# Existence and uniqueness of vacuum of fermion or boson operators

Suppose I have a set of boson (or fermion) annihilation operators $$\{a_i\}$$ defined on a Hilbert space. These operators satisfy the canonical (anti-)commutation rules

$$\text{boson:} \quad [a_i, a^\dagger_j] = \delta_{ij}, \quad [a_i, a_j] = 0$$

$$\text{fermion:} \quad \{a_i, a^\dagger_j\} = \delta_{ij}, \quad \{a_i, a_j\} = 0$$

Define the state $$|0\rangle$$ (called the "vacuum") that is annihilated by all $$a_i$$:

$$a_i |0\rangle = 0$$

My question:

• Can one prove the existence of such $$|0\rangle$$ (I think this part is relatively easy, and will use the number operators $$n_i = a^\dagger_i a_i$$, as in the study of harmonic oscillators)?
• More importantly: if it exists, is it unique (up to normalization)?
• Interesting question, but I think you're going to run in a lot of difficulties before you can answer this. I'm assuming you're talking about an interacting QFT, in which case we don't have a way to construct the annihilation operator (nor does it have a clear meaning), and Haag's theorem says you don't even really know what the Hilbert space is May 19 at 9:37
• @AXensen Do you think CCR/CAR and $a_i |0\rangle = 0$ alone are not enough? I somehow think the question can be tackled in a purely mathematical way without knowledge of the physics background. May 19 at 9:44
• For the fermionic part there is a related question
– LPZ
May 19 at 10:00

Let us consider some (normalized) eigenstate $$|\nu\rangle$$ of the number operator $$\hat{a}^\dagger \hat{a}$$ with eigenvalue $$\nu$$. From commutation relations, we find that the states $$|\psi_{\nu,k}\rangle =\hat{a}^k |\nu\rangle$$ have eigenvalue $$\nu - k$$ for all $$k$$. Also, the norm of these states is $$$$\langle\psi_{\nu,k}|\psi_{\nu,k}\rangle = \langle \nu | \hat{a}^k (\hat{a}^\dagger)^k|\nu\rangle = \nu(\nu-1)\dots(\nu - k + 1).$$$$ From this equation, it directly follows that $$\nu$$ should be non-negative integer. Otherwise, there exist states $$|\psi_{\nu,k}\rangle$$ (with $$k > \nu$$) with negative squared norm in Hilbert space.
Therefore, the state $$|\psi_{\nu,\nu} = \hat{a}^\nu|\nu\rangle$$ is a ground state: it is an eigenvector of $$\hat{a}^\dagger\hat{a}$$ with eigenvalue $$0$$, and it is annihilated by $$\hat{a}$$.
Consider a direct sum of two independent bosonic or fermionic subspaces spanned by the states $$|n\rangle_{1,2}$$, where the subscript index indicates the subspace, and $$n$$ is the number of quanta. Assume that the creation/annihilation operator acts on each of the subspaces independently. Then, there are two vacuum states in the system. An example of such a system is a spin-$$1/2$$ particle in a harmonic potential: both of the states $$|0, \uparrow\rangle$$ and $$|0, \downarrow\rangle$$ are annihilated by $$\hat{a}$$.
• I think your counter-example on the uniqueness is not correct. I think your $\hat{a}$ only refers to the ladder operator of the harmonic potential. But what I want is a vacuum that is annihilated by all annihilation operators, including that of the spin-1/2. So I think only $|0,\downarrow\rangle$ is the vacuum in my sense, which is annihilated by both $\hat{a}$ and the spin ladder operator $\hat{S}^-$. May 20 at 2:40
• In your question, you asked about some pre-defined set of creation/annihilation operators. In particular, this set can consist of only a single pair $\hat{a}, \hat{a}^\dagger$. For this set, you require that 1) they act on a Hilbert space 2) they obey commutation relations, and nothing more. In my example, both conditions are satisfied. $S^{-}$ was not initially included in the set. Maybe you would like to impose some other conditions on your set of creation/annihilation operators, but you should update the question accordingly or ask another question. May 20 at 10:06