Is there a known particle field equation of a similar form $$ \begin{equation} (\Gamma^n \pi_n)^2 \Psi = (mc)^2 \Psi \tag{1} \end{equation} $$ such that by reducing the number of degrees of freedom for the spinor $\Psi$ into a spinor of lesser degrees of freedom, such as a scalar $\psi_0$, two three-vectors $\boldsymbol{\psi}_\pm$ or two two-vectors $\boldsymbol{\phi}_\pm$, it reduces Eq. 1 into either ...

  • a spin zero field equation $$ \begin{equation} \pi^n \pi_n \psi_0 = (mc)^2 \psi_0, \tag{2} \end{equation} $$
  • a spin one field equation $$ \begin{equation} (I\pi_0\pm i \boldsymbol{\pi} \times) (I\pi_0\mp i \boldsymbol{\pi} \times) \boldsymbol{\psi}_ \pm = (mc)^2 \boldsymbol{\psi}_ \pm \tag{3} \end{equation} $$
  • or a spin 1/2 field equation $$ \begin{equation} (I\pi_0\pm\boldsymbol{\sigma}\cdot\boldsymbol{\pi}) (I\pi_0\mp\boldsymbol{\sigma}\cdot\boldsymbol{\pi}) \boldsymbol{\phi}_\pm = (mc)^2 \boldsymbol{\phi}_\pm? \tag{4} \end{equation} $$

In these expressions $\pi_n$ is the four-component momentum operator which includes the electromagnetic four-potential interaction $A_n$ with the particle's charge $q$ written as $$ \begin{equation} \pi_n = i\hbar \partial_n - q A_n , \tag{5} \end{equation} $$ and $$ \begin{equation} \boldsymbol{\pi} = -i\hbar \boldsymbol{\nabla} - q \boldsymbol{A} \tag{6} \end{equation} $$ uses bold to indicate a euclidean vector, specific to 3-components. The three two-by-two matrices $\boldsymbol{\sigma}$ in Eq. 4 are the Pauli Spin Matrices.

  • $\begingroup$ theguardian.com/lifeandstyle/2015/jan/14/… $\endgroup$
    – user16438
    Feb 21, 2015 at 3:51
  • $\begingroup$ @caseyr547 Interesting. How come you you did not post this on my most recent question? $\endgroup$ Feb 21, 2015 at 14:13
  • $\begingroup$ they are strange at cogsci i'm banned but even then had i posted a really great answer they would have downvoted it into oblivion then called the answer low quality haha $\endgroup$
    – user16438
    Feb 21, 2015 at 18:49
  • $\begingroup$ @caseyr547 How come? Wait you have an answer? Please send it to me via email. Thanks. $\endgroup$ Feb 21, 2015 at 20:44
  • $\begingroup$ they are bigots...three users said I was cheating on the Se system and threatened to leave...they didn't like my answers and were down voting the. Because they disagreed not because it was low quality...so I went through and looked at artem's supposed high quality posts and his is mostly bs passing off as science...then the other two guys I looked at theirs too and they were writing a bunch of stuff too like supporting unhealthy behavior...then they are really scared to talk about antidepressants, antipsychotics and stimulants... $\endgroup$
    – user16438
    Feb 21, 2015 at 21:26

1 Answer 1


The known wavefunctions for scalar, spin-1/2 and vector fields follow from the theory of unitary representations of the Poincare group. Especially the theory of induced representations (keyword here is Mackey's machine) and are special cases of a general wavefunction that an be written in the form

$$Q(p)\psi = \psi$$

where $Q(p)$ is a projection operator.

Given a subgroup $K$ of a lie group $G$ we can induce a unitary rep of $G$ via a unitary rep of $K$ (also called the 'little group') by the following: Take a more or less arbitrary (non-unitary) rep of $G$, $D(g)$ whose restriction to $K$, $D(k)$, is unitary. We then construct a unitary rep of $G$, U(g), with

$$U(g)\psi(c) = D(g)\psi(cg)$$

where $c\in C = G/K$ is an element of the right coset space.

Now for the Lorentz/Poincare case the little group is $SU(2)$ and the coset space are the boosts, i.e. momenta. Therefore the above tells us that given a finite dimensional rep of the Lorentz group that is unitary on $SU(2)$, we can have a continuous unitary rep of the Lorentz group, which is the field $\psi(p)$ and that must obey the subsidiary condition

$$Q(p)\psi(p) = 0$$ (this is what will be called the wave equation)

and transformation law

$$(U(\Lambda)\psi)(p) = \psi'(p) = D(g)\psi(p') = D(g) \psi(\Lambda^{-1}p)$$

the projectors $Q(p)$ for the common cases are:

scalar: trivial $\psi'(p) = \psi(\Lambda^{-1}p), p^2\psi = m^2\psi$ (KG)

spin $1/2$: $Q(p) = \frac{1}{2m}(\gamma^\mu p_\mu + m)$, $(\gamma^\mu p_\mu-m)\psi = 0$ (dirac)

vector: $Q(p) = g^\mu_\nu - \frac{p^\mu p\nu}{m^2}$, $p_\mu A^\mu = 0$(proca)

For a generic spin $j$ we get the Bargmann Wigner equation

$$(\gamma^\mu p_\mu - m)_{\alpha_r\alpha'_r}\psi_{\alpha_1\dots\alpha_r\dots,\alpha_{2j}} = 0, r=1\dots2n$$

A good reference on this

Niederer, U. H. and O'Raifeartaigh, L. Realizations of the Unitary Representations of the Inhomogeneous Space-Time Groups II. Covariant Realizations of the Poincaré Group

parts I and II

  • $\begingroup$ I will have to take some time to read the reference you provided, but first what happens if I do check your answer as the accepted answer before the deadline? Do you not receive the bounty? $\endgroup$ Mar 18, 2014 at 17:13
  • 1
    $\begingroup$ @linuxfreebird: I could be mistaken but I don't think you assign the bounty, but namehere does. $\endgroup$
    – JeffDror
    Mar 18, 2014 at 17:25
  • $\begingroup$ I think your answer is generic to any spin and any dimension? Could you provide an example of a field equation that reduces into equations 2, 3, and 4? Thanks. $\endgroup$ Mar 18, 2014 at 17:37
  • $\begingroup$ @linuxfreebird did you just want the corresponding field equations for the vector and fermioni wavefunction? $\endgroup$
    – luksen
    Mar 18, 2014 at 18:22
  • $\begingroup$ @luksen I think so. At least it would make sense for those unfamiliar with high level theory. $\endgroup$ Mar 19, 2014 at 0:32

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