Today’s outline - September 16, 2015csrri.iit.edu/~segre/phys405/15F/lecture_07.pdfToday’s...
Transcript of Today’s outline - September 16, 2015csrri.iit.edu/~segre/phys405/15F/lecture_07.pdfToday’s...
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
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
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
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
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
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
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
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
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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.
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 2 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
Delta function potential well
Consider a potential of the form
where α is a positive constant
the Schrodinger equation is now
and 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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
Delta function potential well
Consider a potential of the form
where α is a positive constant
the Schrodinger equation is now
and 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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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)
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 3 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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
~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 4 / 18
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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
Derivative of the wave function
We mentioned previously that the wavefunction which solves theSchrodinger equation must satisfy two conditions:
ψ is always continuous
dψdx 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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
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)→∞
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 5 / 18
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)→∞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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 6 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 7 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 7 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 7 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 7 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 7 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 7 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 7 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 8 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
=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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 9 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 10 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 10 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 10 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 11 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 11 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 11 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 11 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 11 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 11 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 11 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 11 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 12 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 12 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 12 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 12 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 12 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 12 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 13 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 13 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 13 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 13 / 18
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~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 14 / 18
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~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 14 / 18
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~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 14 / 18
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~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 14 / 18
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~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 14 / 18
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~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 14 / 18
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~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 14 / 18
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~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 14 / 18
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~
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 14 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 15 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 15 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 15 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 15 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 15 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 15 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 15 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 15 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 15 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 16 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 16 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 16 / 18
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
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 16 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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
ψ(x) = Ae ikx + Be−ikx
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 17 / 18
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 18 / 18
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 18 / 18
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 18 / 18
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 18 / 18
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 18 / 18
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 18 / 18
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 18 / 18
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, . . . )
C. Segre (IIT) PHYS 405 - Fall 2015 September 16, 2015 18 / 18