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I Is the Fermion number operator squared equal to itself?

  1. Apr 12, 2017 #1
    What the title says. Acting on a fermionic state with the number operator to a power is like acting with the fermionic operator itself. Does this allow us to define ## \hat{n}^k=\hat{n} ##? Or is there any picky mathematical reason not to do so?
     
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  3. Apr 12, 2017 #2

    stevendaryl

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    If you think about it for a second, you'll see that that can't possibly be true. The whole point of a number operator is that [itex]\langle \Psi|\hat{n}|\Psi\rangle[/itex] gives the expected number of particles in state [itex]|\Psi\rangle[/itex]. If [itex]|\Psi\rangle[/itex] is a state with exactly two particles, then


    [itex]\langle \Psi|\hat{n}|\Psi\rangle = 2[/itex]
    [itex]\langle \Psi|(\hat{n})^2|\Psi\rangle = 4[/itex]

    [Edit]

    I apologize. It depends on what you mean. Do you mean the number operator that gives the total number of particles (anywhere, in any state)? In that case, what I said is true.

    However, for a specific state, there is a corresponding number operator, which counts the number of particles in that state. Since for Fermions, that has to be either 0 or 1, then you're right, the square will be 0 or 1, as well. So the number operator for a particular state is equal to its own square.
     
    Last edited: Apr 12, 2017
  4. Apr 12, 2017 #3
    Yup, I meant number operator for a fermionic state (number of particles equals either 0 or 1). So ##\exp({\hat{n}})=\sum{\frac{\hat{n}^k}{k!}}=\sum{\frac{\hat{n}}{k!}}=\hat{n}\sum{\frac{1}{k!}}=\hat{n} e ##?
     
  5. Apr 12, 2017 #4

    stevendaryl

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    That's not quite right. I would say it's equal to [itex] (1 - \hat{n}) + e \hat{n}[/itex].
     
  6. Apr 12, 2017 #5
    How come?
     
  7. Apr 12, 2017 #6

    stevendaryl

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    Well [itex]e^{\hat{n}} = 1 + \hat{n} + \frac{1}{2} (\hat{n})^2 + \frac{1}{6} (\hat{n})^3 + ...[/itex]. But if [itex]\hat{n}^2 = \hat{n}[/itex], then this becomes:

    Well [itex]e^{\hat{n}} = 1 + \hat{n} + \frac{1}{2} \hat{n} + \frac{1}{6} \hat{n} + ...[/itex]
    [itex]= 1 + \hat{n} (1 + \frac{1}{2} + \frac{1}{6} + ...)[/itex]

    But that series is the series for [itex]e-1[/itex]. The series for [itex]e[/itex] starts [itex]1+1+1/2 + ...[/itex]. So

    [itex]e^{\hat{n}} = 1 + \hat{n} (e-1)[/itex]
     
  8. May 22, 2017 #7
    The fermion number operator is defined in terms of operators, c and c^\dagger, which destroy/create fermionic particles. Because of Pauli's exclusion principle, doubly applying any of those gives you zero, that is, c^2|0> = 0, and also (c^\dagger)(c^\dagger)|0>=0.

    Now consider applying the square of the number operator onto the vacuum state, |0>. Since the number operator is defined as n=c^\dagger c, in this case we have

    nn|0> = c^\dagger c c^\dagger c|0>

    However, in order to produce meaningful answers the operators must be NORMAL ORDERED, that is, all the \daggers to the left. In doing this we end up with

    nn|0> = c^\dagger c c^\dagger c|0>=c^\dagger (1-c^\dagger c) c|0> = c^\dagger c|0> - c^\dagger c^\dagger c c|0> = c^\dagger c|0> = n|0>

    where we have used the fermionic anticommutation relations {c,c^\dagger}=1 and that cc|0>=0.
     
  9. May 22, 2017 #8

    Demystifier

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    You forgot the ##k=0## term.
     
  10. May 22, 2017 #9

    vanhees71

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    Say we are in a box, so that ##\{\hat{C}(\vec{p},\sigma),\hat{C}^{\dagger}(\vec{p}',\sigma')\}=\delta_{\vec{p},\vec{p}'} \delta_{\sigma,\sigma'}## with the ##\delta##'s being Kronecker ##\delta##'s with values ##1## and ##0## rather than Dirac-##\delta## distributions. Then you can just do the calculation: By definition
    $$\hat{N}=\hat{C}^{\dagger}(\vec{p},\sigma) \hat{C}(\vec{p},\sigma)$$
    leading to
    $$\hat{N}^2=\hat{C}^{\dagger} \hat{C} \hat{C}^{\dagger} \hat{C}=\hat{C}^{\dagger} [\{\hat{C},\hat{C}^{\dagger} \}-\hat{C^{\dagger} \hat{C}}] \hat{C}=\hat{C}^{\dagger} \hat{1} \hat{C}=\hat{N},$$
    because ##\{C,C \}=2 \hat{C}^2=0##.

    Another argument is that ##\hat{N}## has the eigenvalues 0 and 1, i.e.,
    $$\hat{N}^2=\sum_{n=0}^1 |n \rangle \langle n|n^2 =\sum_{n=0}^1 |n \rangle \langle n|n=\hat{N},$$
    because both ##0^2=0## and ##1^2=1##, i.e., for both eigenvalues ##n^2=n##.
     
  11. May 22, 2017 #10
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