# Where is the error in my calculation for a simple bosonic system?

Physics Asked on May 9, 2021

$$newcommand{ket}[1]{left|#1rightrangle}$$I was doing some calculations and I came across something that confused me and maybe you can explain to me where my reasoning is wrong. Let’s say I have some bosonic systems described by the tensor product of four independent states:
$$ket{psi} = ket{n_1} otimes ket{n_2} otimes ket{n_3} otimes ket{n_3} tag{1}label{1}$$
The vacuum state is of course
$$ket{varnothing} = ket{0} otimes ket{0} otimes ket{0} otimes ket{0} tag{2}label{2}$$
Let’s define some composite operator
$$hat{A}=sum_i hat{a}_i tag{3}label{3}$$
where $$hat{a}_i$$ is the annihilation operator for the $$i$$-th state of the tensor product. The action of the adjunct of the operator $$(ref{3})$$ onto the vacuum state $$(ref{2})$$ is
begin{align} hat{A}^daggerket{varnothing}&= ket{1} otimes ket{0} otimes ket{0} otimes ket{0} &quad+ket{0} otimes ket{1} otimes ket{0} otimes ket{0} &quad+ket{0} otimes ket{0} otimes ket{1} otimes ket{0} &quad+ket{0} otimes ket{0} otimes ket{0} otimes ket{1} tag{4a}label{4a} &=ket{psi} otimes ket{psi} otimes ket{psi} otimes ket{psi} tag{4b}label{4b} end{align}
where
$$ket{psi} = ket{1} + 3ket{0} tag{5}$$
Now, repeating this identical for the operator $$(ref{3})$$ as written results in
begin{align} hat{A}ket{varnothing}&= 0 otimes ket{0} otimes ket{0} otimes ket{0} &quad+ket{0} otimes 0 otimes ket{0} otimes ket{0} &quad+ket{0} otimes ket{0} otimes 0 otimes ket{0} &quad+ket{0} otimes ket{0} otimes ket{0} otimes 0 tag{6a}label{6a} &=ket{psi} otimes ket{psi} otimes ket{psi} otimes ket{psi} tag{6b}label{6b} end{align}
where
$$ket{psi} = 3ket{0} tag{7}$$
which is obviously not correct, since the result should be the zero-vector. Where is my mistake? Is the step from $$(ref{4a})$$ to $$(ref{4b})$$ actually allowed? Clearly, for $$(ref{6a})$$ to $$(ref{6b})$$ this cannot be the case.

Each term in eq. (6a) is multiplied by zero (not the $$vert 0 rangle$$ vector, but the number zero). Therefore the whole is identically zero (the number again).

I also do not agree with result (4b). The fourfold tensor product of the given $$vert psi rangle$$ (5) contains a term $$vert 1,1,1,1 rangle$$ which is clearly absent from (4a).

Correct answer by Nephente on May 9, 2021