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Easy Question About the Number Operator

  1. May 8, 2015 #1
    Suppose I have a system of fermions in the ground state ##\Psi_0##. If I operate on this state with the number operator, I get
    [tex] \langle \Psi_0 | c_k^{\dagger} c_k | \Psi_0 \rangle = \frac{1}{e^{(\epsilon_k - \mu)\beta} + 1} [/tex]
    which is, of course, the fermi distribution. What if I operate with ##c^{\dagger}_k c_l##, where ##k \neq l##? I.e, what is
    [tex] \langle \Psi_0 | c_k^{\dagger} c_l | \Psi_0 \rangle? [/tex]
    My hunch says that this is zero, but I'm not sure. This might be obvious.
     
  2. jcsd
  3. May 8, 2015 #2

    fzero

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    You can show that this is zero by writing out the ground state ##|\Psi_0\rangle## as a product of creation operators acting on the vacuum ##|0>##. Since ##\{c^\dagger_k,c_l\}=0## for ##k\neq l##, we can anticommute the ##c^\dagger_k## through to the ##c^\dagger_k## appearing in the product. Then we find a factor of ##(c^\dagger_k)^2=0##.
     
  4. May 8, 2015 #3
    Nice! That's really clever. This seems like a trick that I'll be using a lot. :)
     
  5. May 8, 2015 #4
    Okay, another question. Let ##\Psi_{0,\downarrow}## be the ground state for spin down electrons (for example we could have a partially polarized electron gas, with ##\Psi_{0,\downarrow}## representing the filled fermi sphere for down-spin electrons). If I try to act on this with the number operator for spin up particles, like
    [tex] \langle \Psi_{0,\downarrow} | c^{\dagger}_{\uparrow} c_{\uparrow} | \Psi_{0,\downarrow} \rangle [/tex]
    do I get 0 (since there are no spin-up particles in the down-spin ground state), or can I just pull the spin-up operators out, and write
    [tex] \langle \Psi_{0,\downarrow} | c^{\dagger}_{\uparrow} c_{\uparrow} | \Psi_{0,\downarrow} \rangle = c^{\dagger}_{\uparrow} c_{\uparrow}? [/tex]
    Again, this might be obvious, but I don't have a lot of confidence with second quantization yet and am trying to build intuition. Thanks!
     
  6. May 9, 2015 #5

    fzero

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    You will get zero because you can anticommute to get the expression

    $$ c^\dagger_{k\uparrow} \prod_r^\mathcal{N} c^\dagger_{r\downarrow} c_{k\uparrow} | 0 \rangle.$$

    Also, you generally can't pull operators out of an expectation value. If the expectation value you're writing down is physically sensible then the operator acts on the state that you're using to compute the expectation value.
     
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