# When do annihilation operators act - second quantisation

I am confused as to when does an annihilation operator annihilate its creation operator counter-part.

As an intro to second quantisation in my QFT notes we have:

The operators $c_j$ annihilate their counterparts and for this we require that: $c_{j_1}c_{j_2}^\dagger |0\rangle =\delta_{j_1 j_2} |0 \rangle$

Doesn't this mean that an operator $\hat c_{j_1} \hat c_{j_1}^\dagger = 1$? Or do we require the $|0\rangle$ to be able to say that?

I guess that would imply that annihilation may only happen when the term is a state rather than an operator? e.g. $\hat c_{j_1} \hat c_{j_1}^\dagger |0\rangle = |0\rangle$ vs $\hat c_{j_1} \hat c_{j_1}^\dagger = \hat c_{j_1} \hat c_{j_1}^\dagger$

I am asking this question because I am dealing with fermions on sites, and I was curious whether an operator such as $f_j f^\dagger_j$ would simply "collapse" or "annihilate" to 1.

• Are these specifically Fermion operators, or are you asking about the Boson case too? – DanielSank Jan 20 '17 at 2:38
• @DanielSank sorry i removed "number operator" out, I generally mean any operator – Zeeshan Ahmad Jan 20 '17 at 3:04

## 1 Answer

Isn't the order for the number operator you listed incorrect? Doesn't the creation operator lie to the left of the annihilation operator?

Edit: You didn't mention if you are talking about fermions or bosons. For fermions, that isn't the number operator.

Either way, you only showed that it equals 1 when when acting on the vacuum state. To show that it is equal to one always, you have to show that it equals 1 on a basis of the space (so, for every state.)

• So say you have a general operator $c c^\dagger$. Will this term always collapse to one? – Zeeshan Ahmad Jan 20 '17 at 3:07
• I don't think so. For bosons, $\hat{c}_{i}\hat{c}_{i}^{\dagger}|n_{1},....n_{i},....>$ will give $n_{i}|n_{1},....,n_{i},.....>$. For fermions, it will give |0> if state i had a particle, and it will give back the original state if state i didn't have a particle. – Ben S Jan 20 '17 at 3:47
• Saw an update for this due to an upvote. I believe my previous comment should have $n_{i}+1$ and not $n_{i}$. – Ben S Aug 3 '18 at 3:30