19
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ theguardian.com/lifeandstyle/2015/jan/14/… $\endgroup$
    – user16438
    Commented Feb 21, 2015 at 3:51
  • $\begingroup$ @caseyr547 Interesting. How come you you did not post this on my most recent question? $\endgroup$ Commented 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
    Commented Feb 21, 2015 at 18:49
  • $\begingroup$ @caseyr547 How come? Wait you have an answer? Please send it to me via email. Thanks. $\endgroup$ Commented 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
    Commented Feb 21, 2015 at 21:26

1 Answer 1

4
+100
$\begingroup$

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

$\endgroup$
8
  • $\begingroup$ 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? $\endgroup$ Commented 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
    Commented 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. $\endgroup$ Commented Mar 18, 2014 at 17:37
  • $\begingroup$ @linuxfreebird did you just want the corresponding field equations for the vector and fermioni wavefunction? $\endgroup$
    – luksen
    Commented Mar 18, 2014 at 18:22
  • $\begingroup$ @luksen I think so. At least it would make sense for those unfamiliar with high level theory. $\endgroup$ Commented Mar 19, 2014 at 0:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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