Particle Magnetic Field

8/4/2019 Particle Magnetic Field

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Charged Particle in a Magnetic FieldMichael Fowler 1/16/08Introduction

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

v BF q E

c

= +

This velocity-dependent force is quite different from the conservative forces from potentials thatwe have dealt with so far, and the recipe for going from classical to quantum mechanics

replacing momenta with the appropriate derivative operatorshas to be carried out with more

care. We begin by demonstrating how the Lorentz force law arises classically in the Lagrangianand 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 LagrangianL:

( , ) ( , )0, , being coordinates and velocities.i i i i

i i

i i

L q q L q qdq q

dt q q

=

The canonical momentumpi is defined by the equation

i

i

Lpq

=

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

( , ) ( , )i i i i i iH q p p q L q q=

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

,i ii i

H Hq p

p q

= =

These are known asHamiltons Equations. Note that if the Hamiltonian is independent of a

particular coordinate qi, the corresponding momentumpi remains constant. (Such a coordinate is

termed cyclic, because the most common example is an angular coordinate in a sphericallysymmetric Hamiltonian, where angular momentum remains constant.)


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For the conservative forces we have been considering so far, L = TV, H= T+ V, with Tthekinetic energy, Vthe potential energy.

Poisson Brackets

Any dynamical variablefin the system is some function of the qis andpis and (assuming it

does not depend explicitly on time) its development is given by:

( , ) { , }.i i i ii i i i i i

d f f f H f H f q p q p f H

dt q p q p p q

= + = =

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

{ , } .i i i i

A B A BA B

q p p q

=

We have shown from Hamiltons equations that for any variable { , }. f f H =

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

{ , } 0 { , }, { , } .i j i j i j ijq q p p q p = = =

This was the classical mathematical structure that led Dirac to link up classical and quantummechanics: 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:

1, .

A B A E

c t

= =

The right Lagrangian turns out to be:

212

. .q

L mv q v Ac

= +

(Note: if youre familiar with Relativity, the interaction term here looks less arbitrary: the

relativistic version would have the relativistically invariant ( )/q c A dx added to the actionintegral, where the four-potential ( ),A A =

and ( )1 2 3, , ,dx dx dx dx cdt = . This is the simplest

possible invariant interaction between the electromagnetic field and the particles four-velocity.

Then in the nonrelativistic limit, ( )/q c A dx just becomes ( ). / .q v A c dt

)


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The derivation of the Lorentz force from this Lagrangian is given by Shankar on page 84. Wegive the (equivalent) derivation from the Hamilton equations below.

Note that for zero vector potential, the Lagrangian has the usual TV form.

For this one-particle problem, the general coordinates qi are just the Cartesian co-ordinates( 1 2 3, ,i )x x x x= , the position of the particle, and the are the three componentsiq i ix v= of the

particles velocity.

The important new point is that the canonical momentum

i i

i i

i

L L q p mv

q x cA

= = = +

is no longer mass velocitythere is an extra term!

The Hamiltonian is

212

212

( , ) ( , )

.

i i i i i i

i i i

H q p p q L q q

q qmv A v mv q v A

c c

mv q

=

= + +

= +

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

However, to get Hamiltons equations of motion, the Hamiltonian has to be expressed solely in

terms of the coordinates and canonical momenta. That is,

( )2

( , ) / ( , )

2

p qA x t cH q

m

= + x t

where we have noted explicitly that the potentials mean those at the position x

of the particle at

time t.

Let us now consider Hamiltons equations

,i i

i i

H Hx pp x

= =

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

.i i i iq q

p mv A mx Ac c

= + = +i


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The second equation yields the Lorentz force law, but is a little more tricky. The first point to

bear in mind is that dp/dtis notthe acceleration, theA 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,

.ii i i i j j i

Aq q p mx A mx v Ac c t

= + = + +

The right-hand side of the second Hamilton equationi

i

Hp

x

=

is

( )( , ) / ( , ). .

.

i i

j i j i

p qA x t c H q Aq

i

x t

x m c x x

qv A q

c

=

=

Putting the two sides together, the Hamilton equation reads:

.i

i j j i j i j

Aq qmx v A v A q

c t ci

= + +

Using ( ) ( . ) ( . ) , ,v A v A v A B = =

A

and the expressions for the electric and

magnetic fields in terms of the potentials, the Lorentz force law emerges:

v Bmx q E

c

= +

Quantum Mechanics of a Particle in a Magnetic Field

We make the standard substitution:

, so that [ , ] as usual: but now .i j ij i i

p i x p i p= =

mv

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

/i i imv i qA c=

it is easy to check that

2[ , ]x y

iqv v B

m c=

To actually solve Schrdingers equation for an electron confined to a plane in a uniform


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perpendicular magnetic field, it is convenient to use the Landau gauge,

( , , ) ( , 0,0)A x y z By=

giving a constant fieldB in thez direction. The equation is

2

21( , ) ( / ) ( , ) ( , ).

2 2

y

x

p H x y p qBy c x y E x y

m m

= + + =

Note thatxdoes not appear in this Hamiltonian, so it is a cyclic coordinate, andpx is conserved.

In other words, thisHcommutes withpx, soHand px have a common set of eigenstates. We

know the eigenstates ofpx are just the plane waves , so the common eigenstates must have

the form:

/xip xe

/( , ) ( ).x

ip x x y e y =

Operating on this wavefunction with the Hamiltonian, the operatorpx appearing inHsimply

gives its eigenvalue. That is, the px inHjust becomes a number! Therefore, writing

they-component of the wavefunction satisfies:/yp i d d = ,y ( )y

22 2

21022

( ) ( ) ( ) ( )2

d qBy m y y y E y

m dy mc

+ =

Where

0

/ .x

y cp qB=

We now see that the conserved canonical momentumpx in thex-direction is actually thecoordinate of the center of a simple harmonic oscillator potential in they-direction! This simple

harmonic oscillator has frequency = |q|B/mc, so the allowed values of energy for a particle in aplane in a perpendicular magnetic field are:

1 12 2

( ) ( ) | | / E n n q B mc.= + = +

The frequency is of course the cyclotron frequencythat of the classical electron in a circular

orbit in the field (given by ) .2/ / , / / mv r qvB c v r qB mc= = =

Let us confine our attention to states corresponding to the lowest oscillator state, 1

2E = .

How many such states are there? Consider a square of conductor, areax y

A L L= , and, for

simplicity, take periodic boundary conditions. The center of the oscillator wave functiony0 must

lie between 0 andLy. But remember that 0 /x y cp qB= , and with periodic boundary conditions/

1, so 2 / / .x xip L

x xe p n L n= = = xh L This means thaty0 takes a series of evenly-spaced


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discrete values, separated by

0/ .

x y ch qBL =

So the total number of states0/ ,y N L y=

0

. ,x yL L B

N Ahc

qB

= =

where is called the flux quantum. So the total number of states in the lowest energy level0

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E = (usually referred to as the lowest Landau level, orLLL) is exactly equal to the total

number of flux quanta making up the fieldB penetrating the areaA.

It is instructive to findy0 from a purely classical analysis.

Writing in components,q

mv v Bc

=

,

.

qBmx y

c

qBmy x

c

=

=

These equations integrate trivially to give:

0

0

( )

( )

qBmx y yc

qBmy x x

c

=

=

,

.

Here (x0,y0) are the coordinates of the center of the classical circular motion (the velocity vector

( ),r x y= is always perpendicular to ( )0r r

) , and

0r

is given by

0

0

/ /

/ /

x x

y y

y y cmv qB cp qB

. x x cmv qB x cp qB

= =

= + = +

(Recall that we are using the gauge ( , , ) ( ,0,0)A x y z By=

, and x x x

L q p mv A

x c

= = +

, etc.)

Just asy0 is a conserved quantity, so isx0: it commutes with the Hamiltonian since

/ , / 0y x x cp qB p qBy c .+ + =


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However,x0 andy0do not commute with each other:

[ ]0 0, / .x y i c q= B

This is why, when we chose a gauge in whichy0 was sharply defined,x0 was spread over the

sample. If we attempt to localize the point (x0, y0) as well as possible, it is fuzzed out over anarea essentially that occupied by one flux quantum. The natural length scale of the problem istherefore the magnetic length defined by

.c

lqB

=

References: the classical mechanics at the beginning is similar to Shankars presentation, the

quantum mechanics is closer to that in Landau.

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Particle Magnetic Field