4
$\begingroup$

We see evidence of bosonic coherent states everywhere. Lasers and microwave circuits naturally condense into photonic coherent states and resonators do the same except with phonons.

A coherent state is a state with an uncertainty in the number of quanta. This makes sense for bosons, but I cannot think of a situation where the same would happen for fermions. Yet I find a lot of articles that discuss fermionic coherent states with a remarkably similar formula as for bosons.

Is this just a mathematical trick or are there really situations where a fermionic field can exist with an indeterminate number of particles?

$\endgroup$
  • $\begingroup$ Can you give a reference of such an article? $\endgroup$ – Sofia Feb 11 '15 at 12:05
  • $\begingroup$ Here is a theoretical paper in the Journal of Physics A. There are also lots of PDFs (p. 227) available on the internet. $\endgroup$ – mactud Feb 11 '15 at 12:11
  • $\begingroup$ Fermion states can contain more than 1 fermion only if there is some difference in the properties of the fermions. Thus, a coherent fermion state is of the form $A|0\rangle + B|1\rangle$. If additional fermions participate to the state, they must have, each one of them a different value of some quantum number than the rest of the fermions. For instance, we have the spin singlet (not considered a coherent state), comprising two fermions, but they differ in spin-projection. In the He atom we have 2 electrons of opposite spin. $\endgroup$ – Sofia Feb 11 '15 at 13:06
  • $\begingroup$ Note that the formula may be the same, but since the fermionic operators obey anticommuation relations rather than commutation relations, the action of the operator creating a coherent state is very different. Also, how could this be a "mathematical trick"? If it is a valid state in the Hilbert space, it is a valid state of the system, unless quantum mechanics is false. $\endgroup$ – ACuriousMind Feb 11 '15 at 13:58
  • 3
    $\begingroup$ A pure fermionic coherent state violates the superselection rule corresponding to fermion number conservation, so in some sense it is not a valid state (or rather, it can always be replaced by a mixed state with the same measurement statistics in the number basis). However, it is an extremely useful form for evaluating expectation values and constructing path integrals (the exponential form is particularly useful for the latter). $\endgroup$ – Mark Mitchison Feb 11 '15 at 15:49

Your Answer

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

Browse other questions tagged or ask your own question.