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$\def\+{\!\!\kern0.08333em}$ Say I have a spin-1/2 particle in a general, pure superposition state $$ |\psi\rangle=\alpha|\+\uparrow\rangle+\beta|\+\downarrow\rangle, $$ or equivalently $$ \rho=(\alpha|\+\uparrow\rangle+\beta|\+\downarrow\rangle)(\alpha^*\langle\uparrow\+|+\beta^*\langle\downarrow\+|)= \begin{bmatrix} |\alpha|^2 & \alpha\beta^*\\ \alpha^*\beta & |\beta|^2 \end{bmatrix}. $$ When I measure this state in the $\{|\+\uparrow\rangle,|\+\downarrow\rangle\}$ basis, I get $|\+\uparrow\rangle$ with probability $|\alpha|^2$ and $|\+\downarrow\rangle$ with probability $|\beta|^2$.

Now assume that this state was measured, but I do not know the outcome (e.g. my friend measured it but did not tell me the result, and will never do so). In this case, it seems appropriate, from my point of view, to describe the post measurement state as the mixed state $\rho=|\alpha|^2|\+\uparrow\rangle\langle\uparrow\+|+|\beta|^2|\+\downarrow\rangle\langle\downarrow\+|$. After all, I know that the measurement yielded either $|\+\uparrow\rangle\langle\uparrow\+|$ or $|\+\downarrow\rangle\langle\downarrow\+|$ with the given probabilities.

Is this reasoning correct? If no, why not?

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Yes, this reasoning is correct.

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  • $\begingroup$ (In case you don't find this useful, make your question more specific.) $\endgroup$ – Norbert Schuch Oct 23 '18 at 22:34
  • $\begingroup$ One could add that this is called a "nonselective measurement". $\endgroup$ – Noiralef Oct 23 '18 at 22:37
  • $\begingroup$ @Noiralef Never heard that term. $\endgroup$ – Norbert Schuch Oct 23 '18 at 22:37
  • $\begingroup$ I thought it was standard terminology. I just checked and found it used in the book by Breuer and Petruccione as well as in some papers. (Other books I checked: Sakurai only defines "selective measurement" without talking explicitly about the nonselective case, Wiseman and Milburn call it "unconditional measurement".) $\endgroup$ – Noiralef Oct 23 '18 at 22:53
  • $\begingroup$ @Noiralef If you measure & forget I usually wouldn't call this a measurement. It's just a CP map. $\endgroup$ – Norbert Schuch Oct 23 '18 at 22:55

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