Chapter 35
Problem 55
†
Given
U = 12 mω 2 x2
Solution
a) What is Schrodinger’s equation for the harmonic oscillator.
Schrodinger’s equation is
h̄2 ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ
+ 2 + 2 + U (x, y, z)ψ = Eψ
−
2m ∂x2
∂y
∂z
In one dimension it becomes
h̄2 ∂ 2 ψ
−
+ U ψ = Eψ
2m ∂x2
Substituting in the potential energy function gives
h̄2 ∂ 2 ψ
−
+ 12 mω 2 x2 ψ = Eψ
2m ∂x2
b) Show that ψ = A0 e−α
2 x2 /2
satisfies this equation where α2 = mω/h̄.
Take the first derivative of this function with respect to x.
−α2 x2 /2
∂
A
e
0
∂ψ
2 2
=
= A0 e−α x /2 (−α2 2x/2)
∂x
∂x
∂ψ
2 2
= A0 e−α x /2 (−α2 x) = ψ(−α2 x)
∂x
Take the second derivative
−α2 x2 /2
2
∂
A
e
(−α
x)
2
0
∂ ψ
∂(∂ψ/∂x)
=
=
∂x2
∂x
∂x
∂ 2ψ
2 2
2 2
= A0 e−α x /2 (−α2 ) + A0 e−α x /2 (−α2 x)2
2
∂x
∂ 2ψ
2 2
= A0 e−α x /2 (−α2 ) + (−α2 x)2 = (−α2 ) + (−α2 x)2 ψ
2
∂x
Now substitute this into Schrodinger’s equation
−
h̄2 (−α2 ) + (−α2 x)2 ψ +
2m
h̄2 −
(−α2 ) + (−α2 x)2 ψ +
2m
Factor out ψ, energy becomes
1
mω 2 x2
2
ψ = Eψ
1
mω 2 x2
2
ψ = Eψ
h̄2 E=−
(−α2 ) + (−α2 x)2 +
2m
†
1
mω 2 x2
2
Problem from Essential University Physics, Wolfson
Since α2 = mω/h̄, then
h̄2 (−mω/h̄) + (−mωx/h̄)2 + 12 mω 2 x2
2m
h̄2 −mω m2 ω 2 x2
2 2
1
E=−
+
+
mω
x
2
2m
h̄
h̄2
h̄ω 1
− 2 mω 2 x2 + 12 mω 2 x2 = 12 h̄ω
E=
2
This is the ground state of the harmonic oscillator.
E=−
c) Normalize the function and find the value of A0 .
The solution to the harmonic oscillator is a bell curve and normalization involves a subtitle process.
Write out the normalization process for ψ
Z ∞
1=
ψ 2 dx
−∞
Z
∞
1=
Z
2
−α2 x2 /2
2
A0 e
dx = A0
−∞
∞
−α2 x2
e
dx = 1
−∞
Now let’s square the integration portion and take the square root.
s
2
Z ∞
2
2
2
(e−α x ) dx
1 = A0
−∞
Write out each term of the square based on a different variable
sZ
Z ∞
∞
2 y2
2 x2
2
−α
−α
1 = A0
(e
) dy
(e
) dx
−∞
−∞
Since each variable is independent of the other the two individual integrals can be treated as a double
integral.
sZ Z
sZ Z
∞
∞
∞
∞
2 x2 −α2 y 2
2
2
−α
1 = A0
e
e
dxdy = A0
e−α2 (x2 +y2 ) dxdy
−∞
−∞
−∞
−∞
Notice this double integral is over all space in Cartesian coordinates. Change to polar coordinates and
dxdy becomes rdrdθ and x2 + y 2 becomes r2 . Also the limits for θ go from 0 to 2π and the limits for r go
from 0 to ∞.
s
Z 2π Z ∞
2
1 = A0
e−α2 r2 rdrdθ
0
0
Make a u-substitution on r where u = −α2 r2 and du = −2α2 rdr. The limits now go from 0 to −∞.
s
Z 2π Z −∞ u
e
2
1 = A0
dudθ
−2α2
0
0
Solving the integral gives
s
r
Z 2π Z −∞
1
1
2
2
1 = A0
dθ
eu du = A0
(2π − 0) (e−∞ − e0 )
2
−2α
−2α2
0
0
r
1=
A20
1
(2π) (0 − 1) = A20
−2α2
Solving for A0 gives
r
2
4 α
A0 =
π
Since α2 = mω/h̄, then
r
mω
A0 = 4
πh̄
r
2π
= A20
2α2
r
π
α2