Making Movies - Time Development

Recall the rectangular initial wave packet in the infinite square well shownbelow. How does it evolve in time?

V (x) = 0 0 < x < a

= ∞ x ≤ 0 and x ≥ a

|Ψ(x , 0)〉 = 1√d

x0 ≤ x ≤ x1 and d = x0 − x1

= 0 otherwise

x0 x1

V(x)

Ψ(x,0)

0 a x

a = 1.0 Åx0 = 0.3 Åx1 = 0.5 Å

Jerry Gilfoyle Time Development 1 / 13

Probabilities of Different States

a = 1.0 Å

x0 = 0.3 Å

x1 = 0.5 Å

0 500 1000 1500 2000 2500 30000.0

0.1

0.2

0.3

0.4

Energy (eV)

Pro

babili

ty

Rectangular Wave in a Square Well

Jerry Gilfoyle Time Development 2 / 13

Time Development of a Square Wave

0.0 0.2 0.4 0.6 0.8 1.00

2

4

6

8

x

Pro

babili

tyD

ensity

time=0.0000⨯10-16s.

0.0 0.2 0.4 0.6 0.8 1.00

2

4

6

8

x

Pro

babili

tyD

ensity

time=0.0040⨯10-16s.

0.0 0.2 0.4 0.6 0.8 1.00

2

4

6

8

x

Pro

babili

tyD

ensity

time=0.0080⨯10-16s.

0.0 0.2 0.4 0.6 0.8 1.00

2

4

6

8

x

Pro

babili

tyD

ensity

time=0.0120⨯10-16s.

Jerry Gilfoyle Time Development 3 / 13

Comparison of Bound and Free Particles

Particle in a Box

The potential

V =0 0 < x < a

=∞ otherwise

Eigenfunctions and eigenvalues

|φn〉 =√

2

asin(nπx

a

)En = n

2 ~2π2

2ma2

Superposition

|ψ〉 =∞∑n=1

bn|φn〉 〈φm|φn〉 = δm,n

Getting the coefficients

bn = 〈φn|ψ〉 Pn = |bn|2

Free Particle

The potential

V = 0

Eigenfunctions and eigenvalues

|φ(k)〉 = 1√2π

e±ikx E =~2k2

2m

Superposition

|ψ〉 =∫ ∞−∞

b(k)φ(k)dk

〈φ(k ′)|φ(k)〉 =δ(k − k ′)

Getting the coefficients

b(k) = 〈φ(k)|ψ〉 Pn = |b(k)|2

Jerry Gilfoyle Time Development 4 / 13

Time Development of the Initial Gaussian

Recall the Gaussian initial wave packet for the free particle shown below.How does it evolve in time?

V (x) = 0 |Ψ(x , 0)〉 = 1(2πσ2)1/4

e−x2/4σ2

-4 -2 0 2 40.0

0.1

0.2

0.3

0.4

0.5

x

ψ

Jerry Gilfoyle Time Development 5 / 13

Time Development of the Initial Gaussian

0 1 2 3 4 5 60

1

2

3

4

5

x

|ψ2

t=0

0 1 2 3 4 5 60

1

2

3

4

5

x

|ψ2

t=1.5

0 1 2 3 4 5 60

1

2

3

4

5

x

|ψ2

t=3.

0 1 2 3 4 5 60

1

2

3

4

5

x

|ψ2

t=4.5

Jerry Gilfoyle Time Development 6 / 13

Time Development of Nuclear Fusion

Consider a case of one dimensional nuclear ‘fusion’. A neutron is in thepotential well of a nucleus that we will approximate with an infinite squarewell with walls at x = 0 and x = L. The eigenfunctions and eigenvalues are

En =n2~2π2

2ma2φn =

√2

asin(nπx

a

)0 ≤ x ≤ a

= 0 x < 0 and x > a .

The neutron is in the n = 4 state when it fuses with another nucleus thatis the same size, instantly putting the neutron in a new infinite square wellwith walls at x = 0 and x = 2a.

1 What are the new eigenfunctions and eigenvalues of the fused system?

2 How will the initial wave packet evolve in time?

Jerry Gilfoyle Time Development 7 / 13

Time Development of Nuclear Fusion

0 5 10 15 200.00

0.05

0.10

0.15

0.20

x(fm)

|ψ2

t=0×10-21 s

0 5 10 15 200.00

0.05

0.10

0.15

0.20

x(fm)

|ψ2

t=0.3×10-21 s

0 5 10 15 200.00

0.05

0.10

0.15

0.20

x(fm)

|ψ2

t=0.6×10-21 s

0 5 10 15 200.00

0.05

0.10

0.15

0.20

x(fm)

|ψ2

t=0.9×10-21 s

Jerry Gilfoyle Time Development 8 / 13

Comparison of Bound and Free Particles

Particle in a Box

The potential

V =0 0 < x < a

=∞ otherwise

Eigenfunctions and eigenvalues

|φn〉 =√

2

asin(nπx

a

)En = n

2 ~2π2

2ma2

Superposition

|ψ〉 =∞∑n=1

bn|φn〉 〈φm|φn〉 = δm,n

Getting the coefficients

bn = 〈φn|ψ〉 Pn = |bn|2

Time Dependence

Ψ(x , t) =∞∑n=1

bn|φn(x)〉e−iωnt

Free Particle

The potential

V = 0

Eigenfunctions and eigenvalues

|φ(k)〉 =1√

2πe±ikx E =

~2k2

2m

Superposition

|ψ〉 =∫ ∞−∞

b(k)φ(k)dk

〈φ(k ′)|φ(k)〉 =δ(k − k ′)

Getting the coefficients

b(k) = 〈φ(k)|ψ〉 Pn = |b(k)|2

Time Dependence

Ψ(x , t) =

∫ ∞−∞

b(k)φk (x)e−iω(k)tdk

Jerry Gilfoyle Time Development 9 / 13

Liboff 6.4 - 1

Jerry Gilfoyle Time Development 10 / 13

Liboff 6.4 - 2

Jerry Gilfoyle Time Development 11 / 13

Liboff 6.4 - 3

Mathematica result

Jerry Gilfoyle Time Development 12 / 13

Liboff 6.4 - 4

Mathematica result

Jerry Gilfoyle Time Development 13 / 13

Liboff 6.4 - 5

Black Curve - Initial

Probability

Density.

Red Curve - Probability

Density at

t = 10 s.

Chance of being in

the original box: 0.993

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.070

2

4

6

8

10

12

x (m)

|ψ2(m

-1)

Square Wave Time Development

Jerry Gilfoyle Time Development 14 / 13