Question

[Refer to. Problem 4.59for background.] Suppose $\mathit{A}\mathbf{=}\frac{{\mathbf{B}}_{\mathbf{0}}}{\mathbf{2}}\left(X\hat{\varnothing }-y\hat{I}\right)$ and$\mathit{\phi }\mathbf{=}\mathit{K}{\mathit{z}}^{\mathbf{2}}$, where ${\mathit{B}}_{\mathbf{0}}$ and Kare constants.

(a) Find the fields E and B.

(b) Find the allowed energies, for a particle of mass m and charge q , in these fields, Answer: $$\begin{gathered} \mathit{E}\left(n_1,n_2\right)\mathbf{=}\left(n_1+\frac{1}{2}\right)\mathit{ħ}{\mathit{\omega }}_{\mathbf{1}}\mathbf{+}\left(n_2+\frac{1}{2}\right)\mathit{ħ}{\mathit{\omega }}_{\mathbf{2}}\mathbf{}\mathbf{,}\left(n_1,n_2=0,1,2,...\right)\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e} \\ {\mathit{\omega }}_{\mathbf{1}}\mathbf{\equiv }\mathit{q}{\mathit{B}}_{\mathbf{0}}\mathbf{/}\mathit{m}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}\mathbf{}{\mathbf{\omega }}_{\mathbf{2}}\mathbf{\equiv }\sqrt{\mathbf{2}\mathbf{qK}\mathbf{/}\mathbf{m}}\mathbf{}\mathbf{}\mathbf{.} \end{gathered}$$ Comment: If K=0this is the quantum analog to cyclotron motion;${\mathit{\omega }}_{\mathbf{1}}$ is the classical cyclotron frequency, and it's a free particle in the z direction. The allowed energies,$\left(n_1+\frac{1}{2}\right)\mathit{ħ}{\mathit{\omega }}_{\mathbf{1}}$, are called Landau Levels.

Short answer

a) The values of fields are $\overrightarrow{E}=-2kz\hat{k}$and the value of $\overrightarrow{B}=B_0\hat{k}$.

b) The allowed energies are $\left(n_1+\frac{1}{2}\right)ħw_1+\left(n_2+\frac{1}{2}\right)ħw_2$.

Step-by-step solution

Definition of the cyclotron

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

The fields E and B 

(a)

Consider that,

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

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

$\overrightarrow{E}=-\nabla \varphi -\frac{\partial A}{\partial t}$

Substituting the values, and we get,

$$\begin{gathered} \overrightarrow{E}=-\nabla \varphi -\frac{\partial A}{\partial t} \\ =-\left(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z}\right)\left(k{z}^2\right)-0 \\ =-2kz\hat{k} \\ \mathrm{And} \\ \overrightarrow{\mathrm{B}}=\nabla \times \mathrm{A} \end{gathered}$$

Substituting the values, and we get,

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

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

The allowed energies

(b)

According to the question,

$H\psi =E\psi$

For time-independent potentials,

$$\begin{gathered} \frac{1}{2m}\left(-iħ\nabla -qA\right)\left(-iħ\nabla -qA\right)\psi +q\varphi \psi =E \\ \frac{1}{2m}\left(-i{ħ}^2{\nabla }^2\psi +iħq\left[\nabla \left(A\psi \right)+A\left(\nabla \psi \right)\right]+q^2{A}^2\psi \right)+q\varphi \psi =E\psi \end{gathered}$$

However,

$\nabla \left(A\psi \right)=\left(\nabla .A\right)\psi +\left(A.\nabla \psi \right)$

Then the expression can be written as:

$E\psi =\frac{-{ħ}^2}{2m}{\nabla }^2\psi +\frac{iħq}{2m}\left[2A\left(\nabla \psi \right)+\left(\nabla .A\right)\psi \right]+\left(\frac{q^2}{2m}{A}^2+q\varphi \right)\psi$

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

Now,

$\nabla .A=0$

And

$A.\left(\nabla \psi \right)=\frac{B_0}{2}\left[x\frac{\partial \psi }{\partial y}-y\frac{\partial \psi }{\partial x}\right]$

Also,

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

Since,

$$\begin{gathered} L_z=\frac{ħ}{i}\left(x\frac{\partial }{\partial y}-y\frac{\partial }{\partial x}\right) \\ A.\left(\nabla \psi \right)=\frac{B_0}{2}\left(\frac{i}{ħ}\right)L_z\psi \end{gathered}$$

Thus,

$E\psi =\frac{-ħ}{2m}{\nabla }^2\psi -\frac{q{B}_0}{2m}\left(L_z\psi \right)+\left[\frac{q^2{B}_0^2}{8m}\left(x^2+qk{z}^2\right)\right]\psi$

Given, $L_2\psi =\overline{m}ħ\psi \overline{m}=0,\pm 1,\pm 2,...$where $\stackrel{-}{m}$is the magnetic quantum number, then,

$$\begin{gathered} \frac{-{ħ}^2}{2m}{\nabla }^2\psi -\frac{q{\displaystyle \stackrel{-}{m}}}{2m}{B}_0ħ\psi +\left(\frac{q^2{B}_0^2}{8m}\left(x^2+y^2\right)+qk{z}^2\right)\psi =E\psi \\ \left[-\frac{{ħ}^2}{2m}{\nabla }^2+\frac{q^2{B}_0^2}{8m}\left(x^2+y^2\right)+qk{z}^2\right]\psi =\left(E+\frac{q^2{B}_0^2ħ\stackrel{-}{m}}{2m}\right)\psi \end{gathered}$$

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

$\left(-\frac{ħ}{2m}{\nabla }^2+\frac{m{w}_1^2}{8}\left(x^2+y^2\right)+\frac{1}{2}m{w}_2^2{z}^2\right)\psi =\left(E+\frac{1}{2}{w}_1ħ\stackrel{-}{m}\right)\psi$

The above equation may be expressed in cylindrical coordinates as:

$-\frac{{ħ}^2}{2m}\left[\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial \psi }{\partial x}\right)+\frac{1}{r^2}\frac{{\partial }^2\psi }{\partial {\varphi }^2}+\left(\frac{{\partial }^2\psi }{\partial z^2}\right)\right]+\frac{1}{8}m{w}_1^2\left(x^2+y^2\right)\psi +\frac{1}{2}m{w}_2^2{z}^2\psi =\left(E+\frac{1}{2}\stackrel{-}{m}ħw_1\right)\psi$

From $L_z=\frac{ħ}{i}\frac{\partial }{\partial \varphi }$,

$$\begin{gathered} \frac{{\partial }^2\psi }{\partial {\varphi }^2}=\frac{-1}{{ħ}^2}{L}_z^2\psi \\ \frac{{\partial }^2\psi }{\partial {\varphi }^2}=\frac{1}{{ħ}^2}\left({\overline{m}}^2{ħ}^2\right)\psi =-{\overline{m}}^2\psi \end{gathered}$$

Using the separation of the variable $\psi \left(r,\varphi ,z\right)=R\left(r\right)\Phi \left(\varphi \right)Z\left(z\right)$above equation becomes:

$\frac{-{ħ}^2}{2m}\left[\begin{array}{ccc}\Phi Z\frac{1}{r}\frac{d}{dr}& \left(r\frac{dR}{dr}\right)& \\ -\frac{{\displaystyle \stackrel{-}{m}}}{r^2}& R\Phi Z+& R\Phi \frac{d^{2}Z}{d{z}^2}\end{array}\right]+\left(\begin{array}{c}\frac{1}{8}m{w}_1^2{r}^2\\ +\frac{1}{2}m{w}_2^2{z}^2\end{array}\right)R\Phi Z=\left(E+\frac{1}{2}\stackrel{-}{m}ħw_1\right)R\Phi Z$

Dividing the above equation by $R\Phi Z$:

$$\begin{gathered} \frac{-ħ}{2m}\left[\frac{1}{rR}\frac{d}{dr}\left(r\frac{dR}{dr}\right)-\frac{{{\displaystyle \stackrel{-}{m}}}^2}{r^2}+\frac{1}{Z}\frac{d^{2}Z}{d{z}^2}\right]+\frac{1}{8}m{w}_1^2{r}^2+\frac{1}{2}m{w}_2^2{z}^2=E+\frac{1}{2}\stackrel{-}{m}ħw_1 \\ \left\{\frac{-ħ}{2m}\left[\frac{1}{rR}\frac{d}{dr}\left(r\frac{dR}{dr}\right)-\frac{{\displaystyle {\stackrel{-}{m}}^2}}{r^2}\right]+\frac{1}{8}m{w}_1^2{r}^2\right\}+\left\{\frac{-ħ}{2m}\frac{1}{Z}\frac{d^{2}Z}{d{z}^2}+\frac{1}{2}m{w}_2^2{z}^2\right\}=E+\frac{1}{2}\stackrel{-}{m}ħw \end{gathered}$$

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

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

And

$$\begin{gathered} E_{z}Z=-\frac{{ħ}^2}{2m}\left(\frac{d^{2}Z}{d{z}^2}\right)+\frac{1}{2}m{w}_2^2{z}^{2}Z \\ E+\frac{1}{2}\stackrel{-}{m}ħw_1=E_r+E_z \\ E=E_r+E_z-\frac{1}{2}\stackrel{-}{m}ħw_1 \end{gathered}$$

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

$$\begin{gathered} E_z=\left(n_2+\frac{1}{2}\right)ħw_2 \\ {n}_2=0,1,2,... \end{gathered}$$

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{ħ}}^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{ħ}}^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" $$\begin{gathered} E=\left(j_{max}+/+\frac{3}{2}\right)ħ \\ W{\left(l+\frac{1}{2}\right)}^2={\stackrel{-}{m}}^2 \\ \left|\stackrel{-}{m}\right|-\frac{1}{2}=l \\ {E}_r=\left(j_{max}+\left|\stackrel{-}{m}\right|+1\right)\frac{ħW_1}{2}\mathrm{where}{j}_{max}=0,2,4... \\ \mathrm{Such}\mathrm{that}{\mathrm{n}}_1=\frac{{\mathrm{j}}_{\max }}{2}\mathrm{for}\stackrel{-}{\mathrm{m}}\ge 0. \\ \mathrm{And}{\mathrm{n}}_1=\frac{{\mathrm{j}}_{\max }}{2}-\mathrm{for}\stackrel{-}{\mathrm{m}}\le 0. \\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{allowed}\mathrm{energy}\mathrm{is}\left({\mathrm{n}}_1+\frac{1}{2}\right){\mathrm{ħw}}_1+\left({\mathrm{n}}_2+\frac{1}{2}\right){\mathrm{ħw}}_2. \end{gathered}$$