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:

$\chi=c_1\alpha+c_2\beta$

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 90^o, 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 $\vert \beta\rangle_{\hat{\boldsymbol{\mathit{r}}}}$ .

Question

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

Answer

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,

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

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

So,

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

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

$c_2=c_1e^{i\phi}\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}$

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

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

Question

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

Answer

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: