# Charged Particle in a Magnetic Field

*Michael Fowler 1/16/08*

### Introduction

Classically, the force on a charged particle in electric and magnetic fields is given by the Lorentz force law:

$$\overrightarrow{F}=q\left(\overrightarrow{E}+\frac{\overrightarrow{v}\times \overrightarrow{B}}{c}\right)$$

This velocity-dependent force is quite different from the conservative forces from potentials that we have dealt with so far, and the recipe for going from classical to quantum mechanics$\u2014$replacing momenta with the appropriate derivative operators$\u2014$has to be carried out with more care. We begin by demonstrating how the Lorentz force law arises classically in the Lagrangian and Hamiltonian formulations.

### Laws of Classical Mechanics

Recall first (or look it up in Shankar, Chapter 2) that the
Principle of Least Action leads to the Euler-Lagrange equations for the
Lagrangian *L*:

$$\frac{d}{dt}\left(\frac{\partial L({q}_{i},{\dot{q}}_{i})}{\partial {\dot{q}}_{i}}\right)-\frac{\partial L({q}_{i},{\dot{q}}_{i})}{\partial {q}_{i}}=0,\text{}\text{\hspace{0.17em}}{q}_{i},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\dot{q}}_{i}\text{beingcoordinatesandvelocities}\text{.}$$

The canonical momentum *p _{i}* is defined by
the equation

$${p}_{i}=\frac{\partial L}{\partial {\dot{q}}_{i}}$$

and the Hamiltonian is defined by performing a Legendre transformation of the Lagrangian:

$H({q}_{i},{p}_{i})={\displaystyle \sum {p}_{i}{\dot{q}}_{i}-L({q}_{i},{\dot{q}}_{i})}$

It is straightforward to check that the equations of motion can be written:

$${\dot{q}}_{i}=\frac{\partial H}{\partial {p}_{i}},\text{}{\dot{p}}_{i}=-\frac{\partial H}{\partial {q}_{i}}$$

These are known as *Hamilton’s
Equations*. Note that if the
Hamiltonian is independent of a particular coordinate *q _{i}*, the
corresponding momentum

*p*remains constant. (Such a coordinate is termed

_{i}*cyclic*, because the most common example is an angular coordinate in a spherically symmetric Hamiltonian, where angular momentum remains constant.)

For the conservative forces we have been considering so far,
*L* = *T* − *V*,
*H* = *T* + *V*, with *T* the kinetic energy, *V*
the potential energy.

### Poisson Brackets

Any dynamical variable *f* in the system is some
function of the *q _{i}*’s and

*p*’s and (assuming it does not depend explicitly on time) its development is given by:

_{i}$$\frac{d}{dt}f({q}_{i},{p}_{i})=\frac{\partial f}{\partial {q}_{i}}{\dot{q}}_{i}+\frac{\partial f}{\partial {p}_{i}}{\dot{p}}_{i}=\frac{\partial f}{\partial {q}_{i}}\frac{\partial H}{\partial {p}_{i}}-\frac{\partial f}{\partial {p}_{i}}\frac{\partial H}{\partial {q}_{i}}=\{f,H\}.$$

The curly brackets are called *Poisson Brackets*, and
are defined for any dynamical variables as:

$$\{A,B\}=\frac{\partial A}{\partial {q}_{i}}\frac{\partial B}{\partial {p}_{i}}-\frac{\partial A}{\partial {p}_{i}}\frac{\partial B}{\partial {q}_{i}}.$$

We have shown from

It is easy to check that for the coordinates and canonical momenta,

$\{{q}_{i},{q}_{j}\}=0=\{{p}_{i},{p}_{j}\},\text{}\{{q}_{i},{p}_{j}\}={\delta}_{ij}.$

This was the classical mathematical structure that led Dirac to link up classical and quantum mechanics: he realized that the Poisson brackets were the classical version of the commutators, so a classical canonical momentum must correspond to the quantum differential operator in the corresponding coordinate.

### Particle in a Magnetic Field

The Lorentz force is velocity dependent, so cannot be just the gradient of some potential. Nevertheless, the classical particle path is still given by the Principle of Least Action. The electric and magnetic fields can be written in terms of a scalar and a vector potential:

$$\overrightarrow{B}=\overrightarrow{\nabla}\times \overrightarrow{A},\text{}\overrightarrow{E}=-\overrightarrow{\nabla}\phi -\frac{1}{c}\frac{\partial \overrightarrow{A}}{\partial t}.$$

The right Lagrangian turns out to be:

$$L={\scriptscriptstyle \frac{1}{2}}m{\overrightarrow{v}}^{2}-q\phi +\frac{q}{c}\overrightarrow{v}.\overrightarrow{A}.$$

(*Note*: if you’re
familiar with Relativity, the interaction term here looks less arbitrary: the
relativistic version would have the relativistically invariant $\left(q/c\right){\displaystyle \int {A}^{\mu}d{x}_{\mu}}$ added to the action integral, where the four-potential
${A}_{\mu}=\left(\overrightarrow{A},\phi \right)$ and $d{x}_{\mu}=\left(d{x}_{1},d{x}_{2},d{x}_{3},cdt\right)$.
This is the simplest possible invariant
interaction between the electromagnetic field and the particle’s four-velocity.
Then in the nonrelativistic limit, $\left(q/c\right){\displaystyle \int {A}^{\mu}d{x}_{\mu}}$ just becomes $\int q\left(\overrightarrow{v}.\overrightarrow{A}/c-\phi \right)}\text{\hspace{0.17em}}dt.$ )

The derivation of the Lorentz force from this Lagrangian is
given by Shankar on page 84. We give the
(equivalent) derivation from the

Note that for zero vector potential, the Lagrangian has the
usual *T* − *V* form.

For this one-particle problem, the general coordinates *q*_{i} are just the Cartesian
co-ordinates ${x}_{i}=\left({x}_{1},{x}_{2},{x}_{3}\right)$,
the position of the particle, and the ${\dot{q}}_{i}$ are the three components ${\dot{x}}_{i}={v}_{i}$ of the particle’s velocity.

The
important *new* point is that the
canonical momentum

$${p}_{i}=\frac{\partial L}{\partial {\dot{q}}_{i}}=\frac{\partial L}{\partial {\dot{x}}_{i}}=m{v}_{i}+\frac{q}{c}{A}_{i}$$

is no longer mass $\times $ velocity$\u2014$there is an extra term!

The Hamiltonian is

$$\begin{array}{c}H({q}_{i},{p}_{i})={\displaystyle \sum {p}_{i}{\dot{q}}_{i}-L({q}_{i},{\dot{q}}_{i})}\\ ={\displaystyle \sum \left(m{v}_{i}+\frac{q}{c}{A}_{i}\right){v}_{i}}-{\scriptscriptstyle \frac{1}{2}}m{\overrightarrow{v}}^{2}+q\phi -\frac{q}{c}\overrightarrow{v}.\overrightarrow{A}\\ ={\scriptscriptstyle \frac{1}{2}}m{\overrightarrow{v}}^{2}+q\phi \end{array}$$

Reassuringly, the Hamiltonian just has the familiar form of
kinetic energy plus potential energy.
However, to get

$$H=\frac{{\left(\overrightarrow{p}-q\overrightarrow{A}(\overrightarrow{x},t)/c\right)}^{2}}{2m}+q\phi \text{\hspace{0.17em}}(\overrightarrow{x},t)$$

where we have noted explicitly that the potentials mean
those at the position $\overrightarrow{x}$* *of the particle at time *t*.

Let us now consider Hamilton’s equations

$${\dot{x}}_{i}=\frac{\partial H}{\partial {p}_{i}},\text{}{\dot{p}}_{i}=-\frac{\partial H}{\partial {x}_{i}}$$

It is easy to see how the first equation comes out, bearing in mind that

$${p}_{i}=m{v}_{i}+\frac{q}{c}{A}_{i}=m{\dot{x}}_{i}+\frac{q}{c}{A}_{i}.$$

The second equation yields the Lorentz force law,
but is a little more tricky. The first
point to bear in mind is that *dp*/*dt* is *not *the
acceleration, the *A* term also varies in time, and in a quite complicated
way, since it is *the field at a point moving with the particle*. That is,

$${\dot{p}}_{i}=m{\ddot{x}}_{i}+\frac{q}{c}{\dot{A}}_{\text{\hspace{0.05em}}i}=m{\ddot{x}}_{i}+\frac{q}{c}\left(\frac{\partial {A}_{\text{\hspace{0.05em}}i}}{\partial t}+{v}_{j}{\nabla}_{j}A{\text{\hspace{0.05em}}}_{i}\right).$$

The right-hand side of the second Hamilton equation ${\dot{p}}_{i}=-\frac{\partial H}{\partial {x}_{i}}$ is

$$\begin{array}{c}-\frac{\partial H}{\partial {x}_{i}}=\frac{\left(\overrightarrow{p}-q\overrightarrow{A}(\overrightarrow{x},t)/c\right)}{m}.\frac{q}{c}.\frac{\partial \overrightarrow{A}}{\partial {x}_{i}}-q\frac{\partial \phi (\overrightarrow{x},t)}{\partial {x}_{i}}\\ =\frac{q}{c}{v}_{j}{\nabla}_{i}{A}_{j}-q{\nabla}_{i}\phi .\end{array}$$

Putting the two sides together, the Hamilton equation reads:

$$m{\ddot{x}}_{i}=-\frac{q}{c}\left(\frac{\partial {A}_{\text{\hspace{0.17em}}i}}{\partial t}+{v}_{j}{\nabla}_{j}{A}_{i}\right)+\frac{q}{c}{v}_{j}{\nabla}_{i}{A}_{j}-q{\nabla}_{i}\phi \text{\hspace{0.17em}}.$$

$\text{Using}\overrightarrow{v}\times (\overrightarrow{\nabla}\times \overrightarrow{A})=\overrightarrow{\nabla}(\overrightarrow{v}.\overrightarrow{A})-(\overrightarrow{v}.\overrightarrow{\nabla})\overrightarrow{A},\text{}\overrightarrow{B}=\overrightarrow{\nabla}\times \overrightarrow{A},$ and the expressions for the electric and magnetic fields in terms of the potentials, the Lorentz force law emerges:

$$m\ddot{\overrightarrow{x}}=q\left(\overrightarrow{E}+\frac{\overrightarrow{v}\times \overrightarrow{B}}{c}\right)$$

### Quantum Mechanics of a Particle in a Magnetic Field

We make the standard substitution:

$\overrightarrow{p}=-i\hslash \overrightarrow{\nabla},\text{sothat}[{x}_{i},{p}_{j}]=i\hslash {\delta}_{ij}\text{asusual:butnow}{p}_{i}\ne m{v}_{i}.$

This leads to the novel situation that the velocities in
different directions *do not commute*.
From

$m{v}_{i}=-i\hslash {\nabla}_{i}-q{A}_{i}/c$

it is easy to check that

$$[{v}_{x},{v}_{y}]=\frac{iq\hslash}{{m}^{2}c}B$$

To actually solve Schrödinger’s equation for an electron confined to a plane in a uniform perpendicular magnetic field, it is convenient to use the Landau gauge,

$\overrightarrow{A}(x,y,z)=(-By,0,0)$

giving a constant field *B* in the *z* direction. The equation is

$$H\psi (x,y)=\left[\frac{1}{2m}{({p}_{x}+qBy/c)}^{2}+\frac{{p}_{y}^{2}}{2m}\right]\psi (x,y)=E\psi (x,y).$$

Note that *x* *does
not appear in this Hamiltonian*, so it is a cyclic coordinate, and *p _{x}*
is conserved. In other words, this

*H*commutes with

*p*, so

_{x}*H*and

*p*have a common set of eigenstates. We know the eigenstates of

_{x}*p*are just the plane waves ${e}^{i{p}_{x}x/\hslash}$, so the common eigenstates must have the form:

_{x}$\psi (x,y)={e}^{i{p}_{x}x/\hslash}\chi (y).$

Operating on this wavefunction with the Hamiltonian, the
operator *p _{x}* appearing in

*H*simply gives its eigenvalue. That is, the

*p*in

_{x}*H*just becomes a number! Therefore, writing ${p}_{y}=-i\hslash d/dy,$ the

*y*-component $\chi \left(y\right)$ of the wavefunction satisfies:

$$-\frac{{\hslash}^{2}}{2m}\frac{{d}^{2}}{d{y}^{2}}\chi (y)+{\scriptscriptstyle \frac{1}{2}}m{\left(\frac{qB}{mc}\right)}^{2}{(y-{y}_{0})}^{2}\chi (y)=E\chi (y)$$

where

${y}_{0}=-c{p}_{x}/qB.$

We now see that the conserved canonical momentum *p _{x}* in the

*x*-direction is actually the coordinate of the center of a simple harmonic oscillator potential in the

*y*-direction! This simple harmonic oscillator has frequency

*ω*= |

*q*|

*B*/

*mc*, so the allowed values of energy for a particle in a plane in a perpendicular magnetic field are:

$E=(n+{\scriptscriptstyle \frac{1}{2}})\hslash \omega =(n+{\scriptscriptstyle \frac{1}{2}})\hslash \left|q\right|B/mc.$

The frequency is of course the cyclotron frequency$\u2014$that of the classical electron in a circular orbit in the field (given by $m{v}^{2}/r=qvB/c,\text{\hspace{1em}}\omega =v/r=qB/mc$ ) .

Let us confine our attention to states corresponding to the
lowest oscillator state, $E={\scriptscriptstyle \frac{1}{2}}\hslash \omega $. How many such states are there? Consider a square of conductor, area $A={L}_{x}\times {L}_{y}$,
and, for simplicity, take periodic boundary conditions. The center of the oscillator wave function *y*_{0}
must lie between 0 and *L _{y}*.
But remember that ${y}_{0}=-c{p}_{x}/qB$,
and with periodic boundary conditions ${e}^{i{p}_{x}{L}_{x}/\hslash}=1,\text{so}{p}_{x}=2n\pi \hslash /{L}_{x}=nh/{L}_{x}.$ This
means that

*y*

_{0}takes a series of evenly-spaced discrete values, separated by

$\Delta {y}_{0}=ch/qB{L}_{x}.$

So the total number of states $N={L}_{y}/\Delta {y}_{0},$

$$N=\frac{{L}_{x}{L}_{y}}{\left(\frac{hc}{qB}\right)}=A.\frac{B}{{\Phi}_{0}},$$

where ${\Phi}_{0}$ is called the “flux quantum”. So the total number of states in the lowest energy
level $E={\scriptscriptstyle \frac{1}{2}}\hslash \omega $ (usually referred to as the lowest Landau
level, or *LLL*) is exactly equal to
the total number of flux quanta making up the field *B* penetrating the
area *A*.

It is instructive to find *y*_{0} from a purely
classical analysis.

$\text{Writing}m\dot{\overrightarrow{v}}=\frac{q}{c}\overrightarrow{v}\times \overrightarrow{B}\text{incomponents,}$

$$\begin{array}{c}m\ddot{x}=\frac{qB}{c}\dot{y},\text{}\\ m\ddot{y}=-\frac{qB}{c}\dot{x}.\end{array}$$

These equations integrate trivially to give:

$$\begin{array}{l}m\dot{x}=\frac{qB}{c}(y-{y}_{0}),\\ m\dot{y}=-\frac{qB}{c}(x-{x}_{0}).\end{array}$$

Here (*x*_{0}, *y*_{0}) are the
coordinates of the center of the classical circular motion (the velocity vector
$\dot{\overrightarrow{r}}=\left(\dot{x},\dot{y}\right)$ is always perpendicular to $\left(\overrightarrow{r}-{\overrightarrow{r}}_{0}\right)$ ) , and ${\overrightarrow{r}}_{0}$ is given by

$\begin{array}{l}{y}_{0}=y-cm{v}_{x}/qB=-c{p}_{x}/qB\\ {x}_{0}=x+cm{v}_{y}/qB=x+c{p}_{y}/qB.\end{array}$

(Recall that we are using the gauge $\overrightarrow{A}(x,y,z)=(-By,0,0)$, and ${p}_{x}=\frac{\partial L}{\partial \dot{x}}=m{v}_{x}+\frac{q}{c}{A}_{x}$, etc.)

Just as *y*_{0}
is a conserved quantity, so is *x*_{0}:
it commutes with the Hamiltonian since

$\left[x+c{p}_{y}/qB,{p}_{x}+qBy/c\right]=0.$

However, *x*_{0}
and *y*_{0} *do not commute with each other*:

$\left[{x}_{0},{y}_{0}\right]=-i\hslash c/qB.$

This is why, when we chose a gauge in which *y*_{0} was sharply defined, *x*_{0} was spread over the
sample. If we attempt to localize the point (x0, y0) as well as possible, it is
fuzzed out over an area essentially that occupied by one flux quantum. The natural length scale of the problem is therefore
the magnetic length defined by

$$l=\sqrt{\frac{\hslash c}{qB}}.$$

*References:* the
classical mechanics at the beginning is similar to Shankar’s presentation, the
quantum mechanics is closer to that in Landau.