Chapter 4.4, Problem 4.35P

(a)

To determine

The probability that would get $+\frac{\hslash }{2}$ at time $t$, if the component of spin angular momentum is measured along the $x$ direction.

Answer to Problem 4.35P

The probability that would get $+\frac{\hslash }{2}$ at time $t$, if the component of spin angular momentum is measured along the $x$ direction is $P_{+}^{\left(x\right)}\left(t\right)=\frac{1}{2}\left[1+\sin \alpha \cos \left(\gamma B_{0}t\right)\right]$.

Explanation of Solution

Write the expression to find $\chi$, Equation 4.163

  $\chi =\left(\begin{array}{c}\cos \left(\alpha /2\right)e^{i\gamma B_{0}t/2}\\ \sin \left(\alpha /2\right)e^{-i\gamma B_{0}t/2}\end{array}\right)$        (I)

Write the expression for ${\chi }_{+}^{\left(x\right)†}$, Equation 4.151

    ${\chi }_{+}^{\left(x\right)†}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\end{array}\right)$        (II)

Write the expression for $c_{+}^{\left(x\right)}$

    $c_{+}^{\left(x\right)}={\chi }_{+}^{\left(x\right)†}\chi$        (III)

Write the expression to fine the probability along $x$ direction

    $P_{+}^{\left(x\right)}\left(t\right)={\left|c_{+}^{\left(x\right)}\right|}^2$

Substitute equation (I), (II), and (III) in the above equation and solve for $P_{+}^{\left(x\right)}\left(t\right)$

$\begin{array}{c}{P}_{+}^{\left(x\right)}\left(t\right)={\left|c_{+}^{\left(x\right)}\right|}^2\\ ={\left|{\chi }_{+}^{\left(x\right)†}\chi \right|}^2\\ ={\left|\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& 1\end{array}\right)\left(\begin{array}{c}\cos \left(\alpha /2\right)e^{i\gamma B_{0}t/2}\\ \sin \left(\alpha /2\right)e^{-i\gamma B_{0}t/2}\end{array}\right)\right|}^2\\ =\frac{1}{2}\left[\cos \left(\alpha /2\right)e^{i\gamma B_{0}t/2}+\sin \left(\alpha /2\right)e^{-i\gamma B_{0}t/2}\right]\left[\cos \left(\alpha /2\right)e^{i\gamma B_{0}t/2}+\sin \left(\alpha /2\right)e^{-i\gamma B_{0}t/2}\right]\end{array}$

Solving further,

$\begin{array}{c}{P}_{+}^{\left(x\right)}\left(t\right)=\frac{1}{2}\left[{\cos }^2\frac{\alpha }{2}+{\sin }^2\frac{\alpha }{2}+\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\left(e^{i\gamma B_{0}t/2}+e^{-i\gamma B_{0}t/2}\right)\right]\\ =\frac{1}{2}\left[1+2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\cos \left(\gamma B_{0}t\right)\right]\\ =\frac{1}{2}\left[1+\sin \alpha \cos \left(\gamma B_{0}t\right)\right]\end{array}$

Conclusion:

Thus, the probability that would get $+\frac{\hslash }{2}$ at time $t$, if the component of spin angular momentum is measured along the $x$ direction is $P_{+}^{\left(x\right)}\left(t\right)=\frac{1}{2}\left[1+\sin \alpha \cos \left(\gamma B_{0}t\right)\right]$.

(b)

To determine

The probability that would get $+\frac{\hslash }{2}$ at time $t$, if the component of spin angular momentum is measured along the $y$ direction.

Answer to Problem 4.35P

The probability that would get $+\frac{\hslash }{2}$ at time $t$, if the component of spin angular momentum is measured along the $y$ direction is $P_{+}^{\left(y\right)}\left(t\right)=\frac{1}{2}\left[1-\sin \alpha \cos \left(\gamma B_{0}t\right)\right]$.

Explanation of Solution

Write the expression for $c_{+}^{\left(y\right)}$

    $c_{+}^{\left(y\right)}={\chi }_{+}^{\left(y\right)†}\chi$        (IV)

Write the expression to fine the probability along $x$ direction

    $P_{+}^{\left(y\right)}\left(t\right)={\left|c_{+}^{\left(y\right)}\right|}^2$

Substitute equation (I) and (IV) in the above equation and solve for $P_{+}^{\left(y\right)}\left(t\right)$

$\begin{array}{c}{P}_{+}^{\left(y\right)}\left(t\right)={\left|c_{+}^{\left(y\right)}\right|}^2\\ ={\left|{\chi }_{+}^{\left(y\right)†}\chi \right|}^2\\ ={\left|\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& -i\end{array}\right)\left(\begin{array}{c}\cos \left(\alpha /2\right)e^{i\gamma B_{0}t/2}\\ \sin \left(\alpha /2\right)e^{-i\gamma B_{0}t/2}\end{array}\right)\right|}^2\\ =\frac{1}{2}\left[\cos \left(\alpha /2\right)e^{-i\gamma B_{0}t/2}+i\sin \left(\alpha /2\right)e^{i\gamma B_{0}t/2}\right]\left[\cos \left(\alpha /2\right)e^{-i\gamma B_{0}t/2}-i\sin \left(\alpha /2\right)e^{i\gamma B_{0}t/2}\right]\end{array}$

Solving further,

$\begin{array}{c}{P}_{+}^{\left(y\right)}\left(t\right)=\frac{1}{2}\left[{\cos }^2\frac{\alpha }{2}+{\sin }^2\frac{\alpha }{2}+i\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\left(e^{i\gamma B_{0}t/2}-e^{-i\gamma B_{0}t/2}\right)\right]\\ =\frac{1}{2}\left[1-2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\cos \left(\gamma B_{0}t\right)\right]\\ =\frac{1}{2}\left[1-\sin \alpha \cos \left(\gamma B_{0}t\right)\right]\end{array}$

Conclusion:

Thus, the probability that would get $+\frac{\hslash }{2}$ at time $t$, if the component of spin angular momentum is measured along the $y$ direction is $P_{+}^{\left(y\right)}\left(t\right)=\frac{1}{2}\left[1-\sin \alpha \cos \left(\gamma B_{0}t\right)\right]$.

(c)

To determine

The probability that would get $+\frac{\hslash }{2}$ at time $t$, if the component of spin angular momentum is measured along the $z$ direction.

Answer to Problem 4.35P

The probability that would get $+\frac{\hslash }{2}$ at time $t$, if the component of spin angular momentum is measured along the $z$ direction is $P_{+}^{\left(z\right)}\left(t\right)={\cos }^2\frac{\alpha }{2}$.

Explanation of Solution

Write the expression for ${\chi }_{+}^{\left(z\right)†}$

    ${\chi }_{+}^{\left(z\right)†}=\left(\begin{array}{c}1\\ 0\end{array}\right)$        (V)

Write the expression for $c_{+}^{\left(z\right)}$

    $c_{+}^{\left(z\right)}={\chi }_{+}^{\left(z\right)†}\chi$        (VI)

Write the expression to fine the probability along $z$ direction

    $P_{+}^{\left(z\right)}\left(t\right)={\left|c_{+}^{\left(z\right)}\right|}^2$

Substitute equation (I), (V), and (VI) in the above equation and solve for $P_{+}^{\left(z\right)}\left(t\right)$

$\begin{array}{c}{P}_{+}^{\left(z\right)}\left(t\right)={\left|c_{+}^{\left(z\right)}\right|}^2\\ ={\left|{\chi }_{+}^{\left(z\right)†}\chi \right|}^2\\ ={\left|\left(\begin{array}{cc}1& 0\end{array}\right)\left(\begin{array}{c}\cos \left(\alpha /2\right)e^{i\gamma B_{0}t/2}\\ \sin \left(\alpha /2\right)e^{-i\gamma B_{0}t/2}\end{array}\right)\right|}^2\\ =\frac{1}{2}\left[\cos \left(\alpha /2\right)e^{i\gamma B_{0}t/2}\right]\left[\cos \left(\alpha /2\right)e^{i\gamma B_{0}t/2}\right]\end{array}$

Solving further,

$P_{+}^{\left(z\right)}\left(t\right)={\cos }^2\frac{\alpha }{2}$

Conclusion:

Thus, the probability that would get $+\frac{\hslash }{2}$ at time $t$, if the component of spin angular momentum is measured along the $z$ direction is $P_{+}^{\left(z\right)}\left(t\right)={\cos }^2\frac{\alpha }{2}$.