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Second Quantization for Fermions

  1. Nov 21, 2012 #1
    Please let me know if I get this right. Second Quantization for Fermions used the definition of its annihilation and creation operators instead of wavefunctions. We use second quantization to express this many body problem in a hamiltonian. Am I right? Can someone please explain this to me in simple terms? How do I start in understanding this second quantization?
  2. jcsd
  3. Nov 21, 2012 #2
    Not sure, what you mean.second quantization for fermions do use creation and annihilation operator which satisfy anticommutation rule rather than commutation rule.It is used to bring pauli exclusion principle into account i.e. two fermions can not occupy same state.
  4. Nov 22, 2012 #3
    Second quantization is a somewhat misleading term to me (and many others) because it seems to imply that you do two steps of quantisation, which is not correct.

    What happens is this:
    In QM, you have a wave function that assigns a prob. amplitude to each point in space (and time). The Dirac equation or Klein-Gordon equation was initially conceived as an equation of a wave function like the Schroedinger equation.

    Then people realised that this is not correct (because particle number is not conserved due to creation of particles/antiparticles etc.) and that you need in fact a field theory.
    In a field theory, you use the same equation (for example the Klein-Gordon equation), but you now interpret this as the field equation of a classical field. Solutions of this equations look like they did before (when you thought that you were dealing with a prob. amplitude), but now the field has to be interpreted as a classical quantity (for example, the displacement of a membrane as a intuitive example).

    In the second step, you then quantise this field, using the standard rules of quantum theory, converting observables to operators. Since the classical field itself is an observable, it becomes a field operator.

    What makes things confusing is that you now usually do not deal with wave functions anymore - QFT is usually not phrased that way (laudable exception in the book of Hatfield "QFT of point particles and strings"). If you think in terms of wave functions, what QFT does is to assign a probability amplitude to each possible field configuration. (Since this is not easy to do because of the infinity of possible functions, people prefer other ways of describing QFT.)

    The standard way of explaining this quantisation - turning wave functions into creation/annihilation operators - is also confusing for another reason, in my opinion: Before you have a set of possible solutions, with coefficients a and b that can have any possible value. Then you do the second quantisation and these coefficients turn into creation/annihilation operators, which are precisely defined objects. Where did all the possibilities for the solution go?
    Answer: The 2nd quantised solution that contains operators has to be applied to a state vector, and the freedom you have is now in the state vector, not in the solution anymore. Unfortunately, this is not always explained.

    Hope this helps - if you can read German, you can also look at my blog where I explain many aspects of QFT - see here and look for the QFT series:
  5. Nov 23, 2012 #4
    yeah. Got it! Thanks. but how can we say that the slater determinant is asymmetric?
  6. Nov 24, 2012 #5
    Do you even need the slater determinante in QFT? Usually it is used to anti-symmetrise the many-particle wave function, but this is not existing in QFT.
    Perhaps this link may help:
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