Question

(a) If you measured the component of spin angular momentum along the x direction, at time t, what is the probability that you would get $\mathbf{+}\mathbf{h}\mathbf{/}\mathbf{2}$?

(b) Same question, but for the ycomponent.

(c) Same, for the z component.

Short answer

(a) The probability along x direction is $\frac{1}{2}[1+sin\alpha cos(\gamma B_{0}t\left)\right]$.

(b) The probability along y direction is $\frac{1}{2}[1-sin\alpha sin(\gamma B_{0}t\left)\right]$.

(c) The probability along y direction is ${\cos }^2\frac{\alpha }{2}$.

Step-by-step solution

Definition of Eigen spinors and probability

In quantum physics, eigen spinors are considered basis vectors that represent a particle's general spin state. They are not vectors in the strictest sense, but rather spinors.

The probability formula states that the ratio of the number of favorable outcomes to the total number of alternatives equals the likelihood of an event occurring.

(a) Determination of the probability of getting +h/2 along x-direction

Use equations. 4.151 and 4.163 that are mentioned as follows,

$$\begin{gathered} {\chi }_{+}^{\left(x\right)}=\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right),\left(eigenvalue+\frac{\hslash }{2}\right) \\ {\chi }_{+}^{\left(x\right)}\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right),\left(eigenvalue-\frac{\hslash }{2}\right) \end{gathered}$$

The generic spinor x (Equation 4.139) can be written as a linear combination since the eigenvectors of a hermitian matrix, they span the space.

$x=\left(\frac{a+b}{\sqrt{2}}\right)x_{+}^{\left(x\right)}+\left(\frac{a-b}{\sqrt{2}}\right)x_{-}^{\left(x\right)}$

Determine the value of $\overline{)x)}$.

$\overline{)x)=\left(\begin{array}{cc}\cos \frac{\alpha }{2}& e^{i\gamma B_0\frac{1}{2}}\\ \sin \frac{\alpha }{2}& e^{-i\gamma B_0\frac{1}{2}}\end{array}\right)}$

Find the value of probability,$P_{+}^x$.

localid="1659017003900" $$\begin{gathered} P_{+}^x={\left|\left(x_{+}^x|x\right)\right|}^2 \\ =\frac{1}{2}\left|\left(11\right)\left(\begin{array}{cc}\cos \frac{\alpha }{2}& e^{iy{B}_0\frac{1}{2}}\\ \sin \frac{\alpha }{2}& e^{-i7y{B}_0\frac{f}{2}}\end{array}\right)\right| \\ =\frac{1}{2}\left[\cos \frac{\alpha }{2}{e}^{iy{B}_0\frac{t}{2}}+\sin \frac{\alpha }{2}{e}^{iy{B}_0\frac{t}{2}}\right]\times \left[\cos \frac{\alpha }{2}{e}^{iy{B}_0\frac{t}{2}}+\sin \frac{\alpha }{2}{e}^{-iy{B}_0\frac{t}{2}}\right] \\ =\frac{1}{2}\left[{\cos }^2\frac{\alpha }{2}+\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}\sin \frac{\alpha }{2}\left(e^{iy{B}_0}+e^{-iy{B}_0}\right)\right] \end{gathered}$$

Evaluate the above expression further.

$$\begin{gathered} P_{+}^x=\frac{1}{2}\left[1+2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\cos \left(\gamma B_{0}t\right)\right] \\ {P}_{+}^x\left(t\right)=\frac{1}{2}\left[1+\sin \alpha \cos \left(\gamma B_{0}t\right)\right] \end{gathered}$$

Thus, the probability along -direction is $\frac{1}{2}\left[1+\sin \alpha \cos \left(\gamma B_{0}t\right)\right]$.

(b) Determination of the probability of getting +h/2 along y-direction

From part a ofProblem 4.32,

$x_{+}^{\left(y\right)}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ i\end{array}\right)$

Substitute the above value in $c_{+}^{\left(y\right)}=x_{+}^{\left(y\right)┼}x$.

$$\begin{gathered} c_{+}^{\left(y\right)}=x_{+}^{\left(y\right)┼}x \\ =\frac{1}{\sqrt{2}}\left(1-i\right)\left(\begin{array}{c}\cos \frac{\alpha }{2}{e}^{iy{B}_{0}t/2}\\ \sin \frac{\alpha }{2}{e}^{-iy{B}_{0}t/2}\end{array}\right) \\ =\frac{1}{\sqrt{2}}\left[\cos \frac{\alpha }{2}{e}^{iy{B}_{0}t/2}-i\sin \frac{\alpha }{2}{e}^{-iy{B}_{0}t/2}\right] \end{gathered}$$

Determine the value of the probability, $P_{+}^y$.

$$\begin{gathered} P_{+}^x={\left|\left(x_{+}^x\right|x)\right|}^2 \\ =\frac{1}{2}\left|\left(1-i\right)\left(\begin{array}{c}\cos \frac{\alpha }{2}{e}^{i-B_0\frac{1}{2}}\\ \sin \frac{\alpha }{2}{e}^{-17}{B}_0\frac{1}{2}\end{array}\right)\right| \\ =\frac{1}{2}\left[{\cos }^2\frac{\alpha }{2}+{\sin }^2\frac{\alpha }{2}+i\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\left(e^{iy-B_{0}t}-e^{-i-y{B}_{0}t}\right)\right] \\ =\frac{1}{2}\left[1-2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\sin \left(\gamma B_{0}t\right)\right] \\ {P}_{+}^y\left(t\right)=\frac{1}{2}\left[1-\sin \alpha \sin \left(\gamma B_{0}t\right)\right] \end{gathered}$$

Thus, the probability along -direction is $\frac{1}{2}\left[1-\sin \alpha \sin \left(\gamma B_{0}t\right)\right]$.

(c) Determination of the probability of getting +h/2 along z-direction

The normalized eigen spinors for is as follows,

$$\begin{gathered} x_{+}^{\left(z\right)}=\left(\begin{array}{c}1\\ 0\end{array}\right) \\ {c}_{+}^{\left(z\right)}=\left(\begin{array}{cc}1& 0\end{array}\right)\left(\begin{array}{c}\cos \frac{\alpha }{2}{e}^{iy{B}_{0}t/2}\\ \sin \frac{\alpha }{2}{e}^{-iy{B}_{0}t/2}\end{array}\right) \\ =\cos \frac{\alpha }{2}{e}^{-iy{B}_{0}t/2} \end{gathered}$$

Determine the value of the probability,$P_{+}^{\left(z\right)}$.

$$\begin{gathered} P_{+}^{\left(z\right)}\left(t\right)={\left|c_{+}^{\left(z\right)}\right|}^2 \\ ={\cos }^2\frac{\alpha }{2} \end{gathered}$$

Hence, the probability along z-direction is ${\cos }^2\frac{\alpha }{2}$.