a- Using the axioms for inner products, prove (\langle \mathbf A + \langle \mathbf B) | \mathbf C \rangle = \langle \mathbf A | \mathbf C \rangle + \langle \mathbf B | \mathbf C \rangle b- Prove \langle \mathbf A | \mathbf A \rangle is a real number.
Using the axioms for the inner product (that it is linear and that interchanging bra and kets correspond to complex conjugation:
(\langle \mathbf A +\langle \mathbf B) | \mathbf C \rangle = \overline{ \langle \mathbf C | ( \mathbf A \rangle + \mathbf B \rangle )} = \overline{\langle \mathbf C | \mathbf A \rangle} + \overline{\langle \mathbf C | \mathbf B \rangle} = \langle \mathbf A | \mathbf C \rangle + \langle \mathbf B | \mathbf C \rangle
Since interchanging bra and kets correspond to complex conjugation:
\langle \mathbf A | \mathbf A \rangle = \begin{bmatrix} \bar \alpha_1 & \bar \alpha_2 & \cdots & \bar \alpha_n \end{bmatrix} \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \cdots \\ \alpha_n \end{bmatrix} = \bar \alpha_1 \alpha_1 + \bar \alpha_2 \alpha_2 + \cdots + \bar \alpha_n \alpha_n
This is a real number because it is the sum of the product of the component time their complex conjugate, and for a complex number z = a + ib:
\bar z \, z = (x-iy)(x+iy) = x^2 + ixy - iyx -i^2y^2 = x^2 + y^2
which is a real number.
Show that the inner product defined by Eq. 1.2 satisfies all the axioms of inner products.
\langle \mathbf C | ( z\,\mathbf A \rangle + w\,\mathbf B \rangle) = z\,\langle \mathbf C | \mathbf A \rangle + w\, \langle \mathbf C | \mathbf B \rangle
Proof:
\begin{aligned} \langle \mathbf C | ( z\,\mathbf A \rangle + w\, \mathbf B \rangle) & = \begin{bmatrix} \bar \gamma_1 & \bar \gamma_2 & \cdots & \bar \gamma_n \end{bmatrix} \left( z \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \cdots \\ \alpha_n \end{bmatrix} + w\begin{bmatrix} \beta_1 \\ \beta_2 \\ \cdots \\ \beta_n \end{bmatrix} \right) = \begin{bmatrix} \bar \gamma_1 & \bar \gamma_2 & \cdots & \bar \gamma_n \end{bmatrix} \begin{bmatrix} z\, \alpha_1 + \beta_1 \\ z\, \alpha_2 + \beta_2 \\ \cdots \\ z\, \alpha_n + \beta_n \end{bmatrix} \\ & = \bar \gamma_1 \left(z\, \alpha_1 + w\, \beta_1 \right) + \bar \gamma_2 \left(z\, \alpha_2 + w\, \beta_2 \right) + \cdots + \bar \gamma_n \left(z\, \alpha_n + w \, \beta_n \right) \\ & = z \, \bar \gamma_1 \alpha_1 + w \, \bar \gamma_1 \beta_1 + z\, \bar \gamma_2 \alpha_2 + w \, \bar \gamma_2 \beta_2 + \cdots + z\, \bar \gamma_n \alpha_n + w \, \bar \gamma_n z\, \beta_n \\ z \langle \mathbf C | \mathbf A \rangle + w \langle \mathbf C | \mathbf B \rangle & = z\begin{bmatrix} \bar \gamma_1 & \bar \gamma_2 & \cdots & \bar \gamma_n \end{bmatrix} \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \cdots \\ \alpha_n \end{bmatrix} + w \begin{bmatrix} \bar \gamma_1 & \bar \gamma_2 & \cdots & \bar \gamma_n \end{bmatrix} \begin{bmatrix} \beta_1 \\ \beta_2 \\ \cdots \\ \beta_n \end{bmatrix} \\ & = z \, \bar \gamma_1 \alpha_1 + w \, \bar \gamma_1 \beta_1 + z\, \bar \gamma_2 \alpha_2 + w \, \bar \gamma_2 \beta_2 + \cdots + z\, \bar \gamma_n \alpha_n + w \, \bar \gamma_n z\, \beta_n \end{aligned}
\langle \mathbf B | \mathbf A \rangle = \overline{\langle \mathbf A | \mathbf B \rangle}
Proof:
\begin{aligned} \overline{\langle \mathbf A | \mathbf B \rangle} & = \overline{\begin{bmatrix} \bar \alpha_1 & \bar \alpha_2 & \cdots & \bar \alpha_n \end{bmatrix} \begin{bmatrix} \beta_1 \\ \beta_2 \\ \cdots \\ \beta_n \end{bmatrix}} = \overline{\bar \alpha_1 \beta_1 + \bar \alpha_2 \beta_1 + \cdots + \bar \alpha_n \beta_n} \\ & = \alpha_1 \bar \beta_1 + \alpha_2 \bar \beta_2 + \cdots + \alpha_n \bar \beta_n = \langle \mathbf B | \mathbf A \rangle \end{aligned}
Prove that the vector |\mathbf{r}\rangle in Eq. 2.5 is orthogonal to vector |\mathbf{l}\rangle in Eq. 2.6.
\begin{aligned} \langle r | l \rangle & = \left(\frac{1}{\sqrt 2} \langle u | + \frac{1}{\sqrt 2} \langle d | \right)\left( \frac{1}{\sqrt 2}| u\rangle - \frac{1}{\sqrt 2}| d\rangle\right) \\ & = \frac{1}{2} \langle u | u \rangle - \frac{1}{2} \langle u | d \rangle + \frac{1}{2} \langle d | u \rangle - \frac{1}{2} \langle d | d \rangle = \frac{1}{2} - \frac{1}{2} = 0 \end{aligned}
Prove that |\mathbf{i} and |\mathbf{o}\rangle satisfy all of the conditions in Eqs. 2.7, 2.8 and 2.9. Are they unique in that respect?
For 2.7:
\begin{aligned} \langle i | o \rangle & = \left(\frac{1}{\sqrt 2} \langle u | - \frac{i}{\sqrt 2} \langle d \right)\left( \frac{1}{\sqrt 2}| u\rangle - \frac{i}{\sqrt 2}| d\rangle\right) \\ & = \frac{1}{2} \langle u | u \rangle - \frac{1}{2} \langle u | d \rangle - \frac{1}{2} \langle d | u \rangle + \frac{i^2}{2} \langle d | d \rangle = \frac{1}{2} - \frac{1}{2} = 0 \end{aligned}
For 2.8 it will be done for each term.
Considering | o \rangle = \frac{1}{\sqrt{2}}| u \rangle + \frac{i}{\sqrt{2}}| d \rangle and its dual \langle o | = \frac{1}{\sqrt{2}}\langle u | - \frac{i}{\sqrt{2}}\langle d |, for \langle o | u \rangle \langle u | o \rangle:
\langle o | u \rangle \langle d | u \rangle = \left( \frac{1}{\sqrt{2}}\langle u | - \frac{i}{\sqrt{2}}\langle d | \right) | u \rangle \cdot \langle d | \left( \frac{1}{\sqrt{2}}| u \rangle + \frac{i}{\sqrt{2}}| u \rangle \right) = \frac{1}{\sqrt 2} \frac{1}{\sqrt 2} = \frac{1}{2}
For \langle o | d \rangle \langle d | o \rangle:
\langle o | d \rangle \langle d | o \rangle = \left( \frac{1}{\sqrt{2}}\langle u | - \frac{i}{\sqrt{2}}\langle d | \right) | d \rangle \cdot \langle d | \left( \frac{1}{\sqrt{2}}| u \rangle + \frac{i}{\sqrt{2}}| d \rangle \right) = \frac{-i}{\sqrt 2} \frac{i}{\sqrt 2} = \frac{1}{2}
Considering | i \rangle = \frac{1}{\sqrt{2}}| u \rangle - \frac{i}{\sqrt{2}}| d \rangle and its dual \langle i | = \frac{1}{\sqrt{2}}\langle u | + \frac{i}{\sqrt{2}}\langle d |, for \langle i | u \rangle \langle u | i \rangle:
\langle i | u \rangle \langle u | i \rangle = \left( \frac{1}{\sqrt{2}}\langle u | + \frac{i}{\sqrt{2}}\langle d | \right) | u \rangle \cdot \langle u | \left( \frac{1}{\sqrt{2}}| u \rangle - \frac{i}{\sqrt{2}}| d \rangle \right) = \frac{1}{\sqrt 2} \frac{1}{\sqrt 2} = \frac{1}{2}
For \langle i | d \rangle \langle d | i \rangle:
\langle i | d \rangle \langle d | i \rangle = \left( \frac{1}{\sqrt{2}}\langle u | + \frac{i}{\sqrt{2}}\langle d | \right) | d \rangle \cdot \langle d | \left( \frac{1}{\sqrt{2}}| u \rangle - \frac{i}{\sqrt{2}}| d \rangle \right) = \frac{i}{\sqrt 2} \frac{-i}{\sqrt 2} = \frac{1}{2}
For 2.9:
Computing \langle o | r \rangle:
\langle o | r \rangle = \left( \frac{1}{\sqrt{2}}\langle u | + \frac{i}{\sqrt{2}}\langle d | \right)\left( \frac{1}{\sqrt{2}}| u \rangle + \frac{1}{\sqrt{2}}| d \rangle \right) = \frac{1}{2}\langle u | u \rangle + \frac{i}{2}\langle d | u \rangle + \frac{1}{2}\langle u | d \rangle + \frac{i}{2}\langle d | d \rangle = \frac{1}{2} + \frac{i}{2}
Next, compute \langle r | o \rangle, which is the complex conjugate of \langle o | r \rangle, because \langle r | o \rangle = \overline{\langle o | r \rangle}:
\langle r | o \rangle = \frac{1}{2} - \frac{i}{2}
Now, multiply \langle o | r \rangle and \langle r | o \rangle:
\langle o | r \rangle \langle r | o \rangle = \left( \frac{1}{2} + \frac{i}{2} \right)\left( \frac{1}{2} - \frac{i}{2} \right) = \frac{1}{4} - \frac{i^2}{4} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
Computing \langle o | l \rangle:
\langle o | l \rangle = \left( \frac{1}{\sqrt{2}}\langle u | + \frac{i}{\sqrt{2}}\langle d | \right)\left( \frac{1}{\sqrt{2}}| u \rangle - \frac{1}{\sqrt{2}}| d \rangle \right) = \frac{1}{2}\langle u | u \rangle - \frac{1}{2}\langle u | d \rangle + \frac{i}{2}\langle d | u \rangle - \frac{i}{2}\langle d | d \rangle = \langle o | l \rangle = \frac{1}{2} - \frac{i}{2}
Then, for \langle l | o \rangle, which is the complex conjugate of \langle o | l \rangle, we have:
\langle l | o \rangle = \frac{1}{2} + \frac{i}{2}
Multiplying \langle o | l \rangle and \langle l | o \rangle together:
\langle o | l \rangle \langle l | o \rangle = \left( \frac{1}{2} - \frac{i}{2} \right)\left( \frac{1}{2} + \frac{i}{2} \right) = \frac{1}{4} + \frac{i}{4} - \frac{i}{4} - \frac{1}{4}i^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
Computing \langle i | r \rangle:
\langle i | r \rangle = \left( \frac{1}{\sqrt{2}}\langle u | - \frac{i}{\sqrt{2}}\langle d | \right) \left( \frac{1}{\sqrt{2}}| u \rangle + \frac{1}{\sqrt{2}}| d \rangle \right) = \frac{1}{2}\langle u | u \rangle + \frac{1}{2}\langle u | d \rangle - \frac{i}{2}\langle d | u \rangle - \frac{i}{2}\langle d | d \rangle = \frac{1}{2} - \frac{i}{2}
Then, for \langle r | i \rangle, the complex conjugate of \langle i | r \rangle, we have:
\langle r | i \rangle = \frac{1}{2} + \frac{i}{2}
Multiplying \langle i | r \rangle and \langle r | i \rangle together:
\langle i | r \rangle \langle r | i \rangle = \left( \frac{1}{2} - \frac{i}{2} \right)\left( \frac{1}{2} + \frac{i}{2} \right) = \frac{1}{4} + \frac{i}{4} - \frac{i}{4} - \frac{1}{4}i^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
Computing \langle i | l \rangle:
\langle i | l \rangle = \left( \frac{1}{\sqrt{2}}\langle u | - \frac{i}{\sqrt{2}}\langle d | \right) \left( \frac{1}{\sqrt{2}}| u \rangle - \frac{1}{\sqrt{2}}| d \rangle \right)= \frac{1}{2}\langle u | u \rangle - \frac{1}{2}\langle u | d \rangle - \frac{i}{2}\langle d | u \rangle + \frac{i}{2}\langle d | d \rangle = \frac{1}{2} + \frac{i}{2}
Then, for \langle l | i \rangle, the complex conjugate of \langle i | l \rangle, we have:
\langle l | i \rangle = \frac{1}{2} - \frac{i}{2}
Multiplying \langle i | l \rangle and \langle l | i \rangle together:
\langle i | l \rangle \langle l | i \rangle = \left( \frac{1}{2} + \frac{i}{2} \right)\left( \frac{1}{2} - \frac{i}{2} \right) = \frac{1}{4} - \frac{i}{4} + \frac{i}{4} - \frac{1}{4}i^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
These vector are not unique since adding a phase will not change the product since complex conjugating a phase gives e^{-i\theta}e^{\i\theta}=1.
It would be possible to demonstrate (using a similar approach as Exercise 2.3) that these are the only 3 possible basis that can be constructed.
Starting from (for example) the representation on the |u \rangle, |d\rangle basis:
\begin{aligned} & | u \rangle = 1 \, | u \rangle + 0 \, | d \rangle \\ & | d \rangle = 0 \, | u \rangle + 1 \, | d \rangle \\ & | r \rangle = \dfrac{1}{\sqrt 2}| u \rangle + \dfrac{1}{\sqrt 2}| d \rangle \\ & | l \rangle = \dfrac{1}{\sqrt 2}| u \rangle - \dfrac{1}{\sqrt 2}| d \rangle \\ & | i \rangle = \dfrac{1}{\sqrt 2}| u \rangle + \dfrac{i}{\sqrt 2}| d \rangle \\ & | o \rangle = \dfrac{1}{\sqrt 2}| u \rangle - \dfrac{i}{\sqrt 2}| d \rangle \\ \end{aligned}
Trying to find two new vectors |s\rangle and |t\rangle:
\begin{array}{cc} | s \rangle = \alpha \, | u\rangle + \beta \, | d\rangle & | t \rangle = \gamma \, | u \rangle + \delta \, | d\rangle \end{array} and then computing the probabilities, then it would be possible to show that this lead to a mathematical inconsistency and therefore there are no other vectors that can satisfy the requested relationships.
For the moment, forget that Eqs. 2.10 give us working definitions for |\mathbf{i} and |\mathbf{o} in terms of |\mathbf{u}\rangle and |\mathbf{d}\rangle, and assume that the components \alpha, \beta, \gamma and \delta are unknown:
\begin{aligned} &|\mathbf{i}\rangle = \alpha|\mathbf{u}\rangle + \beta|\mathbf{d}\rangle \\ & |\mathbf{o}\rangle = \gamma|\mathbf{u}\rangle + \delta|\mathbf{d}\rangle \end{aligned}
\alpha^*\alpha = \beta^*\beta = \gamma^*\gamma = \delta^*\delta = \frac1{2}
\alpha^*\beta + \alpha\beta^* = \gamma^*\delta + \gamma\delta^* = 0
If \alpha^*\beta is pure imaginary, then \alpha and \beta cannot both be real. The same reasoning applies to \gamma^*\delta.
Assuming that the spin are defined in the | u \rangle, | d \rangle, | r \rangle and | l \rangle as:
\begin{array}{cc} | u \rangle = 1 \, | u \rangle + 0 \, | d \rangle & d \rangle = 0 \, | u \rangle + 1 \, | d \rangle \\[12pt] | r \rangle = \dfrac{1}{\sqrt 2}| u\rangle + \dfrac{1}{\sqrt 2}| d\rangle & | l \rangle = \dfrac{1}{\sqrt 2}| u\rangle - \dfrac{1}{\sqrt 2}| d\rangle \end{array}
For | i \rangle and | o \rangle assuming generic coefficients \gamma, \beta, \gamma, and \delta:
\begin{array}{cc} | i \rangle = \alpha \, | u\rangle + \beta \, | d\rangle & | o \rangle = \gamma \, | u\rangle + \delta \, | d\rangle \end{array}
If the system has been prepared along the y axis, then taking a measurement along the z axis give 50% chances to be up and 50% down or along the x axis 50% to be right and 50% to be left.
\begin{array}{ll} P_{i\,u} = \langle i | u \rangle \langle u | i \rangle = \frac{1}{2} & P_{i\,d} = \langle i | d \rangle \langle d | i \rangle = \frac{1}{2} \\ P_{i\,r} = \langle i | r \rangle \langle r | i \rangle = \frac{1}{2} & P_{i\,l} = \langle i | l \rangle \langle l | i \rangle = \frac{1}{2} \\ P_{o\,u} = \langle o | u \rangle \langle u | o \rangle = \frac{1}{2} & P_{o\,d} = \langle o | d \rangle \langle d | o \rangle = \frac{1}{2} \\ P_{o\,r} = \langle o | r \rangle \langle r | o \rangle = \frac{1}{2} & P_{o\,l} = \langle o | l \rangle \langle l | o \rangle = \frac{1}{2} \end{array}
Starting from the probabilities along | u \rangle, | d \rangle:
\begin{aligned} \langle i | u \rangle \langle u | i \rangle & = \left[ (\bar \alpha \, \langle u | + \bar \beta \,\langle d |)(1 \, | u \rangle + 0 \, | d \rangle)\right] \left[(1 \, \langle u | + 0 \, \langle d | ) (\alpha \, | u\rangle + \beta \, | d \rangle) \right]\\ & = (\bar \alpha \langle u | u \rangle + \bar \beta \langle d | u \rangle) (\alpha \langle u | u \rangle + \beta \langle u | d \rangle) = \bar \alpha \, \alpha = \tfrac{1}{2} \\ \langle i | d \rangle \langle d | i \rangle & = \left[ (\bar \alpha \, \langle u | + \bar \beta \,\langle d |)(0 \, | u \rangle + 1 \, | d \rangle)\right] \left[(0 \, \langle u | + 1 \, \langle d | ) (\alpha \, | u\rangle + \beta \, | d \rangle) \right]\\ & = (\bar \alpha \langle u | d \rangle + \bar \beta \langle d | d \rangle) (\alpha \langle d | u \rangle + \beta \langle d | d \rangle) = \bar \beta \, \beta = \tfrac{1}{2} \\ \langle o | u \rangle \langle u | o \rangle & = \left[ (\bar \gamma \, \langle u | + \bar \delta \,\langle d |)(1 \, | u \rangle + 0 \, | d \rangle)\right] \left[(1 \, \langle u | + 0 \, \langle d | ) (\gamma \, | u\rangle + \delta \, | d \rangle) \right]\\ & = (\bar \gamma \langle u | u \rangle + \bar \delta \langle d | u \rangle) (\gamma \langle u | u \rangle + \delta \langle u | d \rangle) = \bar \gamma \, \gamma = \tfrac{1}{2} \\ \langle o | d \rangle \langle d | o \rangle & = \left[ (\bar \gamma \, \langle u | + \bar \delta \,\langle d |)(0 \, | u \rangle + 1 \, | d \rangle)\right] \left[(0 \, \langle u | + 1 \, \langle d | ) (\gamma \, | u\rangle + \delta \, | d \rangle) \right]\\ & = (\bar \gamma \langle u | d \rangle + \bar \delta \langle d | d \rangle) (\gamma \langle d | u \rangle + \delta \langle d | d \rangle) = \bar \delta \, \delta = \tfrac{1}{2} \end{aligned}
Moving to the probabilities along | l \rangle, | r \rangle:
\begin{aligned} \langle i | r \rangle \langle r | i \rangle & = \left[ (\bar \alpha \, \langle u | + \bar \beta \,\langle d |)(\tfrac{1}{\sqrt 2}| u\rangle + \tfrac{1}{\sqrt 2}| d \rangle)\right] \left[(\tfrac{1}{\sqrt 2} \langle u | + \tfrac{1}{\sqrt 2} \langle d |) (\alpha \, | u\rangle + \beta \, | d \rangle) \right] \\ & = (\tfrac{\bar \alpha}{\sqrt 2} \langle u | u \rangle + \tfrac{\bar \alpha}{\sqrt 2} \langle u | d \rangle + \tfrac{\bar \beta}{\sqrt 2} \langle d | u \rangle + \tfrac{\bar \beta}{\sqrt 2} \langle d | d \rangle)(\tfrac{\alpha}{\sqrt 2} \langle u | u \rangle + \tfrac{\alpha}{\sqrt 2} \langle u | d \rangle + \tfrac{\beta}{\sqrt 2} \langle d | u \rangle + \tfrac{\beta}{\sqrt 2} \langle d | d \rangle) \\ & = (\tfrac{\bar \alpha}{\sqrt 2} + \tfrac{\bar \beta}{\sqrt 2})(\tfrac{\alpha}{\sqrt 2} + \tfrac{\beta}{\sqrt 2}) = \tfrac{1}{2} (\bar \alpha \,\alpha + \bar \alpha \, \beta + \bar \beta \, \alpha + \bar \beta \, \beta ) = \tfrac{1}{2} \\ \langle i | l \rangle \langle l | i \rangle & = \left[ (\bar \alpha \, \langle u | + \bar \beta \,\langle d |)(\tfrac{1}{\sqrt 2}| u\rangle - \tfrac{1}{\sqrt 2}| d \rangle)\right] \left[(\tfrac{1}{\sqrt 2} \langle u | - \tfrac{1}{\sqrt 2} \langle d |) (\alpha \, | u\rangle + \beta \, | d \rangle) \right] \\ & = (\tfrac{\bar \alpha}{\sqrt 2} \langle u | u \rangle - \tfrac{\bar \alpha}{\sqrt 2} \langle u | d \rangle + \tfrac{\bar \beta}{\sqrt 2} \langle d | u \rangle - \tfrac{\bar \beta}{\sqrt 2} \langle d | d \rangle)(\tfrac{\alpha}{\sqrt 2} \langle u | u \rangle + \tfrac{\alpha}{\sqrt 2} \langle u | d \rangle - \tfrac{\beta}{\sqrt 2} \langle d | u \rangle - \tfrac{\beta}{\sqrt 2} \langle d | d \rangle) \\ & = (\tfrac{\bar \alpha}{\sqrt 2} - \tfrac{\bar \beta}{\sqrt 2})(\tfrac{\alpha}{\sqrt 2} - \tfrac{\beta}{\sqrt 2}) = \tfrac{1}{2} (\bar \alpha \,\alpha - \bar \alpha \, \beta - \bar \beta \, \alpha + \bar \beta \, \beta ) = \tfrac{1}{2} \\ \langle o | r \rangle \langle r | o \rangle & = \left[ (\bar \gamma \, \langle u | + \bar \delta \,\langle d |)(\tfrac{1}{\sqrt 2}| u\rangle + \tfrac{1}{\sqrt 2}| d \rangle)\right] \left[(\tfrac{1}{\sqrt 2} \langle u | + \tfrac{1}{\sqrt 2} \langle d |) (\gamma \, | u\rangle + \delta \, | d \rangle) \right] \\ & = (\tfrac{\bar \gamma}{\sqrt 2} \langle u | u \rangle + \tfrac{\bar \gamma}{\sqrt 2} \langle u | d \rangle + \tfrac{\bar \delta}{\sqrt 2} \langle d | u \rangle + \tfrac{\bar \delta}{\sqrt 2} \langle d | d \rangle)(\tfrac{\gamma}{\sqrt 2} \langle u | u \rangle + \tfrac{\gamma}{\sqrt 2} \langle u | d \rangle + \tfrac{\delta}{\sqrt 2} \langle d | u \rangle + \tfrac{\delta}{\sqrt 2} \langle d | d \rangle) \\ & = (\tfrac{\bar \gamma}{\sqrt 2} + \tfrac{\bar \delta}{\sqrt 2})(\tfrac{\gamma}{\sqrt 2} + \tfrac{\delta}{\sqrt 2}) = \tfrac{1}{2} (\bar \gamma \,\gamma + \bar \gamma \, \delta + \bar \delta \, \gamma + \bar \delta \, \delta ) = \tfrac{1}{2} \\ \langle o | l \rangle \langle l | o \rangle & = \left[ (\bar \gamma \, \langle u | + \bar \delta \,\langle d |)(\tfrac{1}{\sqrt 2}| u\rangle - \tfrac{1}{\sqrt 2}| d \rangle)\right] \left[(\tfrac{1}{\sqrt 2} \langle u | - \tfrac{1}{\sqrt 2} \langle d |) (\gamma \, | u\rangle + \delta \, | d \rangle) \right] \\ & = (\tfrac{\bar \gamma}{\sqrt 2} \langle u | u \rangle - \tfrac{\bar \gamma}{\sqrt 2} \langle u | d \rangle + \tfrac{\bar \delta}{\sqrt 2} \langle d | u \rangle - \tfrac{\bar \delta}{\sqrt 2} \langle d | d \rangle)(\tfrac{\gamma}{\sqrt 2} \langle u | u \rangle + \tfrac{\gamma}{\sqrt 2} \langle u | d \rangle - \tfrac{\delta}{\sqrt 2} \langle d | u \rangle - \tfrac{\delta}{\sqrt 2} \langle d | d \rangle) \\ & = (\tfrac{\bar \gamma}{\sqrt 2} - \tfrac{\bar \delta}{\sqrt 2})(\tfrac{\gamma}{\sqrt 2} - \tfrac{\delta}{\sqrt 2}) = \tfrac{1}{2} (\bar \gamma \,\gamma - \bar \gamma \, \delta - \bar \delta \, \gamma + \bar \delta \, \delta ) = \tfrac{1}{2} \end{aligned}
Using the previous results:
\begin{aligned} & \tfrac{1}{2} (\bar \alpha \,\alpha + \bar \alpha \, \beta + \bar \beta \, \alpha + \bar \beta \, \beta ) = \tfrac{1}{2} (\tfrac{1}{2} + \bar \alpha \, \beta + \bar \beta \, \alpha + \tfrac{1}{2}) = \tfrac{1}{2} \\ & \bar \alpha \, \beta + \bar \beta \, \alpha = 0 \\ & \tfrac{1}{2} (\bar \alpha \,\alpha - \bar \alpha \, \beta - \bar \beta \, \alpha + \bar \beta \, \beta ) =\tfrac{1}{2} (\tfrac{1}{2} - \bar \alpha \, \beta - \bar \beta \, \alpha + \tfrac{1}{2}) = \tfrac{1}{2} \\ & \bar \alpha \, \beta + \bar \beta \, \alpha = 0 \\ & \tfrac{1}{2} (\bar \gamma \,\gamma + \bar \gamma \, \delta + \bar \delta \, \gamma + \bar \delta \, \delta ) = \tfrac{1}{2} (\tfrac{1}{2} + \bar \gamma \, \delta + \bar \delta \, \gamma + \tfrac{1}{2}) = \tfrac{1}{2} \\ & \bar \gamma \, \delta + \bar \delta \, \gamma = 0 \\ & \tfrac{1}{2} (\bar \gamma \,\gamma - \bar \gamma \, \delta - \bar \delta \, \gamma + \bar \delta \, \delta ) =\tfrac{1}{2} (\tfrac{1}{2} - \bar \gamma \, \delta - \bar \delta \, \gamma + \tfrac{1}{2}) = \tfrac{1}{2} \\ & \bar \gamma \, \delta + \bar \delta \, \gamma = 0 \end{aligned}
Considering a generic complex number z and write it as:
z= a + i\,b, \quad a,\,b \in \mathbb R
Two relationships that hold is that if a number is equal to its conjugate is real, if equal to the opposite of its conjugate is pure imaginary:
\begin {array}{ll} \bar z = a - i\,b \\ z = \bar z \\ a + i\,b = a - i\,b & \rightarrow b = 0 \\ z = -\bar z \\ a + i\,b = -a + i\,b & \rightarrow a = 0 \end{array}
Taking the conjugate of \bar \alpha \, \beta and \bar \gamma\,\delta:
\begin{aligned} & \overline {\bar \alpha \, \beta} = \alpha \bar \beta \\ & \overline {\bar \gamma \, \delta} = \gamma \bar \delta \end{aligned}
Since it was demonstrated that:
\begin{array}{ll} \bar \alpha \, \beta + \bar \beta \, \alpha = 0 & \bar \alpha \, \beta = - \bar \beta \, \alpha \\ \bar \gamma \, \delta + \bar \delta \, \gamma = 0 & \bar \gamma \, \beta = - \bar \delta \, \gamma \end{array}
Both \bar \alpha \, \beta and \bar \gamma\,\delta are pure imaginary numbers; since the product is pure imaginary, they cannot be both real, so it is necessary to introduce complex numbers with imaginary part to represent one component of spin.
Prove the following: If a vector space in N-dimensional, an orthonormal basis of N vectors can be constructed from the eigenvectors of a Hermitian operator.
Given an N-dimensional vector space, let’s consider a Hermitian operator \hat{H} defined on this space. A Hermitian operator is characterized by the property that \langle \mathbf{u} | \hat{H} \mathbf{v} \rangle = \langle \hat{H} \mathbf{u} | \mathbf{v} \rangle for all vectors \mathbf{u}, \mathbf{v} in the space, where \langle \cdot | \cdot \rangle denotes the inner product.
Theorem: From the eigenvectors of a Hermitian operator \hat{H}, an orthonormal basis of N vectors for the N-dimensional vector space can be constructed.
Proof:
Eigenvalues and Eigenvectors: Since \hat{H} is Hermitian, all its eigenvalues are real, and eigenvectors corresponding to distinct eigenvalues are orthogonal. Let \lambda_i be an eigenvalue and \mathbf{v}_i be a corresponding eigenvector, such that \hat{H} \mathbf{v}_i = \lambda_i \mathbf{v}_i.
Construction of an Orthonormal Set: If all eigenvalues are distinct, the eigenvectors \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_N\} form an orthogonal set. Normalize each eigenvector to obtain the orthonormal set \{\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_N\}, where \mathbf{u}_i = \frac{\mathbf{v}_i}{\|\mathbf{v}_i\|} and \|\mathbf{v}_i\| is the norm of \mathbf{v}_i.
Handling Degenerate Eigenvalues: In case of degenerate eigenvalues, where an eigenvalue \lambda has multiplicity greater than 1, apply the Gram-Schmidt process to the corresponding eigenvectors to obtain an orthonormal set of vectors within the eigenspace of \lambda.
Orthonormal Basis: The collection of normalized eigenvectors (including those processed through Gram-Schmidt for degenerate eigenvalues) forms an orthonormal basis for the N-dimensional vector space. This is because the set spans the vector space and consists of N mutually orthogonal, unit vectors.
This construction ensures that any vector in the N-dimensional space can be expressed as a linear combination of the orthonormal basis vectors derived from the eigenvectors of \hat{H}, thereby proving the theorem.
Prove that Eq. 3.16 is the unique solution to Eqs. 3.14 and 3.15.
The objective is to find a matrix:
\sigma_z = \begin{bmatrix} (\sigma_z)_{11} & (\sigma_z)_{12} \\ (\sigma_z)_{21} & (\sigma_z)_{22} \\ \end{bmatrix} which satisfy the relationships 3.14 and 3.15.
By developing the matrix product and identifying the vector components, we obtain a system of four equations with four unknowns: (\sigma_z)_{11}, (\sigma_z)_{12}, (\sigma_z)_{21}, and (\sigma_z)_{22}:
\begin{aligned} & 1(\sigma_z)_{11} + 0(\sigma_z)_{12} = 1 \\ & 1(\sigma_z)_{21} + 0(\sigma_z)_{22} = 0 \\ & 0(\sigma_z)_{11} + 1(\sigma_z)_{12} = 0 \\ & 0(\sigma_z)_{21} + 1(\sigma_z)_{22} = -1 \\ \end{aligned}
Multiplying:
\begin{aligned} & \sigma z_{11} = 1 \\ & \sigma z_{21} = 0 \\ & \sigma z_{12} = 0 \\ & \sigma z_{22} = -1 \\ \end{aligned}
Which gives the matrix:
\sigma_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
As this is a system of n equations in n unknown (with n=4), the solution is unique.
Calculate the eigenvectors and eigenvalues of \sigma_n. Hint: Assume the eigenvector \lambda_1 has the form:
\begin{pmatrix} \cos\alpha \\ \sin\alpha \\ \end{pmatrix},
where \alpha is an unknown parameter. Plug this vector into the eigenvalue equation and solve for \alpha in terms of \theta. Why did we use a single parameter \alpha? Notice that our suggested column vector must have unit length.
Considering the case where \mathbf n lies in the x-z plane; since it an unit vector, it can be written in polar coordinates as:
\begin{aligned} & n_x = \sin(\theta) \\ & n_y = 0 \\ & n_z = \cos(\theta) \end{aligned}
where \theta is the angle between \mathbf n and the z axis. The operator then becomes:
\sigma_n = \begin{bmatrix} n_z & \left(n_x -i\,n_y \right) \\ \left(n_x + i\,n_y \right) & - n_z \end{bmatrix} = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & - \cos(\theta) \end{bmatrix}
Starting from the eigenvalues:
\begin{aligned} \left| \sigma_n - \lambda\,I\right| & = \begin{vmatrix} \cos(\theta) - \lambda & \sin(\theta) \\ \sin(\theta) & - \cos(\theta) - \lambda \end{vmatrix} = (\cos(\theta) - \lambda)(-\cos(\theta) - \lambda) - \sin(\theta)^2 \\[12pt] & = \lambda_2 - \cos(\theta)^2 - \sin(\theta)^2 = \lambda^2 - 1 = 0 \\ \lambda_1 & = 1 \\ \lambda_2 & = -1 \end{aligned}
For \lambda_1 = 1 the eigenvector is:
(\sigma_n - I)\mathbf x = \begin{bmatrix} \cos(\theta) - 1 & \sin(\theta) \\ \sin(\theta) & - \cos(\theta) - 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
and expanding the system of equations (using trigonometric identities for \frac{\theta}{2} and 2\theta):
\begin{aligned} & (\cos(\theta) - 1)x_1 + \sin(\theta)x_2 = -2\sin\left(\tfrac{\theta}{2}\right)^2x_1 + 2\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right)x_2 = 0 \\ & \sin(\theta)x_1 - (\cos(\theta) + 1)x_2 = 2\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right)x_1 - 2\cos\left(\tfrac{\theta}{2}\right)^2 x_2 = 0 \\ \end{aligned}
Considering the case \theta=0, \mod( \pi), then any vector of the form [x_1, \quad 0]^T is an eigenvector (taking [1, \quad 0]^T so it has unitary length). If \theta \neq 0 it is possible to divide by \sin\left(\tfrac{\theta}{2}\right) the first equation:
-\sin\left(\tfrac{\theta}{2}\right)x_1 + \cos\left(\tfrac{\theta}{2}\right)x_2 = 0
Then is possible to choose as unitary eigenvector:
| v_1 \rangle = \begin{bmatrix} \cos\left(\tfrac{\theta}{2}\right) \\[12pt] \sin\left(\tfrac{\theta}{2}\right) \end{bmatrix}
For \lambda_2 = -1 the eigenvector is:
(\sigma_n - I)\mathbf x = \begin{bmatrix} \cos(\theta) + 1 & \sin(\theta) \\ \sin(\theta) & - \cos(\theta) + 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
and expanding the system of equations (using trigonometric identities for \frac{\theta}{2} and 2\theta):
\begin{aligned} & (\cos(\theta) + 1)x_1 + \sin(\theta)x_2 = 2\cos\left(\tfrac{\theta}{2}\right)^2x_1 + 2\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right)x_2 = 0 \\ & \sin(\theta)x_1 - (\cos(\theta) - 1)x_2 = 2\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right)x_1 + 2\sin\left(\tfrac{\theta}{2}\right)^2 x_2 = 0 \\ \end{aligned}
Considering the case \theta=0, \mod( \pi), then any vector of the form [0, \quad x_2]^T is an eigenvector (taking [0, \quad 1]^T so it has unitary length. If \theta \neq 0 it is possible to divide by \sin\left(\tfrac{\theta}{2}\right) the second equation:
\cos\left(\tfrac{\theta}{2}\right)x_1 + \sin\left(\tfrac{\theta}{2}\right)x_2 = 0
Then is possible to choose as unitary eigenvector:
| v_2 \rangle = \begin{bmatrix} -\sin\left(\tfrac{\theta}{2}\right) \\[12pt] \cos\left(\tfrac{\theta}{2}\right) \end{bmatrix}
There are some fact regarding these findings: the eigenvalues are again +1 and -1, and it is not unexpected as the apparatus can give only these values; the eigenvectors are orthogonal:
\langle v_1 | v_2 \rangle = -\cos\left(\tfrac{\theta}{2}\right)\sin\left(\tfrac{\theta}{2}\right) + \sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right) = 0
That is incidentally true for \theta=0, \mod( \pi), and this would be the trivial case in which \mathbf n is aligned or opposite to the z axis (and therefore \sigma_n = \sigma_z if aligned or \sigma_n = -\sigma_z if opposite); similarly \theta=\frac{\pi}{2}, \mod( \pi) would be the case in which \mathbf n is aligned or opposite to the x axis (and therefore \sigma_n = \sigma_x if aligned or \sigma_n = -\sigma_x if opposite).
Suppose \mathcal A initially points along the z axis and the spin is prepared to be +1 (|u \rangle); then it is rotated to lie along the \mathbf n axis; the probability of observing \sigma_n = +1 is:
P(+1) = \left| \langle u | v_1 \rangle \right|^2 = \cos\left(\tfrac{\theta}{2}\right)^2
Similarly:
P(-1) = \left| \langle u | v_2 \rangle \right|^2 = \sin\left(\tfrac{\theta}{2}\right)^2
The expected value of the measurement:
\langle \mathbf \sigma_n \rangle = \sum_i \lambda_iP\left(v_i \right) = (+1)\cos\left(\tfrac{\theta}{2}\right)^2 + (-1)\sin\left(\tfrac{\theta}{2}\right)^2 = \cos\left(\tfrac{\theta}{2}\right)^2 - \sin\left(\tfrac{\theta}{2}\right)^2 = \cos\left(2\tfrac{\theta}{2}\right) = \cos\left(\theta\right)
The last has been obtained from the formula for the double angle.
Let n_z = \cos\theta, n_x = \sin\theta\cos\phi and n_y = \sin\theta\sin\phi. Angles \theta and \phi are defined according to the usual conventions for spherical coordinates (Fig. 3.2). Compute the eigenvalues and eigenvectors for the matrix of Eq. 3.23.
Considering the case for a generic \mathbf n; since it an unit vector, it can be written in spherical coordinates as:
\begin{aligned} & n_x = \sin(\theta)\cos(\phi) \\ & n_y = \sin(\theta)\sin(\phi) \\ & n_z = \cos(\theta) \end{aligned}
where \theta is the angle between \mathbf n and the z axis and \phi angle between the projection in the x-y plane and the axis x. The operator then becomes, using Euler’s formula: e^{\pm i\,x}=\cos(x) \pm i\,\sin(x):
\begin{aligned} \sigma_n & = \begin{bmatrix} n_z & \left(n_x -i\,n_y \right) \\ \left(n_x + i\,n_y \right) & - n_z \end{bmatrix} = \begin{bmatrix} \cos(\theta) & \left(\sin(\theta)\cos(\phi) - i\,\sin(\theta)\sin(\phi) \right) \\ \left(\sin(\theta)\cos(\phi) + i\,\sin(\theta)\sin(\phi) \right) & - \cos(\theta) \end{bmatrix} \\[12pt] & = \begin{bmatrix} \cos(\theta) & \sin(\theta) e^{-i\,\phi} \\ \sin(\theta) e^{i\,\phi} & - \cos(\theta) \end{bmatrix} \end{aligned}
Starting from the eigenvalues:
\begin{aligned} \left| \sigma_n - \lambda\,I\right| & = \begin{vmatrix} \cos(\theta) - \lambda & \sin(\theta) e^{-i\,\phi} \\ \sin(\theta) e^{i\,\phi} & - \cos(\theta) - \lambda \end{vmatrix} = (\cos(\theta) - \lambda)(-\cos(\theta) - \lambda) - \sin(\theta)^2 e^{-i\,\phi} e^{i\,\phi}\\[12pt] & = \lambda_2 - \cos(\theta)^2 - \sin(\theta)^2 = \lambda^2 - 1 = 0 \\ \lambda_1 & = 1 \\ \lambda_2 & = -1 \end{aligned}
For \lambda_1 = 1 the eigenvector is:
(\sigma_n - I)\mathbf x = \begin{bmatrix} \cos(\theta) - 1 & \sin(\theta)e^{-i\,\phi} \\ \sin(\theta)e^{i\,\phi} & - \cos(\theta) - 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
and expanding the system of equations (using trigonometric identities for \frac{\theta}{2} and 2\theta):
\begin{aligned} & (\cos(\theta) - 1)x_1 + \sin(\theta)e^{-i\,\phi}x_2 = -2\sin\left(\tfrac{\theta}{2}\right)^2x_1 + 2\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right)e^{-i\,\phi}x_2 = 0 \\ & \sin(\theta)e^{i\,\phi}x_1 - (\cos(\theta) + 1)x_2 = 2\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right)e^{i\,\phi}x_1 - 2\cos\left(\tfrac{\theta}{2}\right)^2 x_2 = 0 \\ \end{aligned}
The trivial cases are again not considered, as \theta=0 is the case parallel or opposite to the z axis. If \theta \neq 0 it is possible to divide by \sin\left(\tfrac{\theta}{2}\right) the first equation:
-\sin\left(\tfrac{\theta}{2}\right)x_1 + \cos\left(\tfrac{\theta}{2}\right)e^{-i\,\phi}x_2 = 0
Then is possible to choose as eigenvector:
| v_1 \rangle = \begin{bmatrix} \cos\left(\tfrac{\theta}{2}\right) \\[12pt] e^{i\,\phi} \sin\left(\tfrac{\theta}{2}\right) \end{bmatrix}
This is unitary:
\langle v_1 | v_1 \rangle = \langle \begin{bmatrix} \cos\left(\tfrac{\theta}{2}\right) \\[12pt] e^{-i\,\phi}\sin\left(\tfrac{\theta}{2}\right) \end{bmatrix} | \begin{bmatrix} \cos\left(\tfrac{\theta}{2}\right) \\[12pt] e^{i\,\phi} \sin\left(\tfrac{\theta}{2}\right) \end{bmatrix} \rangle = \cos\left(\tfrac{\theta}{2}\right)^2 + e^{-i\,\phi}e^{i\,\phi}\sin\left(\tfrac{\theta}{2}\right)^2 = 1
For \lambda_2 = -1 the eigenvector is:
(\sigma_n - I)\mathbf x = \begin{bmatrix} \cos(\theta) + 1 & \sin(\theta)e^{-i\,\phi} \\ \sin(\theta)e^{i\,\phi} & - \cos(\theta) + 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
and expanding the system of equations (using trigonometric identities for \frac{\theta}{2} and 2\theta):
\begin{aligned} & (\cos(\theta) + 1)x_1 + \sin(\theta)e^{-i\,\phi}x_2 = 2\cos\left(\tfrac{\theta}{2}\right)^2x_1 + 2\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right)e^{-i\,\phi}x_2 = 0 \\ & \sin(\theta)e^{i\,\phi}x_1 - (\cos(\theta) - 1)x_2 = 2\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right)e^{i\,\phi}x_1 + 2\sin\left(\tfrac{\theta}{2}\right)^2 x_2 = 0 \\ \end{aligned}
If \theta \neq 0 it is possible to divide by \sin\left(\tfrac{\theta}{2}\right) the second equation:
\cos\left(\tfrac{\theta}{2}\right)x_1e^{i\,\phi} + \sin\left(\tfrac{\theta}{2}\right)x_2 = 0
Then is possible to choose as eigenvector:
| v_2 \rangle = \begin{bmatrix} -\sin\left(\tfrac{\theta}{2}\right) \\[12pt] e^{i\,\phi}\cos\left(\tfrac{\theta}{2}\right) \end{bmatrix}
This is also unitary:
\langle v_2 | v_2 \rangle = \langle \begin{bmatrix} -\sin\left(\tfrac{\theta}{2}\right) \\[12pt] e^{-i\,\phi}\cos\left(\tfrac{\theta}{2}\right) \end{bmatrix} | \begin{bmatrix} -\sin\left(\tfrac{\theta}{2}\right) \\[12pt] e^{i\,\phi}\cos\left(\tfrac{\theta}{2}\right) \end{bmatrix} \rangle = \sin\left(\tfrac{\theta}{2}\right)^2 + e^{-i\,\phi}e^{i\,\phi}\cos\left(\tfrac{\theta}{2}\right)^2 = 1
The eigenvalues are again +1 and -1; the eigenvectors are orthogonal:
\langle v_1 | v_2 \rangle = \langle \begin{bmatrix} \cos\left(\tfrac{\theta}{2}\right) \\[12pt] e^{-i\,\phi}\sin\left(\tfrac{\theta}{2}\right) \end{bmatrix} | \begin{bmatrix} -\sin\left(\tfrac{\theta}{2}\right) \\[12pt] e^{i\,\phi}\cos\left(\tfrac{\theta}{2}\right) \end{bmatrix} \rangle = -\cos\left(\tfrac{\theta}{2}\right)\sin\left(\tfrac{\theta}{2}\right) + e^{-i\,\phi}e^{i\,\phi}\sin\left(\tfrac{\theta}{2}\right)\cos\left(\tfrac{\theta}{2}\right) = 0
Suppose \mathcal A initially points along the z axis and the spin is prepared to be +1 (|u \rangle); then it is rotated to lie along the \mathbf n axis; the probability of observing \sigma_n = +1 is:
P(+1) = \left| \langle u | v_1 \rangle \right|^2 = e^{i\,\phi}e^{-i\,\phi} \cos\left(\tfrac{\theta}{2}\right)^2 = \cos\left(\tfrac{\theta}{2}\right)^2
Similarly:
P(-1) = \left| \langle u | v_2 \rangle \right|^2 = e^{i\,\phi}e^{-i\,\phi}\sin\left(\tfrac{\theta}{2}\right)^2 = \sin\left(\tfrac{\theta}{2}\right)^2
The expected value of the measurement:
\langle \mathbf \sigma_n \rangle = \sum_i \lambda_iP\left(v_i \right) = (+1)\cos\left(\tfrac{\theta}{2}\right)^2 + (-1)\sin\left(\tfrac{\theta}{2}\right)^2 = \cos\left(\tfrac{\theta}{2}\right)^2 - \sin\left(\tfrac{\theta}{2}\right)^2 = \cos\left(2\tfrac{\theta}{2}\right) = \cos\left(\theta\right)
Which lead to the exact measurement as in the case where the orientation is on x-z plane.
Suppose that a spin is prepared so that \sigma_m = +1. The apparatus is then rotated to the \hat{n} direction and \sigma_n is measured. What is the probability that the result is +1? Note that \sigma_m = \sigma\cdot\hat{m}, using the same convention we used for \sigma_n.
Consider that a spin is prepared along an \mathbf m direction so that \sigma_m = +1; the apparatus \mathcal A is then rotated to an \mathbf n direction and \sigma_n is measured.
To measure the probability that the result of this experiment is +1 is:
P(+1) = \cos\left(\tfrac{\theta_{mn}}{2}\right)^2
where \theta_{mn} is the angle between \mathbf m and \mathbf n.
This can be immediately proven just observing that if it considered a reference frame which align the \mathbf m direction with the up direction (z' axis), then the probability of finding +1 is the same as the spherical coordinate case, in which the angle \theta' is the angle between the direction \mathbf n and the axis z', which is \theta_{mn}.
It is possible to have a more formal answer without any particular assumptions; considering the standard reference frame it is possible to express the probability using spherical coordinates; the eigenvector corresponding to the eigenvalue +1 are known:
\begin{aligned} & | v_{1m} \rangle = \begin{bmatrix} \cos\left(\tfrac{\theta_m}{2}\right) \\[12pt] e^{i\,\phi_m} \sin\left(\tfrac{\theta_m}{2}\right) \end{bmatrix} \\ & | v_{1n} \rangle = \begin{bmatrix} \cos\left(\tfrac{\theta_n}{2}\right) \\[12pt] e^{i\,\phi_n} \sin\left(\tfrac{\theta_n}{2}\right) \end{bmatrix} \end{aligned}
The probability is therefore:
\begin{aligned} P(+1) = & |\langle v_{1m} | v_{1n} \rangle|^2 = \langle v_{1m} | v_{1n} \rangle \langle v_{1n} | v_{1m} \rangle \\ = & \langle \begin{bmatrix} \cos\left(\tfrac{\theta_m}{2}\right) \\[12pt] e^{-i\,\phi_m} \sin\left(\tfrac{\theta_m}{2}\right) \end{bmatrix} \begin{bmatrix} \cos\left(\tfrac{\theta_n}{2}\right) \\[12pt] e^{i\,\phi_n} \sin\left(\tfrac{\theta_n}{2}\right) \end{bmatrix} \rangle \langle \begin{bmatrix} \cos\left(\tfrac{\theta_n}{2}\right) \\[12pt] e^{-i\,\phi_n} \sin\left(\tfrac{\theta_n}{2}\right) \end{bmatrix} \begin{bmatrix} \cos\left(\tfrac{\theta_m}{2}\right) \\[12pt] e^{i\,\phi_m} \sin\left(\tfrac{\theta_m}{2}\right) \end{bmatrix} \rangle \\ = & \left[ \cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right) + e^{-i\,\phi_m}e^{i\,\phi_n} \sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right)\right] \\ & \times \left[\cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right) + e^{-i\,\phi_n}e^{i\,\phi_m} \sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right)\right] \\ = & \left[ \cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right) + e^{-i\,\phi_m}e^{i\,\phi_n} \sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right)\right] \\ & \times \left[\cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right) + e^{-i\,\phi_n}e^{i\,\phi_m} \sin\left(\tfrac{\theta_m}{2}\right)\left(\tfrac{\theta_n}{2}\right)\right] \\ = & \cos\left(\tfrac{\theta_m}{2}\right)^2\cos\left(\tfrac{\theta_n}{2}\right)^2 + \cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right)e^{-i\,\phi_n}e^{i\,\phi_m} \sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right) \\ & + e^{-i\,\phi_m}e^{i\,\phi_n} \sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right)\cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right) \\ & + e^{-i\,\phi_m}e^{i\,\phi_n} e^{-i\,\phi_n}e^{i\,\phi_m} \sin\left(\tfrac{\theta_m}{2}\right)^2\sin\left(\tfrac{\theta_n}{2}\right)^2 \\ = & \cos\left(\tfrac{\theta_m}{2}\right)^2\cos\left(\tfrac{\theta_n}{2}\right)^2 + \left[e^{-i\,\phi_m}e^{i\,\phi_n} + e^{-i\,\phi_n}e^{i\,\phi_m} \right] \cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right)\sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right) \\ & + 1 \sin\left(\tfrac{\theta_m}{2}\right)^2\sin\left(\tfrac{\theta_n}{2}\right)^2 \\ = & \cos\left(\tfrac{\theta_m}{2}\right)^2\cos\left(\tfrac{\theta_n}{2}\right)^2 \\ & + 2 \left[\cos\left(\phi_m\right)\cos\left(\phi_n\right) + \sin\left(\phi_m\right)\sin\left(\phi_n\right)\right]\cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right)\sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right) \\ & + \sin\left(\tfrac{\theta_m}{2}\right)^2\sin\left(\tfrac{\theta_n}{2}\right)^2 \end{aligned}
as \cos(\alpha -\beta ) =\cos (\alpha) \cos (\beta) +\sin (\alpha) \sin (\beta).
For the phases:
\begin{aligned} & \left[\cos\left(\phi_m\right) - i\sin\left(\phi_m\right)\right]\left[\cos\left(\phi_n\right) - i\sin\left(\phi_n\right)\right] + \left[\cos\left(\phi_n\right) - i\sin\left(\phi_n\right)\right]\left[\cos\left(\phi_m\right) - i\sin\left(\phi_m\right)\right] \\ & \quad = \cos\left(\phi_m\right)\cos\left(\phi_n\right) + \sin\left(\phi_m\right)\sin\left(\phi_n\right) + i \left[\sin\left(\phi_n\right)\cos\left(\phi_m\right) - \sin\left(\phi_m\right)\cos\left(\phi_n\right)\right] \\ & \quad + \cos\left(\phi_n\right)\cos\left(\phi_m\right) + \sin\left(\phi_n\right)\sin\left(\phi_m\right) + i \left[\sin\left(\phi_m\right)\cos\left(\phi_n\right) - \sin\left(\phi_n\right)\cos\left(\phi_m\right)\right] \\ & \quad = 2 \left[\cos\left(\phi_m\right)\cos\left(\phi_n\right) + \sin\left(\phi_m\right)\sin\left(\phi_n\right)\right] \end{aligned}
If \phi_m = \phi_n = \phi then:
2 \left[\cos\left(\phi_m\right)\cos\left(\phi_n\right) + \sin\left(\phi_m\right)\sin\left(\phi_n\right)\right] = 2 \left[\cos\left(\phi\right)^2 + \sin\left(\phi\right)^2\right] = 2
And the probability results:
\begin{aligned} = & \cos\left(\tfrac{\theta_m}{2}\right)^2\cos\left(\tfrac{\theta_n}{2}\right)^2 \\ & + 2 \left[\cos\left(\phi_m\right)\cos\left(\phi_n\right) + \sin\left(\phi_m\right)\sin\left(\phi_n\right)\right]\cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right)\sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right) \\ & + \sin\left(\tfrac{\theta_m}{2}\right)^2\sin\left(\tfrac{\theta_n}{2}\right)^2 \\ = & \cos\left(\tfrac{\theta_m}{2}\right)^2\cos\left(\tfrac{\theta_n}{2}\right)^2 + 2\cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right)\sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right) + \sin\left(\tfrac{\theta_m}{2}\right)^2\sin\left(\tfrac{\theta_n}{2}\right)^2 \\ = & \left[\cos\left(\tfrac{\theta_m}{2}\right)\cos\left(\tfrac{\theta_n}{2}\right) + \sin\left(\tfrac{\theta_m}{2}\right)\sin\left(\tfrac{\theta_n}{2}\right) \right]^2 = \cos\left(\tfrac{\theta_m}{2} - \tfrac{\theta_n}{2}\right)^2 = \cos\left(\tfrac{\theta_{mn}}{2}\right)^2 \end{aligned}
which gives the previous result.