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Why a bosonic background in SUGRA?

  1. Oct 30, 2015 #1
    In this http://www.ulb.ac.be/sciences/ptm/pmif/Rencontres/specgeom.pdf, for example on p.7 and in many other references, people usually say:

    we will concentrate on the bosonic part of the Lagrangian in N=1 (sometimes N=2) supergravity.

    In other incidents, people say, that they are going to consider backgrounds that are considered having vanishing values of fermions, but why? Why do we have to consider bosonic backgrounds and not, for instance, fermionic backgrounds?
     
    Last edited: Oct 30, 2015
  2. jcsd
  3. Oct 30, 2015 #2

    fzero

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    The spin-statistics theorem requires that, in a QFT, fermions are anticommuting objects. A background value for a fermion would have to have components that also anticommute (i.e. it is not enough to have noncommuting matrices, the matrix elements themselves have to be noncommuting). If we are dealing with classical geometry and real numbers, we can't construct such a thing, so we must have vanishing fermion VEVs.

    It may be the case that people that study noncommutative geometry, where the spacetime includes anticommuting variables, have studied some sort of fermionic backgrounds, but I am not familiar with any examples.
     
  4. Oct 30, 2015 #3
    Thank you for your answer @fzero. You said

    I do not know what classical geometry and real numbers mean here and if the case that I'm asking about has the latter two conditions. The case that I'm asking about is for example the case of N=2 supergravity.. So, why is it that fermions must vanish there? Is it because what you've mentioned in your previous answer? If yes, I would be grateful if you can elaborate on the ideas of classical geometry and real numbers there.
     
    Last edited: Oct 30, 2015
  5. Oct 30, 2015 #4

    fzero

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    Yes, it's easy to elaborate. Say that our space time is Minkowski space ##M##, at least locally and we have coordinates ##x^\mu##. Now we have a quantum scalar field ##\Phi(x)## and the classical background value is ##\phi(x) = \langle v|\Phi(x)|v\rangle ## where the expectation value is taken in the vacuum state ##|v\rangle##. For a real field ##\phi(x)## is just a function from Minkowski space to the real numbers.

    Now consider a particular component of a real fermion field ##\Psi(x)##. If you want, think of this as the positive helicity component of a Weyl fermion. Spin-statistics says that we cannot have two fermions in the same state, so for any state ##|\alpha\rangle##, ##\Psi(x) \Psi(x) |\alpha \rangle =0##. Now assume that the classical value of this field, ##\psi(x) = \langle v|\Psi(x)|v\rangle ## is not zero. Then consider
    $$ 0 = \langle v| \Psi(x) \Psi(x)|v\rangle = \sum_{\alpha} \langle v| \Psi(x) |\alpha\rangle\langle \alpha| \Psi(x)|v\rangle,$$
    where I have inserted a complete set of states into the expression. The vacuum state appears in the sum, so we can write
    $$ \begin{split} 0 = & \langle v| \Psi(x) |v\rangle^2 + \sum_{\alpha\neq v} \langle v| \Psi(x) |\alpha\rangle\langle \alpha| \Psi(x)|v\rangle \\
    = & (\psi(x))^2 + \sum_{\alpha\neq v} |\langle v| \Psi(x) |\alpha\rangle|^2. \end{split}$$
    Since both terms are positive definite, we conclude that it must be the case that ##\Psi(x)|v\rangle=0##. It therefore follows that ##\psi(x)=0## by definition.

    The argument I was thinking of previously when I referred to classical geometry and real numbers was that because of statistics, we know that ##(\psi(x))^2=0## even if we assumed that ##\psi(x)\neq 0##. But for a real function of the coordinates ##x^\mu##, this is impossible
     
  6. Oct 30, 2015 #5
    @fzero Great answer. I now understand how the argument works in Minkowski space. I am just still a little unsure about the answer of the second part of my question "the case of N=2 supergravity". I will repeat it so that I can see the full image. So my last question, how is my case similar to your argument in Minkowski space?
     
    Last edited: Oct 30, 2015
  7. Oct 30, 2015 #6

    fzero

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    Nothing about the argument really depends on having Minkowski space. I wrote that before I had put the argument together and it turned out not to be important and didn't think about it again. The argument doesn't really depend on what sort of theory you have, since it applies to any component of any fermion in all theories.

    If you really wanted to define the vacuum or other states in your theory, then the field content and background geometry would be important, but the argument didn't require those specifics.
     
  8. Oct 30, 2015 #7
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