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

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