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Physics Part IA Cambridge University, 2012-2013

Notes for Waves and Quantum Waves, Part 3 Quantum-mechanical wavefunctions: The de Broglie hypothesis states that particles can be thought of as matter waves of wavelength

l=h/ p . They can therefore be described by a wavefunction ps ( x , t) of

the general form:

ps ( x , t )=ps 0 e i (ot-kx ) k

Using the de Broglie equation, constants

k :

Wavenumber

k=

and

o can be written as:

o :

Angular frequency

2p 2pp p

=

=

l

h

o=2pu=

p 2 mE

[?] k= =[?]

[?] o=

2pE E

=

h

E

Thus, the wavefunction for a mono-energetic particle travelling in the positive with energy

E and momentum i

ps ( x , t )=ps 0 e

x direction

p is:

( Et- px )

Physical meaning of the wavefunction: The wavefunction

ps

itself has no physical meaning, however

density function for the particle described by 1-dimension: x+ dx

2 [?] |ps| dx

[?] [?]

x x

3-dimension: x+ dx y+dy z +dz

2 1-dimension:

[?]

2 [?] |ps| dx=1

-[?]

dV =dxdydz

y

centred on

2 [?] [?] [?] |ps| dx dy dz

|ps| dx dy

dx

Is the probability of finding the particle in a line element volume element

is the probability

ps , such that:

2-dimensions: x+ dx y+dy

2 |ps|

y

z

dx , area element dA=dxdy

or

(x , y , z) , with normalisation condition:

2-dimensions:

3-dimension: 2

liint -[?] [?]|ps| dxdy=1

2 liiint -[?] [?] |ps| dxdydz=1

Physics Part IA Cambridge University, 2012-2013 That is, the particle must be found somewhere in space at a given time.

ps

This physical meaning of

can be understood by considering: 2

E [?]|ps|

1. Energy of a classical wave is

2. For a monochromatic light beam, energy is quantised into photons of equal energy

3. For a light beam

4. Therefore

E [?] N , where

is number of photons per unit volume

2 |ps| [?] N

5. In some volume element

6. Hence

N

2 |ps|

dV , the probability of finding a photon is [?] N dV

represents the probability of a photon being observed in

7. By de Broglie hypothesis, particles exhibit both particle and wave behaviour, like photons

8. Hence

2 |ps|

dV

represents the probability of a particle being observed in

Expectation value of position

x :

In the time-independent, one-dimensional case, particle position probability density function

2 |ps ( x )| =ps (x) ps (x)

x is described by the

. Any one measurement of

x will be an

inaccurate description of the nature of the particle, which is in fact described by a probability density distribution. The best single value describing the 'true position' of the particle to greatest accuracy is therefore the expectation value

:

[?]

< x > =[?] x ps ps dx

-[?]

The one-dimensional, time-independent Schrodinger equation: The one-dimensional, time-independent Schrodinger equation describes the behaviour of a mono-energetic particle with energy The particle has a wavefunction varies with

x :

T ps ( x ) + V ps ( x )=Eps ( x )

E travelling along a line with some position

(i / (Et -p x))

ps ( x , t )=ps 0 e and a potential V (x)

x .

which

Physics Part IA Cambridge University, 2012-2013 Where

T

is the kinetic energy operator,

V is the potential energy operator, and

E

is the total energy of the particle: Kinetic energy operator:

-i

ps ( x )=ps 0 e

[?] T =

px

[?] p=

Potential energy operator:

=V (x ) V

- dps

- d

[?] p =

i dx i dx

p2 - 2 d 2

=

2m 2m d x 2

2 d 2 ps

d2 ps 2

[?]2 m ( E-V )

[?]-

+V ( x)ps=Eps[?] 2 +k ps=0 ,k =

2 2m d x

dx The Schrodinger equation can be shown to obey the wave equation and hence describe the wavelike properties of the particle:

x :

Derivative in

Derivative in

[?]2 ps

=-k 2 ps

2 [?]x

ps ( x , t )=ps 0 e

[?]

[?]

t : i (Et- px )

[?]2 ps i 2 E 2

= 2 ps =-o2 ps=-k 2 c 2 ps

2 [?]t

[?]2 ps 1 [?]2 ps

=

[?] x2 c 2 [?] t 2

Where

k =[?]2m ( E-V ) / , or

[?] 2m ( E ) /

two and three dimensions, the wavevector 2

|k| =

2m ( E-V ( x ) )

2 for a zero potential function. Generalising to

k =(k x , k y , k z ) is given by:

Physics Part IA Cambridge University, 2012-2013 Conservation of energy for the Schrodinger equation: Energy is conserved for particles which obey the Schrodinger equation: i

psE=Vps-

( ( E-V ) t - px )

2 [?]2 ps

, ps=ps 0 e

2 2m [?] x

Substituting the wavefunction into the equation: 2

psE=

[?] E=

Where

2 (

2 p p

ps +Vps=ps

+V 2m

2m

()

)

p2

+V 2m 2 p /2m gives the kinetic energy and V

gives the potential energy.

Particles in infinite-potential wells:

[?]

[?]

V =0 0 a x An infinite potential well totally confines a particle such that the Schrodinger equation only has a solution inside the well:

V ( x ) =0,0 < x< a V ( x ) =[?], x <0, x> a Boundary conditions:

ps ( 0 )=ps ( a )=0 Outside the well:

Inside the well:

Physics Part IA Cambridge University, 2012-2013 2

[?] ps 2m

+ 2 ( E-V ( x ) ) ps =0, E-V ( x ) --[?]

2 [?]x

2 [?] ps=0,

2 [?] ps 2m ( )

+ 2 E ps =0 2

[?]x

Which in the time-independent case is an ordinary differential equation with solution:

[?] ps

=0 2

[?]x

ps= A sin ( kx ) + B cos ( kx ) , k 2=

2m ( E-V ) 2

Applying the boundary conditions gives:

B=0, 0=A sin(k a)

[?]k=

pn ,n[?] Z, a 2

[?] ps= A sin kx , E=

2 2

p n 2 2ma

Where the allowed energy levels within the potential well are called 'energy eigenstates'.

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