Is the Fermion number operator squared equal to itself?

In summary, the conversation discusses the possibility of defining the number operator to a power as equal to the number operator itself for fermionic states. However, this is not possible due to the nature of the number operator and the anticommutation relations for fermionic particles. The calculation and eigenvalue arguments presented support this conclusion.
  • #1
voila
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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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  • #2
voila said:
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?

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.
 
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  • #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 ##?
 
  • #4
voila said:
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 ##?

That's not quite right. I would say it's equal to [itex] (1 - \hat{n}) + e \hat{n}[/itex].
 
  • #5
How come?
 
  • #6
voila said:
How come?
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]
 
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  • #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.
 
  • #8
voila said:
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 ##?
You forgot the ##k=0## term.
 
  • #9
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##.
 
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  • #10

1. What is the Fermion number operator?

The Fermion number operator is a mathematical operator used in quantum mechanics to describe the number of fermions (particles with half-integer spin) in a system. It is denoted by the symbol NF.

2. How is the Fermion number operator squared?

The Fermion number operator is squared by multiplying it by itself. So the squared operator is NF2.

3. Is the Fermion number operator squared equal to itself?

Yes, the Fermion number operator squared is equal to itself, as it is a self-adjoint operator. This means that when it operates on a state, it gives the same state back.

4. Why is the Fermion number operator squared important?

The squared Fermion number operator is important because it is a conserved quantity in many physical systems. This means that its value remains constant throughout the system's evolution, making it a useful tool for understanding and predicting the behavior of fermions.

5. In what other contexts is the Fermion number operator squared relevant?

The Fermion number operator squared is also relevant in particle physics, where it is used to classify particles as fermions or bosons. In addition, it is used in condensed matter physics to describe the number of particles occupying different energy levels in a system.

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