Today’s outline - September 16, 2015csrri.iit.edu/~segre/phys405/15F/lecture_07.pdfToday’s outline - September 16, 2015 Delta function well Finite square well Problem 2.39 Problem

Today’s outline - September 16, 2015

• Delta function well

• Finite square well

• Problem 2.39

• Problem 2.46

• Scattering from a well/barrier

Reading Assignment: Chapter 3.1

Homework Assignment #03:Chapter 2: 16, 18, 20, 21, 38, 49due Monday, September 21, 2015

C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 1 / 18




• Problem 2.39

• Problem 2.46








• Problem 2.39

• Problem 2.46








• Problem 2.39

• Problem 2.46








• Problem 2.39

• Problem 2.46








• Problem 2.39

• Problem 2.46








• Problem 2.39

• Problem 2.46








• Problem 2.39

• Problem 2.46





Dirac delta function

The Dirac delta function is a simplepotential to solve and it illustratesbound and scattering states.

It isinfinitely high and narrow and it hasunit area.

When multiplied with a function itpicks out the function at a singlevalue

δ(x) ≡

{0, if x 6= 0

∞, if x = 0∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx = f (a)

This integral works for any limits which include the peak of the deltafunction.



The Dirac delta function is a simplepotential to solve and it illustratesbound and scattering states. It isinfinitely high and narrow

and it hasunit area.


δ(x) ≡

{0, if x 6= 0

∞, if x = 0∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx = f (a)




The Dirac delta function is a simplepotential to solve and it illustratesbound and scattering states. It isinfinitely high and narrow

and it hasunit area.


δ(x) ≡

{0, if x 6= 0

∞, if x = 0

∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx = f (a)




The Dirac delta function is a simplepotential to solve and it illustratesbound and scattering states. It isinfinitely high and narrow and it hasunit area.


δ(x) ≡

{0, if x 6= 0

∞, if x = 0

∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx = f (a)






δ(x) ≡

{0, if x 6= 0

∞, if x = 0∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx = f (a)






δ(x) ≡

{0, if x 6= 0

∞, if x = 0∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx = f (a)






δ(x) ≡

{0, if x 6= 0

∞, if x = 0∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)

∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx = f (a)






δ(x) ≡

{0, if x 6= 0

∞, if x = 0∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx

= f (a)






δ(x) ≡

{0, if x 6= 0

∞, if x = 0∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx = f (a)






δ(x) ≡

{0, if x 6= 0

∞, if x = 0∫ +∞

−∞δ(x)dx = 1

f (x)δ(x − a) = f (a)δ(x − a)∫ +∞

−∞f (x)δ(x − a)dx = f (a)

∫ +∞

−∞δ(x − a)dx = f (a)



Delta function potential well

Consider a potential of the form

where α is a positive constant

the Schrodinger equation is nowand if E < 0, there is a bound state

start with region x < 0 where thepotential is zero

the general solution is

but since the first term is un-bounded as x → −∞, we chooseA = 0

V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0

ψ(x) = ��Ae−κx + Be+κx

= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)





the Schrodinger equation is now

and if E < 0, there is a bound state




V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)





the Schrodinger equation is now

and if E < 0, there is a bound state




V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0

ψ(x) = Ae−κx + Be+κx

= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)









V (x) = −αδ(x)

Eψ = − ~2

2m

d2ψ

dx2− αδ(x)ψ

d2ψ

dx2= −2mE

~2ψ = κ2ψ

κ ≡√−2mE

~> 0


= Be+κx , (x < 0)


Bound state solution

In the region x > 0 we have thesame Schrodinger equation and thesolution must have the same form

this time we must choose G = 0 tohave a bounded wave function, so

but ψ(x) must be continuous, so atx = 0 we see that F = B

the energy of this state is

E = −~2κ2

2m

= −mα2

2~2

normalizing

ψ(x) = Fe−κx +��Ge+κx

ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m

= −mα2

2~2

normalizing

ψ(x) = Fe−κx + Ge+κx

ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m

= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m

= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m

= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m

= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m

= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx

= 2|B|2∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx

=|B|2

κ→ B =

√κ =

√mα

~







E = −~2κ2

2m= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ

→ B =√κ =

√mα

~







E = −~2κ2

2m= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ

=

√mα

~







E = −~2κ2

2m= −mα2

2~2

normalizing


ψ(x) =

{Be+κx , x ≤ 0

Be−κx , x ≥ 0

ψ(x)

x

κ1/2

1 =

∫ +∞

−∞|ψ(x)|2dx = 2|B|2

∫ ∞0

e−2κxdx =|B|2

κ→ B =

√κ =

√mα

~


Derivative of the wave function

We mentioned previously that the wavefunction which solves theSchrodinger equation must satisfy two conditions:

ψ is always continuousdψdx is continuous except where V (x) =∞

what can this second condition tell us in the case of the delta functionpotential? Integrate the Schrodinger equation through the delta functiondiscontinuity at x = 0. Now take the limit ε→ 0.

��

E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0

= − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

The last term is usually zero, unless V (x)→∞




ψ is always continuous

dψdx is continuous except where V (x) =∞


��

E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0

= − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx







��

E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0

= − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx






what can this second condition tell us in the case of the delta functionpotential?

Integrate the Schrodinger equation through the delta functiondiscontinuity at x = 0. Now take the limit ε→ 0.

��

E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0

= − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx






what can this second condition tell us in the case of the delta functionpotential? Integrate the Schrodinger equation through the delta functiondiscontinuity at x = 0.

Now take the limit ε→ 0.

��

E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0

= − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx








E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0

= − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx








E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0

= − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx







E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0

= − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx







��

E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0 = − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx







��

E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0 = − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx







��

E

∫ +ε

−εψ(x)dx = − ~2

2m

∫ +ε

−ε

d2ψ

dx2dx +

∫ +ε

−εV (x)ψ(x)dx

0 = − ~2

2m

dψ

dx

∣∣∣∣ε−ε

+

∫ +ε

−εV (x)ψ(x)dx

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

The last term is usually zero, unless V (x)→∞C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18

Delta function discontinuity

For the delta function, thislimit is non-zero and can becalculated

For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε

−ε−αδ(x)ψ(x)dx

= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ

,dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ

→ ψ(0) = B




For our solution

ψ(x) =

{Be−κx , x ≥ 0

Be+κx , x ≤ 0

and

dψ

dx=

{−Bκe−κx , x > 0

+Bκe+κx , x < 0

Thus

∆

(dψ

dx

)=

2m

~2limε→0

∫ +ε

−εV (x)ψ(x)dx

=2m

~2limε→0

∫ +ε


= −2mα

~2ψ(0) lim

ε→0

∫ +ε

−εδ(x)dx

= −2mα

~2ψ(0) = −2κψ(0)

dψ

dx

∣∣∣∣+

= −Bκ, dψ

dx

∣∣∣∣−

= +Bκ

∆

(dψ

dx

)= −2Bκ → ψ(0) = B


Delta function well summary

V (x) = −αδ(x)

ψ(x) =

√mα

~e−mα|x |/~

2

E =mα2

2~2

∆

(dψ

dx

)= −2mα

~2ψ(0)

ψ(x)

x

κ1/2

Thus, the negative delta-function potential has a single bound state.There is always only one bound state for this potential, independent of thestrength of the potential α.

Now turn to a more complex finite bound system



V (x) = −αδ(x)

ψ(x) =

√mα

~e−mα|x |/~

2

E =mα2

2~2

∆

(dψ

dx

)= −2mα

~2ψ(0)

ψ(x)

x

κ1/2





V (x) = −αδ(x)

ψ(x) =

√mα

~e−mα|x |/~

2

E =mα2

2~2

∆

(dψ

dx

)= −2mα

~2ψ(0)

ψ(x)

x

κ1/2





V (x) = −αδ(x)

ψ(x) =

√mα

~e−mα|x |/~

2

E =mα2

2~2

∆

(dψ

dx

)= −2mα

~2ψ(0)

ψ(x)

x

κ1/2





V (x) = −αδ(x)

ψ(x) =

√mα

~e−mα|x |/~

2

E =mα2

2~2

∆

(dψ

dx

)= −2mα

~2ψ(0)

ψ(x)

x

κ1/2

Thus, the negative delta-function potential has a single bound state.

There is always only one bound state for this potential, independent of thestrength of the potential α.




V (x) = −αδ(x)

ψ(x) =

√mα

~e−mα|x |/~

2

E =mα2

2~2

∆

(dψ

dx

)= −2mα

~2ψ(0)

ψ(x)

x

κ1/2





V (x) = −αδ(x)

ψ(x) =

√mα

~e−mα|x |/~

2

E =mα2

2~2

∆

(dψ

dx

)= −2mα

~2ψ(0)

ψ(x)

x

κ1/2




Finite square well

-a +a x

0

I II III

-Vo

In region I, x < −a

Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx

for the well in region II, |x | ≤ a

Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)

~ψ = C sin(lx) + D cos(lx)

finally, in region III, x > +a

Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~

ψ = ��Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)

~

ψ = C sin(lx) + D cos(lx)


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~

ψ = Fe−κx +��Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx + Ge+κx


Finite square well

-a +a x

0

I II III

-Vo


Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = ��

Ae−κx + Be+κx


Eψ = − ~2

2m

d2ψ

dx2− V0ψ

− l2ψ =d2ψ

dx2, l ≡

√2m(E + V0)



Eψ = − ~2

2m

d2ψ

dx2

κ2ψ =d2ψ

dx2, κ ≡

√−2mE

~ψ = Fe−κx +��

Ge+κx


Boundary conditions

Both the wave function and its derivative must be continuous at theboundaries of the three regions, x = −a,+a.

Let’s consider the evensolutions initially, where C ≡ 0.

ψ(x) =

Be+κx , x < −aD cos(lx), |x | < a

Fe−κx , x > +a

let z ≡ la, and z0 ≡ a~√

2mV0

from our defining equations

κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have

Fe−κa = D cos(la)

− κFe−κa = −lD sin(la)

dividing the two equations:

κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions

Both the wave function and its derivative must be continuous at theboundaries of the three regions, x = −a,+a. Let’s consider the evensolutions initially, where C ≡ 0.

ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

=2mV0

~2

=z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√

z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z

√(z0z

)2− 1= tan z


Boundary conditions


ψ(x) =


Fe−κx , x > +a


2mV0


κ2 + l2 =−2mE

~2+

2m(E + V0)

~2

κ2 +z2

a2=

2mV0

~2=

z20a2

κa =√z20 − z2

at x = +a we have




κ = l tan(la)

1

a

√z20 − z2 = l tan z

1

z

√z20 − z2 = tan z√(z0z

)2− 1= tan z


Even solutions to finite well

tan z =

√(z0z

)2− 1

This is a transcendental equationwhich defines the discrete ener-gies which are allowed as stationarystates.

0 π/2 π 3π/2 2π 5π/2 3π

z

z0=2z0=4

z0=8



tan z =

√(z0z

)2− 1


0 π/2 π 3π/2 2π 5π/2 3π

z

z0=2z0=4

z0=8



tan z =

√(z0z

)2− 1


0 π/2 π 3π/2 2π 5π/2 3π

z

z0=2z0=4

z0=8


Limiting case: deep (wide) well

and the even solutions approach

z = la =

√2m(E + V0)

~a

Since En + V0 is just the energyabove the bottom of the well andthe width is 2a, the even solutions(and the odd ones, of course) ap-proach those of the infinite squarewell.

If V0 →∞ then z0 →∞

zn →nπ

2, n = 1, 3, 5, · · ·

nπ

2a~ ≈

√2m(En + V0)

n2π2~2

(2a)2≈ 2m(En + V0)

En + V0 ≈n2π2~2

2m(2a)2




z = la =

√2m(E + V0)

~a



zn →nπ

2, n = 1, 3, 5, · · ·

nπ

2a~ ≈

√2m(En + V0)

n2π2~2

(2a)2≈ 2m(En + V0)

En + V0 ≈n2π2~2

2m(2a)2




z = la =

√2m(E + V0)

~a



zn →nπ

2, n = 1, 3, 5, · · ·

nπ

2a~ ≈

√2m(En + V0)

n2π2~2

(2a)2≈ 2m(En + V0)

En + V0 ≈n2π2~2

2m(2a)2




z = la =

√2m(E + V0)

~a



zn →nπ

2, n = 1, 3, 5, · · ·

nπ

2a~ ≈

√2m(En + V0)

n2π2~2

(2a)2≈ 2m(En + V0)

En + V0 ≈n2π2~2

2m(2a)2




z = la =

√2m(E + V0)

~a



zn →nπ

2, n = 1, 3, 5, · · ·

nπ

2a~ ≈

√2m(En + V0)

n2π2~2

(2a)2≈ 2m(En + V0)

En + V0 ≈n2π2~2

2m(2a)2




z = la =

√2m(E + V0)

~a



zn →nπ

2, n = 1, 3, 5, · · ·

nπ

2a~ ≈

√2m(En + V0)

n2π2~2

(2a)2≈ 2m(En + V0)

En + V0 ≈n2π2~2

2m(2a)2




z = la =

√2m(E + V0)

~a



zn →nπ

2, n = 1, 3, 5, · · ·

nπ

2a~ ≈

√2m(En + V0)

n2π2~2

(2a)2≈ 2m(En + V0)

En + V0 ≈n2π2~2

2m(2a)2




z = la =

√2m(E + V0)

~a



zn →nπ

2, n = 1, 3, 5, · · ·

nπ

2a~ ≈

√2m(En + V0)

n2π2~2

(2a)2≈ 2m(En + V0)

En + V0 ≈n2π2~2

2m(2a)2


Limiting case: shallow (narrow) well

As the well becomes more shallow, V0 → 0 and z0 → 0 as well.

The number of states decreases until the lowest odd bound state vanishes.However, the ground state (lowest even state) will never vanish. There isalways a bound state no matter how shallow the well!

0 π/2 π 3π/2 2π 5π/2 3π

z

z0=2z0=4

z0=8




The number of states decreases until the lowest odd bound state vanishes.

However, the ground state (lowest even state) will never vanish. There isalways a bound state no matter how shallow the well!

0 π/2 π 3π/2 2π 5π/2 3π

z

z0=2z0=4

z0=8






0 π/2 π 3π/2 2π 5π/2 3π

z

z0=2z0=4

z0=8






0 π/2 π 3π/2 2π 5π/2 3π

z

z0=2z0=4

z0=8






0 π/2 π 3π/2 2π 5π/2 3π

z

z0=2z0=4

z0=8




The number of states decreases until the lowest odd bound state vanishes.However, the ground state (lowest even state) will never vanish. There isalways a bound state no matter how shallow the well!

0 π/2 π 3π/2 2π 5π/2 3π

z

z0=2z0=4

z0=8


Problem 2.39

(a) Show that the wave function of a particle in the infinite square wellreturns to its original form after a quantum revival timeT = 4ma2/π~. That is: Ψ(x ,T ) = Ψ(x , 0) for any state (not just astationary state).

(b) What is the classical revival time, for a particle of energy E bouncingback and forth between the walls?

(c) For what energy are the two revival times equal?

(a) Start with a generalized time-dependent wave function

Ψ(x , t) =∞∑n=1

cnψn(x)e−iEnt/~

where En = n2π2~2/2ma2


Problem 2.39





Ψ(x , t) =∞∑n=1

cnψn(x)e−iEnt/~



Problem 2.39





Ψ(x , t) =∞∑n=1

cnψn(x)e−iEnt/~



Problem 2.39





Ψ(x , t) =∞∑n=1

cnψn(x)e−iEnt/~



Problem 2.39(a) (cont.)

If we postulate that there is a revival time T , we can write that

Ψ(x , t) = Ψ(x , t + T )∞∑n=1

cnψn(x)e−iEnt/~ =∞∑n=1

cnψn(x)e−iEn(t+T )/~

=∞∑n=1

cnψn(x)e−iEnt/~−in2π2~T/2ma2

=∞∑n=1

cnψn(x)e−iEnt/~e−in2π2~T/2ma2

The revival condition can only be met for all elements in the sum when thelast term in the exponent is 2π, leading to

T = 2π2ma2

π2~=

4ma2

π~




Ψ(x , t) = Ψ(x , t + T )

∞∑n=1



=∞∑n=1


=∞∑n=1



T = 2π2ma2

π2~=

4ma2

π~




Ψ(x , t) = Ψ(x , t + T )∞∑n=1



=∞∑n=1


=∞∑n=1



T = 2π2ma2

π2~=

4ma2

π~




Ψ(x , t) = Ψ(x , t + T )∞∑n=1



=∞∑n=1


=∞∑n=1



T = 2π2ma2

π2~=

4ma2

π~




Ψ(x , t) = Ψ(x , t + T )∞∑n=1



=∞∑n=1


=∞∑n=1



T = 2π2ma2

π2~=

4ma2

π~




Ψ(x , t) = Ψ(x , t + T )∞∑n=1



=∞∑n=1


=∞∑n=1



T = 2π2ma2

π2~=

4ma2

π~




Ψ(x , t) = Ψ(x , t + T )∞∑n=1



=∞∑n=1


=∞∑n=1



T = 2π2ma2

π2~=

4ma2

π~




Ψ(x , t) = Ψ(x , t + T )∞∑n=1



=∞∑n=1


=∞∑n=1



T = 2π2ma2

π2~

=4ma2

π~




Ψ(x , t) = Ψ(x , t + T )∞∑n=1



=∞∑n=1


=∞∑n=1



T = 2π2ma2

π2~=

4ma2

π~


Problem 2.39(b & c)

The classical revival time is thetime for a round trip of the par-ticle in the well, Tc = 2a/v .

Thevelocity can be calculated from theenergy

Equating the two revival times

which is simply one quarter of theground state energy

E =1

2mv2

v =

√2E

m

Tc = a

√2m

E

4ma2

π~= a

√2m

E

E =π2~2

8ma2=

E1

4


Problem 2.39(b & c)

The classical revival time is thetime for a round trip of the par-ticle in the well, Tc = 2a/v . Thevelocity can be calculated from theenergy



E =1

2mv2

v =

√2E

m

Tc = a

√2m

E

4ma2

π~= a

√2m

E

E =π2~2

8ma2=

E1

4


Problem 2.39(b & c)




E =1

2mv2

v =

√2E

m

Tc = a

√2m

E

4ma2

π~= a

√2m

E

E =π2~2

8ma2=

E1

4


Problem 2.39(b & c)




E =1

2mv2

v =

√2E

m

Tc = a

√2m

E

4ma2

π~= a

√2m

E

E =π2~2

8ma2=

E1

4


Problem 2.39(b & c)




E =1

2mv2

v =

√2E

m

Tc = a

√2m

E

4ma2

π~= a

√2m

E

E =π2~2

8ma2=

E1

4


Problem 2.39(b & c)




E =1

2mv2

v =

√2E

m

Tc = a

√2m

E

4ma2

π~= a

√2m

E

E =π2~2

8ma2=

E1

4


Problem 2.39(b & c)




E =1

2mv2

v =

√2E

m

Tc = a

√2m

E

4ma2

π~= a

√2m

E

E =π2~2

8ma2=

E1

4


Problem 2.39(b & c)




E =1

2mv2

v =

√2E

m

Tc = a

√2m

E

4ma2

π~= a

√2m

E

E =π2~2

8ma2

=E1

4


Problem 2.39(b & c)




E =1

2mv2

v =

√2E

m

Tc = a

√2m

E

4ma2

π~= a

√2m

E

E =π2~2

8ma2=

E1

4


Problem 2.46

Imagine a bead of mass m that slides frictionlessly around a circular wirering of circumference L. (This is just like a free particle, except thatψ(x + L) = ψ(x).) Find the stationary states (with appropriatenormalization) and the corresponding allowed energies. Note that thereare two independent solutions for each energy En – corresponding toclockwise and counter-clockwise circulation; call them ψ+

n (x) and ψ−n (x).

If we define the x coordinate as thedirection along the circumferenceof the ring with the positive direc-tion as the clockwise direction andthe negative direction as counter-clockwise, we have

d2ψ

dx2= k2ψ

k =

√2mE

~

ψ(x) = Ae ikx + Be−ikx


Problem 2.46




d2ψ

dx2= k2ψ

k =

√2mE

~



Problem 2.46




d2ψ

dx2= k2ψ

k =

√2mE

~



Problem 2.46




d2ψ

dx2= k2ψ

k =

√2mE

~



Problem 2.46 (cont)

Now we can applythe boundary condi-tion that the wave-function must be iden-tical when x → x + L

We apply it at two val-ues: x = 0 and x =π/2k

Adding the two equa-tions


Ae ikx + Be−ikx = Ae ik(x+L) + Be−ik(x+L)

= Ae ikxe ikL + Be−ikxe−ikL

A + B= Ae ikL + Be−ikL

Ae iπ/2 + Be−iπ/2 = Ae iπ/2e ikL + Be−iπ/2e−ikL

iA− iB = iAe ikL − iBe−ikL

A− B= Ae ikL − Be−ikL

A +��B + A−��B = Ae ikL +��Be−ikL + Ae ikL −��

Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)


We apply it at two val-ues: x = 0

and x =π/2k










Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)


We apply it at two val-ues: x = 0

and x =π/2k










Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)











A + B + A− B = Ae ikL + Be−ikL + Ae ikL − Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A

−→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)












Be−ikL

2Ae ikL = 2A −→ A = 0 or e ikL = 1

thus kL = 2nπ (n = 0,±1,±2,±3, . . . )


Problem 2.46 (cont)

Note that if A = 0, we can obtainthe same result by subtracting thetwo equations and putting a con-straint on B

Because we get positive and nega-tive solutions in any case, the fullsolution can be written

normalizing gives

and the full solution is thus

ψ+n (x) = Ae+i(2nπx/L)

ψ−n (x) = Be−i(2nπx/L)∫ L

0|ψ±|2 dx = 1

A = B =1√L

ψ±n (x) =1√Le±i(2nπx/L); En =

2n2π2~2

mL2(n = 0, 1, 2, 3, . . . )


Problem 2.46 (cont)



normalizing gives




0|ψ±|2 dx = 1

A = B =1√L


2n2π2~2

mL2(n = 0, 1, 2, 3, . . . )


Problem 2.46 (cont)



normalizing gives




0|ψ±|2 dx = 1

A = B =1√L


2n2π2~2

mL2(n = 0, 1, 2, 3, . . . )


Problem 2.46 (cont)



normalizing gives



ψ−n (x) = Be−i(2nπx/L)

∫ L

0|ψ±|2 dx = 1

A = B =1√L


2n2π2~2

mL2(n = 0, 1, 2, 3, . . . )


Problem 2.46 (cont)



normalizing gives




0|ψ±|2 dx = 1

A = B =1√L


2n2π2~2

mL2(n = 0, 1, 2, 3, . . . )


Problem 2.46 (cont)



normalizing gives




0|ψ±|2 dx = 1

A = B =1√L


2n2π2~2

mL2(n = 0, 1, 2, 3, . . . )


Problem 2.46 (cont)



normalizing gives




0|ψ±|2 dx = 1

A = B =1√L


2n2π2~2

mL2(n = 0, 1, 2, 3, . . . )


Problem 2.46 (cont)



normalizing gives




0|ψ±|2 dx = 1

A = B =1√L


2n2π2~2

mL2(n = 0, 1, 2, 3, . . . )


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Today’s outline - September 16, 2015csrri.iit.edu/~segre/phys405/15F/lecture_07.pdfToday’s outline - September 16, 2015 Delta function well Finite square well Problem 2.39 Problem