I know basically the difference between Klein-Gordon and Dirac field is spin. But I am not sure where we need to implement this info.

The solutions of both equations are the wave packets which includes the sum of creation and annihilation operators.

$$ \Psi(x) = \int [a_pe^{-ipx} + a_p^\dagger e^{ipx}]\,d^3 p. $$

where we use the spin information? Why should I use Klein Gordon for spin-0 and Dirac equation for 1/2?

  • $\begingroup$ That mode expansion does not hold for solutions of the Dirac equation. Therein the plane waves are spinors of the form $u(p)e^{ipx}$ where $u(p)$ is a 4-component spinor and encodes the information about the spin of the plane wave. The mode expansion then includes a sum over these plane waves and the usual creation and annihilation operators. $\endgroup$ Commented Jan 4, 2015 at 9:25

1 Answer 1


The momentum decomposition you wrote is valid only for a scalar (spinless), real field, satisfying the Klein-Gordon equation.

When considering a field with spin, like a spin-$1/2$ field satisfying the Dirac equation, you must include the polarization vectors, obtaining something of the form $$ \psi_\alpha(x) = \sum_{\textbf{p},s} N_{s,\textbf{p}}[ c_s(\textbf{p}) u_\alpha(s,\textbf{p}) e^{-ipx} + d_s^\dagger(\textbf{p}) v_\alpha(s,\textbf{p}) e^{ipx} ] $$ $$ \bar \psi_\alpha(x) = \sum_{\textbf{p},s} N_{s,\textbf{p}}[ c_s^\dagger(\textbf{p}) \bar u_\alpha(s,\textbf{p}) e^{ipx} + d_s(\textbf{p}) \bar v_\alpha(s,\textbf{p}) e^{-ipx} ] $$ where $c,c^\dagger$ and $d,d^\dagger$ are respectively annihilation/creation operators of particles and corresponding antiparticles, $u,v$ are the polarization vectors, which encode the information about the spin, and $N_{s,\textbf{p}}$ normalization factors depending on the convention used. The sum is extended over all momentum ($\textbf{p}$) and spin ($s$) eigenstates. The objects I denoted with $ue^{-ipx}$ and $ve^{ipx}$ are called Dirac spinors. They have four components, reflecting the fact that a Dirac field describes both electron and positron, each having 2 spin degrees of freedom (for a total of $2+2=4$ degrees of freedom).

In other words, the spin information is encoded in the additional degrees of freedom of the field, in this case the spin-index which I denoted with $\alpha$. This additional degrees of freedom do not evolve independently, as you can see from the (free) Dirac equation, which showing explicitly the spinor indices reads $$ ( i \gamma^\mu_{\alpha\beta} \partial_\mu - m \delta_{\alpha\beta})\psi_\beta(x) = 0, \quad \forall \alpha=1,2,3,4,$$ where $\gamma^\mu$ are the gamma matrices, and a sum over the spin-index $\beta$ is implicit.

For another example you can look at spin-1 fields (e.g. photons). In this case the quantum field is denoted by $A_\mu(x)$ with $\mu$ a vector index, which is the spin-index for spin-1 fields. The decomposition has now the form: $$ A_\mu(x) = \sum_{\textbf{p},s} N_{s,\textbf{p}} [ a(s,\textbf{p}) \varepsilon_\mu(s,\textbf{p}) e^{-ipx} + a^\dagger(s,\textbf{p}) \varepsilon_\mu^*(s,\textbf{p}) e^{ipx} ]. $$ Again, $s$ denotes the spin states (which are usually called polarization states in this context), and $\varepsilon$ are the polarization vectors.

Finally, note that each spin component of the Dirac field satisfied the Klein-Gordon equation: $$ (\square + m^2)\psi_\alpha(x) = 0, \forall \alpha $$ An excellent explanation of this can be found here.

Related discussions:

  • $\begingroup$ So for spin 1/2 α should be 1 or 2, I mean spin up and down conditions right? and 'u' will be the linear independent vector including up and down conditions? Thanks.. $\endgroup$
    – aQuestion
    Commented Jan 4, 2015 at 9:39
  • $\begingroup$ @Major_Tom I edited the answer to address that question. $u_\alpha(s,\textbf{p})e^{-ipx}$ is the $\alpha$ component of a Dirac spinor describing an electron with momentum $\textbf{p}$ and spin $s$. This are four-components object, despite the fact that they describe two spin degrees of freedom. Look at the wikipedia article I linked to see their explicit expressions. $\endgroup$
    – glS
    Commented Jan 4, 2015 at 9:53

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