Fermions and Coherent States: What's the Connection?

[ShayanJ]

I know coherent states are minimal uncertainty states and can provide a link from quantum to classical physics.But when I hear fermions can't have coherent states,or at least are limited in this sense,I can't see any relationship!
What's the point?

And...another thing...is there sth called Classical Coherent State?

Thanks

 

[DrDu]

Due to the univalence superselection rule, superpositions of states with different numbers of fermions don't exist, hence there are also no fermion coherent states made up from states with different numbers of particles.
But for a connection with classical physics this isn't important:For photons, position and momentum operators are non-diagonal in particle number while for fermions, position and momentum are diagonal in particle number.
Hence the coherent states relevant for fermions for discussing the classical limit are wave packets with sharp particle numbers.

 

[Demystifier]

Coherent states contain many particles in the same state, which is impossible for fermions.

 

[DrDu]

Coherent states contain many particles in the same state, which is impossible for fermions.

That depends. You can also construct coherent states for e.g. a harmonic oscillator.

 

[vanhees71]

There indeed exists a definition for coherent states for fermions which is quite important for the path-integral formulation of (non-relativistic as well as relativistic) quantum field theory. The complication noted by Demystifier needs to be overcome by introducing Grassmann valued c-number fields. See, e.g., my QFT manuscript (Section 4.5):

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

 

[ShayanJ]

Due to the univalence superselection rule, superpositions of states with different numbers of fermions don't exist, hence there are also no fermion coherent states made up from states with different numbers of particles.
But for a connection with classical physics this isn't important:For photons, position and momentum operators are non-diagonal in particle number while for fermions, position and momentum are diagonal in particle number.
Hence the coherent states relevant for fermions for discussing the classical limit are wave packets with sharp particle numbers.

In (http://dirac.fciencias.unam.mx/papers/limitations.pdf) Univalence superselection rule,is said to be because of the different behavior of bosons and fermions under a $2 \pi$ rotation.Considering that line of thought,I can only accept that it is not possible to superimpose a state with even number of fermions with a state with odd number of fermions!But its OK to superimpose states which their number of fermions are of the same parity!
Also,in a one particle problem,the issues you mentioned can't arise!

Coherent states contain many particles in the same state, which is impossible for fermions.

As I said,What about one particle(fermion) problems?

There indeed exists a definition for coherent states for fermions which is quite important for the path-integral formulation of (non-relativistic as well as relativistic) quantum field theory. The complication noted by Demystifier needs to be overcome by introducing Grassmann valued c-number fields. See, e.g., my QFT manuscript

Well,I was looking for the reason that why it seems impossible at first and then a change of view makes it OK!

 

[DrDu]

Yes, you can have coherent states formed from only even or odd numbers of fermions. Excitations in superconductors are an example and are observable as coherence factors in all kinds of measurements in superconductors.

The Grassmannian coherent states mentioned by van Hees are indeed an important formal concept in QFT, but they can't be prepared as actual states. So you can't think of classical states of fermions as limits of Grassmannian coherent states.

 

[strangerep]

I was looking for the reason that why it seems impossible at first and then a change of view makes it OK!

Taking the definition of "coherent state" as one of "minimal uncertainty" is too restrictive. Generalized coherent states can be constructed group-theoretically, and the construction is applicable to a surprisingly large number of cases.

Try this book:

J-P. Gazeau, "Coherent States in Quantum Physics",
Wiley 2009, ISBN 978-3-527-40709-5

https://www.amazon.com/dp/352740709X/?tag=pfamazon01-20

Chapter 11 discusses fermionic coherent states.

 

[Demystifier]

Coherent states contain many particles in the same state, which is impossible for fermions.

That depends. You can also construct coherent states for e.g. a harmonic oscillator.

Also,in a one particle problem,the issues you mentioned can't arise!

As I said,What about one particle(fermion) problems?

The original question in the first post, referring to fermions, suggested that one had the many-particle field-theoretic notion of coherent state in mind.

By the way, except those two meanings of the term "coherent state", there is also a third meaning: any pure state, i.e., state in the Hilbert space with a well defined phase.

 

[Demystifier]

The Grassmannian coherent states mentioned by van Hees are indeed an important formal concept in QFT, but they can't be prepared as actual states. So you can't think of classical states of fermions as limits of Grassmannian coherent states.

True. Or from a mathematical point of view, Grassmannian coherent states are not states in the (physical) complex Hilbert space.