1. No category

Physics 220
Homework #3
Spring 2015
Due Wednesday 4/22/15
1.
A particle of mass m is moving in one dimension in a potential V ( x,t ) . The wave
function for the particle is Ψ ( x,t ) = Axe
and k are constants.
⎛ km ⎞ 2
⎛3 k ⎞
−⎜
x −i⎜
t
⎝ 2! ⎟⎠
⎝ 2 m ⎟⎠
e
for −∞ < x < ∞ , where A
a. Show that V is independent of t , and determine V ( x ) .
b. Normalize the wave function and determine the constant A .
c. Using the normalized wave function, calculate x , x 2 , p , and p 2 .
a. To determine the potential we use the SWE.
Ψ = Axe
−
3 k
km 2
t
x −i
2 m
2!
e
! d Ψ
+ VΨ = EΨ
2m dx 2
−
dΨ
d⎛
EΨ = i!
= i! ⎜ Axe
dt
dt ⎝
−
2
2
−
dΨ d ⎛
= ⎜ Axe
dx dx ⎝
3 k
km 2
t
x −i
2 m
2!
e
3 k
km 2
t
x −i
2 m
2!
e
⎞
−
⎟ = Ae
⎠
⎞ 3! k
⎟ = 2 mΨ
⎠
3 k
km 2
t
x −i
2 m
2!
e
⎛ 2x km −
+ Ax ⎜ −
e
2!
⎝
3 k
km 2
t
x −i
2 m
2!
e
d 2 Ψ d ⎛ dΨ ⎞
km
km
= ⎜
6Ψ + 2 4x 2 Ψ
⎟⎠ = −
2
⎝
dx
dx dx
2!
4!
−
⎞
!2 d 2Ψ
!2 ⎛
km
km
3! k
+
VΨ
=
EΨ
→
−
−
6Ψ + 2 4x 2 Ψ ⎟ + VΨ =
Ψ
2
⎜
2m dx
2m ⎝ 2!
4!
2 m
⎠
⎛3 k 1 k
k⎞ 1 2 1 2
∴V = ! ⎜
−
−
+ kx = kx
m ⎟⎠ 2
2
⎝2 m 2 m
⎞
⎟
⎠
b. We normalize the wave function:
⎛
⎞
⎜
⎟
∞
2 km 2
−
x
π
⎟
*
2 2
2⎜
2!
dx = A
3
∫ Ψ Ψdx = 1 → 1 = −∞∫ A x e
⎜
2 ⎟
⎜ 2 ⎛ km ⎞ ⎟
⎜⎝ ⎜⎝ ! ⎟⎠ ⎟⎠
3
4
2 ⎛ km ⎞
∴A = 1 ⎜
π 4 ⎝ ! ⎟⎠
c. The expectation values:
∞
x =
∞
∫ Ψ xΨ dx = A ∫ x e
*
2
−∞
3
−
2 km 2
x
2!
dx = 0
−∞
∞
2 km 2
⎡ ∞ − 2 2!km x 2
⎤
−
x
d⎞
2 km
*⎛
2
3
2!
Ψ
−i!
Ψ
dx
=
A
xe
dx
−
x
e
dx ⎥ = 0
⎢∫
∫−∞ ⎜⎝ dx ⎟⎠
∫
2! −∞
⎣ −∞
⎦
∞
p =
∞
x
2
=
∞
∫ Ψ x Ψ dx = A ∫ x e
* 2
−∞
2
4
−∞
∞
p2 =
−
2 km 2
x
2!
5
3 π ⎛ ! ⎞ 2 3 km
dx = A
⎜
⎟ =
4 ⎝ km ⎠
20!
2
∞
∞
2
2
d ⎞⎛
d⎞
*⎛
2
* d Ψ
2
* d Ψ
Ψ
−i!
−i!
Ψ
dx
=
!
Ψ
dx
=
!
Ψ
∫−∞ ⎜⎝ dx ⎟⎠ ⎜⎝ dx ⎟⎠
∫−∞ dx 2
∫−∞ dx 2 dx
=
2.
Determine which of the following one-dimensional wave functions represent state of
definite momentum. For each wave function that does correspond to a state of
definite momentum, determine the momentum.
a. ψ ( x ) = eikx
b. ψ ( x ) = xeikx
c. ψ ( x ) = sin ( kx ) + i cos ( kx )
d. ψ ( x ) = eikx + e−ikx
dψ
d
= −i! ( eikx ) = −i! ( ik ) eikx = !kψ = pψ ∴ p = !k
dx
dx
dψ
d
b. −i!
= −i! ( xeikx ) = −i! ( eikx + ikxeikx ) ≠ pψ ∴ NO
dx
dx
dψ
d
c. −i!
= −i! ( sin kx + i cos kx ) = −i! ( k cos kx − ik sin kx ) = −!kψ ∴ p = −!k
dx
dx
dψ
d
d. −i!
= −i! ( eikx + e−ikx ) = −i! ( ikeikx − ike−ikx ) = !k ( eikx − e−ikx ) ≠ pψ ∴ NO
dx
dx
a. −i!
3. G1.4
a. Normalizing the wave function we have:
⎛ A (b − x ) ⎞
⎛ Ax ⎞
1 = ∫ ψ ψ dx = ∫ ⎜ ⎟ dx + ∫ ⎜
dx
⎝ a ⎠
⎝ b − a ⎟⎠
0
a
2
a
2
b
*
a
1=
A2 2
A2
x
dx
+
a 2 ∫0
( b − a )2
b
∫ (b − x )
2
dx
a
3
⎛
b − a) ⎞
A ⎛a ⎞
A
(
⎛ a b − a⎞
2 b
1= 2 ⎜ ⎟ −
= A2 ⎜ +
⎟⎠ = A
2 ⎜0−
⎟
⎝
a ⎝ 3 ⎠ (b − a ) ⎝
3 ⎠
3
3
3
2
∴A =
3
2
3
b
b.
c. From the graph the most probably location is at x = a .
d. To calculate the probabilities:
2
a
A 2 ⎛ a 3 ⎞ 3a 3 a
⎛ Ax ⎞
*
P = ∫ ψ ψ dx = ∫ ⎜ ⎟ dx = 2 ⎜ ⎟ =
= .
⎝ a ⎠
a ⎝ 3 ⎠ 3ba 2 b
0
We check the case that a = b , and we see that the probability is Unity.
We check the case that 2a = b an isosceles triangle, and we see that the
a a 1
probability is P = =
= .
b 2a 2
e. To calculate the expectation value of the position, we evaluate:
b
⎧ a 3
⎫⎪
1
2
*
2⎪ 1
x = ∫ ψ xψ dx = A ⎨ 2 ∫ x dx +
2 ∫ x ( b − x ) dx ⎬
(b − a ) a
⎩⎪ a 0
⎭⎪
a
b
3 ⎧⎪ 1 ⎛ x 2 ⎞
1 ⎛ 2 x2
x 3 x 4 ⎞ ⎫⎪
x = ⎨ 2⎜ ⎟ +
b
− 2b + ⎟ ⎬
b ⎪ a ⎝ 4 ⎠ 0 ( b − a )2 ⎜⎝ 2
3 4 ⎠ a⎪
⎩
⎭
4
⎛b
⎞ 2a + b
1
x =
− 3a 2b + 2a 3 ⎟ =
2 ⎜
4
⎠
4 (b − a ) ⎝ 3
4. G1.15
a. To show this result, we start with the definition of probability.
∞
P=
∞
∫ Ψ Ψ dx = ∫ Ψ
*
−∞
2
dx . Now if is normalized does it stay normalized and what
−∞
happens if the potential function is not real but contains an imaginary part? So, to
see if it stay normalized we take the time derivative. If the time derivative is zero
then the probability function will stay normalized. Let’s take the time derivative
(and note that the integral and time derivative commute):
∞
∞
∞
⎞ ∞ ∂
⎡ ∂Ψ ∂Ψ * ⎤
dP d ⎛
∂
2
2
= ⎜ ∫ Ψ dx ⎟ = ∫
Ψ dx = ∫
Ψ *Ψ dx = ∫ ⎢ Ψ *
+
Ψ ⎥ dx .
dt dt ⎝ −∞
∂t
∂t
∂t
⎠ −∞ ∂t
⎦
−∞
−∞ ⎣
Now we need expressions for the terms in the brackets.
! 2 ∂2 Ψ
∂Ψ
∂Ψ i! ∂ 2 Ψ iV
−
+
VΨ
=
i!
→
=
− Ψ and the complex conjugate
2m ∂x 2
∂t
∂t 2m ∂x 2
!
*
2
*
*
∂Ψ
i! ∂ Ψ iV *
=−
+
Ψ , assuming that the potential can be complex. Back to
∂t
2m ∂x 2
!
the expression in brackets.
2
⎡ * ∂Ψ ∂Ψ * ⎤
iV ⎞ ⎛ i! ∂ 2 Ψ iV * * ⎞
* ⎛ i! ∂ Ψ
Ψ
+
Ψ
=
Ψ
−
Ψ + −
+
Ψ ⎟Ψ
⎢
⎥
⎜⎝ 2m ∂x 2
∂t
∂t
! ⎟⎠ ⎜⎝ 2m ∂x 2
!
⎠
⎣
⎦
( )
(
)
i! ⎛ * ∂ 2 Ψ ∂ 2 Ψ * ⎞ i
=
Ψ
−
Ψ ⎟ + ( −VΨ *Ψ + V *Ψ *Ψ )
2
2
⎜
2m ⎝
∂x
∂x
⎠ !
If the potential function is given as V = V0 + iΓ and taking the complex conjugate
V * = V0 − iΓ we can evaluate the last term on the right and we find
i
i
2Γ *
2Γ 2
−V + V * ) Ψ *Ψ = (V0 − iΓ − V0 + iΓ ) Ψ *Ψ = −
Ψ Ψ=−
Ψ . Now
(
!
!
!
!
finishing the problem.
∞
∞
⎡ i! ⎛ * ∂ 2 Ψ ∂ 2 Ψ * ⎞ 2Γ 2 ⎤
⎡ ∂Ψ ∂Ψ * ⎤
dP
= ∫ ⎢Ψ*
+
Ψ ⎥ dx = ∫ ⎢
Ψ
−
Ψ −
Ψ ⎥ dx
dt −∞ ⎣
∂t
∂t
2m ⎜⎝
∂x 2
∂x 2 ⎟⎠ !
⎦
⎦
−∞ ⎣
∞
∞
∞
dP
∂ ⎡ i! ⎛ * ∂Ψ ∂Ψ * ⎞ ⎤ 2Γ
i! ⎛ * ∂Ψ ∂Ψ * ⎞
2Γ
2
=∫
Ψ
−
Ψ
−
Ψ
dx
=
Ψ
−
Ψ⎟ −
P
⎢
⎥
∫
⎜
⎟
⎜
dt −∞ ∂x ⎣ 2m ⎝
∂x
∂x ⎠ ⎦ ! −∞
2m ⎝
∂x
∂x ⎠ −∞ !
dP
2Γ
=−
P
dt
!
where evaluating the expression in parenthesis is zero since the wave function is
normalized it has to vanish at ±∞ .
b. From part a, we can solve for the probability as a function of time. We have:
dP
2Γ
dP
2Γ
2Γ
=−
P→
=−
dt → ln P = −
t
dt
!
P
!
!
.
2Γ
∴ P ( t ) = P0 e
−
!
t
Comparing this with P(t) given in the problem, we calculate the lifetime and
initial probability of the of the state. We have:
P ( t ) = P0 e
−
2Γ
t
!
=e
−
t
τ
⎧
P0 = 1
.
⎪
∴ ⎨ 2Γ 1
!
⎪ ! = τ → τ = 2Γ
⎩
!
5. Suppose that a wave function Ψ ( r,t ) is normalized. Show that the wave function
!
eiθ Ψ ( r,t ) , where θ is an arbitrary real number, is also normalized.
!
!
If Ψ ( r,t ) is normalized then P = ∫ Ψ *Ψd 3r = 1 . To show that eiθ Ψ ( r,t ) is also
!
!
!
!
normalized we compute P' = ∫ e−iθ Ψ * ( r,t ) eiθ Ψ ( r,t ) d 3r = ∫ Ψ * ( r,t ) Ψ ( r,t ) d 3r = 1 .
!
Therefore for eiθ Ψ ( r,t ) , the wave function is still normalized.
6. Griffith’s problem 1.17 parts a – e.
a. The normalization is
a
1= A
∫ (a
2
−a
2
−x
)
2 2
a
⎡
2a 2 x 3 x 5 ⎤
dx = 2A ∫ ( a − 2a x + x ) dx = 2A ⎢ a 4 x −
+ ⎥
3
5 ⎦0
⎣
0
a
2
4
2
2
4
2
16 5 2
15
a A →A=
15
16a 5
b. The expectation value of the position is:
1=
a
x = ∫ ψ xψ dx =
*
−a
a
∫ x (a
2
−a
− x 2 ) dx = 0 on Mathematica .
2
c. The expectation value of the momentum is:
a
a
d⎞
*⎛
2
p = ∫ ψ ⎜ −i! ⎟ ψ dx = −i!A ∫ ( a 2 − x 2 ) ( −2x ) dx = 0 on Mathematica
⎝
dx ⎠
−a
−a
d. The expectation value of the position squared is:
a
a
a
x 2 = ∫ ψ * x 2ψ dx = A 2 ∫ x 2 ( a 2 − x 2 ) dx = 2A 2 ∫ ( a 4 x 2 − 2a 2 x 4 + x 6 ) dx
−a
x2 =
2
a
7
−a
2
0
e. The expectation value of the momentum squared is:
2
a
a
a
d⎞
2
*⎛
2 2
2
2
2 2
p = ∫ ψ ⎜ −i! ⎟ ψ dx = −! A ∫ ( a − x ) ( −2 ) dx = 4! A ∫ ( a 2 − x 2 ) dx
⎝
dx ⎠
−a
−a
0
p2 =
5! 2
2a 2
7. Griffith’s problem 2.3
There are two ways to answer this problem.
d 2ψ −2mE
=
ψ
2
!2
Method I: The SWE reads: dx
.
The two cases we have are:
Case 1:
d 2ψ
E = 0 → 2 = 0 → ψ = A + Bx
dx
BC ' s : ψ ( 0 ) = 0 = A + B ( 0 ) → A = 0
ψ (a) = 0 = B(a) → B = 0
∴ψ = 0
Case 2:
d 2ψ
2mE
= − 2 ψ = k 2ψ → ψ = Aekx + Be− kx
2
dx
!
BC ' s : ψ ( 0 ) = 0 = A + B → A = −B → ψ = B ( ekx − e− kx )
E<0→
⎧⎪
B=0
ψ ( a ) = 0 = B ( eka − e− ka ) → ⎨ − ka
ka
2 ka
⎪⎩ e = e → 1 = e → ln(1) = 0 = 2ka → k = 0
∴ψ = 0
Method II: The lowest energy in the infinite square well is E1 =
π 2!2
. Why cannot the
2ma 2
p2
< 0 implies that the
energy be zero or negative? For the case of E =
2m
momentum is imaginary and a measurement would not yield a real value. For the
case of E = 0 , the uncertainty principle, the uncertainty in the position of the
!
!
particle is Δx ~ a and therefore ΔxΔp ≥ → Δp ≥
. But E = 0 if then
2
2a
p2
E=0=
→ Δp = 0 , which violates the uncertainty principle.
2m
8. Suppose that ψ 1 and ψ 2 are two different solutions of the time-independent
Schrodinger wave equation with the same energy E .
a. Show that ψ 1 + ψ 2 is also a solution with energy E .
b. Show that cψ 1 is also a solution of the Schrodinger equation with energy E .
a. To show ψ 1 + ψ 2 that is solutions to the SWE:
!2 2
ψ1 : −
∇ ψ 1 + Vψ 1 = Eψ 1
2m
!2 2
ψ2 :−
∇ ψ 2 + Vψ 2 = Eψ 2
2m
!2 2
Add : −
∇ (ψ 1 + ψ 2 ) + V (ψ 1 + ψ 2 ) = E (ψ 1 + ψ 2 )
2m
!2 2
∴−
∇ ψ + Vψ = Eψ ; ψ = ψ 1 + ψ 2
2m
Thus since ψ satisfies the SWE and is a solution, the sum ψ 1 + ψ 2 must also satisfy
the SWE.
b. To show that cψ 1 is solutions to the SWE:
!2 2
ψ1 : −
∇ ψ 1 + Vψ 1 = Eψ 1
2m
!2 2
!2
ψ 2 = cψ 1 : −
∇ ψ 2 + Vψ 2 = Eψ 2 → −
c∇ 2ψ 1 + Vcψ 1 = Ecψ 1
2m
2m
!2 2
∴−
∇ ψ 1 + Vψ 1 = Eψ 1
2m
which is a solution to the wave equation, so cψ 1 is a solution to the wave equation.
9. A particle of mass m is moving in one dimension near the speed of light so that the
p2
relation for the kinetic energy E =
is no longer valid. Instead, the total energy is
2m
given by E 2 = p 2 c 2 + m 2 c 4 . So, we can no longer use the Schrodinger equation.
Suppose that the wave function for the particle Ψ ( x,t ) is an eigenfunction of the
energy operator and an eigenfunction of the momentum operator, and also assume
there is no potential energy V . Derive a linear differential equation for Ψ ( x,t ) .
E 2 = p2c2 + m 2c4
i!
dψ
d ⎛ dψ ⎞
d
⎛ dψ ⎞
2
= Eψ → i! ⎜ i!
⎟⎠ = i! ( Eψ ) = E ⎜⎝ i!
⎟⎠ = E ψ
⎝
dt
dt
dt
dt
dt
d 2ψ
= E 2ψ
2
dt
dψ
d ⎛
dψ ⎞
d
2
−i!
= pψ → −i! ⎜ −i!
⎟⎠ = −i! ( pψ ) = p ψ
⎝
dx
dx
dx
dx
∴−! 2
∴−! 2
d 2ψ
= p 2ψ
2
dx
2
d 2ψ
2 2 d ψ
SWE : E ψ = p c ψ + m c ψ → −!
= −! c
+ m 2 c 4ψ
2
2
dt
dx
2
2 2
2 4
2
1 d 2ψ d 2ψ ⎛ mc ⎞
= 2 −⎜
ψ
⎝ ! ⎟⎠
c 2 dt 2
dx
This equation is called the Dirac equation and it forms the basis for the study of
relativistic quantum mechanics. It’s the relativistic form of the Schrodinger wave
equation.
2
∴
10. Determine the odd solutions to the finite square well. Determine the energy of the
single bound state with E < V0 . Normalize your solutions in each region to determine
the unknown coefficient A in each region. Plot your solution for ψ 2 (x) .
The plot of this wave function and the energy of the state are below in question #2.
11. Determine the normalization coefficients for the second energy state of the even
solutions to the finite square well. That is, renormalize the solutions and determine
B in each region for E3 . Plot your solution for ψ 3 (x) , along with the solutions for
ψ 2 (x) from above and ψ 1 (x) from class.
⎛ ka ⎞
Below is the plot of k ' = −k cot ⎜ ⎟ for the case that r = 4 . The coefficients for the
⎝ 2⎠
ka
mEa 2
ma 2V0
=
plot are plotted using α =
and r =
. The intersection was
2
2! 2
2! 2
determined using Mathematica’s find roots command and α = 2.475 . Therefore the
energy of this bound state is determined from α where
2
2
mEa 2 Er 2
⎛α⎞
⎛ 2.475 ⎞
α2 =
=
→
E
=
V
=
⎜⎝ ⎟⎠ 0 ⎜⎝
⎟ V0 = 0.383V0 .
2! 2
V0
r
4 ⎠
10
5
1
2
3
4
5
-5
Below is the plot for the wave functions of the bound states (ψ 1 ,ψ 2 , & ψ 3 ) of this
particular finite square well using r = 4 .
The wave functions are:
⎧
17.9 7.6a x
⎪
e
a
⎪
⎪⎪ 1.25
⎛ 2.5x ⎞
ψ1 = ⎨
cos ⎜
⎝ a ⎟⎠
a
⎪
⎪
17.9 − 7.6a x
e
⎪
a
⎪⎩
⎧
18.6 6.32a x
⎪
−
e
a
⎪
⎪⎪ 1.23 ⎛ 4.96x ⎞
ψ2 = ⎨
sin ⎜
⎝ a ⎟⎠
⎪ a
⎪
18.6 − 6.32a x
e
⎪
a
⎪⎩
⎧
5.8 3.48a x
⎪
−
e
a
⎪
⎪⎪ 1.25
⎛ 7.2x ⎞
ψ3 = ⎨
cos ⎜
⎝ a ⎟⎠
a
⎪
⎪
5.8 − 3.48a x
−
e
⎪
a
⎪⎩
ψ2
ψ1
1.0
0.5
ψ3
- 1.0
- 0.5
0.0
0.5
1.0
- 0.5
- 1.0
12. G2.40
⎧ Asin ( kx ) + B cos ( kx ) 0 ≤ x ≤ a
⎪
The solutions to the wave equation are: ψ ( x ) = ⎨
Ce k ' x + De− k ' x x > a
⎪⎩
where k =
2m ( E + V )
−2mE
and k ' =
.
!
!
Boundary Conditions:
C = 0 based on ψ having to vanish as x → ∞ .
Continuity of the wave function at x = 0 says that B = 0 .
Continuity of the wave function at x = a gives
Asin ( ka ) = De− k 'a → D = Ae k 'a sin ( ka ) .
Continuity of the first derivative of the wave function at x = a gives
Ak cos ( ka ) = −k ' De− k 'a = −k ' Ae k 'a sin ( ka ) e− k 'a . This yields −k cot(ka) = k ' .
−2mEa 2 2mVa 2
+
. In the
!2
!2
expression for k 'a we’ll pull a mathematical trick, namely adding and subtracting the
2mVa 2
same quantity,
. We get
!2
Multiply the last result by a and define z = ka =
−ka cot(ka) = k 'a → −z cot(z) = k 'a =
−z cot(z) = −z 2 +
2mVa
!2
The potential is V =
−2mEa 2 2mVa 2 2mVa 2
+
−
!2
!2
!2
2
.
32! 2
and our equation for the energies reduces to
ma 2
2ma 2 ⎛ 32! 2 ⎞
= 64 − z 2 . Plotting this (on Mathematica) shows
2
2 ⎟
⎜
! ⎝ ma ⎠
the intersection of the two curves and from the FindRoot command we numerically
find the solutions, z .
−z cot(z) = −z 2 +
The roots are z = {2.7859,5.5215, 7.9573} . Thus there are 3 bound states. The
2
⎛ z ⎞
energies of each state are given by E = ⎜ ⎟ V0 and we find the energies of the
⎝ 64 ⎠
bound states to be E = {0.1213V0 ,0.4764V0 ,0.9894V0 } .
Next we normalize the solution to determine the unknown constant A . We have
a
∞
0
a
1 = ∫ A 2 sin 2 ( kx )dx + ∫ A 2 e2 k 'a sin 2 ( ka ) e−2 k ' x dx
⎡ e−2ak ' ⎤
⎡ a sin ( ka ) cos ( ka ) ⎤
2 2 k 'a
2
1= A ⎢ −
⎥ + A e sin ( ka ) ⎢ 2k ' ⎥
2k
⎣2
⎦
⎣
⎦
2
Multiply both sides by k ' , multiply the middle term on the right hand side by
a
and
a
k 'asin ( ka ) cos ( ka )
A2 ⎡
⎤
k 'a −
+ sin 2 ( ka ) ⎥ . From our
⎢
2 ⎣
ka
⎦
k 'a
energy equation we have
= − cot(ka) = − cot(z) . Thus we have
ka
let z = ka . We have k ' =
A2
A2
2
⎡ k 'a − (− cot(z)sin ( z ) cos ( z )) + sin ( z ) ⎤⎦ =
k' =
[ k 'a + 1] . Thus the
2 ⎣
2
2k '
normalization constant is A =
.
1+ k 'a
Then the probability of finding the particle outside of the well is given by
∞
⎛ 2k ' ⎞ 2 k 'a 2
−2 k ' x
P = ∫ ψ *ψ dx = ∫ ⎜
dx
⎟⎠ e sin ( ka ) e
⎝
1+
k
'a
a
⎡ e−2 k 'a ⎤ sin 2 ( ka )
⎛ 2k ' ⎞ 2 k 'a 2
P=⎜
e sin ( ka ) ⎢
⎥=
⎝ 1+ k 'a ⎟⎠
⎣ 2k ' ⎦ 1+ k 'a
∴ P3 =
sin 2 ( z )
1+ 64 − z 2
=
sin 2 ( 7.9573)
1+ 64 − ( 7.9573)
2
.
= 0.5426
Or 54.6% of finding the particle in the highest energy state outside of the well.
13. G2.47
I. For b = 0 we have an ordinary finite square well. Exponential decay outside;
sinusoidal inside (cos for ψ 1 , sin for ψ 2 . No nodes forψ 1 , one node for ψ 2 .
ψ1
ψ2
II. Ground state is even. Exponential decay outside, sinusoidal inside the wells,
hyperbolic cosine in barrier. First excited state is odd – hyperbolic sine in barrier. No
nodes for ψ 1 , one node for ψ 2 .
ψ1
ψ2
III. For b >> a we same as (ii), but wave function very small in barrier region.
Essentially two isolated finite square wells; ψ 1 and ψ 2 are degenerate (in energy);
they are even and odd linear combinations of the ground states of the two separate
wells.
ψ1
ψ2
b,c.
There is a mixture of states here given by ψ 1 + ψ 2 and ψ 1 − ψ 2 . In the (even)
ψ 1 − ψ 2 ground state the energy is lowest in configuration (i), with b → 0 , so the
electron tends to draw the nuclei together, promoting bonding of the atoms. In the
(odd) ψ 1 + ψ 2 first excited state, by contrast, the electron drives the nuclei apart.
E+V0 E2+V0 E1+V0