The Aharonov-Bohm Effect in Relativistic Quantum Theory

“I think I can safely say that nobody understands Quantum Mechanics.” – R P Feynman

The Aharonov-Bohm Effect in Relativistic Quantum Theory.

Craig Millar

200405021

12521 Research Project

Abstract In this report I will be introducing Relativistic Quantum Mechanics in order to lead to a discussion

on the Klein paradox and the Aharonov-Bohm effect. Along the way will be a reminder of the

Schrödinger equation, before moving onto the beginnings of relativistic quantum mechanics with

the Klein-Gordon equation then onto the Dirac equation. Gauge theory plays a crucial part in the

prediction of the Aharonov-Bohm effect, hence this will be introduced.

Craig Millar 200405021 Page 2

Contents Abstract ................................................................................................................................................... 1

Contents .................................................................................................................................................. 2

Introduction ............................................................................................................................................ 3

Review of Non-Relativistic Quantum Mechanics .................................................................................... 3

The Schrödinger Equation ................................................................................................................... 3

Observables and Operators (3) ............................................................................................................. 4

Relativistic Quantum Mechanics ............................................................................................................. 7

The Klein-Gordon Equation ................................................................................................................. 7

Derivation ........................................................................................................................................ 7

Problems with the Klein-Gordon Equation ..................................................................................... 8

Recovering the Schrödinger Equation from the Klein-Gordon Equation ...................................... 10

Dirac Equation ................................................................................................................................... 11

Energy-Momentum Relationship .................................................................................................. 12

Probability Density ........................................................................................................................ 13

Lorentz Covariance ........................................................................................................................ 14

Klein Paradox ................................................................................................................................. 14

E-M Theory and Potentials .................................................................................................................... 19

Gauge Theory .................................................................................................................................... 19

Minimal Coupling .............................................................................................................................. 21

Gauge Transform of the Dirac Equation ........................................................................................... 23

Aharonov-Bohm (A-B) Effect ................................................................................................................. 24

Conclusion ............................................................................................................................................. 29

Acknowledgements ............................................................................................................................... 29

References ............................................................................................................................................. 30


Introduction The Aharonov-Bohm effect was first indicated by Ehrenberg and Siday in their paper on the

refractive index in electron optics in 1948(1) but little attention was paid to it at the time. The

phenomenon was rediscovered independently in 1959 by Aharonov and Bohm(2). In brief the

theory says that in quantum mechanics potentials can act on charged particles even where the

field lines are excluded, this will be discussed in detail in the last part of the report. This report

begins at the beginning with a simple reminder of the Schrödinger equation in Non-Relativistic

Quantum Mechanics (N.R.Q.M.) and operators and observables. We shall then follow the

historical route to the Dirac Equation for Relativistic Quantum Mechanics (R.Q.M.) via the Klein-

Gordon equation pointing out its flaws along the way. Once we have derived the Dirac Equation

we will look at the Klein Paradox in which we are in the odd situation where you seem to get out

a larger current than you put in from an electron incident on a potential step. We will then look

at Gauge theory and the coupling of the Dirac equation to the electromagnetic field and proving

that the Dirac equation is invariant under a gauge transform. Once we have covered all of these

points we are in a position to discuss the Aharonov-Bohm effect in Quantum Mechanics in detail.

Review of Non-Relativistic Quantum Mechanics

The Schrödinger Equation The Schrodinger equation allows us to predict the probability of finding a particle at a specific

point in time or space so long as we know its wave function, . The Schrodinger equation can be

justified as a suitable equation by taking a simple plane wave and using Planck’s law and the de

Broglie’s hypothesis. (3)

If we take the standard wave equation for a plane wave to be

expA i k x t (1.1)

Where A is the complex normalisation constant, k is the wave vector, x is the position vector, ω is

the angular frequency and t is time.

Planck’s Law is E and the de Broglie hypothesis states that p k .

For the moment it is assumed that we are working in one dimension. Taking the 2nd derivative of

(1.1) with respect to x we get

2

2

2k

x

(1.2)


Taking the kinetic energy and multiplying by (1.1) and then using the de Broglie hypothesis we get

2 2 2

2 2

p kE

m m

(1.3)

Now replacing 2k with (1.2), we are lead to

2 2

22E

m x

(1.4)

Now the derivative of (1.1) with respect to time is taken.

it

i

dt

(1.5)

Multiplying Planck’s Law by the wave function

E (1.6)

Now inserting (1.5) into (1.6)

E it

(1.7)

Since E must be the same in each case (1.4) and (1.7) can be set equal to each other to get

2 2

22i

t m x

(1.8)

This is the Schrödinger equation for a free particle.

The Schrödinger equation can easily be changed to three dimensions and also changed for a

particle in a potential to form the time dependant Schrödinger equation below.

2

2 ,2

i V x tt m

(1.9)

Where 2 2 2

2

2 2 2x y z

Observables and Operators (3) In Quantum mechanics physical quantities such as energy, position and momentum are referred

to as observables. Each observable has an operator associated with it to allow it to be calculated


from the wave function. An observable usually represented by A will be represented by the

operator A and the expectation values and mean square can be calculated by

*ˆ ˆ( , ) ,A x t A x t dx

(1.10)

2 * 2ˆ ˆ( , ) ,A x t A x t dx

(1.11)

Taking a generalised wave equation relating the position wave function, , and the momentum

wave function, , through a Fourier transform we have (1.12) in one dimension.

1

, , exp2

ipxx t p t dp

(1.12)

The inverse of equation (1.12) is given by

1

, , exp2

ipxp t x t dp

(1.13)

In both equations (1.12) and (1.13), p is the momentum and the rest of the symbols have the

same definitions as in the Schrödinger equation.

Calculating the expectation values of the momentum in the usual way

2

,p p p t dp

(1.14)

Where 2 *( , )p t

If we multiply p into both sides of equation (1.12) we get

1, , exp

2

1, exp

2

,

ipxp x t p p t dp

ipxi p t dp

x

i x tx

(1.15)

So now the expectation value of p becomes


* , ,p x t i x t dxx

(1.16)

By taking a similar approach with the expectation value of the mean square of the momentum

2 2

22

2

1, , exp

2

1, exp

2

1, exp

2

,

ipxp x t p p t dp

ipxi p p t dp

x

ipxi i p t dp

x x

x tx

(1.17)

This leads to the expectation value of

2

2 * 2

2, ,p x t x t dx

x

(1.18)

So from(1.10), (1.11) and (1.16), (1.18) we can see that the momentum observable , p, is

represented by the momentum operator

p ix

(1.19)

Which becomes

p i (1.20)

In 3 dimensions.

There is also an operator associated with the energy. The energy observable is given by

2

2

pE V x

m . To change this into an operator we have to put in the momentum operator

(1.19) and the position operator ˆx x giving


2

2

2 2

2

ˆˆ ˆ,2

,2

,2

pH V x t

m

ix

V x tm

V x tm x

(1.21)

H is known as the Hamiltonian operator and the eigenvalues of it are the energy of the system

such that ˆ , ,H x t E x t .(4)

Relativistic Quantum Mechanics The Schrödinger equation is not invariant under a general Lorentz transformation, i.e. each side

of the equation transforms differently under a Lorentz transform. Therefore it is only useful for

describing phenomena for velocities very much less than the speed of light. This is where we

need to start using Relativistic Quantum Mechanics.

The first attempt at a relativistic quantum theory was in 1926(5) by Klein and Gordon with the

Klein-Gordon Equation but there were problems with their equation some of which will be

discussed below. This was then followed in 1928 by the Dirac Equation which correctly describes

elementary spin ½ particles, such as the electron. (6)(7)

The Klein-Gordon Equation Derivation (8)(9)

In 1926 Klein and Gordon attempted a relativistic version of the Schrödinger equation by first

finding a Lorenz invariant version of the Hamiltonian(1.21). They used

2 2 2 2 4H p c m c (2.1)

Using (1.20), (2.1) becomes

2 2 2 2 2 4H c m c (2.2)

When we return to the free particle Schrödinger equation with our new Hamiltonian we see that


1/22 2 2 2 4

22 2 2 2 2 4

2

i c m ct

c m ct

(2.3)

Rearranging (2.3) we are lead to

22 2 2 2 2 4

2

2 2 22 2 2

22 2

2 2 2 2

2 2 20

2 22

2

0

0

0

0

c m ct

m cc t x

m c

x x

m c

(2.4)

Which is the Klein-Gordon Equation, where2

2 2

2 2

1

c t

.

Problems with the Klein-Gordon Equation

The first and most obvious problem with the Klein-Gordon equation is that when we try to find

the energy from equation (2.1) we see that we have introduced a negative energy term such that

2 2 2 4 2 2 2 4 OR H p c m c p c m c (2.5)

The second problem with the Klein-Gordon equation is that of a positive definite probability

density, ρ. In order to find the probability density we have to find the conservation equation for

the Klein-Gordon equation. For this we need the complex conjugate of equation (2.4)

2 2

2 *

20

m c

(2.6)

Taking equation (2.6) back a few steps in its derivation we return to

2 * 2 * 2 2*

22 2 2

10

m c

c t x

(2.7)

If we now take the complex conjugate wave function multiplied by (2.4) and subtract the wave

function times (2.7) and set it equal to 0.


2 2 2 2 2 2 2 2* *

2 22 2 2 2 2 2

1 10

m c m c

c t c tx x

(2.8)

This can be re-written as

*

* * *

20

2 2

i i

t mc t t m

(2.9)

After some manipulation(8). If we then compare (2.9) with the standard continuity equation

0jt

(2.10)

Where is the probability density and j is the probability density current, we would like to

interpret

*

*

22

i

mc t t

(2.11)

as the probability density. However a probability density has to be a positive definite expression

which this is not.

To find out what (2.11) is, we use a plane wave of the form

exp i k x t (2.12)

And

22

2

22

2

and appropriate complex conjugates

it t

ik kx x

(2.13)

Substituting (2.12)and (2.13) into (2.11) we get

2e e e e

2

i k x t i k x t i k x t i k x tii i

mc

(2.14)

* *

22

ii i

mc


*

22

2

ii

mc

*

2mc

(2.15)

Which is an energy density since E and 2E mc so we have a ratio or energies multiplied

by a probability density giving us an energy density.

Using equations (2.4) and (2.13) to find the allowed frequencies of (2.12) we find that

2 2 22

2 2 2

2 2 22

2 2

2 42 2 2

2

10

0

m c

c t

m ck

c

m cc k

(2.16)

If we take the non-relativistic limit where k ,equation (2.16) becomes

2mc

(2.17)

If we now substitute (2.17)into (2.15) we can recover the probability density

* (2.18)

for the Schrödinger equation as we would expect.

Recovering the Schrödinger Equation from the Klein-Gordon Equation

If the Klein-Gordon equation is to be a candidate description for relativistic quantum mechanics it

would be sensible for it to collapse down to the Schrödinger equation in the non-relativistic limit.

As we have already seen from (2.18) the probability density collapses to the correct form so what

about the rest of the equation?

Sticking with our plane wave

exp i k x t (2.19)

And taking the first derivative with respect to time we get

it

(2.20)


Taking the square root of (2.16) and rearranging

12 2 2 2

2 21

mc k

m c

(2.21)

Now using a Taylor expansion to the first two terms to approximate (2.21)

2 2 2

2 2

121

1!

mc k

m c

2 21

2

mc k

m

(2.22)

Substituting (2.22)into (2.20)

2 2

2 22

2 22

2

1

2

1

2

2

mc ki

t m

ki mc

t m

i mct m x

(2.23)

Which we have already seen is the Schrödinger equation with a potential 2V mc .

Dirac Equation Much of the following is adapted from Bjorken and Drell. (8)

In 1928 Dirac took the search for a relativistically covariant version of the Schrödinger equation

down a different route to Klein and Gordon. He noticed that the Schrödinger equation is linear in

the time derivative and decided that the Hamiltonian should be linear in the space derivatives

too. He suggested an equation of the form

2

1 2 31 2 3i i c mc H

t x x x

(3.1)

For this to be a suitable form the constants and must be square matrices and must be a

column matrix, this will led to a set of coupled first order equations


2

1 2 31 2 3

1

N N

N

i i c mct x x x

H

(3.2)

In order for it to be a candidate it must satisfy the same energy momentum relationship as the

Klein-Gordon Equation, give a continuity equation with a positive definite probability density and

be Lorentz Covariant.

Energy-Momentum Relationship

In order for (3.1) to be a suitable equation it must first give the correct energy momentum

relationship (2.3) as in the Klein-Gordon equation. This can be done by taking the derivative of

(3.1) with respect to time and multiplying by i which gives

22

2

23 32 2 3 2 2 4

, 1 , 12

j i i j

i ii j ii j i j

i it t t

c i mc m cx x x

(3.3)

In order for (3.3) to be equal to (2.3) the following must be true about the coefficient matrices

2 2

2

0

1

i j j i ij

i i

i

(3.4)

Where 1ij for i j and 0 for i j . The other restriction on the matrices is that they must

be hermitian so that the Hamiltonian of equation (3.1)is also hermitian. From the third equation

in (3.4) we know that the eigenvalues of the matrices are 1 and from the anti-commutation

properties the sum of the diagonal elements, the trace, must be zero. From this we know that

the matrices must also be even dimensional matrices because if trace is zero and the diagonal

elements are the eigenvalues then there must be an equal number of positive and negative

eigenvalues which can only occur if the matrix is evenly dimensional.

The first suitable matrix is a 4x4 matrix, this is because a 2x2 matrix can only contain the 3 Pauli

matrices because they are anti-commutating, and the identity matrix which can commute with

the Pauli matrices which is not allowed by (3.4). The usual choice of the first matrix combination

is


0 1 0

0 0 1

i

i

i

(3.5)

Where i are the Pauli matrices,

1

2

3

0 1

1 0

0

0

1 0

0 1

x

y

z

i

i

(3.6)

1 is the 2x2 identity matrix and 0 is the 2x2 zero matrix. By using equations(3.4), (3.5)and (3.6)

we can recover the correct energy-momentum relationship,(2.3) from (3.3).

Probability Density

The second criterion which (3.1) must satisfy is to give a positive definite probability density. To

do this we need the hermitian conjugates of the wave function and of(3.1). The hermitian

conjugate is the complex conjugate of the transpose of the matrix denoted by † . The hermitian

conjugate of is † and the hermitian conjugate of (3.1) is

† † † †

† † † † 2 †

1 2 31 2 3i i c mc

t x x x

(3.7)

We know that † † and i i so (3.7) can be simplified.

Left multiplying (3.1) by † and subtracting (3.7)right multiplied by gives

† †3 3† † 2 † 2 †

1 1

3† †

1

3† †

1

0

k kk kk k

k

kk

k

kk

i i i c mc i c mct t x x

i i ct x

i i ct x

(3.8)

Comparing this with equation (2.10) we can make the identification that

† (3.9)

Which is what we would expect for a probability density.


Lorentz Covariance

First off we need to write the Dirac equation in a covariant form. To do this we rearrange (3.1) to

equal zero then multiply by c

to get

2

1 2 31 2 3

1 2 31 2 3

0

0

i mcc t x x x

i mcc t x x x

(3.10)

Since 2 1 from the last equation in (3.4).

We know that 0

1

c t x

and if we set 0 and i i where 1,2,3i we get to

0 1 2 3

0 1 2 30

0

i mcx x x x

i mcx

(3.11)

In this representation there is a slight change to the matrices in (3.5) to

0

0 1 0

0 0 1

ii

i

(3.12)

Equation (3.11) is Lorentz covariant and this is proved in detail in Bjorken and Drell chapter 2. (8)

Klein Paradox

Let’s have a look at what happens to an electron incident on a potential barrier in the z direction.

For and electron in region 1 with an energy E and a momentum p=pz we have

2

2 2 2Ep m c

c

(3.13)


Figure 1 - Diagram of an electron incident on a potential barrier.(10)

However in region 2 where it is now in a constant potential V0 equation (3.13) becomes

2

2 2 20E Vp m c

c

(3.14)

Where p’ is the momentum of the electron in region 2.

As we are only working in one dimension we can reduce the Dirac equation (3.1) to

3 2

2ˆ

i i c mct z

c p mc E

(3.15)

For region 1 with α=α3 and p=pz and for region 2 we get

2

0ˆc p mc E V (3.16)

If we now look at the possible solutions for the incident, reflected and transmitted waves. The

incident wave in region 1, travelling left to right has a solution of (8)

1

1

2

1

0

0

ik z

inc ae ck

E mc

(3.17)

Remembering that 1p k .


The reflected wave in region 1 travelling right to left has a solution of

1 1

1

2

1

2

1 0

0 1

0

0

ik z ik z

ref be b eck

E mc ck

E mc

(3.18)

The wave transmitted into region 2 is subject to a constant potential so 2 2

0E mc E V mc

giving

2 2

2

2

0 2

2

0

1 0

0 1

0

0

ik z ik z

trans de d eck

E V mc ck

E V mc

(3.19)

In (3.19) 2p k and the wave is again travelling left to right.

In order for the solutions above to be correct the wave must be continuous across the boundary

so that means that

inc ref trans (3.20)

By choosing a coordinate system where z=0 exactly at the boundary and using equations (3.17),

(3.18), (3.19) and (3.20)we can calculate a, b, b’, d and d’ for the wave functions. Using the

matrices we get 4 simultaneous equations.

1 1 2

1 2

1 1 2

1 2

1 1 2

2 2 2

0

1 2

2 2

0

ik z ik z ik z

ik z ik z

ik z ik z ik z

ik z ik z

ae be de

b e d e

ck ck ckae be e

E mc E mc E V mc

ck ckb e d e

E mc E V mc

(3.21)

Now using the fact that at the boundary 0 1ikze e and letting 2

2

2

1 0

k E mcr

k E V mc


a b d

b d

a b rd

b rd

(3.22)

From the 2nd and 4th equations in (3.22) we can see that the wave function is only continuous if

0b d . This also tells us that there is no spin flip at the boundary.

Solving the simultaneous equations in (3.22) we get

12

12

1

1

2

1

da r

db r

rb

a r

d

a r

(3.23)

These will be useful in a moment once we have calculated the particle currents. The particle

current is given by †

3( ) ( ) ( )j z c z z . This has to be calculated for the incident, reflected

and transmitted waves. So

1 1† 1

2 1

2

1

2

† 1

2

† 1 1

2 2

2† 1

2

10 0 1 0

00 0 0 1

1 0 01 0 0 0

0 1 0 00

01 0 0

1

0

2

ik z ik z

inc

ckj ca e ae ck

E mcE mc

ck

E mcck

ca aE mc

ck ckca a

E mc E mc

c ka a

E mc

(3.24)


Completing the same for the reflected and transmitted currents we get

2

† 1

22ref

c kj b b

E mc

(3.25)

and

2

† 1

2

0

2trans

c kj d d

E V mc

(3.26)

Now if we compute the ratio of the reflected and transmitted currents to the incident current and

using the equations in (3.23) we get

2

4

1

trans

inc

j r

j r

(3.27)

And

2

2

1

1

ref

inc

j r

j r

(3.28)

Remembering that 2

2

2

1 0

k E mcr

k E V mc

. Looking at the limits of equations (3.27)and (3.28) as

2

0V E mc , (i.e. 0r ).

Looking at (3.27) if 0r then the whole equation is negative which is implying that the

transmitted current is travelling in the opposite direction to the incident current. This is rather

counter intuitive as you would expect the transmitted current to be travelling the same direction.

Leaving (3.27) for a moment to look at(3.28). When 0r the equation is negative, this implies

that the reflected current is travelling in the opposite direction to the incident wave as expected

but 1ref

inc

j

j which implies that the reflected current is greater than the incident current which is

counter intuitive. Together the above analysis point to a current being produced in region 2

travelling from right to left. But where is this current coming from? We have assumed up until

this point that region 2 is empty, clearly this cannot be the case as electrons are leaving it

producing the current. So in order to make sense of this we have to reassess this assumption. If

we change the assumption to be that all electron states with energy 2E mc must all be

occupied with electrons, then in region 2 when a potential is applied the electron energy is raised.


If the energy is raised enough there will be an overlap between the filled negative continuum of

region 2 and the positive continuum of region 1. This then allows electron coming in from the left

to knock extra electrons out of the vacuum in region 2 increasing the reflected current and giving

the negative transmitted current we see from equations (3.27) and (3.28). If we are knocking

electrons out of region 2, what is left behind? Originally it could only be one thing, the proton but

this didn’t fit with the physics so Dirac postulated the existence of a particle identical to the

electron only with a positive charge. It took the discovery of the positron in 1932 by Carl D.

Anderson(11) to complete the picture and prove Dirac’s postulation correct.

E-M Theory and Potentials (Note: In all the remaining sections 1c for simplicity)

Gauge Theory Gauge theories are a general class of quantum field theories which are used to describe

elementary particles and their interactions. Electromagnetic (EM) theory closely relates the

behaviour of the forces to the symmetry principle. The principle of symmetry in the real world is

basically an operation in which the object looks the same before and after. This can be applied to

the laws of physics too, if we can perform some mathematical operation on an equation and

leave it looking as it did before the equation is said to be invariant under a transformation. A very

important fact that we need to know is the difference between a global and local invariance. A

global invariance is one in which a transformation occurs simultaneously throughout space and

time and a local invariance is one in which different transforms are carried out at different space-

time points. Unfortunately many globally invariant theories, such as EM, are not locally invariant

but by a clever choice of fields acting on a particle a local invariance can be restored to the

system. To demonstrate this I will use Maxwell’s equations.

Maxwell’s equations are(12)

E (4.1)

B

Et

(4.2)

0B (4.3)

E

B jt

(4.4)


In Heavyside-Lorentz units where ε0 and μ0 are unity by choice of suitable units for the charge and

current and also rationalising to remove 4π(12).

Maxwell modified (4.4) to include the E

t

term otherwise, from the continuity equation, (2.10),

the charge density would have to be constant in time. The extra term also allows the charge

density to be locally conserved. As we already know the continuity equation tells us that charge

must be locally conserved which also means that it is globally conserved. Why must it be locally

conserved? In order to globally conserve it but not locally we would have to be able to transmit

signals instantaneously across any distance which is not possible because it would violate special

relativity.

Generally E and B are replaced by functions of a vector potential ( )A x as follows(12)

B A (4.5)

And

A

E Vt

(4.6)

Where V is the scalar potential and A is the3-vector potential.

Using (4.5) and (4.6) we can see that (4.2)and (4.3)are satisfied straight away.

A major point to note about A and V is that they are not unique for a certain value of B or E which

means that they can be transformed while leaving E and B unchanged. So what are the

transformations which allow this to be the case?

Using equation (4.5)we can see that

A A A (4.7)

Where χ is an arbitrary function of r and t. Putting (4.7)into (4.5) we see

B A

A

A

A

(4.8)


Since 0 . Now if we look at (4.6) it is not quite so simple. This time V will have to

transform as well as A. Let us now look at inserting (4.7)into (4.6)to find the transformation of V.

AE V

t

V At

AV

t t

AV

t t

(4.9)

If we now compare (4.9) to (4.6) we see that in order for it to be invariant V Vt

which

implies that V Vt

.

So (4.9) now becomes

AE V

t t t

AV

t

(4.10)

This can all be re-written in a compact form using the 4-vector potential, ,A V A (12). This

transforms as

A Ax

(4.11)

Where ,x t

.

Minimal Coupling So now, how is the Dirac equation connected to electromagnetism?

Remember that the Dirac equation is

i i mt

(4.12)


With 1c but (4.12) is not invariant under gauge transformation. In order to make it gauge

invariant we must now look at the Gauge Principle.(13)

If we start by considering a gauge transform which has the form

iqe (4.13)

Where q is charge and χ and ψ are functions of r and t. If we now take the gradient of (4.13)

iq

iq

e

e iq

(4.14)

We see that we have now gained the term iqe iq . The method of resolving this problem

is to introduce the field A into the gradient in such a way as to be able to cancel out this term. So

if we introduce the field using

iqA (4.15)

Now we get

iq

iq iq

iq

iqA iqA e

e iqA e

e iq A

(4.16)

Then if we make it so that A A the theory is now invariant under the gauge

transformation.

Now if we do the same for the time derivative of (4.13)

iq

iq iq

iq

et t

iq e et t

e iqt t

(4.17)

We have gained the term iqe iqt

, if we now introduce the scalar potential V such that so


iqVt t

(4.18)

now the time derivative of (4.13) becomes

iq

iq iq

iq

iqV iqV et t

e iqV et

e iq Vt t

(4.19)

In the same was as for the gradient we can make it that V Vt

. So taking what we know

from (4.15)and (4.18)and putting it into the Dirac equation, (4.12)we can get it in the gauge

invariant form

i iqV i iqA mt

i qV i qA mt

i i qA m qVt

(4.20)

For an electron with q e this then becomes

i i eA m eVt

(4.21)

This equation expresses the minimal coupling of the Dirac equation to an electromagnetic field

when the Dirac particle is taken to be a point charge, in this case an electron.(8)

Gauge Transform of the Dirac Equation Now let us prove that the Dirac Equation in the form of (4.20)is gauge invariant.

Taking

i i qA m qVt

(4.22)

As our new Dirac equation and by using the transforms


A A (4.23)

V Vt

(4.24)

And

iqe (4.25)

Let’s now see if we can get back to equation (4.20). Putting (4.23), (4.24)and (4.25)into (4.22)we

get

iq iq

iq iq

i e i q A m q V et t

i q V e i q A m et t

(4.26)

Now multiplying out and performing the derivatives we get

iq iqq i qV q e q i q A q m et t t

(4.27)

Through cancellation of terms we are led to

iq iqi qV e i q A m et

(4.28)

Upon dividing through by the phase factor and taking out common factors we get

i qV i qA mt

i i qA m qVt

(4.29)

Which we can see is identical to the original equation (4.20) hence the Dirac Equation coupled to

the electromagnetic field in this way is gauge invariant.

Aharonov-Bohm (A-B) Effect If we think back to classical physics for a moment we will remember that only E and B are

physically significant and the vector potential, A, and the scalar potential, V, can be changed by a


gauge transform without altering the physics of E and B. However this doesn’t hold in quantum

physics as we will see shortly. In their paper(2) Aharonov and Bohm suggested two different

effects, the magnetic A-B effect and the electrical A-B effect, both are dependent on their

respective gauge transforms. We shall first look at the magnetic A-B effect as this is the better

known of the two.

If we start with the same gauge transforms as in (4.23), (4.24)and (4.25) but with a slight notation

change to introduce the phase factor and replacing V with φ, to avoid confusion with V for the

potential in the electrical A-B effect, we get

( , )A A r t (4.30)

t

(4.31)

and

exp ( , )i r t (4.32)

Where the phase factor ( , ) ( , )r t q r t .

If we now take the first case where 1

( ) ( ) ( )A r r rq

then from (4.5)we have

1

0

B A

q

(4.33)

Since 0S for all values of S and we also get(14)

( ) ( )

b

a

r q A r dr (4.34)

Which is a line integral between points a and b, so by putting (4.34)into (4.32) we get that(14)

exp ( )

b

a

iq A r dr

(4.35)


This now suggests that even if B=0, A can be non-zero which implies from (4.35)that the particle

undergoes a phase change despite there being no force acting on it.

Figure 2- Charged particle moving along a closed loop

If we now look at a charged particle which travels from a to b along a path then back to a along

another path as in Figure 2.

The phase change along each path will be (4.34) so the total phase change will be

1 2

( ) ( )

( )

path path

loop

q A r dr q A r dr

q A r dr

(4.36)

By Stokes theorem (12) (4.36) can be rewritten as

loop

enclosedsurface

enclosedsurface

q A dr

q A dS

q B dS q

(4.37)

Where is the magnetic flux within the closed loop. So now we know from (4.37)that the phase

change, , around the loop is the charge times the magnetic flux, q . Obviously if there is no

magnetic flux inside the loop there will be no phase change as expected but if a charged particle

is moving in a region where A≠0 there will be a phase change even if B=0. Aharonov and Bohm

suggest an experiment based on the Youngs slits experiment in order to test this interesting

phenomenon, it is shown in Figure 3. The electron beam is split by a set of Youngs slits positioned

at A, these beams then pass either side of a tight wound solenoid and the brought back together

at F where an interference pattern will be observed. If the B field is changed this will have an


effect on the interference pattern at F despite the fact the magnetic field is contained within the

solenoid and the electrons are only passing through an area where 0B .

This effect was first shown in experiment by Chambers(15) in 1960 using a set up similar the one in

Figure 3.

Figure 3 - Schematic experiment to demonstrate interference with a time-independent vector potential(2)

More recently in 1986 Tonomura et. al.(16) used a toroidal magnet coated with super-conducting

material, to confine the magnetic field by the Meissner effect, and electron holography to see the

effect. Figure 4 shows one of their sets of results, you will see in (a) and (c) that the interference

lines are in different places inside and outside of the toroid which proves that there is a relative

phase shift between electrons passing through the toroid and those passing around it.

Meanwhile in (b) there appears to be no shift, this is because the phase has shifted my 2n thus

making it appear to have not moved. The white dashed lines indicate where we would expect the

interference lines to be if the toroid had no effect.

Figure 4 - evidence of the A-B effect.(16)

The second type of A-B effect is the Electric A-B effect. Starting from the same equations as for

the magnetic A-B effect and following the same process but this time with

( )

b

a

t q dt (4.38)


Which leads us to having

exploop

iq dt

(4.39)

Aharonov and Bohm suggested an experiment as in Figure 5 where A, B, C, D, E are devices for

splitting and redirecting beams, W1 and W2 are wave packets, M1 and M2 are cylindrical metal

beams and F is the interference region. The idea behind this is to have wave packets travelling

along the system and have M1 and M2 change in potential only when the wave packets are well

inside them, this protects the wave packets from the electric field because they are acting as a

Faraday Cage. This means that the wave packet is never exposed to the potential but there will

still be a change the scalar potential around the loop which will cause the interference in (4.38).

Figure 5 - Schematic of an experiment to demonstrate interference with a time-dependent scalar potential.(2)

This experiment was difficult to realise so an alternative was sought and eventually in 1998

Oudenaarden et. al.(17) showed the electric A-B effect in an experiment designed to show both the

electric and magnetic A-B effect using metal rings with tunnelling junctions in them so that a well

defined voltage can be applied to either side of the ring.

The A-B effect was originally worked out for the non-relativistic case so how would it change

when taken to the relativistic case? Well it doesn’t, so long as you use the fully relativistic form

for the vector and scalar potentials. In both the non-relativistic and relativistic cases all we get is

a phase change which does not affect the physics of the wave function in any way as we have

seen from the gauge transforms earlier in the report.


Conclusion The relativistic quantum world is very different from the macroscopic world we live in. It allows

things to happen which seem impossible, from the Klein paradox where we appear to get a larger

current out than we put in, to the Aharonov-Bohm effect where wave functions undergo a phase

change despite the fact that there is no force field acting on them, the change is entirely due to

the vector potential or the scalar potential. In this report I have led you from the Schrödinger

equation, through the Klein-Gordon equation and the Dirac equation, including the Klein paradox,

and eventually ending up at the Aharonov-Bohm effect via Gauge theory.

Acknowledgements First of all I must thank Prof. Barnett for all his help and patience throughout this project, even on

the 3rd attempt at explaining something to me. I would also like to thank Dr Jeffers for his help

whilst Prof. Barnett was unavailable; you saved me becoming overly stressed. I would also like to

thank Laura and Nick for helping proof read my report and finding the mistakes I just couldn’t see.

Lastly I would like to thank all my friends for keeping me sane when things just didn’t seem to be

going right.


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The Aharonov-Bohm Effect in Relativistic Quantum Theory