# Wavefunction collapse in Stern-Gerlach experiment

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?

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

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.
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.