4
$\begingroup$

It is known that there is a set of gamma matrices that can be purely imaginary (called Majorana basis), thus one can solve the 1st quantized Majorana wave function in terms of real wave function.

However, I am confused by the implications for this real Majorana wave function.

  1. Isn't that the wave function should still be complex under time evolution? If so, what is the big deal to call this real Majorana wave function?

  2. What is the meaning for this set of real Majorana wave function, when we go from 1st quantized to 2nd quantization language?

p.s. One should clarify 1st quantized and 2nd quantization languages. The Ref below seems to mix two up.

Below from Wilczek on Majorana returns:

enter image description here

$\endgroup$
6
$\begingroup$

Neat question. The short answer is no, the real spinor does not turn complex under time evolution. That's at the very heart of the "big deal".

To see this explicitly, go to the rest frame, as you may always Lorentz-transform yourself back. So, the Dirac equation reduces to just $$ (i\tilde \gamma ^0 \partial_t - m)\psi =0, $$ hence $$ (\partial_t +im\tilde \gamma^0)\psi=0 . $$ Note in this representation $i\tilde \gamma ^0$ is real and antisymmetric, and thus antihermitean, as it should be!

Since it does not mix up the real and imaginary parts of the Dirac spinor $\psi$, we may consistently take the imaginary part to vanish, so the spinor is real: A Majorana spinor in this, Majorana, representation. (In the somewhat off-mainstream basis you display, you may take $C=-i\tilde \gamma^0=-C^T=-C^\dagger=-C^{-1} $.)

The evident solution, then, is $$ \psi (t) = \exp (-itm\tilde \gamma ^0 ) ~~\psi(0)=\exp (-itm ~\tilde \sigma_2\otimes\sigma_1 ) ~~\psi(0)~~, $$ where, as seen, the exponential is real, so the time-evolved spinor is real, as well, forever and ever. Thus, $$ \psi (t) = (\cos (tm)-i\tilde \gamma^0 \sin (tm) ) ~\psi(0) . $$

Now, thruth be told, this is an "existence proof" of the consistency of the split. Few of my friends actually use the Majorana representation in their daily lives. It just reminds you that a Dirac spinor is resolvable to a Majorana spinor plus i times another Majorana spinor.

The properties of the Dirac equation are the same in both first and second quantization, so all the moves and points made also hold for field theory as well, without departure from the standard textbook transition of the Dirac equation.


  • Edit in response to @annie marie hart 's question.

In effect, the complex nature of Schroedinger's equation is replicated with real matrix quantities, given $\Gamma\equiv i\tilde \gamma^0 ~\leadsto ~\Gamma^2=-I$. As a result, all complex unitary propagation features of Schroedinger's wavefunctions are paralleled here by the real unitary matrices acting on spinor wave functions, normalized in the technically analogous sense.

$\endgroup$
7
  • $\begingroup$ thanks very much for the nice answer. vote up $\endgroup$ Jul 24 '19 at 15:17
  • $\begingroup$ Another way of saying this could be to think of the free Hamiltonian as a matrix $\chi^T H \chi$, where $\chi$ are Majorana fields. If this is to be nonzero then by fermionic statistics it must be antisymmetric, and since $H$ is Hermitian we can conclude that the Hamiltonian for a collection of Majoranas must be purely imaginary. This means that the time evolution $e^{-iHt}$ acts as a real matrix. $\endgroup$ Aug 28 '19 at 20:05
  • $\begingroup$ How do we appreciate better that the $\exp(-iEt)$ type of time evolution does not get involved? In the naive picture we have $H |\psi_E> = i \hbar \partial_t |\psi_E>$ with a textbook solution $ |\psi_E>= \exp(-iEt) |\psi_0>$ and $H |\psi_0>=E |\psi_E>$. But $\exp(-iEt)$ complexes the solution... $\endgroup$ Oct 2 '20 at 0:39
  • $\begingroup$ Also are all the real solutions, (such as $\psi (t) = (\cos (tm)-i\tilde \gamma^0 \sin (tm) ) ~\psi(0) $) normalizable as in the Hilbert space requires $\int |\psi (t)|^2 d^3x=1$? $\endgroup$ Oct 2 '20 at 0:41
  • $\begingroup$ See this question physics.stackexchange.com/q/345106/42982 - also comment below - "An Introduction to Quantum Field Theory" by Peskin and Schroeder at page 41 it says: "In fact the Lorentz group, being non compact, has no faithful, finite-dimensional representations that are unitary. But that does not matter to us, since 𝜓 is not a wavefunction; it is a classical field." "--> I think this P&S statement seems wrong -- but it says how the wavefunction (1st quantized) and field (2nd quantized) may differ? $\endgroup$ Apr 28 '21 at 1:04

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.