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Griffiths Quantum Mechanics 3e: Problem 1.11 Page 1 of 4
Problem 1.11
[This problem generalizes Example 1.2.] Imagine a particle of mass m and energy E in a potentialwell V (x), sliding frictionlessly back and forth between the classical turning points (a and b inFigure 1.10).
Classically, the probability of finding the particle in the range dx (if, for example, you took asnapshot at a random time t) is equal to the fraction of the time T it takes to get from a to b thatit spends in the interval dx:
ρ(x) dx =dt
T=
(dt/dx) dx
T=
1
v(x)Tdx, (1.41)
where v(x) is the speed, and
T =
ˆ T
0dt =
ˆ b
a
1
v(x)dx. (1.42)
Thus
ρ(x) =1
v(x)T. (1.43)
This is perhaps the closest analog22 to |Ψ|2.
(a) Use conservation of energy to express v(x) in terms of E and V (x).
(b) As an example, find ρ(x) for the simple harmonic oscillator, V (x) = kx2/2. Plot ρ(x), andcheck that it is correctly normalized.
(c) For the classical harmonic oscillator in part (b), find 〈x〉, 〈x2〉, and σx.
Solution
By conservation of energy, the sum of potential and kinetic energies must be equal to E, the totalmechanical energy.
E = PE + KE = V (x) +1
2mv2
22If you like, instead of photos of one system at random times, picture an ensemble of such systems, all with thesame energy but with random starting positions, and photograph them all at the same time. The analysis is identical,but this interpretation is closer to the quantum notion of indeterminacy.
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Griffiths Quantum Mechanics 3e: Problem 1.11 Page 2 of 4
Solve for v = v(x).
v(x) = ±√
2
m[E − V (x)]
For a simple harmonic oscillator, V (x) = kx2/2, which means the classical turning points arex = ±A, where A is the amplitude of oscillation.
The formula for velocity becomes
v(x) = ±
√2
m
(E − 1
2kx2)
= ±√
2E − kx2m
,
and the probability distribution becomes
ρ(x) =1
v(x)T=
1
v(x)
ˆ A
−A
1
v(x)dx
=1√
2E − kx2m
ˆ A
−A
√m
2E − kx2dx
=1√
2E − kx2m
√m
2E
ˆ A
−A
dx√1− k
2Ex2
=1
2√
1− k2Ex
2
ˆ A
0
dx√1− k
2Ex2
.
Make the following trigonometric substitution.
x =
√2E
ksin θ → 1− k
2Ex2 = 1− sin2 θ = cos2 θ
dx =
√2E
kcos θ dθ
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Griffiths Quantum Mechanics 3e: Problem 1.11 Page 3 of 4
As a result,
ρ(x) =1
2√
1− k2Ex
2
ˆ sin−1
(√k2EA
)0
√2Ek cos θ dθ
cos θ
=1
2√
1− k2Ex
2√
2Ek
ˆ sin−1(1)
0dθ
=1
2√
2Ek − x2
(π2
)=
1
π√A2 − x2
.
The domain of this function reveals where it’s valid: |x| < A. For |x| ≥ A, ρ(x) = 0 because it’simpossible for the particle to be here with this energy E.
Observe that the probability amplitude is lowest at the bottom of the well, where the particle ismoving fastest, and that the probability amplitude is highest near the top, where the particlecomes to a stop. Of course, the probability distribution is normalized because
ˆ A
−Aρ(x) dx =
ˆ A
−A
1
v(x)Tdx =
1
T
ˆ A
−A
1
v(x)dx =
1H
HHHHHH
ˆ A
−A
1
v(x)dx
HHHHHHH
ˆ A
−A
1
v(x)dx = 1.
Check it anyway using the formula found for ρ(x).
ˆ A
−Aρ(x) dx =
ˆ A
−A
1
π√A2 − x2
dx =1
π
ˆ A
−A
dx√A2 − x2
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Griffiths Quantum Mechanics 3e: Problem 1.11 Page 4 of 4
Make the following trigonometric substitution.
x = A sin θ → A2 − x2 = A2(1− sin2 θ) = A2 cos2 θ
dx = A cos θ dθ
As a result,
ˆ A
−Aρ(x) dx =
1
π
ˆ sin−1(1)
sin−1(−1)
A cos θ dθ
A cos θ
=1
π
ˆ π/2
−π/2dθ
=1
π(π)
= 1.
Use the probability distribution to calculate the expectation values of x and x2.
〈x〉 =
ˆ A
−Axρ(x) dx
ˆ A
−Aρ(x) dx
=
ˆ A
−Axρ(x) dx
1=
ˆ A
−Axρ(x) dx
=
ˆ A
−A
x
π√A2 − x2
dx
= 0
This integral is zero because the integrand is an odd function and the interval of integration issymmetric.
〈x2〉 =
ˆ A
−Ax2ρ(x) dx
ˆ A
−Aρ(x) dx
=
ˆ A
−Ax2ρ(x) dx
1=
ˆ A
−Ax2ρ(x) dx
=
ˆ A
−A
x2
π√A2 − x2
dx
=
ˆ sin−1(1)
sin−1(−1)
A2 sin2 θ
π(A cos θ)(A cos θ dθ)
=A2
π
ˆ π/2
−π/2sin2 θ dθ
=A2
π
ˆ π/2
−π/2
1
2(1− cos 2θ) dθ
=A2
2
The trigonometric substitution at the top of this page was used. Finally, the standard deviation is
σx =√〈x2〉 − 〈x〉2 =
√A2
2=
A√2≈ 0.707A.
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