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I Eigenvalues of Fermionic field operator

  1. Feb 22, 2017 #1
    Hello,

    I'm a bit confused about the eigenvalues of the second quantized fermionic field operators [itex] \psi(x)_a[/itex]. Since these operators satisfy the condition [itex] \{\psi(x)_a, \psi(y)_b\} = 0 [/itex] the eigenvalues should also anti-commute? Does this mean that the eigenvalues of [itex] \psi(x)_a[/itex] are Grassmann-numbers?

    Thanks for your help
     
  2. jcsd
  3. Feb 23, 2017 #2

    DrDu

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    The Pauli matrices ##\sigma_x## and ##\sigma_y## also anti-commute. Would you also conclude that they have Grassmann valued eigenvalues?
     
  4. Feb 23, 2017 #3
    No, but the eigenvalues are not matrices? And in the path integral formalism we use Grassmann valued fields.
     
  5. Feb 23, 2017 #4

    Demystifier

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    There is one important difference:
    $$ \psi_a(x)\psi_a(x)=0$$
    but
    $$\sigma_x\sigma_x \neq 0$$
    A fermionic field operator cannot be diagonalized and it does not have eigenvalues. A Pauli matrix can be diagonalized and it does have eigenvalues. Fermionic field operator is not hermitian, Pauli matrix is hermitian.
     
  6. Feb 23, 2017 #5
    But isn't there a self-adjoint extension with generalized eigenstates like for the QM operators [itex] x,p [/itex]?
     
  7. Feb 23, 2017 #6

    Demystifier

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    No.
     
  8. Feb 23, 2017 #7

    DrDu

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    I just wanted to give a counter example of anti-commuting operators which have eigenvalues. Even more field operator like are the matrices ##\sigma_\pm=\sigma_x\pm i\sigma_y##. They anticommute and are non-hermitian. They are not diagonisable.
     
  9. Feb 23, 2017 #8

    Demystifier

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    Yes, ##\sigma_\pm## is a much better analog of fermionic field. ##\sigma_\pm## does not have (non-zero) eigenvalues.
     
  10. Feb 23, 2017 #9

    stevendaryl

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    In some developments of the quantum mechanics of fermions (an example is "Quantum Field Theory of Point Particles and Strings" by Brian Hatfield), Feynman's path integral is extended to include integration over non-commuting Grassman variables. I personally find it hard to understand, because I'm used to thinking of noncommutation as being a property of operators, not variables.
     
  11. Feb 23, 2017 #10

    Demystifier

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    A long time ago I have tried to overcome this
    https://arxiv.org/abs/hep-th/0210307
    but a smart referee found an error for the case of many fermionic degrees, which I couldn't repair. (If someone is interested, Eq. (21), which would be needed in a consistent approach, is not really satisfied by my construction. A more complicated construction is needed to satisfy (21), but then the formalism is too complicated to be useful.)

    But in the case of one fermionic degree of freedom my approach is perfectly fine. The final result in this case is the functional integral (14). It's nice from a conceptual point of view, but probably not too useful.
     
    Last edited: Feb 23, 2017
  12. Feb 23, 2017 #11
    So the only possibility to get hermitian operators are the number-operators? But could the Lagrangian (it should also be hermitian) be rewriten in terms of number-operators?
     
  13. Feb 24, 2017 #12

    DrDu

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    No, you can also express your Hamiltonian in terms of e.g. ##\psi+\psi^\dagger## and ## i(\psi-\psi^\dagger)## which are hermitian.
     
  14. Feb 24, 2017 #13
    Thanks. So the eigenvalues of this operators are real?

    But shouldn't the hermitian operators [itex] \psi + \psi^\dagger[/itex] and [itex] i(\psi - \psi^\dagger)[/itex] you gave, satisfy the commutation relations if and only if [itex] \psi, \psi^\dagger [/itex] satisfy the anti-commutation relations?
     
    Last edited: Feb 24, 2017
  15. Feb 26, 2017 #14

    Demystifier

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    No. See Eq. (4) in the paper I linked above.
     
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