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Consider a projective measurement of a superposition of states:

$$\frac{1}{\sqrt{2}} |\Psi_1\rangle + \frac{1}{\sqrt{2}} |\Psi_2\rangle \xrightarrow[measurement]{projective} |\Psi_1\rangle, \qquad (|\Psi_1\rangle \neq |\Psi_2\rangle)$$

Assuming an unitary evolution of the "universe = quantum system + measure apparatus" the state evolves as:

$$\left(\frac{1}{\sqrt{2}} |\Psi_1\rangle + \frac{1}{\sqrt{2}} |\Psi_2\rangle\right)\otimes |\chi_0\rangle \xrightarrow[measurement]{unitary} |\Psi_1\rangle \otimes |\chi_1\rangle$$

But:

  • Can we somehow prove that the initial state information is contained in the final state, or is it just an assumption we make, difficult to prove in practice?
  • Would it be possible to construct a very simple measuring device where this would be easier to verify?
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Can we somehow prove that the initial state information is contained in the final state, or is it just an assumption we make, difficult to prove in practice?

Not only can we not prove this, but it is not true. If the final state is indeed $|\Psi_1 \rangle$, as it would be in textbook Quantum Mechanics, then there is no way of inferring the initial state from the final state. After all, any superposition with a non-zero coefficient in front of $|\Psi_1 \rangle$ could have led to the final state $|\Psi_1 \rangle$. Thus there are infinitely many initial states consistent with just 1 final state.

Would it be possible to construct a very simple measuring device where this would be easier to verify?

There is no way to verify that unitary evolution would give you the non-superposition state $|\Psi_1 \rangle$. That is because unitary evolution does not give you that state; it is inconsistent with the linearity of the Schrödinger equation. Here is how we know:

First, we know that measurement of an eigenstate just gives you back that eigenstate. If we assume that measurement happens through unitary evolution, then

$$|\Psi_1 \rangle \otimes |\chi_0 \rangle \xrightarrow[measurement]{unitary} |\Psi_1 \rangle \otimes |\chi_1 \rangle $$

and

$$|\Psi_2 \rangle \otimes |\chi_0 \rangle \xrightarrow[measurement]{unitary} |\Psi_2 \rangle \otimes |\chi_2 \rangle $$

Then, since the unitary evolution in quantum mechanics requires the sum of solutions to be a solution, we can add those solutions together to find out that

$$\alpha |\Psi_1 \rangle \otimes |\chi_0 \rangle + \beta |\Psi_2 \rangle \otimes |\chi_0 \rangle\xrightarrow[measurement]{unitary}\alpha |\Psi_1 \rangle \otimes |\chi_1 \rangle + \beta |\Psi_2 \rangle \otimes |\chi_2\rangle $$

Taking the example of $\alpha = \beta = \frac{1}{\sqrt{2}}$, this reads $$ \left(\frac{1}{\sqrt{2}}|\Psi_1 \rangle + \frac{1}{\sqrt{2}}|\Psi_2 \rangle \right)\otimes |\chi_0 \rangle\xrightarrow[measurement]{unitary}\frac{1}{\sqrt{2}} |\Psi_1 \rangle \otimes |\chi_1 \rangle + \frac{1}{\sqrt{2}}|\Psi_2 \rangle \otimes |\chi_2 \rangle $$

...which is in direct contradiction with the statement you wanted to prove, which I'll copy below:

$$\left(\frac{1}{\sqrt{2}} |\Psi_1\rangle + \frac{1}{\sqrt{2}} |\Psi_2\rangle\right)\otimes |\chi_0\rangle \xrightarrow[measurement]{unitary} |\Psi_1\rangle \otimes |\chi_1\rangle$$

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  • $\begingroup$ But what if the state $|\bar{\chi}_0\rangle$ in the evolution $|\Psi_1\rangle\otimes|\bar{\chi}_0\rangle \to |\Psi_1\rangle\otimes|\chi_1\rangle$, and the state $|\chi_0\rangle$ in the evolution $(\alpha|\Psi_1\rangle+\beta\Psi_2)\otimes|\chi_0\rangle \to |\Psi_1\rangle\otimes|\chi_1\rangle$ are not equal? I am not sure why you are not assuming that $|\bar{\chi}_0\rangle \neq |\chi_0\rangle$? $\endgroup$
    – Davius
    Commented Aug 25, 2022 at 15:57
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    $\begingroup$ Because starting from the initial pre-measurement state of the detector $|\chi_0 \rangle$, we could choose to measure either a particle in superposition or not. The state of the detector can be chosen independently of the state of the particle since before the measurement happens, they are just completely separate systems. $\endgroup$
    – don't train ai on me
    Commented Aug 25, 2022 at 16:01
  • $\begingroup$ I am not sure, assume that initially we have $|\Psi_1\rangle \otimes |\chi_0\rangle$, maybe, then, the unitary evolution leads to $|\Psi_1\rangle \otimes |\chi_0\rangle \to |\Psi_1\rangle \otimes |\bar{\chi}_1\rangle$ with $|\bar{\chi}_1\rangle \neq |\chi_1\rangle$. The problem is that we do not know the details of an hypothetical unitary evolution (under the assumption of an interpretation without "objective collapse" of wave function, which underlie "quantum information paradox": en.wikipedia.org/w/…). $\endgroup$
    – Davius
    Commented Aug 25, 2022 at 21:38
  • $\begingroup$ How do you define $|\chi_1 \rangle$ vs $|\bar{\chi}_1\rangle$? $\endgroup$
    – don't train ai on me
    Commented Aug 25, 2022 at 22:00
  • $\begingroup$ Also, looking back at the definition of $|\chi_0 \rangle$ vs $\bar{\chi}_0 \rangle$, we only need to say that $|\chi_0 \rangle$ is one possible initial state of the detector. Then, we could consider a system measuring states other than $|\Psi_1 \rangle$, which start with the detector in the same state $|\chi_0 \rangle$. Physically, we have detectors which are capable of detecting many different results... e.g. the same stern-gerlach apparatus can detect either spin up or spin down. So, by choosing one reference starting state for the detector, the other ones are just defined to be the same. $\endgroup$
    – don't train ai on me
    Commented Aug 25, 2022 at 22:04

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