Sequential Stern-Gerlach experiments

A sequential Stern-Gerlach experiment involves passing a spin-\frac{1}{2} particle through multiple inhomogeneous magnetic fields, each of a certain orientation. In the first example, a beam of spin-\frac{1}{2} particles, e.g. silver atoms, undergoes a first measurement as it passes through an inhomogeneous magnetic field, which is parallel to the z-axis (see diagram below).

The beam is split into two, and the S_z=\frac{\hbar}{2} beam is allowed to travel through another inhomogeneous magnetic field, which is also z-directional (the S_z=-\frac{\hbar}{2} beam is blocked). The general state \chi of the valence silver electron prior to passing through the first magnetic field is given by eq171:


where \alpha and \beta are basis vectors representing the electron spin eigenstates of \vert s=1/2,m_s=1/2\rangle and  \vert s=1/2,m_s=-1/2\rangle respectively; \vert c_1\vert^{2} is the probability of finding the electron with the state \vert s=1/2,m_s=1/2\rangle, and \vert c_2\vert^{2} is the probability of finding the electron with the state \vert s=1/2,m_s=-1/2\rangle, with \vert c_1\vert^{2}+\vert c_2\vert^{2}=1.

Since the direction of the first magnetic field is arbitrary assigned and the beam of silver atoms emerging from the source is not polarised, we have equal probability of silver atoms with valence electrons in each eigenstate emerging, i.e. \vert c_1\vert^{2}=\vert c_2\vert^{2}=0.5. Noting the orthonormality of basis vectors, the expectation value of S_z with reference to eq168 is:

\langle S_z\rangle=\langle\chi\vert\hat{S}_z\vert\chi\rangle=\langle\sqrt{0.5}\alpha+\sqrt{0.5}\beta\vert\sqrt{0.5} \frac{\hbar}{2}\alpha+\sqrt{0.5} \left ( -\frac{\hbar}{2}\right )\beta\rangle=0

The state \chi of the valence silver electron entering the second magnetic field is \chi=c_1\alpha+c_2\beta, where \vert c_1\vert^{2}=1 and \vert c_2\vert^{2}=0. Therefore, the expectation value of S_z is \langle S_z\rangle=\frac{\hbar}{2}.

For the second example, the S_z=\frac{\hbar}{2} beam emerging from the first magnetic field is passed through a second magnetic field that is rotated 90o, i.e. in the x-direction (see diagram below).

To determine the result of the second measurement, we must express the state of the valence electron in the S_z=\frac{\hbar}{2} beam in terms of \vert\alpha_x\rangle and \vert\beta_x\rangle, or in general where the second magnetic field is rotated parallel to an arbitrary direction \hat{\boldsymbol{\mathit{r}}} (\hat{\boldsymbol{\mathit{r}}} is the unit vector in spherical coordinates, i.e. \hat{\boldsymbol{\mathit{r}}}=sin\theta cos\phi\boldsymbol{\mathit{i}}+sin\theta sin\phi\boldsymbol{\mathit{j}}+cos\theta \boldsymbol{\mathit{k}}), in terms of \vert \alpha\rangle_{\hat{\boldsymbol{\mathit{r}}}} and \vert \beta\rangle_{\hat{\boldsymbol{\mathit{r}}}}.



How do we derive the unit vector \hat{\boldsymbol{\mathit{r}}} in spherical coordinates?


Substitute eq77 in the definition of a unit vector,

\hat{\boldsymbol{\mathit{r}}}=\frac{\boldsymbol{\mathit{r}}}{\left |\boldsymbol{\mathit{r}}\right |}=\frac{x\boldsymbol{\mathit{i}}+y\boldsymbol{\mathit{j}}+z\boldsymbol{\mathit{k}}}{r}=sin\theta cos\phi\boldsymbol{\mathit{i}}+sin\theta sin\phi\boldsymbol{\mathit{j}}+cos\theta\boldsymbol{\mathit{k}}\; \; \; \; \; \; \; \; 182b


In other words, we need eq171 to be in the form:

\chi_\alpha=c_1\alpha_{\hat{\boldsymbol{\mathit{r}}}}+c_2\beta_{\hat{\boldsymbol{\mathit{r}}}}\; \; \; \; \; \; \; \; 183

This implies that we have to construct the operator \hat{S}_{\hat{\boldsymbol{\mathit{r}}}}, which acts on the component of spin angular momentum along \hat{\boldsymbol{\mathit{r}}}. To do so, we take the projection of \boldsymbol{\mathit{S}} onto \hat{\boldsymbol{\mathit{r}}}:

\boldsymbol{\mathit{S}}\cdot\hat{\boldsymbol{\mathit{r}}}=(\boldsymbol{\mathit{i}}S_x+\boldsymbol{\mathit{j}}S_y+\boldsymbol{\mathit{k}}S_z)\cdot(sin\theta cos\phi\boldsymbol{\mathit{i}}+\sin\theta sin\phi\boldsymbol{\mathit{j}}+cos\theta\boldsymbol{\mathit{k}})

\boldsymbol{\mathit{S}}\cdot\hat{\boldsymbol{\mathit{r}}}=S_xsin\theta cos\phi+S_y\sin\theta sin\phi+S_zcos\theta

Hence, the operator is

\hat{S}_{\hat{\boldsymbol{\mathit{r}}}}=\hat{S}_xsin\theta cos\phi+\hat{S}_y\sin\theta sin\phi+\hat{S}_zcos\theta

Substituting eq174, eq177 and eq178 in the above equation to further construct \hat{S}_{\hat{\boldsymbol{\mathit{r}}}} in matrix form,

\hat{S}_{\hat{\boldsymbol{\mathit{r}}}}=\frac{\hbar}{2}\begin{pmatrix} cos\theta &e^{-i\phi}sin\theta \\ e^{i\phi}sin\theta &-cos\theta \end{pmatrix}\; \; \; \; \; \; \; \; 184

The eigenvalue equation is:

\frac{\hbar}{2}\begin{pmatrix} cos\theta &e^{-i\phi}sin\theta \\ e^{i\phi}sin\theta &-cos\theta \end{pmatrix}\chi_\alpha=\begin{pmatrix} \lambda &0 \\ 0 &\lambda \end{pmatrix}\chi_\alpha\; \; \; \; \; \; \; \; 185

where \lambda is the eigenvalue; and the corresponding characteristic equation is:

\frac{\hbar}{2}\begin{vmatrix} cos\theta-\lambda &e^{-i\phi}sin\theta \\ e^{i\phi}sin\theta &-cos\theta-\lambda \end{vmatrix}=0

\lambda=\pm\frac{\hbar}{2}\; \; \; \; \; \; \; \; 186

Substitute eq183 and eq186 in eq185,

\frac{\hbar}{2}\begin{pmatrix} cos\theta &e^{-i\phi}sin\theta \\ e^{i\phi}sin\theta &-cos\theta \end{pmatrix}\begin{pmatrix} c_1\\c_2 \end{pmatrix}=\pm\frac{\hbar}{2}\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}\begin{pmatrix} c_1\\c_2 \end{pmatrix}

\begin{pmatrix} c_1cos\theta +c_2e^{-i\phi}sin\theta \\ c_1e^{i\phi}sin\theta -c_2cos\theta \end{pmatrix}=\pm\begin{pmatrix} c_1\\c_2 \end{pmatrix}\; \; \; \; \; \; \; \; 187


c_1cos\theta +c_2e^{-i\phi}sin\theta =\pm c_1\; \; \; \; \; \Rightarrow \; \; \; \; \; c_2=c_1\frac{\pm1-cos\theta}{e^{-i\phi}sin\theta}

Since 1-cos\theta=2sin^{2}\frac{\theta}{2} and sin\theta=2sin\frac{\theta}{2}cos\frac{\theta}{2},  we have


Substituting the above equation in \vert c_1\vert^{2}+\vert c_2\vert^{2}=1 and using \vert e^{i\phi}\vert ^{2}=1

\vert c_1\vert^{2}=cos^{2}\frac{\theta}{2}\; \; \; \; \; \; \; \; 188



Show that \vert e^{i\phi}\vert^{2}=1.


If a,b\in \mathbb{R}, we have \vert a+ib\vert=\sqrt{a^{2}+b^{2}}.

\vert e^{i\phi}\vert^{2}=\vert cos\phi+isin\phi\vert^{2}=1


Substitute eq188 in \vert c_1\vert^{2} +\vert c_2\vert^{2}=1

\vert c_2\vert^{2}=sin^{2}\frac{\theta}{2}\; \; \; \; \; \; \; \; 189

Therefore, eq183 becomes:

\chi_\alpha=cos\frac{\theta}{2}\alpha_{\hat{\boldsymbol{\mathit{r}}}}+sin\frac{\theta}{2}\beta_{\hat{\boldsymbol{\mathit{r}}}}\; \; \; \; \; \; \; \; 190

Note that we could have equally used c_1e^{i\phi}sin\theta-c_2cos\theta=\pm c_2 in eq187 to arrive at eq188 and eq189. Using eq188, the probability of measuring the \alpha_{\hat{\boldsymbol{\mathit{r}}}=90^{\circ} } state is

\vert c_1\vert^{2}=cos^{2}\frac{90^{\circ}}{2}=0.5\; \; \; \; \; \; \; \; 191

Using eq189, the probability of measuring the \beta_{\hat{\boldsymbol{\mathit{r}}}=90^{\circ} } state is

\vert c_2\vert^{2}=sin^{2}\frac{90^{\circ}}{2}=0.5\; \; \; \; \; \; \; \; 192

Therefore, we observe that the S_z=\frac{\hbar}{2} beam is split equally into two beams by the second magnetic field. Using the same logic, the S_x=\frac{\hbar}{2} beam will again split into two beams by the third magnetic field:

\vert c_1\vert^{2}=cos^{2}\frac{270^{\circ}}{2}=0.5\; \; \; \; \; \; and\; \; \; \;\; \; \vert c_2\vert^{2}=sin^{2}\frac{270^{\circ}}{2}=0.5

In general, the probability of observing the \alpha_{\hat{\boldsymbol{\mathit{r}}}} state is depicted in the graph below.



Do eq188 and eq189 apply to the output of the  beam passing through the second magnetic field in the first example?


Yes. In the first example, \vert c_1\vert^{2}=cos^{2}\frac{0^{\circ}}{2}=1 and \vert c_2\vert^{2}=sin^{2}\frac{0^{\circ}}{2}=0.



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