Easy Question About the Number Operator

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In summary, the conversation discusses the operation of number and creation operators on the ground state ##\Psi_0## of a system of fermions. It is shown that operating with ##c^{\dagger}_k c_l##, where ##k \neq l##, results in a zero value. Additionally, it is explained that the same is true for the number operator acting on a ground state ##\Psi_{0,\downarrow}## for spin down particles.
  • #1
metapuff
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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.
 
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  • #2
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##.
 
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  • #3
Nice! That's really clever. This seems like a trick that I'll be using a lot. :)
 
  • #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!
 
  • #5
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.
 

FAQ: Easy Question About the Number Operator

1. What is the number operator?

The number operator is a mathematical operator that represents the number of elements in a set or the quantity of a certain attribute. It is often denoted as "n" or "#".

2. How is the number operator used in science?

The number operator is used in various scientific fields, such as chemistry, physics, and biology, to represent the number of atoms, particles, or organisms in a given sample or system. It is also used in statistical analysis to represent the number of occurrences of a certain event or outcome.

3. What are some examples of using the number operator in science?

Examples of using the number operator in science include counting the number of cells in a tissue sample, determining the number of protons in an atom, and measuring the number of molecules in a chemical reaction.

4. Can the number operator be negative?

No, the number operator cannot be negative. It represents a count or quantity, so it must be a positive integer or zero.

5. How is the number operator different from other mathematical operators?

The number operator is unique because it is used to represent a specific quantity, rather than performing a mathematical operation. It is also different from other operators in that it is often used in conjunction with other mathematical symbols, such as in equations or formulas.

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