I Is the Fermion number operator squared equal to itself?

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1. Apr 12, 2017

voila

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?

2. Apr 12, 2017

stevendaryl

Staff Emeritus
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 $\langle \Psi|\hat{n}|\Psi\rangle$ gives the expected number of particles in state $|\Psi\rangle$. If $|\Psi\rangle$ is a state with exactly two particles, then

$\langle \Psi|\hat{n}|\Psi\rangle = 2$
$\langle \Psi|(\hat{n})^2|\Psi\rangle = 4$

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
3. Apr 12, 2017

voila

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. Apr 12, 2017

stevendaryl

Staff Emeritus
That's not quite right. I would say it's equal to $(1 - \hat{n}) + e \hat{n}$.

5. Apr 12, 2017

voila

How come?

6. Apr 12, 2017

stevendaryl

Staff Emeritus
Well $e^{\hat{n}} = 1 + \hat{n} + \frac{1}{2} (\hat{n})^2 + \frac{1}{6} (\hat{n})^3 + ...$. But if $\hat{n}^2 = \hat{n}$, then this becomes:

Well $e^{\hat{n}} = 1 + \hat{n} + \frac{1}{2} \hat{n} + \frac{1}{6} \hat{n} + ...$
$= 1 + \hat{n} (1 + \frac{1}{2} + \frac{1}{6} + ...)$

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

$e^{\hat{n}} = 1 + \hat{n} (e-1)$

7. May 22, 2017

Marcello Neto

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. May 22, 2017

Demystifier

You forgot the $k=0$ term.

9. May 22, 2017

vanhees71

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
$$\hat{N}=\hat{C}^{\dagger}(\vec{p},\sigma) \hat{C}(\vec{p},\sigma)$$
$$\hat{N}^2=\hat{C}^{\dagger} \hat{C} \hat{C}^{\dagger} \hat{C}=\hat{C}^{\dagger} [\{\hat{C},\hat{C}^{\dagger} \}-\hat{C^{\dagger} \hat{C}}] \hat{C}=\hat{C}^{\dagger} \hat{1} \hat{C}=\hat{N},$$
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 $n^2=n$.

10. May 22, 2017