91影视

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.

$H=-纬B鈰�S$

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

role="math" localid="1658120332344" $$\begin{gathered} H=-{\mathrm{纬叠}}_0\cos \left(\mathrm{蝇迟}\right)S_z \\ =\frac{-{\mathrm{纬叠}}_0鈩�}{2}\cos \left(\mathrm{蝇迟}\right)\left(\frac{1}{0}\frac{0}{-1}\right) \end{gathered}$$

Thus, the Hamiltonian matrix is $\frac{-{\mathrm{纬叠}}_0鈩�}{2}\cos \left(\mathrm{蝇迟}\right)\left(\frac{1}{0}\frac{0}{-1}\right)$ .

It is known that $\left|\mathrm{蠂}\right(\mathrm{t})>=\left(\begin{array}{c}\mathrm{a}\left(\mathrm{t}\right)\\ \mathrm{尾}\left(\mathrm{t}\right)\end{array}\right)\mathrm{and}\mathrm{i}鈩�\frac{鈭�\mathrm{伪}}{鈭�\mathrm{t}}=\mathrm{贬蠂}$ and

Determine the spin state in the following way,

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

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

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

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

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

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

Determine the probability.

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

Consider the probability,

$$\begin{gathered} P_{-}^x-1鈫� \\ {\sin }^2\frac{aB}{2蝇}\sin \left(蝇t\right)=1 \\ \sin \frac{纬B}{2蝇}\sin \left(蝇t\right)=卤1 \\ \frac{纬B}{2蝇}\sin \left(蝇t\right)=卤\frac{n蟺}{2}鈭赌n鈭�Z \end{gathered}$$

For $\sin \left(蝇t\right)=卤1$,

$B_0=\frac{蝇蟺}{纬}$for$n=1$.

Thus, the minimum field $B_0$ is $\frac{蝇蟺}{纬}$.