Explaining Fermionic Fields & Classical Coherent States

In summary, a coherent state is a superposition of states with definite numbers of particles. In the case of fermions, a coherent state is not possible due to the exclusion principle, but a coherent state of pairs of fermions can be created. There are at least three definitions of coherent states, including being an eigenstate of an annihilation operator, having a minimum uncertainty product, and being obtained by acting on the vacuum with a "displacement" operator. In the case of relativistic systems, generalized coherent states may be used, which involve a Lie algebra of generators that preserve the canonical anti-commutation relations.
  • #1
sadegh4137
72
0
hi
i know what is coherent state, but i read this text in an article and i don't understand this
"
if we wish
to describe long range macroscopic forces, only bosonic fi elds will do, since fermionic fi elds
cannot build up classical coherent states. "

can you explain it for me, how fermionic fields can't build a classical coherent state?
and what's classical coherent state?
i know something about quantum coherent state which has minimum uncertainly.

thanks
 
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  • #2
You can't create a superposition of states containing even and odd numbers of fermions (so called univalence superselection rule) as such a state wouldn't be invariant under a 2π rotation (i.e. identity).
Hence a coherent state is not possible. However, you can create a coherent state of pairs of fermions (like e.g. Cooper pairs) which effectively behave as bosons, though.
 
  • #3
DrDu said:
You can't create a superposition of states containing even and odd numbers of fermions (so called univalence superselection rule) as such a state wouldn't be invariant under a 2π rotation (i.e. identity).
Hence a coherent state is not possible
why we shouldn't create a superposition of fermions?
i can't understand that what is relationship between coherent state and superposition of fermions!
is it possible to explain it for me?
thanks.
 
  • #4
sadegh4137 said:
why we shouldn't create a superposition of fermions?
i can't understand that what is relationship between coherent state and superposition of fermions!
is it possible to explain it for me?
thanks.

A coherent state is a superposition of states with definite numbers of particles. Because in coherent state the number of particles is not definite.
 
  • #5
why!
isn't definition of coherent state that
the states [itex]|a>[/itex] defined by [itex]a|\alpha>=\alpha|\alpha> [/itex], [itex]<\alpha|\alpha>=1[/itex]

and there is a theorem that say
coherent states have a minimum uncertainty relation
 
  • #6
sadegh4137 said:
isn't definition of coherent state that
[itex]<\alpha|\alpha>=1[/itex]

wrong for that.
the average number of particles in a coherent state is [itex]|\alpha|^2[/itex]
 
  • #7
what's exactly definition of coherent state?
 
  • #8
As you wrote it, it is the eigenstate of the annihilation operator (with complex eigenvalue).
wiki's article seems good. look at the relationship with Fock states.
 
  • #9
You can formally define "fermionic coherent states" by analogy to bosonic coherent states. Say I have a bosonic creation operator ##a^\dagger##. Then I can define a bosonic coherent state

##|\alpha\rangle = e^{\alpha a^\dagger} | 0 \rangle##

Which obeys ##a|\alpha\rangle = \alpha | \alpha \rangle## since ##[a, a^\dagger] = 1##.

By increasing ##\alpha## we increase the mean number of particles in the coherent state. For example, classical electromagnetic waves correspond to coherent photon states with a very large number of photons (very large ##\alpha##).

If I have a fermionic creation operator ##b^\dagger##, I can define the analogous fermionic coherent state:

##|\beta\rangle = e^{\beta b^\dagger} | 0 \rangle = (1 + \beta b^\dagger) | 0 \rangle##

This obeys ##b | \beta \rangle = \beta | \beta \rangle## if ##\{b, b^\dagger\} = 1##.

So there is a formal similarity to the case of bosonic coherent states. In this case, though, no matter how large we make ##\beta## the mean number of particles in the fermionic coherent state is never greater than one. Because of the exclusion principle, you can never put more than one fermion in the same state, so you can never build up a coherent wave containing a very large number of fermions all with the same wavelength.
 
  • #10
sadegh4137 said:
what's exactly definition of coherent state?
There are at least 3 definitions that I know of:

1) Eigenstate of an annihilation operator (assuming bosons here),

2) A state of minimum uncertainty product,

3) A state obtained by acting on the vacuum with a (so-called) "displacement" operator.
In the bosonic case this is something like:
$$
e^{za - \bar z a^\dagger} \, |0\rangle
$$

You can study the following tome for a more complete introduction to bosonic coherent state in terms of "displacement" operators:

L. Mandel and E. Wolf, "Optical coherence and quantum optics",
Cambridge Univ. Press, Cambridge 1995.


In the non-relativistic case, it turns out that these 3 definitions are equivalent, but not so in the relativistic case. Hence the question arises: what is the "best" definition of "coherent state"?
One has several possibilities for generalization.

In fact, the cases described above are a special case of so-called group-theoretic "generalized coherent states". In the above, the group is the usual Weyl--Heisenberg group.

More generally, however, one starts with a dynamics given by a Hamiltonian ##H## (and probably some other operators such as rotations, etc). The vacuum state is assumed to be invariant under the action of the Hamiltonian and other "symmetry generators" such as rotations and spatial translations. However, the full dynamical group (which maps solutions of the equations or motion into other solutions) often involves other generators as well. Call this set of generators ##L##. Typically, ##L## together with ##H## form a Lie algebra such that ##[L,H] \in L##. Then (modulo some technical details), one can construct states of the form
$$
\psi(z) ~=~ e^{L(z)} |0\rangle
$$
where now the exponentiated ##L(z)## is shorthand for some particular combination of the generators in ##L##, determined by a set of constant coefficients ##z##.

It turns out that such generalized coherent states ##\psi(z)## form an overcomplete basis for a Hilbert space that carries a representation of the full dynamics, and many calculations of physical properties are more convenient using them.

A classic review article on generalized coherent states is this one:
W.-M. Zhang, D.H. Feng and R. Gilmore, "Coherent states: Theory and some applications",
Rev. Mod. Phys. 62 (1990), 867--927.

Also the following books:

A. Perelomov, "Generalized Coherent States and Their Applications",
Springer-Verlag, 1986, ISBN 3-540-15912-6

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

Anyway... back to fermionic coherent states...

Clearly, exponentiating a fermionic creation operator isn't very useful, since ##(a^\dagger)^2 = 0##, unlike the bosonic case.

However, one can approach it via a different route. Just as the Weyl--Heisenberg emerges when one tries to find the group which preserves the canonical commutation relations, one can also investigate which group preserves the canonical anti-commutation relations. It turns out to involve the group SO(2n+m), iirc, where (I think) there are n paired degrees of freedom and m unpaired ones. My memory might be a bit faulty on this point, so check the Zhang-Feng-Gilmore paper. It all depends on the details of the Hamiltonian for the system under consideration.
 

FAQ: Explaining Fermionic Fields & Classical Coherent States

1. What are fermionic fields?

Fermionic fields are quantum fields that describe the behavior of fermions, which are particles with half-integer spin. These fields are fundamental to the Standard Model of particle physics and are essential for describing the properties of matter at the subatomic level.

2. How do fermionic fields differ from bosonic fields?

Fermionic fields are characterized by the Pauli exclusion principle, which states that no two fermions can occupy the same quantum state simultaneously. This is in contrast to bosonic fields, which do not have this restriction and allow for multiple particles to occupy the same state.

3. What are classical coherent states?

Classical coherent states are quantum states that have properties similar to classical waves. They are represented by a wavefunction that is a Gaussian distribution in phase space, with a well-defined position and momentum. These states are often used to describe the behavior of macroscopic systems.

4. How are fermionic fields related to classical coherent states?

Fermionic fields can be represented as a superposition of classical coherent states. This means that fermionic fields can be described using classical wave-like behavior, even though they are fundamentally quantum in nature. This allows for a more intuitive understanding of fermionic fields and their properties.

5. What is the significance of explaining fermionic fields and classical coherent states?

Understanding fermionic fields and classical coherent states is crucial for advancing our understanding of the fundamental laws of nature. These concepts play a central role in the Standard Model of particle physics and are essential for describing the behavior of matter at the subatomic level. Additionally, developments in these areas have led to important technological advancements, such as the development of quantum computers.

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