Suppose we are given the vector $$|n_1, n_2, \ldots \rangle \in H^{\otimes n}_s$$ where $H^{\otimes n}_s$ is the n-fold symmetric tensor product of a Hilbert space $H$, and $|n_1, n_2, \ldots \rangle$ represents the state where $n_i$ particles are in the $i$th state one (of course we require that $\sum n_k$ = n).

The creation and annihilation operators, $A^\dagger$ and $A$, increase/decrease the total number of particles by one. More explicitly, $$A: H^{\otimes n}_s \rightarrow H^{\otimes (n-1)}_s\\ A^\dagger: H^{\otimes n}_s \rightarrow H^{\otimes (n+1)}_s.$$

It is my current understanding that if the set $\{e_i\}$ form a basis for $H$, then for any $k$ we have (omitting any normalization constants): $$A(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k-1,\ldots\rangle\\A^\dagger(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k+1,\ldots\rangle$$ In other words, when applied to the $k$th basis vector these operators create/annihilate a single particle in the $k$th state.

I am now curious what happens if I apply this operator to any $\psi \in H$, and not just a basis vector. How do the total number of particles still increase/decrease by 1 if $\psi = \sum \psi_i e_i$ is now a linear combination of states? What is the physical interpretation of this?

Suppose we are given the vector $$|n_1, n_2, \ldots \rangle \in H^{\otimes n}_s$$ where $H^{\otimes n}_s$ is the $n$-fold symmetric tensor product of a Hilbert space $H$, and $|n_1, n_2, \ldots \rangle$ represents the state where $n_i$ particles are in the $i$th state one (of course we require that $\sum n_k$ = n).

The creation and annihilation operators, $A^\dagger$ and $A$, increase/decrease the total number of particles by one. More explicitly, $$A: H^{\otimes n}_s \rightarrow H^{\otimes (n-1)}_s\\ A^\dagger: H^{\otimes n}_s \rightarrow H^{\otimes (n+1)}_s.$$

It is my current understanding that if the set $\{e_i\}$ form a basis for $H$, then for any $k$ we have (omitting any normalization constants): $$A(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k-1,\ldots\rangle\\A^\dagger(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k+1,\ldots\rangle$$ In other words, when applied to the $k$th basis vector these operators create/annihilate a single particle in the $k$th state.

I am now curious what happens if I apply this operator to any $\psi \in H$, and not just a basis vector. How do the total number of particles still increase/decrease by 1 if $\psi = \sum \psi_i e_i$ is now a linear combination of states? What is the physical interpretation of this?

Suppose we are given the vector $$|n_1, n_2, \ldots \rangle \in H^{\otimes n}_s$$ where $H^{\otimes n}_s$ is the n-fold symmetric tensor product of a Hilbert space $H$, and $|n_1, n_2, \ldots \rangle$ represents the state where $n_i$ particles are in the $i$th state one (of course we require that $\sum n_k$ = n).

The creation and annihilation operators, $A^\dagger$ and $A$, increase/decrease the total number of particles by one. More explicitly, $$A: H^{\otimes n}_s \rightarrow H^{\otimes (n-1)}_s\\ A^\dagger: H^{\otimes n}_s \rightarrow H^{\otimes (n+1)}_s.$$

It is my current understanding that if the set $\{e_i\}$ form a basis for $H$, then for any $k$ we have (omitting any normalization constants): $$A(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k-1,\ldots\rangle\\A^\dagger(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k+1,\ldots\rangle$$ In other words, when applied to the $k$th basis vector these operators create/annihilate a single particle in the $k$th state.

I am now curious what happens if I apply this operator to any $\psi \in H$, and not just a basis vector. How do the total number of particles still increase/decrease by 1 if $\psi = \sum \psi_i e_i$ is now a linear combination of states?

Suppose we are given the vector $$|n_1, n_2, \ldots \rangle \in H^{\otimes n}_s$$ where $H^{\otimes n}_s$ is the n-fold symmetric tensor product of a Hilbert space $H$, and $|n_1, n_2, \ldots \rangle$ represents the state where $n_i$ particles are in the $i$th state one (of course we require that $\sum n_k$ = n).

The creation and annihilation operators, $A^\dagger$ and $A$, increase/decrease the total number of particles by one. More explicitly, $$A: H^{\otimes n}_s \rightarrow H^{\otimes (n-1)}_s\\ A^\dagger: H^{\otimes n}_s \rightarrow H^{\otimes (n+1)}_s.$$

It is my current understanding that if the set $\{e_i\}$ form a basis for $H$, then for any $k$ we have (omitting any normalization constants): $$A(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k-1,\ldots\rangle\\A^\dagger(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k+1,\ldots\rangle$$ In other words, when applied to the $k$th basis vector these operators create/annihilate a single particle in the $k$th state.

I am now curious what happens if I apply this operator to any $\psi \in H$, and not just a basis vector. How do the total number of particles still increase/decrease by 1 if $\psi = \sum \psi_i e_i$ is now a linear combination of states? What is the physical interpretation of this?