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denotes the complex conjugate. Derive the di erential equation that

by taking the complex conjugate of Schrodinger's equation (6.13).

Problem c: Use this result to derive that for a volume V that is xed in time:

@Z ih Z2 dV = 2 2 )dV :

( (6.14)

@t 2m V

j j r ; r

V

Problem d: Use Gauss's law to rewrite this expression as:

@Z ih I 2 dV = ) dS :

2m ( (6.15)

@t j j r ; r

V

Hint, spot the divergence in (6.14) rst.

The left hand side of this expression gives the time-derivative of the probability that the

particle is within the volume V . The only way the particle can enter or leave the volume

is through the enclosing surface S. The right hand side therefore describes the \ ow" of

probability through the surface S. More accurately, one can formulate this as the ux of

the probability density current.

Problem e: Show from (6.15) that the probability density current J is given by:

ih

J = 2m ( ) (6.16)

r ; r

Pay in particular attention to the sign of the terms in this expression.

6.4. FLOWING PROBABILITY 57

As an example let us consider a plane wave:

(r t) = A exp i(k r !t) (6.17)

;

where k is the wavevector and A an unspeci ed constant.

Problem f: Show that the wavelength is related to the wavevector by the relation

=2 = In which direction does the wave propagate?

jkj.

Problem g: Show that the probability density current J for this wavefunction satis es:

J = hk 2 : (6.18)

m j j

This is a very interesting expression. The term 2 gives the probability density of the

j j

particle, while the probability density current J physically describes the current of this

probability density. Since the probability current moves with the velocity of the particle

(why?), the remaining terms in the right hand side of (6.18) must denote the velocity of

the particle:

v = hk : (6.19)

m

Since the momentum p is the mass times the velocity, equation (6.19) can also be written

as p = hk. This relation was proposed by de Broglie in 1924 using completely di erent

arguments than we have used here 13]. Its discovery was a major step in the development

of quantum mechanics.

Problem h: Use this expression and the result of problem f to compute your own wave-

length while you are riding your bicycle. Are quantum-mechanical phenomena im-

portant when you ride you bicycle? Use your wavelength as an argument. Did you

know you possessed a wavelength?

CHAPTER 6. THE THEOREM OF GAUSS

58

Chapter 7

The theorem of Stokes

In section 6 we have noted that in order to nd the gravitational eld of a mass we have

to integrate the eld equation (4.17) over the mass. Gauss's theorem can then be used to

compute the integral of the divergence of the gravitational eld. For the curl the situation

is similar. In section (5.5) we computed the magnetic eld generated by a current in a

straight in nite wire. The eld equation

B = 0J (5:12) again

r

was used to compute the eld away from the wire. However, the solution (5.13) contained

an unknown constant A. The reason for this is that the eld equation (5.12) was only used

outside the wire, where J = 0. The treatment of section (5.5) therefore did not provide

us with the relation between the eld B and its source J. The only way to obtain this

relation is to integrate the eld equation. This implies we have to compute the integral of

the curl of a vector eld. The theorem of Stokes tells us how to do this.

7.1 Statement of Stokes' law

The theorem of Stokes is based on the principle that the curl of a vector eld is the closed

line integral of the vector eld per unit surface area, see section (5.1). Mathematically

this statement is expressed by equation (5.2) that we write in a slightly di erent form as:

I

v dr = (r v) n dS = (r v) dS :

^ (7.1)

dS

The only di erence with (5.2) is that in the expression above we have not aligned the z-

axis with the vector r v. The in nitesimal surface therefore is not necessarily con ned

to the x y-plane and the z-component of the curl is replaced by the component of the

curl normal to the surface, hence the occurrence of the terms n dS in (7.1). Expression

^

(7.1) holds for an in nitesimal surface area. However, this expression can immediately be

integrated to give the surface integral of the curl over a nite surface S that is bounded

by the curve C: I Z

v dr = (r v) dS : (7.2)

C S

59

CHAPTER 7. THE THEOREM OF STOKES

60

This result is known as the theorem of Stokes (or Stokes' law). The line integral in the

left hand side is over the curve that bounds the surface S. A proper derivation of Stokes'

law can be found in ref. 38].

^

n

or

^

n

Figure 7.1: The relation between the sense of integration and the orientation of the surface.

Note that a line integration along a closed surface can be carried out in two directions.

What is the direction of the line integral in the left hand side of Stokes' law (7.2)? To

see this, we have to realize that Stokes' law was ultimately based on equation (5.2). The

orientation of the line integration used in that expression is de ned in gure (5.1), where

it can be seen that the line integration is in the counterclockwise direction. In gure (5.1)

the z-axis points out o the paper, this implies that the vector dS also points out of the

paper. This means that in Stokes' law the sense of the line integration and the direction

of the surface vector dS are related through the rule for a right-handed screw.

There is something strange about Stokes' law. If we de ne a curve C over which we

carry out the line integral, we can de ne many di erent surfaces S that are bounded by

the same curve C. Apparently, the surface integral in the right hand side of Stokes' law

does not depend on the speci c choice of the surface S as long as it is bounded by the

curve C.

z

S2

y

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