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)?
  • $\begingroup$ 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 $\endgroup$
    – AXensen
    Commented May 19, 2023 at 9:37
  • $\begingroup$ @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. $\endgroup$ Commented May 19, 2023 at 9:44
  • $\begingroup$ For the fermionic part there is a related question $\endgroup$
    – LPZ
    Commented May 19, 2023 at 10:00

1 Answer 1


To prove the existence of the ground state, I will follow the derivation made in Richard Feynman "Statistical Mechanics: A Set Of Lectures", Chapter 6, "A Simple Mathematical Problem".

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 \begin{equation} \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). \end{equation} 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 $\hat{a}$.

The uniqueness of the vacuum state can't be proven just from the commutation relations.

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 $\hat{a}$.

  • $\begingroup$ 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 $\hat{S}^-$. $\endgroup$ Commented May 20, 2023 at 2:40
  • $\begingroup$ 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. $\endgroup$
    – E. Anikin
    Commented May 20, 2023 at 10:06

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