Particle in a 2-D Potential well = ( , ) ( , ) H x y - - PowerPoint PPT Presentation

Particle in a 2-D Potential well

Hamiltonian

µ µ µ

x y

H H H m x m y

2 2 2 2

2 2 ∂ ∂ = + = − − ∂ ∂ h h ψ is a product of the eigenfunctions of the parts of Ĥ E is sum of the eigenvalues of the parts of Ĥ

µ

n

H x y E x y ( , ) ( , ) ψ ψ × = ×

y x x x y y

n n x y x y x y L L L L 2 2 ( , ) ( ) ( ) sin sin π π ψ ψ ψ = × = ×

= = +

,

x y x y

n n n n n

E E E E V=0 Lx Ly

Square Box ⇒ Lx = Ly = L

µ µ µ

( )

  Ψ = + Ψ Ψ = + Ψ Ψ = Ψ  

,

x y x y

x y x y n n x y n n

H H H E E E

Degeneracy is manifestation of symmetry

Energy

2-D Potential Well - Wavefunctions

What Quantum Numbers in x and y do this wavefunction correspond to?

Number of nodes = ni-1

Expectation Values and Probability

ψ ψ π π π

∗

= × × × = × × × = × ×

∫ ∫ ∫

2

2 2 sin sin 2 sin

L L

x x dx n n x x x dx L L L L n x x dx L L ψ ψ π π π π π

∗

∂   = × − × ×  ÷ ∂   ∂ = − × × × ∂ − = × ×

∫ ∫ ∫

h h h

2

2 2 sin sin 2 sin cos

x L L

p i dx x n n i x x dx L L x L L i n n n x x dx L L L

π π   = Ψ Ψ = =  ÷     ≈ ∆ ∆ = −  ÷  

∫ ∫

2 2 1 1

* 2 1 2 2 1 2 2 1

2 ( , ) ...( ) 2 ( , ) . if ( ) is small

x x x x

n x P x x dx Sin solve L L n x P x x Sin x x x x L L

Probability of finding the particle in a given small interval (x 1,x2)

Particle in a 2-D Well: Energies

2D Well: 2 Quantum Numbers are required to describe system V=0

2 2 2 , 2 2

, 8 , 1,2,3,4,....

x y

y x n n x y x y

n n h E m L L n n   = +  ÷  ÷   =

V=0 Lx Ly

Square Box ⇒ Lx = Ly = L

Particle in a 3D-Box:Wavefunctions

3-D Box: 3 Quantum Numbers

Cubic Box: Wavefunctions

Importance in Chemistry/Spectroscopy

• Electronic spectra of conjugated molecules (loose p-electrons): 1D-PIB

V=0 V=∞ V=∞

Region I Region II Region III X X=0 X=L

2 2 2 2

1 : ( ) 8 ,

i f f i f i

Spectroscopy of PIB in D n n h h E E E n n mL Longer the box smaller the energy of hv ν → = ∆ = − = −

Increasing length of the box (i.e. the conjugation length) reduces the energy gaps and hence lower energy photons are absorbed or emitted

Nanoscience: Quantum Confinement

Band gap changes due to confinement, and so will the color of emitted light

CB VB CB VB CB VB

Quantum Dots – Particle (excitons) in a Sphere!

Quantum Dots have a huge application In chemistry, biology, and materials science For photoemission imaging purpose, As well as light harvesting/energy science

• +
• +
• +

The Hydrogen Atom The Hydrogen Atom

A A Completely Solvable Completely Solvable problem!! problem!! (kind of rare, in QM!) (kind of rare, in QM!)

CH1037Lecture 5 AC

2 2

sin m θ

What we learnt from solving PIB?

Formulate a correct Hamiltonian (energy) Operator H Impose boundary conditions for eigenfunctions (restriction) and obtain Quantum Numbers

Probability and Average Values

Solve HΨ=EΨ (2nd order PDE) by separation of variable and intelligent trial/guess solutions

Energies of states Corresponding to Quantum Numbers

Eigenstates or Wavefunctions: Should be “well behaved” - Normalization of Wavefunction

Quantum Numbers that specify the “state” of the system

H-Atom: Constructing H=T+V

µ

µ µ

µ

2 2 2 2 2 2 2 2 1

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

N i i i i i

P x y z H KE PE V x y z t V x y z t m m x y z P i and V x y z t Potential energy x y z

=

  ∂ ∂ ∂ = + = + = − + + +  ÷ ∂ ∂ ∂     ∂ ∂ ∂ = − + + =  ÷ ∂ ∂ ∂  

∑

h Q h

·

¶

2 2

2 2

2 2

Nucleus Electron

Electron Nucleus Nucleus Electron

m m

H V

−

+

∇ ∇

= − −

h h

Hydrogenic Atoms: 2-Particle System 1 electron moving around a (massive) central nucleus (+ve)

¶

µ

2 2 2 2 2 2

2 2 2 1

, ( ) ;

2

i i i

N i i i i

where Laplacian i particles x y z

H V m

=

∂ ∂ ∂ = + + → ∂ ∂ ∂

= − ∇ + ∇

∑ h

r

·

¶

2 2

2 2

2 2

Electron Nucleus Nucleus Electron

Nucleus Electron

m m

H V

−

+

∇ ∇

= − −

h h

·

( ) ( ) ( )

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

2

2 2

N N N N e e e e e N e N e N

m x y z m x y z where r x x y y z z

Ze r

H

    ∂ ∂ ∂ ∂ ∂ ∂ + + + +  ÷  ÷ ∂ ∂ ∂ ∂ ∂ ∂     = − + − + −

−

= − −

h h

2 2 2

2 2

,

, 2 2 ( , , , , )

e e e N N N Total e N

N e Total

e N Total Total Total Total Total

Ze m m r where x y z x y z and E E E

E

=

− = Ψ Ψ = +

∇ Ψ ∇ Ψ Ψ Ψ

− −

h h

µ

2 2

: ( )

( ) ~ 4

Coulomb Potential

Ze Ze Potential Energy V U r r r πε = = − −

r

Potential Energy: Coulomb Potential

:

Total

If Complete Wavefunction for H Atom TISE becomes Ψ = −

Reduced Form of TISE for H-Atom: Separation of Variables

Reduced Mass:

e e N N e N CM e e N N e N CM e e N N e N CM e N

m x m x x x x x M m y m y y y y y M m z m z z z z z M m m M µ + = − = + = − = + = − = =

2 2 2

2 2

2 2

in terms of CM and electronic coordinates ( , ) ( ) ( )

, , , , , , , ,

N N N

Total e e e N N N e e e e N N N N

CM e e e e

E M Ze r

Separate H

and E

x y z x y z x y z x y z

µ = − =

Ψ = Ψ

• Ψ

   ÷  

∇ Ψ Ψ ∇ Ψ Ψ

− −

h h

Free Particle: movement of The whole atom: You solved it! Relative motion of the electron and With respect to the Nucleus

2 2 2 2 2 2 2 2 2 2 2

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

e e e e i i i

e e

Ze x y z x y z x y z x y z x y z x y z x y z

E

µ

=

  ∂ ∂ ∂ − Ψ + Ψ + Ψ − Ψ  ÷ ∂ ∂ ∂ + +  

Ψ

h

Problem: 2nd order PDE with 3 variables - can not be separated!

Relative motion of electron wrt nucleus:

: . Movement of electronmuch faster than heavy nucleus Separate translational motion relative frame ⇒

Spherical Polar Coordinates

Used for spherically symmetric systems Conversion from Cartesian coordinates

Hamiltonian: Spherical Polar Coordinates

Solve this PDE need to separate variables r, θ, φ: POSSIBLE

Looks can be deceiving! Looks can be deceiving!