Problem Set 11 W h t tt f I ld lik ( d t d) i We have not gotten as far as I would like (and expected) in lectures (for a variety of reasons). I apologize for this. As one consequence, you can turn in PS 11 anytime before 1 PM on Friday on Friday For help on homework, come to office hours or to the help room on Wednesday (especially if problems I or II are an issue)(especially if problems I or II are an issue) Today’s class will definitely help on problems III and IV,along with reading Sec.11.6 in text.

Total Energy is Sum of Kinetic Energy & Potential EnergyEnergy & Potential Energy

E = Ek + Ep

H = E = px2/2m + Ep(x)

2 2 2 2ˆH E (x) V(x)

p2 2H E (x) V(x)2m 2mx x

is called the Hamiltonian operatorH is called the Hamiltonian operatorH2

2ˆ2m

In 3 dimensions:2 2 2

22 2 2

2

x y z

m

x y z

For two particles: m1, x1, y1, z1 and m2, x2, y2, z2

Schrödinger EquationSchrödinger Equation

Time-independent:2

2 V(x y z) (x y z) E (x y z)

V(x,y,z) (x,y,z) E (x,y,z)

2m

H E

Separability of Solutions

1 2 31 2 3 n 1 2 3 n

If the multi coordinate Hamiltonian operator has the special form( , , ) = ( ) + ( ) + ( )+ + ( ) with each component q q q q q q q qnh h h h

31 2 3 n 1 2 3 n

having the form n

i i i i i ih q T q V q

then there exists a simple, general solution

1 2 31 2 31 2 n( , , , ) q q q q q q i ii=1

( ) ( ) ( ) ( )qn

with each satisfying a 1D Schrödinger like equation h E

Thus whenever the Hamiltonian can be written as a sum then

with each k satisfying a 1D Schrödinger-like equation .k k k kh E

and the total energy is the sum of individual components, .1

n

ii

E E

Thus whenever the Hamiltonian can be written as a sum, then becomes a product of one variable terms and E becomes a sum of energies associated with each degree of freedom.

The difficulty of obtaining a separable solution scales linearly with The difficulty of obtaining a separable solution scales linearly with the number of degrees of freedom, and any 1D problem is very easy!

We will do (almost) anything to achieve separability!

Additional Dimensions/Particles

2 2 2 2

1 2 1 2( , ) 0 for 0 x a and 0 x a otherwise

V x x

1 2

2 2 2 2

1 2 1 22 21 2

2 2 2 2

ˆ , ,2 2

x x V x xm mx x

Two non-interacting 1-D particles in a box of width a

1 2

1 22 21 22 2

V x V xm mx x

Two non-interacting 1-D particles in a box of width a Separable, since we neglected gravitational potential attraction, (m1m2)/(x2-x1)2, in V

2 2 2 2 2 2n n 1 2total 1 22 2

1 2

n nE n 1,2,3 n 1,2,32 2mm a a

21 1 22 2 i i n x xn

Two quantum numbers

1 2

21 1 21 2

1 2

, sin sin

for 0 x 0

n nnx x

a a a aa and x a

Mass m particle in a two-dimensional (x,y) boxV(x, y) = , for x < 0 or x > a or for y < 0 or y > b, and

0 f 0 d 0 b a

= 0 , for 0 x a and 0 y b

2 2 2 2

22

, ,2 2

yxx y

ppT T V x y V x ym m

b m

2 2 2 2

2 2ˆ , 0 inside the rectangle (separable!)

2 2x y

m mx y

22 ( ) ( )xd xE

22

2

( )- ( )2 y

yd yE

2 ( )

2 x xEm dx

2 221

12

n E n 1,2,3,2mx a

2 ( )2 y yEm dy

2 222

y 22 n 1,2,3,E 2mbn

1

1sin

for 0 x 0 otherwise

nn x2(x)

a aa and

2n

2(y) sinb b

for 0 b 0 otherwise

n y

y and

f y

 

1 2

212 2 n x ynx, y sin sin a b a b

f 0 0

n n

d b

  2 2 2 2 2 21 2

total 1 22 2

n nE n 1,2,3 n 1,2,32 2 m bma

for 0 x 0 a and y b

Particle in a two-dimensional (x,y) box

1 2

21n x y2 2 nx, y sin sin a b a b

for 0 x 0

n n

a and y b

a

b m

2 2 2 2 2 21 2

total 1 22 2

n nE n 1,2,3 n 1,2,3

2 2mbma

for 0 x 0 a and y b

2 2mbmaNote especially!!There are two degrees of freedom with constrained motion and the result is that there are two quantum numbers n1 and n2 What do the solutions look like?that there are two quantum numbers, n1 and n2. What do the solutions look like?

Ψ21,2

Nodal line ψ=0 (2D)(points in 1D)(surface in 3D)

Ψ1,2

1,2

y xy x

Particle in a two-dimensional (x,y) box

Consider the ψ2,1 state as well.Now what is the total energy of the 2,1 state? It is the same as the 1,2 state.

Particle in a two-dimensional (x,y) box

The two state functions correspond to the same energy but they are veryThe two state functions correspond to the same energy, but they are verydifferent spatially. This is called degeneracy! Here, twofold degenerate.In fact, these two functions are orthogonal!

More on degeneraciesConsider a square box a by a in extent.

2 2

2 2,2

2 with , sin sin in the square2x y

yxn n x y x yn n

n yn xE n n x ya a ama

2 a a ama

nx , ny

nx

2 + ny2

N d t1,1 2 11

1,2

5

12 2 1

5

Non degenerate

Doubly degenerate2,1 5 21 3,1

10

31

1 3

10

13 Doubly degenerate

1,3 10 13 2,2

8

22

2,3

13

23

Non degenerate

Doubly degenerate 3,2

13

32

Doubly degenerate

The ψij are all orthogonal: 0 unless i j i j

Particle in Three Dimensional (a,b,c) Box

2 2 2 2

2 2 2ˆ , ,

2V x y z

m x y z

b

2

, , 0 inside the box and outsidem x y z

V x y z

Separable into three 1D particle in a box problems!

a cb

Separable into three 1D particle in a box problems!22 22 2

2 2 22yx z

n n n

nn nEm b

2 2 22

1,2, , 1,2, , 1,2, ,

x y zn n n

x y z

m a b cn n n

2 21n x y2 2 2 n nx, y, sin sin sin a b a bx y zn n n

zzc c

a b a b for 0 x , 0 0

x y z c ca y b and z c

Energy Degeneracy for a particle in 3D cubep

2 2

2 2 21 2 32E n n n

1 2 322

E n n nma

ConcepTest #1ConcepTest #

In a cubic box what is the degree of the In a cubic box, what is the degree of the degeneracy if the three quantum numbersn1, n2, n3 can have the values 1, 2, 3?

A. 2-fold

B. 4-foldB. 4 fold

C. 6-fold

D 8 foldD. 8-fold

ConcepTest #1ConcepTest #

In a cubic box what is the degree of the In a cubic box, what is the degree of the degeneracy if the three quantum numbersn1, n2, n3 can have the values 1, 2, 3?

A. 2-fold

B. 4-foldB. 4 fold

C. 6-fold

D 8 foldD. 8-fold

ConcepTest #2ConcepTest #2

In a cubic box which of the following In a cubic box, which of the following states is lowest in energy?

A. n1=3, n2=1, n3=1

B. n1=2, n2=2, n3=2B. n1 2, n2 2, n3 2

C. n1=3, n2=2, n3=1

D. n1=2, n2=2, n3=3

ConcepTest #2ConcepTest #2

In a cubic box which of the following In a cubic box, which of the following states is lowest in energy?

A. n1=3, n2=1, n3=1

B. n1=2, n2=2, n3=2B. n1 2, n2 2, n3 2

C. n1=3, n2=2, n3=1

D. n1=2, n2=2, n3=3

Three Dimensional Box for Electrons: a Quantum Dot

Quantum dot is "Particle in a Box”quantum systemquantum system

Displays size-dependent electronicp y pand optical properties

Useful in nanotechnologyUseful in nanotechnology

Sharp emission of lightp gmakes them useful light "tags"

Energy Levels of Quantum Dot Depend on Dot DiameterDepend on Dot Diameter

Three dimensional cube with an electron inside

2 2

2 2 21 2 322

E n n nm a

Electron transitions between the lowest two l l h ΔE h d d di

2 em a

energy levels have ΔE that depends on diameter

2 2 2 2

2 32 1 1 1 1 1E E E 2,1,1 1,1,1 2 22 1 1 1 1 1

2 2e e

E E Em a m a

hcE h Photon emitted (or absorbed)

2

E h

or a

Photon emitted (or absorbed)with energy E, wavelength

Emission Spectrum ofQuantum Dots

Increasing Quantum Dot Diameter

Quantum Dot “Problem”Quantum Dot ProblemAssume that light emission from quantum dot particles can be described by a cubic (a=b=c) "particle in a box"We observe light is emitted with wavelengths ranging from red (= 650 nm) to green (= 530 nm) from red (= 650 nm) to green (= 530 nm). If the observed light emission arises from the transition (n1=3, n2=3, n3=3) (n1=2, n2=2, n3=2), trans t on (n1 , n2 , n3 ) (n1 , n2 , n3 ), what is the diameter of the quantum dots giving rise to the red and green emissions?

Answer: a = 1.72 nm for red emissiona = 1.55 nm for green emissiong

Micrograph of InGaAs Quantum Micrograph of InGaAs Quantum Dots (20 nm wide, 8 nm high)

Fluorescence Microscopy Imageof mouse in vivo capillary structurep y

using CdSe-ZnS Quantum Dots