# On which states do creation and annihilation operators act?

The Fock space is defined as the direct sum of all $$n$$-particle Hilbert spaces $$H_n$$

$$F = H_0 \oplus H_1 \oplus H_2 \oplus ...$$

Do creation and annihilation operators act (in second quantization) on Fock-states or on states, that are elements of the $$n$$-particle Hilbert spaces $$H_n$$?

Edit:

To be more precise about my question: Let $$|\Psi_2\rangle$$ be a state, which describes two particles and is therefore an element of $$H_2$$. As far as I understand, $$|\Psi_2\rangle$$ is not a Fock state, since a general Fock state would look like this: $$|\Psi\rangle = |\Psi_0\rangle \oplus |\Psi_1\rangle \oplus |\Psi_2\rangle \oplus ...$$

Now I am wondering whether creation and annihilation operators act on $$n$$-particle Hilbert-spaces, such as $$H_2$$ or on elements of the entire Fock space.

In the first case $$c^{\dagger}: H_n \rightarrow H_{n+1}$$ should be valid, whereas in the second case $$c^{\dagger}: F \rightarrow F$$ should be true.

I am just interested on which objects these operators act.

• They can operate on any state, in other words, on any element of the entire Hilbert space. Oct 27, 2021 at 6:49
• Am I right in assuming that each of the $H_n$ only contains one element? However, the Hilbert space also contains all possible superpositions of all the different elements. Therefore, it is not really valid to write is as a direct sum. Oct 27, 2021 at 7:32
• No, $H_n$ is supposed to be the Hilbert space of $n$ particles and therefore contains all possible $n$-particle states. Oct 27, 2021 at 7:51
• OK but then you need some other degrees of freedom to distinguish them, which is not apparent in your notation. Oct 27, 2021 at 10:44

Each $$H_n$$ is a subspace of the Fock space, as you can identify $$|\Psi_n\rangle \in H_n$$ with $$0\oplus \ldots \oplus |\Psi_n\rangle \oplus 0~ \oplus ... \in F$$

The way the creation operators $$c^\dagger$$ are usually defined is the following : define first operators $$c_n^\dagger : H_n \to H_{n+1}$$ and, by linearity, define a $$c^\dagger : F \to F$$ by : $$c^\dagger \Big(|\Psi_0\rangle \oplus |\Psi_1\rangle \oplus \ldots\oplus |\Psi_n\rangle\oplus\ldots\Big) = 0\oplus \Big(c_0^\dagger |\Psi_0\rangle\Big) \oplus \Big(c_1^\dagger |\Psi_1\rangle\Big) \oplus \ldots\oplus\Big( c_n^\dagger |\Psi_n\rangle \Big)\oplus \ldots$$

Edit : Given an operator $$O: F\to F$$, we can define its restrictions $$O_m : H_m\to F$$ by : $$O_m |\Psi_m\rangle = O(0\oplus 0 \oplus \ldots\oplus |\Psi_m\rangle\oplus 0\oplus \ldots)\in F$$

In general, $$O_m |\Psi_m\rangle$$ will not lie in any one $$H_n$$ (ie it will not be of the form $$0\oplus \ldots \oplus |\Psi_n\rangle \oplus \ldots$$, as was the case for $$O = c^\dagger$$). However, we can define operators $$O^{n}_{~~~m}: H_m\to H_n$$ by : $$O_m |\Psi_m \rangle = \Big(O^{0}_{~~~m} |\Psi_m\rangle\Big)\oplus \Big(O^{1}_{~~~m} |\Psi_m\rangle\Big) \oplus \ldots \oplus \Big(O^{n}_{~~~m} |\Psi_m\rangle\Big)\oplus \ldots$$

By linearity, we have : $$O \Big(|\Psi_0\rangle \oplus |\Psi_1\rangle \oplus \ldots\oplus |\Psi_n\rangle\oplus\ldots\Big) = \Big(\sum_{m} O^{0}_{~~~m} |\Psi_m\rangle\Big) \oplus \ldots\oplus\Big(\sum_{m} O^{n}_{~~~m}|\Psi_m\rangle \Big) \oplus \ldots \tag{1}$$

Conversely, given a family of operators $$O^{n}_{~~~m}:H_m \to H_n$$, equation $$(1)$$ defines an operator $$O:F\to F$$.

• Is the same also true for any other operator, e.g. the Hamiltonian? Oct 29, 2021 at 16:59