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I am reading the paper "The quantum-to-classical transition and decoherence" by Maximilian Schlosshauer (https://arxiv.org/abs/1404.2635). I doubt the most basic claims in this paper.

The story is as follows: an isolated microscopic system $S$ (Hilbert space $H_S$) is put into a superposition $\phi=\phi_1+\phi_2$ of states $\phi_1,\phi_2\in H_S$ and then brought into contact with an environment $E$ (Hilbert space $H_E$) which initially is in state $e\in H_E$. Before interaction the composite system $(S,E)$ is in state $$ \Phi=\phi\otimes e = \phi_1\otimes e+\phi_2\otimes e\in H_S\otimes H_E. $$ Now an interaction between $S$ and $E$ takes place and it is claimed that this puts the combined system $(S,E)$ into a state of the form $$ \Psi = \phi_1\otimes e_1+\phi_2\otimes e_2\in H_S\otimes H_E $$ (cited paper, equation (1), p2). This is surprising, since we do not know anything about the interaction (no assumptions made).

I believe that such a principle cannot be maintained: we can write $\phi$ in other ways as a superposition $\phi=\psi_1+\psi_2$ and it must then follow that $\Psi$ is also of the form $$ \Psi = \psi_1\otimes f_1 + \psi_2\otimes f_2\in H_S\otimes H_E, $$ for some vectors $f_1,f_2\in H_E$.

In other words the vectors $e_1,e_2\in H_E$ have the following interesting property: $$ \psi_1+\psi_2=\phi_1+\phi_2\ \Rightarrow\ \exists f_1,f_2\in H_E:\thinspace \phi_1\otimes e_1+\phi_2\otimes e_2 = \psi_1\otimes f_1 + \psi_2\otimes f_2. $$ We will now exhibit an example where such vectors $e_1,e_2$ do not exist. Let $H_S=H_E=L^2([0,1])$. Then the tensor product $H_S\otimes H_E$ is isomorphic to the Hilbert space $L^2([0,1]^2)$ with a canonical isomorphism mapping $f(x)\otimes g(y)$ to the function $f(x)g(y)\in L^2([0,1]^2)$.

Now let $\phi_1(x)=1$, $\phi_2(x)=x$, $\psi_1(x)=1-\sqrt x$ and $\psi_2(x)=x+\sqrt x$. Then $\phi_1+\phi_2=\psi_1+\psi_2$ but there are no functions $e_1(y),e_2(y)\in H$ such that the function $\phi_1(x)e_1(y)+\phi_2(x)e_2(y)$ can be written in the form $$ \phi_1(x)e_1(y)+\phi_2(x)e_2(y) = \psi_1(x)f_1(y)+\psi_2(x)f_2(y), $$ for some functions $f_1,f_2\in H$ since this amounts to the equality $$ e_1(y)+xe_2(y) = (1+\sqrt x)f_1(y)+(x-\sqrt x)f_2(y), $$ for almost all $x,y\in[0,1]$. Simply evaluate the equality at enough different values of $x$ to obtain an overdetermined system of equations in $e_i(y),f_j(y)$ with no solution. Clearly this same argument works if we replace $H=L^2([0,1])$ with a finite dimensional Euclidean space of dimension at least 5.

Why is this not a problem?

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  • $\begingroup$ From the paper, $S$ assumed to be two-dimensional. Thus, any state will be of the form $\phi_1\otimes e_1 + \phi_2\otimes e_2$, if $\phi_1$ and $\phi_2$ for a basis for $S$, which means the initial state must evolve into a state of that form. So whatever calculations you've done can't be a problem. $\endgroup$
    – march
    Commented Dec 30, 2023 at 18:23
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    $\begingroup$ @march OK, thanks, I overlooked this, since it is not explicitly stated in the paper. If you make it an official answer it will be accepted. $\endgroup$
    – gcc
    Commented Dec 30, 2023 at 20:14

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In the paper, the states that OP labels $\phi_1$ and $\phi_2$ correspond to the states $\lvert s_1\rangle$ and $\lvert s_2 \rangle$, which correspond to passage through one of two slits in a double-slit experiment. For that reason, the Hilbert space $S$ is two-dimensional, and so any state in the tensor product space of $S$ and $E$ can be written in the form $$ \phi_1\otimes e_1 + \phi_2\otimes e_2\,, $$ where $s_j$ are some arbitray states in $E$. In particular, the state after some time-evolution must be write-able in this way, so there can be no contradiction.

The issue with the OP's "counterexample" is that they are implicitly assuming that $S$ is of dimension greater than 2, because they are using the four states $\phi_1(x)=1$, $\phi_2(x)=x$, $\psi_1(x)=1-\sqrt x$, and $\psi_2(x)=x+\sqrt x$, which spans a space of dimension 3.

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  • $\begingroup$ It should be noted that the paper is very confusing as a position variable x is postulated on the 2 dimensional space for S, see equation (3) and first line of p3 (the inner product (x,s_i) is defined. This would make the Hilbert space H_S infinite dimensional. $\endgroup$
    – gcc
    Commented Dec 31, 2023 at 19:23

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