Given the Uncertainty Principle, how do you write an equation of motion for a particle?

•First, remember that a particle is only a particle sort of, and a wave sort of, and it’s not quite like anything you’ve encountered in classical physics.

We need to use Fourier’s Theorem to represent the particle as the superposition of many

waves.

dkekax ikx)()0,(

wavefunction of the electron

amplitude of wave with wavenumber k=2/

adding varying amounts of an infinite number of

waves

sinusoidal expression for harmonics

•We saw a hint of probabilistic behavior in the double slit experiment. Maybe that is a clue about how to describe the motion of a “particle” or “wavicle” or whatever.

dxprobability for an electron to be found between x and x+dx

(x,t)2

We can’t write a deterministic equation of motion as in Newtonian Mechanics, however, we know that a large

number of events will behave in a statistically predictable way.

1),(2

dxtx

dxtxPb

a 2),( dxtxP

b

a 2),(

Assuming this particle exists, at any given time it must be somewhere. This requirement can be expressed

mathematically as:

If you search from hither to yon

you will find your particle once, not twice (that would be two particles) but once.

For a linear homogeneous ordinary differential equation, if        and        are solutions, then so is               .

When two or more wave moving through the same region of space, waves will superimpose and produce a well defined combined effect.

When two waves of equal amplitude and frequency but opposite directions of travel superimpose, you get a standing wave- a wave that appears not to move-its nodes and anti-nodes stay in the same place. This happens when traveling waves on a guitar string get to the end of the string and are reflected back.

Conceptual definition

Mathematical definition

Let’s try a typical classical wavefunction:

)sin(),( tkxAtx

We also know that if 1 and 2 are both allowed waves, then 1+2 must also be allowed (this is called the superposition principle).

Similarly, for a particle propagating in the –x direction:

For a particle propagating in the +x direction.

)sin(),( tkxAtx

tkx

tkxAtkxA

txtx

cossin2

)sin()sin(

),(2),(1

21

Oops, the particle vanishes at integer multiples of /2, 2/3, etc.

and we know our particle is somewhere.

sincos

sincos

equationsEuler'

ie

iei

i

Graphical representation

of a complex number z as a point in the complex plane. The horizontal and vertical Cartesian components give the real and imaginary parts of z respectively.

)}sin(){cos(),( )( tkxitkxAAetx tkxi

i

ee

ee

ii

ii

2sin

2cos

:functions ricTrigonomet

22

222

* ))((

:then,if

)()(conjugate

1

ba

bia

ibaibazz

ibaz

ibaibadx

duee

dx

d

ii

uu

Note that we can construct a wavefunction only if the momentum is not precisely defined. A plane wave is unrealistic since it is not normalizable.

dxdxe tkxi 12

)(

You can’t get around the uncertainty principle!

Alright, we think we might have an acceptable wavefunction. Let’s give it a whirl… If we think we know what our wavefunction looks like now, how do we propogate it through time and space?

The Schrodinger Equation:

•Describes the time evolution of your wavefunction.

•Takes the place of Newton’s laws and conserves energy of the system.

•Since “particles” aren’t particles but wavicles, it won’t give us a precise position of an individual particle, but due to the statistical nature of things, it will precisely describe the distribution of a large number of particles.

If you know the position of a particle at time t=0, and you constructed a localized wave packet, superimposing waves of different momenta, then the wave will disperse because, by definition, waves of different momenta travel at different speeds.

time

the position of the real part of the wave…

the probability

density

the potential

dx

dUF

kxkxdx

d

dx

dUF

kxU

springclassicalaforei

2

2

2

12

1

:.,.

kxkxdx

d

dx

dUF

kxU

springclassicalaforei

2

2

2

12

1

:.,.

expression for kinetic energy

kinetic plus potential energy gives the total energy

kpm

pKE ;

2

2kp

m

pKE ;

2

2

Remember our guitar string? We had the boundary condition that the ends of the string were fixed.

Quantum mechanical version- the particle is confined by an infinite potential on either side. The boundary condition- the probability of finding the particle outside of the box is ZERO!

LxattL

xatt

0),(

00),0(

For the guitar string:For the quantum mechanical case:

Assume a general case:

...3,2,12

/2

0)/2sin(2

0

0

0

0

),(2),(1

2

222

/2/2/

)//2(2

)//2(1

21)/(

2)/(

1

)//2(2

)//2(1

)(2

)(1

nmL

nE

nLmE

LmEiA

eeAe

eAeA

Lxat

AAeAeA

xat

eAeA

eAeA

txtx

n

LmEiLmEiiEt

EtLmEiEtLmEi

EtiEti

EtxmEiEtxmEi

tkxitkxi

)()(),(

)(

txtx

eee tiikxtkxi

The wavefunction can be written as the sum of two parts.

Note that:

222

0*

)(|)()(||),(|

1)()(

xtxtx

eeett titi

For a “stationary state” the probability of finding a particle is static!

Note that while =0 solves the problem mathematically, it does not satisfy the conditions for the quantum mechanical wavefunction, because we know that at any given time, the particle must be found SOMEWHERE. N=0 is not a solution.

Appealing to the uncertainty principle can give us a clue.

2

22

22 mL

h

m

pE

L

hp

hxp

2/

...3,2,1

h

nnvrme

an integer number of wavelengths fits into the circular orbit

rn 2

where

p

h

is the de Broglie wavelength