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Coherent state

  1. Aug 9, 2013 #1
    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.

  2. jcsd
  3. Aug 10, 2013 #2


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    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.
  4. Aug 10, 2013 #3
    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?
  5. Aug 10, 2013 #4
    A coherent state is a superposition of states with definite numbers of particles. Because in coherent state the number of particles is not definite.
  6. Aug 10, 2013 #5
    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
  7. Aug 10, 2013 #6


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    wrong for that.
    the average number of particles in a coherent state is [itex]|\alpha|^2[/itex]
  8. Aug 10, 2013 #7
    what's exactly definition of coherent state?
  9. Aug 10, 2013 #8


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    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.
  10. Aug 10, 2013 #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.
  11. Aug 11, 2013 #10


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    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.
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