2014-12-19 chang-kui duan, institute of modern physics, cupt 1 1d systems solve the tise for various...

23/4/11 Chang-Kui Duan, Institute of Modern Physics, CUPT 1

1D systems

Solve the TISE for various 1D potentials

• Free particle

• Infinite square well

• Finite square well

• Particle flux

• Potential stepTransmission and reflection coefficients

• The barrier potentialQuantum tunnellingExamples of tunnelling

• The harmonic oscillator


A Free Particle

2 2

2

d

2 dE

m x

Free particle: no forces so potential energy independent of position (take as zero)

Time-independent Schrödinger equation:

Linear ODE with constant coefficients so try

exp( )x

Combine with time dependence to get full wave function:

General solution:


Notes

• Plane wave is a solution (just as well, since our plausibility argument for the Schrödinger equation was based on this assumption).

• Note signs in exponentials: – Sign of time term (-iωt) is fixed by sign adopted in

time-dependent Schrödinger Equation– Sign of position term (±ikx) depends on

propagation direction of wave. +ikx propagates towards +∞ while -ikx propagates towards –∞

• There is no restriction on k and hence on the allowed energies. The states form a continuum.

( ) i kx tx e


Particle in a constant potential

2 2

2

d

2 dV E

m x

General solutions we will use over and over again


22 2

2 2

2d0,

d

m E VK K

x

22 2

2 2

2d0,

d

m V Eq q

x

Case 1: E > V(includes free particle with V = 0 and K = k)

Case 2: E < V(classically particle can not be here)

Solution:

Solution:


Infinite Square Well

Consider a particle confined to a finite length –a<x<a by an infinitely high potential barrier

xNo solution in barrier region (particle would have infinite potential energy).

In the well V = 0 so equation is the same as before

Boundary conditions:

Continuity of ψ at x = a:

Continuity of ψ at x = -a:

Note discontinuity in dψ/dx allowable, since potential is infinite

V V V(x)

-a a

0V

2 2

2

d

2 dE

m x

General solution:


Add and subtract these conditions:

Even solution: ψ(x) = ψ(-x)

Odd solution: ψ(x) = -ψ(-x)

Infinite Square Well (2)

Energy

We have discrete states labelled by an integer quantum number


Infinite Square Well (3) Normalization

Normalize the solutions

Calculate the normalization integral2

( , ) dN x t x

Normalized solutions are


Sketch solutions


Wavefunctions Probability density

Note: discontinuity of gradient of ψ at edge of well.OK because potential is infinite there.

( )x 2( )x

3

1


Relation to classical probability distribution


Classically particle is equally likely to be anywhere in the box1

( )2clP x

a

so the high energy quantum states are consistent with the classical result when we can’t resolve the rapid oscillations.This is an example of the CORRESPONDENCE PRINCIPLE.

Quantum probability distribution is

2 2

2 2

1cos , 1,3,5

2

1sin , 2,4,6

2

n

n

nx x n

a a

nx x n

a a

2 2cos sin 1/ 2 But


• Energy can only have discrete values: there is no continuum of states anymore. The energy is said to be quantized. This is characteristic of bound-state problems in quantum mechanics, where a particle is localized in a finite region of space.

• The discrete energy states are associated with an integer quantum number.

• Energy of the lowest state (ground state) comes close to bounds set by the Uncertainty Principle:

• The stationary state wavefunctions are even or odd under reflection. This is generally true for potentials that are even under reflection. Even solutions are said to have even parity, and odd solutions have odd parity.

• Recover classical probability distribution at high energy by spatial averaging.

• Warning! Different books differ on definition of well. E.g.– B&M: well extends from x = -a/2 to x = +a/2.

Our results can be adapted to this case easily (replace a with a/2).– May also have asymmetric well from x = 0 to x = a.

Again can adapt our results here using appropriate transformations.

Infinite Square Well (5) – notes


Finite Square WellNow make the potential well more realistic by making the barriers a finite height V0

V(x)

x

-a a

V0

I II III

Region I: Region II: Region III:

0Assume 0 E V

i.e. particle is bound


Finite Square Well (2)

x a

Boundary conditions: match value and derivative of wavefunction at region boundaries:

Solve:

x a

Match ψ:

Match dψ/dx:

Now have five unknowns (including energy) and five equations(including normalization condition)



Cannot be solved algebraically. Solve graphically or on computer

Even solutions when tanq k ka

cotq k kaOdd solutions when

We have changed the notation into q


Finite Square Well (4) Graphical solution

Even solutions at intersections of blue and red curves (always at least one)Odd solutions at intersections of blue and green curves

k0 = 4a = 1


Sketch solutions


Wavefunctions Probability density

Note: exponential decay of solutions outside well

( )x 2( )x


Finite Square Well (6): Notes•Tunnelling of particle into “forbidden” region where V0 > E

(particle cannot exist here classically).

•Amount of tunnelling depends exponentially on V0 – E.

•Number of bound states depends on depth of well,

but there is always at least one (even) state

•Potential is even, so wavefunctions must be even or odd

•Limit as V0→∞:

We recover the infinite well solutions as we should.

0 Solutions at 2

nk ka


Example: the quantum well

Quantum well is a “sandwich” made of two different semiconductors in which the energy of the electrons is different, and whose atomic spacings are so similar that they can be grown together without an appreciable density of defects:

Now used in many electronic devices (some transistors, diodes, solid-state lasers)

Electron energy

Position

Material A (e.g. AlGaAs) Material B (e.g. GaAs)


Summary of Infinite and Finite Wells

Infinite well Infinitely many solutions

Finite well Finite number of solutionsAt least one solution (even parity)Evanescent wave outside well.

tanq k ka cotq k ka

Even parity solutions Odd parity solutions 202

2mq V E

22

2mEk

1cos , 1,3,5

2

1sin , 2,4,6

2

n

n

nx x n

aa

nx x n

aa

Even parity

Odd parity

2 2 2

28n

nE

ma


Particle Flux

2 2exp[ ( )] d dA i kx t x A x

In order to analyse problems involving scattering of free particles, need to understand normalization of free-particle plane-wave solutions.

This problem is related to Uncertainty Principle:

Momentum is completely defined

Position completely undefined; single particle can be anywhere from -∞ to ∞, so probability of finding it in any finite region is zero

Conclude that if we try to normalize so that

2d 1x

we get A = 0.

Solutions: Normalize in a finite boxUse wavepackets (later)Use a flux interpretation


Particle Flux (2)

More generally: what is the rate of change of probability that a particle is in some region (say, between x=a and x=b)?

xa b*

* *Prd d

b bab

a a

d dx x

dt dt t t

Use time-dependent Schrödinger equation: 2 2

2i ( , )

2V x t

t m x


Particle Flux (3)

xa b

Interpretation:Flux entering at x=a

Flux leaving at x=b

minus

**

Particle flux at position

i( , )

2

x

j x tm x x

Note: a wavefunction that is real carries no current

Prabd

dt


Particle Flux (4)

Check: apply to free-particle plane wave.

# particles passing x per unit time = # particles per unit length × velocity

Makes sense:

So plane wave wavefunction describes a “beam” of particles.

**i

( , )2

j x tm x x


Particle Flux (5): Notes

• Particle flux is nonlinear

• Time-independent case: replace

• 3D case,

• Can use this argument to prove CONSERVATION OF PROBABILITY.Put a = -∞, b = ∞, then

**i

( , )2

j x tm x x

, with x t x

x

Pr( , ) ( , ) ( , ) ( , )abd dN

j a t j b t j t j tdt dt

Prab N

If 0, then 0dN

dt

and


Potential Step

Consider a potential which rises suddenly at x = 0:

x

Case 1: E > V0 (above step)

x < 0, V = 0

Boundary condition: particles only incident from left

V(x)

0

V0Case 1

x > 0, V = V0


Potential Step (2)

Continuity of ψ at x = 0:

dContinuity of at 0 :

dx

x

Solve for reflection and transmission amplitudes:

2,

k K kr t

k K k K


Potential Step (3) Transmission and Reflection Fluxes

**i

( )2

j xm x x

x < 0 x > 0

1( ) eikx ikxx e r 2 ( ) iKxx te

Check: conservation of particles 2 21

k qr t

m m

Calculate transmitted and reflected fluxes

(cf classical case: no reflected flux)


Potential Step (4)

Case 2: E < V0 (below step)

Solution for x > 0 is now evanescent wave

Matching boundary conditions:

Transmission and reflection amplitudes:

V(x)

0

V0

Solution for x < 0 same as before

Transmission and reflection fluxes:

This time we have total reflected flux.


Potential Step (5): Notes

• Some tunnelling of particles into classically forbidden region even for energies below step height (case 2, E < V0).

• Tunnelling depth depends on energy difference

• But no transmitted particle flux, 100% reflection, like classical case.

• Relection probability is not zero for E > V0 (case 1). Only tends to zero in high energy limit, E >> V (correspondence principle again).

00

1,

2V E

q m V E


Rectangular Potential Barrier

Now consider a potential barrier of finite thickness:

xb0

V(x)

I II III

V0

Boundary condition: particles only incident from left

Region I: Region II: Region III:

0Assume 0 E V

u = exp(ikx) + B exp(−ikx) u = C exp(Kx) + D exp(−Kx) u = F exp(ikx)


Rectangular Barrier (2)

0x

Match value and derivative of wavefunction at boundaries:

x bMatch ψ:

Match dψ/dx:

Eliminate wavefunction in central region:

1 + B = C + D

1 − B = K/(ik)(C − D)

C exp(Kb) + D exp(−Kb) = F exp(ikb)

C exp(Kb) − D exp(−Kb) = ik/K F exp(ikb)


Rectangular Barrier (3)Transmission and reflection amplitudes:

For very thick or high barrier:

Non-zero transmission (“tunnelling”) through classically forbidden barrier region. Exponentially sensitive to height and width of barrier.

F

|F|2 =


Examples of TunnellingTunnelling occurs in many situations in physics and astronomy:

1. Nuclear fusion (in stars and fusion reactors)

2

0 nucleus

7

( )Barrier height ~ ~ MeV

4

thermal energies (~keV) at 10

Ze

r

T K

V

Nuclear separation x

Repulsive Coulomb interaction

Incident particles

Strong nuclear force (attractive)

Assume a Boltzmann distribution for the KE,

Probability of nuclei having MeV energy is

/E kTP E e1000e

Fusion (and life) occurs because nuclei tunnel through the barrier


Examples of Tunnelling

2. Alpha-decay

Distance of α-particle from nucleus

V

Initial α-particle energy

α-particle must overcome Coulomb repulsion barrier.

2 2qat eTunnelling rate depends sensitively on barrier width and height.

Explains enormous range of α-decay rates, e.g. 232Th, t1/2 = 1010 yrs, 218Th, t1/2 = 10-7s. Difference of 24 orders of magnitude comes from factor of 2 change in α-particle energy!


Examples of Tunnelling

3. Scanning tunnelling microscope

STM image of Iodine atoms on platinum. The yellow pocket is a missing Iodine atom

A conducting probe with a very sharp tip is brought close to a metal. Electrons tunnel through the empty space to the tip. Tunnelling current is so sensitive to the metal/probe distance (barrier width) that even individual atoms can be mapped.

0 4eVE V 02

2

2m E Vq

11 Aq

2 2qat e

If a changes by 0.01A (~1/100th of the atomic size) then current changes by a factor of 0.98,i.e. a 2% change, which is detectable

Tunnelling current proportional to

and

so

Vacuum

Material

Vx

Probea


Particles can tunnel through classically forbidden regions.Transmitted flux decreases exponentially with barrier height and width

Summary of Flux and Tunnelling

The particle flux density is*

*i( , )

2j x t

m x x

022

2m E Vq

2 2qat e

We get transmission and reflection at potential steps.There is reflection even when E > V0. Only recover classical limit for E >> V0 (correspondence principle)

0

0

2,

2,

k K kE V r t

k K k Kk iq k

E V r tk iq k iq

2 2 2 2 2 2

0 0 0 0, ( ), ( )2 2 2

k K qE E V E V V E E V

m m m


Simple Harmonic Oscillator

0

2 2 20

Force

Angular frequency =

1 1Potential energy ( ) 2 2

F kx

k

m

V x kx m x

Example: particle on a spring, Hooke’s law restoring force with spring constant k:

Mass m

x


Problem: still a linear differential equation but coefficients are not constant.

Simplify: change to dimensionless variable

1/ 2

00, =2E/( )

my x

V(x)

x


Simple Harmonic Oscillator (2)

2Suggests we substitute ( ) ( ) exp( / 2)y H y y

Asymptotic solution in the limit of very large y:

2( ) exp( / 2)y y

Try it:

Equation for H(y):

2

22 1 0

d H dHy H

dy dy



Solve this ODE by the power-series method (Frobenius method):

0

( ) pp

p

H y a y

2

22 1 0

d H dHy H

dy dy

Find that series for H(y) must terminate for a normalizable solution

Can make this happen after n terms for either even or odd terms in series (but not both) by choosing

Hence solutions are either even or odd functions (expected on parity considerations)

Label normalizable functions H by the values of n (the quantum number)

Hn is known as the nth Hermite polynomial.

0



EXAMPLES OF HERMITE POLYNOMIALSAND SHO WAVEFUNCTIONS

2

2

2

2

/ 20 0 0

/ 21 1 1

2 2 / 22 2 2

3 3 / 23 3 3

1

2 2

4 2 4 2

8 12 8 12

y

y

y

y

H y y N e

H y y y N ye

H y y y N y e

H y y y y N y y e

0 1 2, , nN N N N are normalization constants

nH y nyis a polynomial of degree


wavefunction Probability density

High n state (n=30)

• Decaying wavefunction tunnels into classically forbidden region• Spatial average for high energy wavefunction gives classical result: another example of the CORRESPONDENCE PRINCIPLE

Simple Harmonic Oscillator (5) Wavefunctions

2 2

2 20

1( )

1

2

clP xa x

E m a


Summary of Harmonic Oscillator1) The quantum SHO has discrete energy levels because of the normalization requirement

2) There is ‘zero-point’ energy because of the uncertainty principle.

3) Eigenstates are Hermite polynomials times a Gaussian

4) Eigenstates have definite parity because V(x) = V(-x). They can tunnel into the classically forbidden region.

5) For large n (high energy) the quantum probability distribution tends to the classical result. Example of the correspondence principle.

6) Applies to any SHO, eg: molecular vibrations, vibrations in a solid (phonons), electromagnetic field modes (photons), etc

00 2

E

0

1, 0,1,2,3

2nE n n

2( ) ( ) exp( / 2)y H y y


Example of SHOs in Atomic Physics:Bose-Einstein Condensation

87Rb atoms are cooled to nanokelvin temperatures in a harmonic trap. de Brogliewaves of atoms overlap and form a giant matter wave known as a BEC. All the atoms go into the ground state of the trap and there is only zero point energy (at T=0). This is a superfluid gas with macroscopic coherence and interference properties.Signature of BEC phase transition:The velocity distribution goes from classical Maxwell-Boltzmann form to the distribution of the quantum mechanical SHO ground state.


Example of SHOs: Molecular vibrations

VIBRATIONAL SPECTRA OF MOLECULESUseful in chemical analysis and in astronomy(studies of atmospheres of cool stars and interstellar clouds).

SHO very useful because any potential is approximately parabolic near a minimum

V(x)

Nuclear separation x

H2 molecule

SHO levels

H H

x

2014-12-19 chang-kui duan, institute of modern physics, cupt 1 1d systems solve the tise for various...

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