# Square of annihilation operator

Sorry about this question, but due to my limited background in quantum physics I wasn't able to figure out why in ALL the refs I have searched they state it as iffor sure: Why is the square of a creation/annihilation operator zero? Given that this is for a Fock space system.

$$\hat{a}^\dagger~\propto~(\omega\hat{q} - i\hat{p}) \tag{2.14}$$ $$\hat{a}^{\dagger 2} = 0$$

The 2 means that the operator is in power two. The same holds true for the annihilation operator. I found some phoney proof (based on the fact that this describes a harmonic oscillator) which I don't like and wish to get feedback from experienced people.

References:

1. C.C. Gerry & P.L. Knight, Introductory Quantum Optics, 2004; eq. (2.45).
• Fermionic operators satisfy $(A^\dagger)^2=0$ since you cannot put two fermions in the same state. Commented Oct 25, 2017 at 17:59
• Hi @Mazem: Is this from a textbook? Which page? Which formula? Commented Oct 25, 2017 at 18:03
• @Qmechanic Yeah. “Introductory Quantum Optics ”, Jerry & Knight. ~p. 15 There is a load of other handouts here and there from the net. They all state the same thing. Commented Oct 25, 2017 at 18:34
• @Qmechanic It has been the case to find the variance of the electric field where the creation + annihilation operators are raised to power 2. I can completely understand that this is not a commutative operation (I.e. a.b =/= b.a) but prefer to approach the creation square(and annihilation square) from a mathematical view point of possible) If it is not & better thought of as an axiom as ZeroTheHero refers, I’ll be quite grateful to cite the ref for this. Commented Oct 25, 2017 at 18:45
• Are you talking about eq. (2.45)? Commented Oct 25, 2017 at 18:51

1. Without context, it first sounded like you were considering fermionic creation & annihilation operators, which square to zero.

2. However, Ref. 1 is considering bosonic creation & annihilation operators, so they definitely do not square to zero.

3. If the $\hat{a}^{2}$ or $\hat{a}^{\dagger 2}$ are sandwiched between a bra vector $\langle m|$ and a ket vector $|n\rangle$, then the result is zero if $m-n\neq \pm 2$. Comparison with Ref. 1 reveals that this is exactly what's going on.

References:

1. C.C. Gerry & P.L. Knight, Introductory Quantum Optics, 2004; eq. (2.45).
• In equation (2.45) as shown ${\hat{a}}^{\dagger 2}$ disappears and the same holds true for ${\hat{a}}^2$. Probably, it is their addition what goes to zero. BTW, their addition gives you $\frac{1}{\hbar \omega}(\omega^2 \hat{q}^2 - \hat{p}^2)$. How could this value be zero? Does it have to do eigen propoerty of vectors? Commented Oct 26, 2017 at 10:37
• They disappear independently! Commented Oct 26, 2017 at 10:42
• is there a way to prove that mathematically? Commented Oct 26, 2017 at 10:45
• Yes. You need: 1. $\hat{a}|0\rangle=0$; 2. $\langle 0|\hat{a}^{\dagger }=0$; 3. the CCR; 4. and a couple of definitions. Commented Oct 26, 2017 at 10:49

There are two different types of creation and annihilation operators:

• One is the bosonic set, which obeys the commutation relation $[a,a^\dagger]=1$ (and trivially $[a,a]=0=[a^\dagger,a^\dagger]$), and for which you can define quadratures via $q=\frac12(a+a^\dagger)$ and $p=\frac{1}{2i}(a-a^\dagger)$ which obey $[q,p]=i$, i.e. $a=q+ip$ and $a^\dagger=q-ip$ can be seen as the ladder operators for a harmonic oscillator with hamiltonian $H=\frac12(p^2+q^2)$.

• The other is the fermionic set, which obeys the anticommutation relations $\{c,c^\dagger\}=1$, $\{c,c\}=0=\{c^\dagger,c^\dagger\}$, from which follows that $c^2=0=(c^\dagger)^2$.

The question incorrectly conflates properties of the two sets, i.e. the question as written asserts that 'the' creation operator obeys both $(a^\dagger)^2=0$ and $a^\dagger = q-ip$ where $q$ and $p$ are reasonably-behaved quadratures. There is no such operator: it's either fermionic, with the first property, or bosonic, with the second property.

However, OP's reference for that first property makes a much weaker claim: Gerry & Knight's eq. (2.45) does not require that $(a^\dagger)^2=0$ or that $a^2=0$; instead, it only requires that their expectation values over Fock states vanish, i.e. $$⟨n|a^2|n⟩=0=⟨n|(a^\dagger)^2|n⟩.$$ These two properties are complex conjugates of each other, and they can be rigorously proved from the facts that $$a|n⟩=\sqrt{n}|n-1⟩$$ so therefore $$a^2|n⟩=\sqrt{n(n-1)}|n-2⟩$$ and thus $$⟨n|a^2|n⟩=\sqrt{n(n-1)}⟨n|n-2⟩=0.$$ However, just because the expectation values of an operator vanish in the Fock basis (like they do for $a$ and $a^\dagger$ themselves) does not mean that the operator is identically zero, so e.g. $⟨0|a^2|2⟩ = \sqrt{2}\neq 0$ as a simple example.

• Many thanks for your clarification. Now that is a complete answer! Honestly, I managed to prove that $<n|a^{\dagger 2}|n> = 0$ just before your post and was just about to answer my own question. But this is more comprehensive and covers the subtlty between fermionic and bosonic operators. Commented Oct 29, 2017 at 13:34