In quantum field theory, one-particle states in the Hilbert space in which a given representation of the Lorentz group acts can be completely characterized by specifying the numbers $|m,\mathbf{p};s,m_s\rangle$. This Hilbert space has dimension $2s+1$ if $m\neq0$ and dimension 2 if $m=0$.

This is usually stated intuitively as follows,

Massive particles propagate $2s+1$ degrees of freedom, while massless particles propagate $2$.

How do fields enter the discussion here? Could someone clarify the mathematical structure intuitively, without getting too technical? In particular,

  1. Why do we represent spin-$0$ particles as scalar fields, spin-$1/2$ particles as spinors and spin-$1$ particles as 4-vectors? I understand that the number of degrees of freedom in each case matches, but how are these fields related to the one-particle states of the Hilbert space?
  2. Spinor and tensor fields seem like completely different objects mathematically. Moreover, to represent a charged spin-$0$ particle we need a complex scalar field! How do we determine what type of object we should use? Is it just a matter of matching the number of components of the object to the degrees of freedom of the particle we want to represent, or is there a more systematical method? Why can't we simply use a two component vector for a charged spin-$0$ particle, for example?
  • 2
    $\begingroup$ See extensive discussion in Weinberg I $\endgroup$
    – Blazej
    Commented Jul 29, 2017 at 6:53

1 Answer 1


I shall attempt to only answer question 1 about the relation between the operator fields and the single particle states. I've been mainly interested in understanding QED and so I've not thought much about charged scalars so I'm unable to answer question 2 at present.

Why do we represent spin-0 particles as scalar fields, spin-1/2 particles as spinors and spin-1 particles as 4-vectors?

The laws of physics are the same in all inertial frames and so classical fields have to transform under the homogeneous Lorentz group SO(1,3). The simplest fields should therefore be irreducible representations of SO(1,3). It is very messy to find the irreps of SO(1,3), however it's much easier to get the irreps of the double cover of the Lorentz group which is the special (unity determinant) linear group of $2\times 2$ complex matrices SL(2,C). Happily, Nature also likes SL(2,C).The finite-dimensional irreps of SL(2,C) are completely symmetric Weyl spinors of the form $\psi^{AB\ldots C}$ where the indices $A,B,\ldots,C$ take values ${1,2}$. The finite-dimensional irreps of SL(2,C) are also complex conjugate fields which are written with dotted indices (Van der Waerden notation) $\chi_{\dot{A}\dot{B}\dots\dot{C}}$ to remind us their transformation matrices are the complex conjugate ones.

A completely symmetric Weyl spinor with $m$ indices has $m+1$ independent components. So, a Weyl spinor field $\psi^{A}$ has two components and represents a spin 1/2, a symmetric spinor field $\psi^{AB}$ has three components and represents a spin 1. The Weyl spinors are not invariant under parity, so to make parity invariant fields one needs the direct sum of a field and one transforming under the complex conjugate transformation. The parity-invariant electron field is of the form $\psi^{A}\oplus\chi^{\dot{B}}$. This has four independent components. When stacked as a column vector with four components it is a Dirac spinor field. The parity invariant photon field is of the form $\psi^{AB}\oplus\psi^{*}_{\dot{A}\dot{B}}$ . The three independent complex components of $\psi^{AB}$ are linearly related to the six components of the electric and magnetic fields $E^{i}$ and $B^{i}$.

It is very easy to set up the equations of motion for these classical fields - there really is no room for guesses. The operator of 4-momentum is $\hat{p}_{\mu}=i\frac{\partial}{\partial x^{\mu}}$. The 4-momentum operator in Weyl spinor notation is $\hat{p}^{\dot{A}}_{B}=\hat{p}^{\mu}[\sigma_{\mu}]^{\dot{A}}_{B}$ where the four $2\times 2$ matrices $\sigma_{\mu}$ are the Pauli matrices. The Dirac equation for the electron field is, \begin{eqnarray*} \hat{p}^{\dot{A}}_{B}\psi^{B}=m\chi^{\dot{A}}\\ \hat{p}^{A}_{\dot{B}}\chi^{\dot{B}}=m\psi^{A} \end{eqnarray*} Maxwell's equations for the photon field are, \begin{equation} \hat{p}^{\dot{A}}_{B}\psi^{BC}=0 \end{equation} The purpose of Dirac's equation is to force the momentum to be on-shell $p^{\mu}p_{\mu}=m^{2}$. The purpose of Maxwell's equations is to reduce the six real degrees of freedom of the photon field $\psi^{AB}$ to the two helicity degrees of freedom of a photon.

The short answer to the question is that the fields have to be made from irreducible representations of the double cover of the homogeneous Lorentz group SL(2,C).

How are these fields related to the one-particle states of the Hilbert space?

Let's write the single particle states as $|p,\lambda\rangle$ where $p$ is momentum and $\lambda$ is helicity. For a massive particle, helicity $\lambda$ is the component of spin measured along the 3-momentum vector of the particle. For a massive spin 1/2 particle the helicity takes values $\lambda=-1/2,+1/2$. For a massive spin 1 particle the helicity takes values $\lambda=-1,0,+1$. Helicity is nicer to work with than z-component of spin in the rest frame of the particle because helicity is invariant under a rotation of the particle and also helicity is invariant under a boost along the direction of the 3-momentum of the particle.

I've thought about this question for electrons, so I'll consider this case. The classical electron field is a Dirac spinor $\psi^{a}(x^{\mu})$ where the index $a=1,2,3,4$ labels the four components of the Dirac spinor. When we go over to quantum theory, this field becomes an operator field $\hat{\psi}^{a}(x^{\mu})$. This operator field has an expansion in terms of emission and absorption operators for electrons and positrons, \begin{equation} \hat{\psi}^{a}(x^{\mu})=\int \frac{d^{3}p}{(2\pi)^{3/2}}\frac{\sqrt{m}}{2\omega}\left(\phi^{a}_{\lambda}(p)\hat{\eta}^{\dagger}_{p\lambda}(0)e^{-i\omega t}+\text{ positron emission terms}\right)e^{ip^{r}x^{r}} \end{equation} where the $\hat{\eta}^{\dagger}_{p\lambda}(0)$ are absorption operators for electrons of 3-momentum $p$ and helicity $\lambda$ in the Schrodinger representation (time independent operators) and $\phi^{a}_{\lambda}(p)$ is a plane wave solution of the Dirac equation and $\omega=p^{0}$ (energy). We can bring in the single particle states by letting the operator $\hat{\psi}^{\dagger}(x)$ act on the vacuum state $|S\rangle$. \begin{equation} \hat{\psi}^{\dagger a}(x^{\mu})|S\rangle=\int \frac{d^{3}p}{(2\pi)^{3/2}}\frac{\sqrt{m}}{2\omega}\phi^{\dagger a}_{\lambda}(p)|p,\lambda\rangle e^{ip_{\mu}x^{\mu}} \end{equation} Take the Hermitian conjugate for convenience to get back to the electron field, \begin{equation} \langle S|\hat{\psi}^{a}(x^{\mu})=\int \frac{d^{3}p}{(2\pi)^{3/2}}\frac{\sqrt{m}}{2\omega}\phi^{a}_{\lambda}(p)e^{-ip_{\mu}x^{\mu}}\langle p,\lambda | \end{equation} Introduce the basis states $\langle S|\hat{\psi}^{a}(x^{\mu})=\langle x^{\mu}, a|$. The above equation is a coordinate transformation between the single particle basis states $\langle p,\lambda|$ and the new basis states $\langle x^{\mu},a|$ labeled by the space-time point $x^{\mu}$ and the Dirac spinor index $a$. The coordinate transformation can be written more transparently as, \begin{equation} \langle x,a|=[T^{-1}]^{(x,a)}_{\ \ \ \ \ (p,\lambda)}\langle p,\lambda| \end{equation} where $(x,a)$ and $(p,\lambda)$ are composite labels and $T^{-1}$ is the transformation matrix from single particle states basis to spinor field basis. The matrix elements of the transformation are, \begin{equation} \langle x,a|p,\lambda\rangle=[T^{-1}]^{(x,a)}_{\ \ \ \ \ (p,\lambda)}=\frac{\sqrt{m}}{(2\pi)^{3/2}}\phi^{a}_{\lambda}(p)e^{-ip_{\mu}x^{\mu}} \end{equation} In the transformation, the sum over the momentum labels uses the Lorentz invariant measure $d^{3}p/(2p^{0})$.

In summary, the electron operator field is, essentially, a basis of states which is related to the single particle basis states by a coordinate transformation. In other words, the electron operator field and the single particle states are equivalent representations of the Poincare group. I learnt this stuff from section 10.5.3, page 207 of "Group Theory in Physics" by Wu-Ki Tung.

After quantization, fields are promoted to operators and the vector space in which they act now consists of the one-particle states?

Julian Rey asked the above question in the comments section. I decided to expand my answer because it is difficult to answer this additional point in a comment.

The emission operator for an electron in a momentum state $p$ and helicity $\lambda$ is $\hat{\eta}_{p,\lambda}$. In other words, when the operator acts on the vacuum $|S\rangle$ we get the single particle state $|p,\lambda\rangle=\hat{\eta}_{p,\lambda}|S\rangle$. These emission operators don't do anything else other than act on the vacuum, so we may as well drop the vacuum state and just register the equivalence $\hat{\eta}_{p,\lambda}\sim |p,\lambda\rangle$. The electron emission operators are synonymous with the single particle electron states. Consequently, the expansion of the electron operator field, given earlier, in terms of the emission and absorption operators, \begin{equation} \hat{\psi}^{a}(x^{\mu})=\int \frac{d^{3}p}{(2\pi)^{3/2}}\frac{\sqrt{m}}{2\omega}\left(\phi^{a}_{\lambda}(p)\hat{\eta}^{\dagger}_{p\lambda}(0)e^{-i\omega t}+\text{ positron emission terms}\right)e^{ip^{r}x^{r}} \end{equation} is really saying that the electron operator field is a linear combination of single particle states. The electron operator field $\hat{\psi}^{a}(x)$ is really an infinite set of basis vectors $\hat{\psi}^{a}(x)\sim |x,a\rangle$ (ignoring positrons to simplify the argument) that span the same space as the single particle states. The answer to the question is negative; the operator field does not act on the single particle states, the operator field is just another set of basis vectors which span the space of single particle states.

The operator fields have another role in the theory - they can be multiplied together to form polynomials which produce multiparticle states.

  • $\begingroup$ Thank you very much, your answer is very clear. There's still something I'd like to clarify, though. Before quantization, fields form the basis of an infinite-dimensional vector space in which a representation of the Lorentz group acts. After quantization, fields are promoted to operators and the vector space in which they act now consists of the one-particle states. This seems very weird to me, since it looks as if fields have gone from being a basis to being a representation acting on another basis! Is this correct? $\endgroup$ Commented Jul 30, 2017 at 15:25
  • $\begingroup$ @JuliánRey : I've extended my answer to try to answer your question. $\endgroup$
    – user7154
    Commented Jul 30, 2017 at 18:50

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