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Consider silver atoms coming from an S(X) apparatus, after S(X) apparatus we place an S(Z) apparatus

$$ |\mathrm{SX+}\rangle= \frac{1}{2}|\mathrm{SZ+}\rangle + \frac{1}{2}|\mathrm{SZ-}\rangle, $$

Action of $\hat{S}_z$ operator collapses the wave state to $|\mathrm{SZ+}\rangle$ or $|\mathrm{SZ-}\rangle$, but action of operator on $|\mathrm{SX+}\rangle$ can be mathematically represented as

\begin{align} \hat{S}_z|\mathrm{SX+}\rangle & = \hat{S}_z\frac{1}{2}|\mathrm{SZ+}\rangle +\hat{S}_z\frac{1}{2}|\mathrm{SZ-}\rangle, \\ \Rightarrow \qquad \hat{S}_z|\mathrm{SX+}\rangle & = \frac{h}{4\pi}\frac{1}{2}|\mathrm{SZ+}\rangle -\frac{h}{4\pi}\frac{1}{2}|\mathrm{SZ-}\rangle, \end{align} which can be normalised as $|\mathrm{SX-}\rangle$ state.

So my doubt is action of operator on a wave function collapses wave function to any of the eigen states or transforming the wavefunction to $|\mathrm{SX-}\rangle$ state?

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    $\begingroup$ see here math.meta.stackexchange.com/questions/5020/… $\endgroup$ Apr 21, 2019 at 17:39
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    $\begingroup$ measurements are not unitary or deterministic operations, and hence they objectively make information about the previous quantum state irretrievable, even in principle $\endgroup$
    – lurscher
    Apr 21, 2019 at 17:46
  • $\begingroup$ I've done a starting round of formatting edits to get your post marginally readable, as far as allowed by your initial notation. $\endgroup$ Apr 21, 2019 at 17:47
  • $\begingroup$ thanks for the help Emilo Pisanty $\endgroup$
    – user212422
    Apr 21, 2019 at 17:49

2 Answers 2

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Your core misconception is here:

Action of $\hat{S}_z$ operator collapses the wave state to $|\mathrm{SZ+}\rangle$ or $|\mathrm{SZ-}\rangle$

This is incorrect. Measurement of the observable $S_z$ will collapse the wavefunction to one of its eigenstates, but that measurement is not modelled by acting with the operator $\hat{S}_z$ on the wavefunction. Instead, if you want to model the collapse, you use one of the two hermitian projection operators \begin{align} \hat{\Pi}_{z,+} & = |\mathrm{SZ+}\rangle \langle \mathrm{SZ+}|, \\ \hat{\Pi}_{z,-} & = |\mathrm{SZ-}\rangle \langle \mathrm{SZ-}| \end{align} (in the sub-optimal notation used in v2 of your answer as the cleanest rendition I could do of your initial text; for clarity, here $\hat{S}_z |\mathrm{SZ\pm}\rangle = \pm \frac12 \hbar |\mathrm{SZ\pm}\rangle$.) If the measurement outcome is $S_z = +\frac12\hbar$, the projective measurement is the replacement $|\psi \rangle \mapsto \hat{\Pi}_{z,+} |\psi⟩$ (with a suitable normalization constant), and similarly for $S_z = -\frac12\hbar$.

Your observation that $$ \hat{S}_z |\mathrm{SX+}\rangle = \frac12 \hbar |\mathrm{SX-}\rangle $$ is correct but it is in no way contradictory or in conflict with any other parts of the formalism.

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As indicated elsewhere, not all operators transform a state into a linear combinations of eigenstates. Instead of $\sigma_z$ consider the operator $\hat \Pi_{+,z}=\vert {+}\rangle_z {_z\langle} {+}\vert$. This will transform the $\vert {+}\rangle_x$ eigenstate to $$ \vert {+}\rangle_z {_z\langle} {+}\vert {+}\rangle_x=\frac{1}{\sqrt{2}}\vert {+}\rangle_z\, , \tag{1} $$ which is a single state and not a linear combinations of eigenstates of $\sigma_z$. Indeed $(1)$ shows how the state collapses to a particular eigenket.

Note that $$ \Pi_{+,z}=\left(\begin{array}{cc} 1&0\\ 0&0\end{array}\right) $$ is still a perfectly valid operator.

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