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I'm trying to learn the fundamentals of quantum theory, lately mostly using the lectures by Prof. Susskind which Stanford University made available on Youtube.

Here is one thing I have trouble following:

  • On the one hand, we can prepare the state of an electron to have its spin point to the right, $\left|r\right> = (\left|u\right>+\left|d\right>)/\sqrt{2}$. It may or may not emit a photon during this process.
  • On the other hand, if we start with two (non-entangled) electrons, one being in state $\left|u\right>$ and the other in state $\left|d\right>$, their states are orthogonal and therefore will remain orthogonal. Even if I prepare both of them to point to the right.

How does this process keep orthogonality? I assume it has to do with the density matrix that captures the possibly emitted photon?

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    $\begingroup$ I'm not sure what you mean when you say "their states are orthogonal". Certainly, $\lvert r \rangle$ is not orthogonal to $\lvert r \rangle$. However, since the two electrons live in different spaces of states, you can't really compare their states - it's not sensible to say that the state of one electron is orthogonal to the other, because they don't belong to the same space. $\endgroup$ – ACuriousMind Mar 30 '15 at 14:06
  • $\begingroup$ True. I guess I hadn't paid attention to that detail at youtu.be/C1erB9p5414?t=1h17m. Then again, time evolution is linear and invertible, so given that I find an electron prepared to $\left|r\right>$, and assuming I found it emitted no photon, does the entanglement with the measurement/preparation device still contain the information about the previous state of the electron? $\endgroup$ – Christopher Creutzig Mar 30 '15 at 14:53

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