An electron is at rest in an oscillating magnetic field

$\mathbf{B}\mathbf{=}{\mathbf{B}}_{\mathbf{0}}\mathbf{cos}\mathbf{\left(}\mathbf{\omega t}\mathbf{\right)}\hat{\mathbf{k}}$

where${\mathit{B}}_{\mathbf{0}}$ and$\mathbf{\omega }$ are constants.

(a) Construct the Hamiltonian matrix for this system.

(b) The electron starts out (at t=0 ) in the spin-up state with respect to the x-axis (that is:$\mathbf{\chi }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{\chi }}_{\mathbf{+}}^{\left(x\right)}\mathbf{}\mathbf{)}$. Determine $\mathbf{X}\left(t\right)$at any subsequent time. Beware: This is a time-dependent Hamiltonian, so you cannot get in the usual way from stationary states. Fortunately, in this case you can solve the timedependent Schrödinger equation (Equation 4.162) directly.

(c) Find the probability of getting$\mathbf{-}\mathbf{h}\mathbf{/}\mathbf{2}$ , if you measure ${\mathbf{S}}_{\mathbf{x}}$. Answer:

${\mathbf{sin}}^{\mathbf{2}}\left(\frac{{\mathrm{\gamma B}}_0}{2\omega }\sin \left(\mathrm{\omega t}\right)\right)$

(d) What is the minimum field$\left(B_0\right)$ required to force a complete flip in${\mathit{S}}_{\mathbf{x}}$ ?

The Hamiltonian of a system expresses its total energy that is, the sum of its kinetic (motion) and potential (position) energy in terms of the Lagrangian function developed from prior studies of dynamics and the position and momentum of individual particles.

Write the expression for the Hamiltonian for a charged particle in an external magnetic field.

$H=-\gamma B\cdot S$

Substitute the$B_0\cos \left(\mathrm{\omega t}\right)$ for B androle="math" localid="1658120380642" $\cos \left(\mathrm{\omega t}\right)S_z$ for S in the above expression.

role="math" localid="1658120332344" $$\begin{gathered} H=-{\mathrm{\gamma B}}_0\cos \left(\mathrm{\omega t}\right)S_z \\ =\frac{-{\mathrm{\gamma B}}_0\hslash }{2}\cos \left(\mathrm{\omega t}\right)\left(\frac{1}{0}\frac{0}{-1}\right) \end{gathered}$$

Thus, the Hamiltonian matrix is $\frac{-{\mathrm{\gamma B}}_0\hslash }{2}\cos \left(\mathrm{\omega t}\right)\left(\frac{1}{0}\frac{0}{-1}\right)$ .

It is known that $\left|\mathrm{\chi }\right(\mathrm{t})>=\left(\begin{array}{c}\mathrm{a}\left(\mathrm{t}\right)\\ \mathrm{\beta }\left(\mathrm{t}\right)\end{array}\right)\mathrm{and}\mathrm{i}\hslash \frac{\partial \mathrm{\alpha }}{\partial \mathrm{t}}=\mathrm{H\chi }$ and

Determine the spin state in the following way,

$$\begin{gathered} \mathrm{i}\left(\begin{array}{c}\hat{\mathrm{a}}\left(\mathrm{t}\right)\\ \hat{\mathrm{t}}\left(\mathrm{t}\right)\end{array}\right)=\frac{-{\mathrm{\gamma B}}_0}{2}\cos \left(\mathrm{\omega t}\right)\left(\frac{1}{0}\frac{0}{-1}\right)\left(\begin{array}{c}\mathrm{a}\left(\mathrm{t}\right)\\ \mathrm{\beta }\left(\mathrm{t}\right)\end{array}\right) \\ \mathrm{i}\left(\begin{array}{c}\hat{\mathrm{a}}\left(\mathrm{t}\right)\\ \hat{\mathrm{t}}\left(\mathrm{t}\right)\end{array}\right)=\frac{-2{\mathrm{B}}_0}{2}\cos \left(\mathrm{\omega t}\right)\left(\begin{array}{c}\mathrm{a}\\ -\mathrm{\beta }\end{array}\right) \\ \mathrm{a}\left(0\right)=\mathrm{\beta }\left(0\right) \\ =\frac{1}{\sqrt{2}} \end{gathered}$$

Use $A=B=\frac{1}{\sqrt{2}}$in the above epression.

$\alpha \left(t\right)=\frac{1}{\sqrt{2}}{e}^{\frac{i-B_e}{2}sin\left(\omega t\right)}and\beta \left(t\right)=\frac{1}{\sqrt{2}}{e}^{\frac{-h_2{R}_0}{2a}sin\left(\omega t\right)}$

Substitute the above values in $\left|\chi \right(t)>=\left(\begin{array}{c}a\left(t\right)\\ \beta \left(t\right)\end{array}\right)$.

role="math" localid="1658120705766" $IX\left(t\right)>\frac{1}{\sqrt{2}}\left(\begin{array}{c}{e}^{\frac{i^{2}R{^}_2}{2}sin\left(\omega t\right)}\\ e^{\frac{-i_2{B}_2}{2\omega }}sin\left(\omega t\right)\end{array}\right)$

Thus, the value at any subsequent time is$\frac{1}{\sqrt{2}}\left(\begin{array}{c}{e}^{\frac{i^{2}R{^}_2}{2}\sin \left(\omega t\right)}\\ e^{\frac{-i_2{B}_2}{2\omega }}\sin \left(\omega t\right)\end{array}\right)$ .

Determine the probability.

Thus, the probability of getting $-h/2$ is role="math" localid="1658122276455" ${\sin }^2\left(\frac{{\mathrm{\gamma B}}_0}{2\mathrm{\omega }}\sin \left(\mathrm{\omega t}\right)\right)$ .

Consider the probability,

$$\begin{gathered} P_{-}^x-1\to \\ {\sin }^2\frac{aB}{2\omega }\sin \left(\omega t\right)=1 \\ \sin \frac{\gamma B}{2\omega }\sin \left(\omega t\right)=\pm 1 \\ \frac{\gamma B}{2\omega }\sin \left(\omega t\right)=\pm \frac{n\pi }{2}\forall n\in Z \end{gathered}$$

For $\sin \left(\omega t\right)=\pm 1$,

$B_0=\frac{\omega \pi }{\gamma }$for$n=1$.

Thus, the minimum field $B_0$ is $\frac{\omega \pi }{\gamma }$.