Question

In classical electrodynamics the force on a particle of charge q

moving with velocity through electric and magnetic fields E and B is given

by the Lorentz force law:$\mathbf{F}\mathbf{=}\mathbf{q}\left(E+v\times B\right)$

This force cannot be expressed as the gradient of a scalar potential energy

function, and therefore the Schrödinger equation in its original form (Equation 1.1)

cannot accommodate it. But in the more sophisticated form $\mathit{i}\sout{\mathbf{h}}\frac{\mathbf{\partial }\mathbf{\psi }}{\mathbf{\partial }\mathbf{t}}\mathbf{=}\mathit{H}\mathit{\psi }$

there is no problem; the classical Hamiltonian is$\mathbf{H}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{m}}{\left(p-\mathrm{qA}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{q\psi }$where A

is the vector potential$\left(B=\nabla \times A\right)$and $\mathbf{\psi }$is the scalar potential $\left(E=-\nabla \psi -\partial A/\partial t\right)$,

so the Schrödinger

equation (making the canonical substitution$\mathbf{p}\mathbf{\to }\left(\sout{h}/i\right)\mathbf{\nabla }\mathbf{)}$becomes$\mathbf{i}\sout{\mathbf{h}}\frac{\mathbf{\partial }\mathbf{\psi }}{\mathbf{\partial }\mathbf{t}}\mathbf{=}\left[\frac{1}{2m}{\left(\frac{\sout{h}}{i}\nabla -\mathrm{qA}\right)}^2+\mathrm{q\psi }\right]\mathbf{\psi }$

(a) Show that $\frac{\mathbf{d}\mathbf{<}\mathbf{r}\mathbf{>}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{m}}\mathbf{<}\left(p-\mathrm{qA}\right)\mathbf{>}$

(b) As always (see Equation ) we identify$\mathbf{d}\mathbf{<}\mathbf{r}\mathbf{>}\mathbf{/}\mathbf{dt}$with$\mathbf{<}\mathbf{v}\mathbf{>}$. Show that

$\mathbf{m}\frac{\mathbf{d}\mathbf{<}\mathbf{v}\mathbf{>}}{\mathbf{dt}}\mathbf{=}\mathbf{q}\mathbf{<}\mathbf{E}\mathbf{>}\mathbf{+}\frac{\mathbf{q}}{\mathbf{2}\mathbf{m}}\mathbf{<}\left(p\times B-B\times p\right)\mathbf{>}\mathbf{-}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{m}}\mathbf{<}\left(A\times B\right)\mathbf{>}$

(c) In particular, if the fields and are uniform over the volume of the wave packet,

show that$\mathbf{m}\frac{\mathbf{d}\mathbf{<}\mathbf{v}\mathbf{>}}{\mathbf{dt}}\mathbf{=}\mathbf{q}\left(E+<V>\times B\right)$so the expectation value of $\left(v\right)$moves

according to the Lorentz force law, as we would expect from Ehrenfest's theorem.

Short answer

a) Hence it is proved that$\frac{\mathrm{d}<\mathrm{r}>}{\mathrm{dt}}=\frac{1}{\mathrm{m}}<(\mathrm{p}-\mathrm{qA})>$.

b) Hence it is proved that$\mathrm{m}\frac{\mathrm{d}<\mathrm{v}>}{\mathrm{dt}}=\mathrm{q}<\mathrm{E}>+\frac{\mathrm{q}}{2\mathrm{m}}<(\mathrm{p}\times \mathrm{B}-\mathrm{B}\times \mathrm{p})>-\frac{{\mathrm{q}}^2}{\mathrm{m}}<(\mathrm{A}\times \mathrm{B})>$.

c) Hence it is proved that $\mathrm{m}\frac{\mathrm{d}<\mathrm{v}>}{\mathrm{dt}}=\mathrm{q}(\mathrm{E}+<\mathrm{V}>\times \mathrm{B})$.

Step-by-step solution

Definition of Lorentz force.

Lorentz force is described as the result of electromagnetic fields combining magnetic and electric forces on a point charge.

Show that d<r>dt=1m<(p-qA)>

(a)

The following equation expresses the energy uncertainty principle.

In the above expression, the Hamiltonianis expressed as follows:

$$\begin{gathered} \mathrm{H}=\frac{1}{2\mathrm{m}}{\left(\mathrm{p}-\mathrm{qA}\right)}^2+\mathrm{q\varphi } \\ =\frac{1}{2\mathrm{m}}\left[\left(\mathrm{p}-\mathrm{qA}\right)\left(\mathrm{p}-\mathrm{qA}\right)\right]+\mathrm{q\varphi } \\ =\frac{1}{2\mathrm{m}}\left[{\mathrm{p}}^2-\mathrm{q}\left(\mathrm{p}.\mathrm{A}+\mathrm{A}.\mathrm{p}\right)+{\mathrm{q}}^2{\mathrm{A}}^2\right]+\mathrm{q\varphi } \end{gathered}$$

Using the given formula for Hamiltonian, obtain the commutator connection between Hamiltonian and position as follows:

$\left[\mathrm{H},\mathrm{x}\right]=\frac{1}{2\mathrm{m}}\left[{\mathrm{p}}^2,\mathrm{x}\right]-\frac{\mathrm{q}}{2\mathrm{m}}\left[\left(\mathrm{p}.\mathrm{A}+\mathrm{A}.\mathrm{p}\right),\mathrm{x}\right].................\left(1\right)$

Obtain the commutator relation$\left[p^2,x\right]$as follows:

$$\begin{gathered} \left[p^2,x\right]=\left[p_x^2+p_y^2+p_z^2,x\right]\left[\therefore \left[p_y^2,x\right]=\left[p_z^2,x\right]=0\right] \\ \left[p^2,x\right]=\left[p_x^2,x\right] \end{gathered}$$

Obtain the momentum in x direction and position commutator relation.

$$\begin{gathered} \left[p_x^2,x\right]=p_x\left[p_x,x\right]+\left[p_x,x\right]p_x \\ =p_x\left[-i\sout{h}\right]+\left[-i\sout{h}\right]p_x.................\left(2\right) \\ =2i\sout{h}{p}_x \end{gathered}$$

Obtain the commutator relationlocalid="1656073559313" $\left[\mathrm{p}.\mathrm{A},\mathrm{x}\right]$as follows:

$$\begin{gathered} \left[\mathrm{p}.\mathrm{A},\mathrm{x}\right]=\left[{\mathrm{p}}_{\mathrm{x}},{\mathrm{A}}_{\mathrm{x}}+{\mathrm{p}}_{\mathrm{y}},{\mathrm{A}}_{\mathrm{y}}+{\mathrm{p}}_{\mathrm{z}}{\mathrm{A}}_{\mathrm{z}},\mathrm{x}\right] \\ =\left[{\mathrm{p}}_{\mathrm{x}}{\mathrm{A}}_{\mathrm{x},}\mathrm{x}\right] \\ ={\mathrm{p}}_{\mathrm{x}}\left[{\mathrm{A}}_{\mathrm{x},}\mathrm{x}\right]+\left[{\mathrm{p}}_{\mathrm{x}},\mathrm{x}\right]{\mathrm{A}}_{\mathrm{x}} \\ =-\mathrm{i}\sout{\mathrm{h}}{\mathrm{A}}_{\mathrm{X}}.......................\left(3\right) \end{gathered}$$

Obtain the commutator relationas follows:

$$\begin{gathered} \left[\mathrm{A}.\mathrm{p},\mathrm{x}\right]=\left[{\mathrm{A}}_{\mathrm{x}},{\mathrm{p}}_{\mathrm{x}}+{\mathrm{A}}_{\mathrm{y}},{\mathrm{p}}_{\mathrm{y}}+{\mathrm{A}}_{\mathrm{z}}{\mathrm{p}}_{\mathrm{z}},\mathrm{x}\right] \\ =\left[{\mathrm{A}}_{\mathrm{x}}{\mathrm{p}}_{\mathrm{x},}\mathrm{x}\right] \\ ={\mathrm{A}}_{\mathrm{x}}\left[{\mathrm{p}}_{\mathrm{x},}\mathrm{x}\right]+\left[{\mathrm{A}}_{\mathrm{x}},\mathrm{x}\right]{\mathrm{p}}_{\mathrm{x}} \\ =-\mathrm{i}\sout{\mathrm{h}}{\mathrm{A}}_{\mathrm{X}}.......................\left(4\right) \end{gathered}$$

Equations $\left(2\right)$, $\left(3\right)$, and $\left(4\right)$should be substituted into equation and solve for

$\left[\mathrm{H},\mathrm{x}\right]$

$$\begin{gathered} \left[\mathrm{H},\mathrm{x}\right]=\frac{1}{2\mathrm{m}}\left(-2\sout{\mathrm{h}}{\mathrm{p}}_{\mathrm{x}}\right)-\frac{\mathrm{q}}{2\mathrm{m}}\left(-2\sout{\mathrm{h}}{\mathrm{p}}_{\mathrm{x}}\right) \\ =-\frac{\mathrm{i}\sout{\mathrm{h}}}{\mathrm{m}}{\mathrm{p}}_{\mathrm{x}}+\frac{\mathrm{i}\sout{\mathrm{h}}{\mathrm{qA}}_{\mathrm{x}}}{\mathrm{m}} \\ =-\frac{\mathrm{i}\sout{\mathrm{h}}}{\mathrm{m}}\left({\mathrm{p}}_{\mathrm{x}}-{\mathrm{qA}}_{\mathrm{x}}\right) \end{gathered}$$

The expression for $\left[\mathrm{H},\mathrm{r}\right]$is derived by generalising the following equation.

$\left[\mathrm{H},\mathrm{r}\right]=\frac{\mathrm{i}\sout{\mathrm{h}}}{\mathrm{m}}\left(\mathrm{p}-\mathrm{qA}\right)$

Now from equation

Show thatmd<v>dt=q<E>+q2m<(p×B-B×p)>-q2m<(A×B)>

(b)

The following is the equation for the predicted value of velocity:

For the situation of $v$, use the energy uncertainty principle.

From equation $\left(5\right)$,

$\frac{\partial \mathrm{v}}{\partial \mathrm{t}}=\frac{-\mathrm{q}}{\mathrm{m}}\frac{\partial \mathrm{A}}{\partial \mathrm{t}}..........\left(7\right)$

Obtain the following expression for $\left[\mathrm{H},\mathrm{v}\right]$:

$$\begin{gathered} \left[\mathrm{H},\mathrm{v}\right]=\left[\frac{1}{2}{\mathrm{mv}}^2+\mathrm{q\varphi },\mathrm{v}\right] \\ =\frac{\mathrm{m}}{2}\left[{\mathrm{v}}^2,\mathrm{v}\right]+\mathrm{q}\left[\mathrm{\varphi },\mathrm{v}\right] \end{gathered}$$

The commutator relation of $\left[\mathrm{\varphi },\mathrm{v}\right]$as follows:

$\left[\mathrm{\varphi },\mathrm{v}\right]=\frac{1}{\mathrm{m}}\left[\mathrm{\varphi },\mathrm{p}\right]$

And the commutator relation $\left[\mathrm{\varphi },{\mathrm{p}}_{\mathrm{x}}\right]$as follows:

$\left[\mathrm{\varphi },{\mathrm{p}}_{\mathrm{x}}\right]=\mathrm{i}\sout{\mathrm{h}}\frac{\partial \mathrm{\varphi }}{\partial \mathrm{x}}$

Generalizing, this gives

$$\begin{gathered} \left[\mathrm{\varphi },\mathrm{p}\right]=\mathrm{i}\sout{\mathrm{h}}∆\mathrm{\varphi } \\ \left[\mathrm{\varphi },\mathrm{p}\right]=\mathrm{i}\sout{\mathrm{h}}∆\mathrm{\varphi }............\left(8\right) \end{gathered}$$

Now solve for commutator $\left({\mathrm{v}}^2,{\mathrm{v}}_{\mathrm{x}}\right)$as follows:

$$\begin{gathered} \left[{\mathrm{v}}^2,{\mathrm{v}}_{\mathrm{x}}\right]=\left[{\mathrm{v}}_{\mathrm{x}}^2,{\mathrm{v}}_{\mathrm{x}}\right]+\left[{\mathrm{v}}_{\mathrm{y}}^2,{\mathrm{v}}_{\mathrm{x}}\right]+\left[{\mathrm{v}}_{\mathrm{z}}^2,{\mathrm{v}}_{\mathrm{z}}\right] \\ =0+\left[{\mathrm{v}}_{\mathrm{y}}^2,{\mathrm{v}}_{\mathrm{x}}\right]+\left[{\mathrm{v}}_{\mathrm{z}}^2,{\mathrm{v}}_{\mathrm{x}}\right] \\ ={\mathrm{v}}_{\mathrm{y}}\left[{\mathrm{v}}_{\mathrm{y}},{\mathrm{v}}_{\mathrm{x}}\right]+\left[{\mathrm{v}}_{\mathrm{y}},{\mathrm{v}}_{\mathrm{x}}\right] \\ ={\mathrm{v}}_{\mathrm{y}}\left[{\mathrm{v}}_{\mathrm{y}},{\mathrm{v}}_{\mathrm{x}}\right]+\left[{\mathrm{v}}_{\mathrm{y}},{\mathrm{v}}_{\mathrm{x}}\right]{\mathrm{v}}_{\mathrm{y}},{\mathrm{v}}_{\mathrm{z}}\left[{\mathrm{v}}_{\mathrm{z}},{\mathrm{v}}_{\mathrm{x}}\right]+\left[{\mathrm{v}}_{\mathrm{z}},{\mathrm{v}}_{\mathrm{x}}\right]{\mathrm{v}}_{\mathrm{z}} \end{gathered}$$

Now use equationto (5) obtain the expression of $\left[v_y,v_x\right]$:

$$\begin{gathered} \left[v_y,v_x\right]=\left[\frac{1}{m}\left(p_y-q{A}_y\right),\frac{1}{m}\left(p_x-q{A}_x\right)\right] \\ =\frac{1}{m^2}\left\{\left[p_y,p_x\right]-\left[p_y,q{A}_x\right]-\left[q{A}_y,p_x\right]+q^2\left[A_y,A_x\right]\right\} \\ =\frac{1}{m^2}\left[-q\left\{\left[p_y,A_x\right]+\left[A_y,p_x\right]\right\}\right]=\frac{-q}{m^2}\left\{-\sout{h\frac{\partial A_x}{\partial y}}+\left(\sout{h\frac{\partial A_y}{\partial x}}\right)\right\} \\ =\frac{-q}{m^2}\left\{\sout{h\frac{\partial A_y}{\partial x}}-\sout{h\frac{\partial A_x}{\partial y}}\right\} \\ =\frac{-\sout{h}q}{m^2}{\left[\nabla \times A\right]}_Z \\ =\frac{-\sout{h}q}{m^2}{B}_z \end{gathered}$$

Obtain the commutator relation role="math" localid="1656133338816" $\left[v_z,v_x\right]$as follows:

$$\begin{gathered} \left[v_z,v_x\right]=\left[\frac{1}{m}\left(p_z-q{A}_z\right),\frac{1}{m}\left(p_x-q{A}_x\right)\right] \\ =\frac{1}{m^2}\left\{\left[p_z,p_x\right]-q\left[p_z,A_x\right]-\left[q{A}_z,p_x\right]+q^2\left[A_z,A_x\right]\right\} \\ =\frac{1}{m^2}\left\{-q\left\{\left[p_z,A_x\right]+\left[A_z,p_x\right]\right\}\right\} \\ =\frac{-q}{m^2}\left\{\sout{h\frac{\partial Ax}{\partial z}}+\left(\sout{h\frac{\partial Az}{\partial x}}\right)\right\} \\ =\frac{-\sout{h}q}{m^2}\left[\frac{\partial A_x}{\partial z}-\frac{\partial A_x}{\partial x}\right] \\ =\frac{-\sout{h}q}{m^2}\left[-{\left(\nabla \times A\right)}_y\right] \\ =\frac{-\sout{h}q}{m^2}{B}_y \end{gathered}$$

Obtain the commutator relation $\left[v^2,v_x\right]$as follows:

$$\begin{gathered} \left[{\mathrm{v}}^2,{\mathrm{v}}_{\mathrm{x}}\right]={\mathrm{v}}_{\mathrm{y}}\left(\frac{-\sout{\mathrm{h}}\mathrm{q}}{{\mathrm{m}}^2}{\mathrm{B}}_{\mathrm{Z}}\right)+\left(\frac{-\sout{\mathrm{h}}\mathrm{q}}{{\mathrm{m}}^2}\right){\mathrm{B}}_{\mathrm{z}}{\mathrm{v}}_{\mathrm{y}}+{\mathrm{v}}_{\mathrm{z}}\left(\frac{-\sout{\mathrm{h}}\mathrm{q}}{{\mathrm{m}}^2}{\mathrm{B}}_{\mathrm{y}}\right)+\left(\frac{-\sout{\mathrm{h}}\mathrm{q}}{{\mathrm{m}}^2}{\mathrm{B}}_{\mathrm{y}}\right){\mathrm{v}}_{\mathrm{z}} \\ =\frac{-\sout{\mathrm{h}}\mathrm{q}}{{\mathrm{m}}^2}\left[-{\mathrm{v}}_{\mathrm{y}}{\mathrm{B}}_{\mathrm{Z}}-{\mathrm{B}}_{\mathrm{Z}}{\mathrm{v}}_{\mathrm{y}}+{\mathrm{v}}_{\mathrm{z}}{\mathrm{B}}_{\mathrm{y}}+{\mathrm{B}}_{\mathrm{y}}{\mathrm{v}}_{\mathrm{z}}\right] \\ =\frac{-\sout{\mathrm{h}}\mathrm{q}}{{\mathrm{m}}^2}\left[-\left({\mathrm{v}}_{\mathrm{y}}{\mathrm{B}}_{\mathrm{Z}}-{\mathrm{v}}_{\mathrm{Z}}{\mathrm{B}}_{\mathrm{y}}\right)+\left({\mathrm{B}}_{\mathrm{y}}{\mathrm{v}}_{\mathrm{Z}}-{\mathrm{B}}_{\mathrm{z}}{\mathrm{v}}_{\mathrm{y}}\right)\right] \\ =\frac{-\sout{\mathrm{h}}\mathrm{q}}{{\mathrm{m}}^2}\left[-{\left(\mathrm{v}\times \mathrm{B}\right)}_{\mathrm{x}}+{\left(\mathrm{B}\times \mathrm{v}\right)}_{\mathrm{x}}\right] \end{gathered}$$

Now generalize the above equation to obtain the expression $\left({\mathrm{v}}^2,\mathrm{v}\right)$as follows:

$\left[{\mathrm{v}}^2,\mathrm{v}\right]=\frac{\mathrm{i}\sout{\mathrm{h}}\mathrm{q}}{{\mathrm{m}}^2}\left[\left(\mathrm{B}\times \mathrm{v}\right)-\left(\mathrm{v}-\mathrm{B}\right)\right]..........\left(9\right)$

Now using equation(6),

The above expression may be expressed as, using equations (7), (8), and (9).

Now calculate $\left(\mathrm{v}\times \mathrm{B}\right)-\left(\mathrm{B}-\mathrm{v}\right)$as follows:0

$$\begin{gathered} \left(\mathrm{v}\times \mathrm{B}\right)-\left(\mathrm{B}-\mathrm{v}\right)=\frac{1}{m}\left[\left\{\left(p-qA\right)\times B\right\}-\left\{B\times \left(P-qA\right)\right\}\right] \\ =\frac{1}{m}\left[\left(p\times B\right)-\left(B\times p\right)\right]-\frac{q}{m}\left[\left(A\times B\right)-\left(B\times A\right)\right] \end{gathered}$$

Using $A\times B=-\left(B\times A\right)$rewrite the above equation as follows:

Show thatmd<v>dt=q(E+<V>×B)

(c)

Use equation$\left(10\right)$,

$\mathrm{m}\frac{\mathrm{d}<\mathrm{v}>}{\mathrm{dt}}=\frac{\mathrm{q}}{2}\left(<(\mathrm{v}\times \mathrm{B}\right)-(\mathrm{B}\times \mathrm{v})>)+\mathrm{q}<\mathrm{E}>$

Consider that$<\mathrm{E}>=\mathrm{E}$and$<\mathrm{v}\times \mathrm{B}>=<\mathrm{v}>\times \mathrm{B}$and rewrite the above equation as follows:

$\mathrm{m}\frac{\mathrm{d}}{\mathrm{dt}}<\mathrm{v}>=\mathrm{q}\left[\mathrm{E}+<\mathrm{v}>\times \mathrm{B}\right]$