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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.

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  • $\begingroup$ They can operate on any state, in other words, on any element of the entire Hilbert space. $\endgroup$
    – flippiefanus
    Oct 27, 2021 at 6:49
  • $\begingroup$ 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. $\endgroup$
    – flippiefanus
    Oct 27, 2021 at 7:32
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    $\begingroup$ No, $H_n $ is supposed to be the Hilbert space of $n$ particles and therefore contains all possible $n$-particle states. $\endgroup$
    – maxxam
    Oct 27, 2021 at 7:51
  • $\begingroup$ OK but then you need some other degrees of freedom to distinguish them, which is not apparent in your notation. $\endgroup$
    – flippiefanus
    Oct 27, 2021 at 10:44

1 Answer 1

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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 $O:F\to F$.

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  • $\begingroup$ Is the same also true for any other operator, e.g. the Hamiltonian? $\endgroup$
    – maxxam
    Oct 29, 2021 at 16:59

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