91Ó°ÊÓ

Charged particles (such as protons, deuterons, or ions) are pushed by an alternating electric field in a constant magnetic field in a cyclotron.

(a)

Consider that,

$$\begin{gathered} \stackrel{→}{A}=\frac{B_0}{2}\left(x\hat{j}-x\hat{i}\right) \\ \mathrm{ϕ}=k{z}^2 \end{gathered}$$

In terms of scalar and vector potentials, the electric and magnetic fields E and B are defined as:

$\stackrel{→}{E}=-∇\mathrm{ϕ}-\frac{∂A}{∂t}$

Substituting the values, and we get,

$$\begin{gathered} \stackrel{→}{E}=-∇\mathrm{ϕ}-\frac{∂A}{∂t} \\ =-\left(\hat{i}\frac{∂}{∂x}+\hat{j}\frac{∂}{∂y}+\hat{k}\frac{∂}{∂z}\right)\left(k{z}^2\right)-0 \\ =-2kz\hat{k} \\ \mathrm{And} \\ \stackrel{→}{\mathrm{B}}=∇×\mathrm{A} \end{gathered}$$

Substituting the values, and we get,

$$\begin{gathered} \stackrel{→}{B}=\frac{B_0}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ -y& x& 0\end{array}\right| \\ =\frac{B_0}{2}\left[\hat{i}\left(-\frac{∂x}{∂y}\right)-\hat{j}\left(0+\frac{∂x}{∂y}\right)+\hat{k}\left(1-\left(-1\right)\right)\right] \\ \stackrel{→}{B}=\frac{B_0}{2}2\hat{k} \\ =B_0\hat{k} \end{gathered}$$

$\mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\stackrel{→}{\mathrm{E}}=-2\mathrm{kz}\hat{\mathrm{k}}\mathrm{and}\mathrm{the}\mathrm{value}\mathrm{of}\stackrel{→}{\mathrm{B}}={\mathrm{B}}_0\hat{\mathrm{k}}.$

(b)

According to the question,

$H\mathrm{ψ}=E\mathrm{ψ}$

For time-independent potentials,

$$\begin{gathered} \frac{1}{2m}\left(-i\mathrm{Ä\S }∇-qA\right)\left(-i\mathrm{Ä\S }∇-qA\right)\mathrm{ψ}+q\mathrm{Ï•}\mathrm{ψ}=E \\ \frac{1}{2m}\left(-i{\mathrm{Ä\S }}^2{∇}^2\mathrm{ψ}+i\mathrm{Ä\S }q\left[∇\left(A\mathrm{ψ}\right)+A\left(∇\mathrm{ψ}\right)\right]+q^2{A}^2\mathrm{ψ}\right)+q\mathrm{Ï•}\mathrm{ψ}=E\mathrm{ψ} \end{gathered}$$

However,

$∇\left(A\mathrm{ψ}\right)=\left(∇.A\right)\mathrm{ψ}+\left(A.∇\mathrm{ψ}\right)$

Then the expression can be written as:

$E\mathrm{ψ}=\frac{-{\mathrm{Ä\S }}^2}{2m}{∇}^2\mathrm{ψ}+\frac{i\mathrm{Ä\S }q}{2m}\left[2A\left(∇\mathrm{ψ}\right)+\left(∇.A\right)\mathrm{ψ}\right]+\left(\frac{q^2}{2m}{A}^2+q\mathrm{Ï•}\right)\mathrm{ψ}$

For electrodynamics, this is a time-independent Schroedinger equation.

Now,

$∇.A=0$

And

$A.\left(∇\mathrm{ψ}\right)=\frac{B_0}{2}\left[x\frac{∂\mathrm{ψ}}{∂y}-y\frac{∂\mathrm{ψ}}{∂x}\right]$

Also,

$A^2=\frac{B_0^2}{4}\left(x^2+y^2\right)$

Since,

$$\begin{gathered} L_z=\frac{\mathrm{Ä\S }}{i}\left(x\frac{∂}{∂y}-y\frac{∂}{∂x}\right) \\ A.\left(∇\mathrm{ψ}\right)=\frac{B_0}{2}\left(\frac{i}{\mathrm{Ä\S }}\right)L_z\mathrm{ψ} \end{gathered}$$

Thus,

$E\mathrm{ψ}=\frac{-\mathrm{Ä\S }}{2m}{∇}^2\mathrm{ψ}-\frac{q{B}_0}{2m}\left(L_z\mathrm{ψ}\right)+\left[\frac{q^2{B}_0^2}{8m}\left(x^2+qk{z}^2\right)\right]\mathrm{ψ}$

Given, $L_2\mathrm{ψ}=\overline{m}\mathrm{Ä\S }\mathrm{ψ}\overline{m}=0,Â\pm 1,Â\pm 2,...$where $\stackrel{-}{m}$is the magnetic quantum number, then,

$$\begin{gathered} \frac{-{\mathrm{Ä\S }}^2}{2m}{∇}^2\mathrm{ψ}-\frac{q{\displaystyle \stackrel{-}{m}}}{2m}{B}_0\mathrm{Ä\S }\mathrm{ψ}+\left(\frac{q^2{B}_0^2}{8m}\left(x^2+y^2\right)+qk{z}^2\right)\mathrm{ψ}=E\mathrm{ψ} \\ \left[-\frac{{\mathrm{Ä\S }}^2}{2m}{∇}^2+\frac{q^2{B}_0^2}{8m}\left(x^2+y^2\right)+qk{z}^2\right]\mathrm{ψ}=\left(E+\frac{q^2{B}_0^2\mathrm{Ä\S }\stackrel{-}{m}}{2m}\right)\mathrm{ψ} \end{gathered}$$

Let $W_1=\frac{q{B}_0}{m}$, and $W_2=\sqrt{\frac{2qk}{m}}$the above expression can be written as:

$\left(-\frac{\mathrm{Ä\S }}{2m}{∇}^2+\frac{m{w}_1^2}{8}\left(x^2+y^2\right)+\frac{1}{2}m{w}_2^2{z}^2\right)\mathrm{ψ}=\left(E+\frac{1}{2}{w}_1\mathrm{Ä\S }\stackrel{-}{m}\right)\mathrm{ψ}$

The above equation may be expressed in cylindrical coordinates as:

$-\frac{{\mathrm{Ä\S }}^2}{2m}\left[\frac{1}{r}\frac{∂}{∂r}\left(r\frac{∂\mathrm{ψ}}{∂x}\right)+\frac{1}{r^2}\frac{{∂}^2\mathrm{ψ}}{∂{\mathrm{Ï•}}^2}+\left(\frac{{∂}^2\mathrm{ψ}}{∂z^2}\right)\right]+\frac{1}{8}m{w}_1^2\left(x^2+y^2\right)\mathrm{ψ}+\frac{1}{2}m{w}_2^2{z}^2\mathrm{ψ}=\left(E+\frac{1}{2}\stackrel{-}{m}\mathrm{Ä\S }{w}_1\right)\mathrm{ψ}$

From $L_z=\frac{\mathrm{Ä\S }}{i}\frac{∂}{∂\mathrm{Ï•}}$,

$$\begin{gathered} \frac{{∂}^2\mathrm{ψ}}{∂{\mathrm{Ï•}}^2}=\frac{-1}{{\mathrm{Ä\S }}^2}{L}_z^2\mathrm{ψ} \\ \frac{{∂}^2\mathrm{ψ}}{∂{\mathrm{Ï•}}^2}=\frac{1}{{\mathrm{Ä\S }}^2}\left({\overline{m}}^2{\mathrm{Ä\S }}^2\right)\mathrm{ψ}=-{\overline{m}}^2\mathrm{ψ} \end{gathered}$$

Using the separation of the variable $\mathrm{ψ}\left(r,\mathrm{ϕ},z\right)=R\left(r\right)\mathrm{Φ}\left(\mathrm{ϕ}\right)Z\left(z\right)$above equation becomes:

$\frac{-{\mathrm{Ä\S }}^2}{2m}\left[\begin{array}{ccc}\mathrm{Φ}Z\frac{1}{r}\frac{d}{dr}& \left(r\frac{dR}{dr}\right)& \\ -\frac{{\displaystyle \stackrel{-}{m}}}{r^2}& R\mathrm{Φ}Z+& R\mathrm{Φ}\frac{d^{2}Z}{d{z}^2}\end{array}\right]+\left(\begin{array}{c}\frac{1}{8}m{w}_1^2{r}^2\\ +\frac{1}{2}m{w}_2^2{z}^2\end{array}\right)R\mathrm{Φ}Z=\left(E+\frac{1}{2}\stackrel{-}{m}\mathrm{Ä\S }{w}_1\right)R\mathrm{Φ}Z$

Dividing the above equation by $R\mathrm{Φ}Z$:

$$\begin{gathered} \frac{-\mathrm{Ä\S }}{2m}\left[\frac{1}{rR}\frac{d}{dr}\left(r\frac{dR}{dr}\right)-\frac{{{\displaystyle \stackrel{-}{m}}}^2}{r^2}+\frac{1}{Z}\frac{d^{2}Z}{d{z}^2}\right]+\frac{1}{8}m{w}_1^2{r}^2+\frac{1}{2}m{w}_2^2{z}^2=E+\frac{1}{2}\stackrel{-}{m}\mathrm{Ä\S }{w}_1 \\ \left\{\frac{-\mathrm{Ä\S }}{2m}\left[\frac{1}{rR}\frac{d}{dr}\left(r\frac{dR}{dr}\right)-\frac{{\displaystyle {\stackrel{-}{m}}^2}}{r^2}\right]+\frac{1}{8}m{w}_1^2{r}^2\right\}+\left\{\frac{-\mathrm{Ä\S }}{2m}\frac{1}{Z}\frac{d^{2}Z}{d{z}^2}+\frac{1}{2}m{w}_2^2{z}^2\right\}=E+\frac{1}{2}\stackrel{-}{m}\mathrm{Ä\S }w \end{gathered}$$

The first term is solely dependent on r , whereas the second term is only dependent on z.

$\mathrm{Let}E,R=\frac{-\mathrm{Ä\S }}{2m}\left(\frac{1}{r}\frac{d}{dr}\left(r\frac{dR}{dr}\right)-\frac{{\stackrel{-}{m}}^2}{r^2}R\right)+\frac{1}{8}m{w}_1^2{r}^{2}R$

And

$$\begin{gathered} E_{z}Z=-\frac{{\mathrm{Ä\S }}^2}{2m}\left(\frac{d^{2}Z}{d{z}^2}\right)+\frac{1}{2}m{w}_2^2{z}^{2}Z \\ E+\frac{1}{2}\stackrel{-}{m}\mathrm{Ä\S }{w}_1=E_r+E_z \\ E=E_r+E_z-\frac{1}{2}\stackrel{-}{m}\mathrm{Ä\S }{w}_1 \end{gathered}$$

The equation in z indicates a one-dimensional harmonic oscillator, and thus,

$$\begin{gathered} E_z=\left(n_2+\frac{1}{2}\right)\mathrm{Ä\S }{w}_2 \\ {n}_2=0,1,2,... \end{gathered}$$

The equation represents a two-dimensional harmonic oscillator in r is:

Let $u\left(r\right)=\sqrt{r}R$

Then,

$$\begin{gathered} R=\frac{u\left(r\right)}{\sqrt{r}} \\ \frac{dR}{dr}=\frac{1}{\sqrt{r}}\left(\frac{du}{dr}\right)-\frac{u\left(r\right)}{2r^{3/2}} \\ r\frac{dR}{dr}=\sqrt{r}\left(\frac{du}{dr}\right)-\frac{u\left(r\right)}{2\sqrt{r}} \\ \mathrm{Now}, \\ \frac{\mathrm{d}}{\mathrm{dr}}\left(\mathrm{r}\frac{\mathrm{dR}}{\mathrm{dr}}\right)=\sqrt{\mathrm{r}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}+\frac{1}{2\sqrt{\mathrm{r}}}\frac{\mathrm{du}}{\mathrm{dr}}-\frac{1}{2\sqrt{\mathrm{r}}}\frac{\mathrm{du}}{\mathrm{dr}}+\frac{\mathrm{u}\left(\mathrm{r}\right)}{4{\mathrm{r}}^{{\displaystyle \frac{3}{2}}}} \\ =\sqrt{\mathrm{r}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}+\frac{\mathrm{u}\left(\mathrm{r}\right)}{4{\mathrm{r}}^{{\displaystyle \frac{3}{2}}}} \\ \frac{1}{\mathrm{r}}\left(\frac{\mathrm{d}}{\mathrm{dr}}\left(\mathrm{r}\frac{\mathrm{dR}}{\mathrm{dr}}\right)\right)=\frac{1}{2\sqrt{\mathrm{r}}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}+\frac{\mathrm{u}\left(\mathrm{r}\right)}{4{\mathrm{r}}^{{\displaystyle \frac{5}{2}}}} \\ \mathrm{Thus},\mathrm{the}\mathrm{equation}\mathrm{becomes}: \\ -\frac{{\mathrm{Ä\S }}^2}{2\mathrm{m}}\left[\frac{1}{\sqrt{\mathrm{r}}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}+\frac{\mathrm{u}\left(\mathrm{r}\right)}{4{\mathrm{r}}^{{\displaystyle \frac{5}{2}}}}-\frac{{{\displaystyle \stackrel{-}{\mathrm{m}}}}^2}{{\mathrm{r}}^2}\frac{\mathrm{u}\left(\mathrm{r}\right)}{\sqrt{\mathrm{r}}}\right]+\frac{1}{8}{\mathrm{mw}}_1^2{\mathrm{r}}^2\frac{\mathrm{u}\left(\mathrm{r}\right)}{\sqrt{\mathrm{r}}}={\mathrm{E}}_{\mathrm{r}}\frac{\mathrm{u}\left(\mathrm{r}\right)}{\sqrt{\mathrm{r}}} \\ -\frac{{\mathrm{Ä\S }}^2}{2\mathrm{m}}\left\{\left[\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}\right]+\left(\frac{1}{4}-{\stackrel{-}{\mathrm{m}}}^2\right)\frac{\mathrm{u}\left(\mathrm{r}\right)}{{\mathrm{r}}^2}\right\}+\frac{1}{8}{\mathrm{mw}}_1^2{\mathrm{r}}^2\mathrm{u}\left(\mathrm{r}\right)=\mathrm{E},\mathrm{u}\left(\mathrm{r}\right) \end{gathered}$$

This is similar to the equation for a three-dimensional harmonic oscillator.

As,

localid="1658401704122" $$\begin{gathered} E=\left(j_{max}+/+\frac{3}{2}\right)\mathrm{Ä\S } \\ W{\left(l+\frac{1}{2}\right)}^2={\stackrel{-}{m}}^2 \\ \left|\stackrel{-}{m}\right|-\frac{1}{2}=l \\ {E}_r=\left(j_{max}+\left|\stackrel{-}{m}\right|+1\right)\frac{\mathrm{Ä\S }{W}_1}{2}\mathrm{where}{j}_{max}=0,2,4... \\ \mathrm{Such}\mathrm{that}{\mathrm{n}}_1=\frac{{\mathrm{j}}_{\max }}{2}\mathrm{for}\stackrel{-}{\mathrm{m}}â‰\yen 0. \\ \mathrm{And}{\mathrm{n}}_1=\frac{{\mathrm{j}}_{\max }}{2}-\mathrm{for}\stackrel{-}{\mathrm{m}}≤0. \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{allowed}\mathrm{energy}\mathrm{is}\left({\mathrm{n}}_1+\frac{1}{2}\right){\mathrm{Ä\S ·É}}_1+\left({\mathrm{n}}_2+\frac{1}{2}\right){\mathrm{Ä\S ·É}}_2. \end{gathered}$$