4-One dimensional potentials PDF

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4 Lecture notes series
Dr Fouzia Ouali ...

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4: One dimensional potentials Additional Reading • Schrodinger equation H-FLAP Module P10.4: Link: http://www.met.reading.ac.uk/pplato2/h-flap/phys10_4.html • Reflection and transmission at steps and barriers H-FLAP Module P11.1: Link: • http://www.met.reading.ac.uk/pplato2/h-flap/phys11_1.html Hyperphysics: For general definitions and summary: • One dimensional box: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/pbox.html • Tunnelling: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html • Eigenvalues, Eigenfunction: http://hyperphysics.phyastr.gsu.edu/hbase/quantum/eigen.html 1

One dimensional potentials Aims and plans: This section aims to apply the Schrödinger equation to simple one dimensional potentials to give an appreciation on how it can be solved to obtained the allowed energy levels and corresponding wavefunctions. We will also be studying quantum mechanical tunnelling. • • • • • •

Free particles Energy levels and wavefunctions of a particle in a box Potential steps Quantum tunnelling through a rectangular potential Application of Quantum tunnelling The Simple Harmonic oscillator’s energy levels and wavefunctions

2

Task and general approach • • •

To use Schrödinger equation to determine the wavefunctions (energy , momentum …where appropriate) for various systems. In this section, motion in one dimension (x) Time dependent Schrödinger equation subject to a potential U(x).

 2  2  ( x, t )  ( x, t )   U ( x ) ( x, t )  i 2 2m x t •

or

 ( x, t ) Hˆ  ( x, t )  i t

If a particle subject to potential a definite energy E and since U(x) is independent of time  time dependent solution

(x,t )  (x)ei Et  Where the time independent wavefunction (x) is solution of time independent Schrödinger equation (SE).

 2 d 2 ( x)  U ( x)  E ( x)   2 m dx2  KE

PE

or

Hˆ  ( x)  E ( x)

2 2  d  Hˆ   U ( x) 2m dx 2

Total energy

3

Task and general approach ctd Approach: 1. Solve time independent SE above equation and obtain a general solution of differential equation 2. Define and apply the boundary conditions for the problem 2 boundary conditions: a. Wavefunction (x) must be continuous: No jumps b. Derivation of  (x) must also be continuous: No kinks 3. Calculate energy eigenvalues and wavefunctions 4. Normalize the wavefunctions using probability argument where appropriate. 5. Time dependent wavefunction

 ( x, t )   ( x )e i Et 

4

4.1: Free particle • Definition of free particle: Particle subject to no force ie F=0, but

F 

dU  0  Potential energy U  constant dx

Take U=0

• Particle with given energy E (total energy) • Potential energy U independent of time, so use time independent SE

  2 d 2 (x )  U ( x) ( x )  E  ( x )  2 m dx 2  0 since U  0

Rearrange 

2

d  (x) dx

2

  2 d 2 ( x )   E ( x)  0 2 m dx2 eq 4.1



2mE

Let k 2 



2

 ( x)  0

2mE  2  5

Free particle ctd 

d 2 ( x ) dx

2

2

 k  ( x)  0

Second order linear differential equation with constant coefficients, same as eq 1.1 (page 12 lecture1)

eq 4.2 General solution

 ( x)  Aeikx  Be ikx with k  k 2 

2mE 2

A and B constants

eq 4.3

eq 4.4

Combine with time dependence to give full wavefunction

 ( x, t )   ( x)eiEt /   Aei kx Et /    Bei  kx  Et   From left to right (+ve x)

From right to left (-ve x)

6

Free particle wavefunction ctd • Plane wave solution to SE for free particle as expected. • Signs and convention: Wavefunction in eq 4.3

 ( x)  Aeikx  Beikx

is superposition of

two waves. By convention:

• Aeikx: Plane Wave with amplitude A travelling in +x-direction (from left to right), with momentum of magnitude p=ħk

Aeikx

Motion from left to right

• Be-ikx : Plane Wave with amplitude A travelling in –x direction (from right to left) with momentumof magnitude p= -ħk

Be-ikx Motion from right to left • No restriction on the particle energy. All energies are allowed Continuum of energy 7

4.2: Particle in a box (infinitely deep square well) Consider particle confined to a finite length L by two infinitely high potential energy barriers (two rigid walls ) and cannot escape from it Particle bound to the box. U I U=

II

III

U=0

U=

Potential energy: U=0 0 ≤x≤ L U=∞ xL x

x=0

L

Procedure: 1. Divide x axis into 3 regions of potential energy and work out SE in each region 2. Write down and solve SE for each region Potential energy U independent of time, so use time independent SE 3. Apply boundary and normalisation conditions Boundary: place where potential energy changes

8

Particle in a box ctd Barriers (regions I and III ) “Well” (region II)

U= ∞ U=0

xL 0≤x≤L

Barrier regions: Particle not allowed because it would have infinite potential energy, hence its wavefunction is zero.

  (x )  0 in barrier regions ( x  0 and x  L)  I (x )  0 for x  0 and III (x )  0 for x  L Well region: 0  x  L SE

Potential energy U=0KE=E-U≥0

  2 d 2 II ( x )  U ( x) II ( x)  E II ( x)  2 m dx 2  0 since U  0

  2 d 2 II ( x )   E II ( x)  0 2 m dx 2

Same equation as that for free particle (eq 4.1, page 4), hence same solution

II ( x)  Aeikx  Be ikx

0 xL

with k 2 

2mE 2

Plane wave solution if particle kinetic energy KE≥0 (positive)

9

Particle in Box ctd So general solution:

I (x )  0  ikx ikx  II (x)  Ae  Be   ( x)  0  III

x 0 0x L xL

Boundary conditions: Two boundaries where potential energy changes: at x=0 and x=L

Continuity of  at x=0 “no jumps” Replace A=-B in II

 I  II  0  Aei 0  Bei 0  A  B  0  A  B  II ( x)  Ae ikx  Ae ikx Aeikx  eikx   2A sin kx  C sin kx

Continuity of  at x=L

 II  I II  0  C sin kL  0 either: C=0 or

Note: d/dx discontinuous at boundaries . Allowed because potential change infinite.

sin kL  0 10

Particle in Box ctd

C=0: not a physical solution, since this would make  II (x )  0 So C#0 and sin kL  0 kL  n Physical solution for n =1, 2, 3…

k  kn 

n L

for all x.

n : integer

n  1, 2, 3

eq 4.5

•

Wavevector kn quantised: takes only specific (discrete) values.

•

n=1, 2, 3..:: quantum number of the system.

but

2mE n 2 2  2  2    L 2 2 h2 2   2 E  En  n n 2mL2 8mL2

2mE k n2  2

n  1, 2, 3 

eq .4.6

•

 Total energy E is also quantised.

•

Particle is bound by the potential En also known as bound states (bound states are quantised) Lowest allowed energy is ground state (for n=1). Note: E=0 is not possible (reason later page 15)

• •

11

Particle in box: Energy n2 eq 4.6 shows that En  2 mL

 16E 1

n =4

9E 1

n =3

4E 1

n =2

E1

n =1

h2  2 2 2  E1  1  2mL2 8mL2  2 2 2 En  n  E1n 2 2 2mL

E

• Quadratic increase with n

Energy inversely proportional to mass of particle • Increases strongly with decreasing well width (characteristic of quantum systems).  Confining particle in a small dimension requires a lot of energy.

12

Particle in Box : wavefunctions •

Normalisation of :



L



 ( x ) 2 dx  C 2 sin 2

space

L

 0

0

1 but sin 2   (1  cos 2 ) 2

L

L

C2  C2 L  2nx    2nx   C sin  x sin  1  cos  dx   1   dx  1  L L n L 2 2 2       0  2

2  nx 

nx dx  1 L

 0

C 2L   1 C  2

2 L

Hence, normalised particle wavefunction

 (x )   (x )  0

2  n sin L  L

 x 

n  1, 2, 3

for 0  x  L (well) elsewhere (barriers) 13

Particle in box: Wavefunctions  (x)

 (x)

• •

•

•

2

Particle likely to be found where probability density is maximum ( shown by

Wavefunctions are standing waves for the box of width L (well). Wavefunctions are either symmetric (cosine like) or asymmetric (sine like) with respect to the centre of well. Reason: Potential is symmetric with respect to the centre of the well. Wavefunctions are non-degenerate. For each level of energy En, there is only only one possible eigenfunction. Time dependent  ( x , t ) 

Where En given by eq 4.6

( x, t) 0

2  n sin  L  L

 iE t /  x  e n 

n  1, 2, 3

for 0  x  L elsewhere

14

Example: Find energy levels of: a) An electron in a 1D 1Å wide box. b) A neutron in a 1D 110-14 m wide box (mn=1.6710-27 kg). c) A pool ball (m=0.2 kg) bouncing between cushions of a frictionless, perfectly elastic pool table (L~1.5 m) Solution a)

15

Particle in a box and uncertainty principle Questions: • Assume particle confined in box of width L. • Use uncertainty principle to determine minimum energy that particle can have. • Compare with ground state of the particle. Answers: • Uncertainty principle: Take •

px x 

   px  2 2 x

x ~ L, p x ~ p

Minimum momentum

Emin

pmin

 p2 2 ~ ~  Emin  2L 2 m 8 mL2

2 1 h2 E1 ~   0 2 2 2 2 8mL 4 8mL 4

E1: ground state of the well

The uncertainty principle is the reason why the minimum energy of any particle in any box cannot be zero.

16

Bohr correspondence principle • • •

Quantum theory applied for sub-microscopic systems Quantum theory must produce same results for macroscopic systems as those predicted by classical mechanics. Macrocsopic systems have large quantum numbers n. The behaviour of quantum mechanical systems reduce to classical physics in the limit of very large quantum numbers.

Bohr correspondence principle Worked example: An electron is confined to a box of width L. 1. Write down the bound energies En, n=1, 2, 3… 2. Calculate the frequency  of the radiation emitted when the electron makes a transition from the nth to (n-1)th energy level. 3. In the limit n  show that  is the same as the classical frequency in which a particle of particle En bounces between the two walls. 17

Solution

18

Finite square potential: notes • • •

Particle in a box model: not realistic Gives some insight of order of magnitudes for confinement energy. Made more realistic by making barriers of finite height U0.

• •

Write down and solve SE in the three regions of potential energy. Consider bound states with EU0KE>0

Region II: x≥0

 2 d 2 II ( x) ( x) II ( x)  E II ( x) U  2m dx2 U 0

  d  II ( x )  U 0  E  II ( x)  0 2 m dx 2 2

2

d 2 II ( x) 2m  II ( x)  0     E U 0 dx 2 2 d 2 II ( x)   2 (x )  0 2 dx 2m E  U 0  μ2  0 2 General solution: plane wave (KE>0) C, D i x  ix  II (x )  Ce  De 26 constants

Case 1: E>U0 ctd Region I: xU0 ctd Boundary conditions:

Solutions in two regions must be joined together at boundary

Continuity of  at x=0  I ( x  0)   II ( x  0)  Ae ik 0  Be  ik 0  Ce i 0  A  B  C d I d II Continuity of d/dx at (x  0)  ( x  0)  ik  A  B   iC dx dx x=0 A  B   k  A  B  Ak    Bk   B  k   and C  2 k  A k  A k  2 2 Jr B v r B  2  2 since v r  vi Reflection coefficient R : defined as R  Ji A vi A k  2 2 vi   vr, vt  2   E E U   B k 0 m m   R E  U0 A k  E  E U 0

Eq 4.9 2

Transmission coefficient T: Note: R+T as expected

2

C v C  J 4k  T t  2 t  2 Ji A vi A k k   2

28

Case 2: Potential step with E...