started with; so the Hamiltonian changes under chirality.

  • $\begingroup$ Because Lorentz Symmetry, the halmitonian need to behave as a zero component of a four vector. $\endgroup$
    – Nogueira
    Commented Jun 24, 2015 at 20:07

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