2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – FenderLesPaul Jan 4 '15 at 9:25
3
$\begingroup$

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:

$\endgroup$
  • $\begingroup$ 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 Jan 4 '15 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 Jan 4 '15 at 9:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.