### The rms deviation

#### Problem:

An electron is moving freely in the x-direction.  At t = 0 the electron is described by the wave function (neglect spin)
Ψ(x,0) = Aexp{-x2/2b2}exp{ip0x/ħ}.
(a)  Compute the constant A such that ∫-∞+∞|ψ(x,0)|2 dx = 1.
(b)  Compute ∆x at t = 0.
(c)  Compute ∆p at t = 0 and show that  for the electron ∆x∆p = ħ/2.
(d)  Assume that the electron has a position uncertainty of ∆x = 10-10 m.
Compute its velocity uncertainty compared to the speed of light.
(me = 9.1*10-31 kg,  ħ = 1.05*10-34 J s,  c = 3*108 m/s).
Hint:  ∫-∞+∞dx exp(-(ax2 + bx + c)) = (π/a)½ exp((b2 - 4ac)/(4a)).
To obtain, for example, ∫-∞+∞dx x exp(-ax2), differentiate with respect to b and then set b = c = 0.

Solution:

• Concepts:
The mean value and the root mean square deviation of an observable
• Reasoning:
The expression for the mean value of an observable A in the normalized state |Ψ> is <A> = <Ψ|A|Ψ>.  If |Ψ> is not normalized then  <A> = <Ψ|A|Ψ>/<Ψ|Ψ>.
The root mean square deviation ΔA characterizes the dispersion of the measurement around <A>.
ΔA = (<(A - <A>2)>)½ = (<A2> - <A>2))½.
• Details of the calculation:
(a)  ∫-∞+∞ dx|Ψ(x)|2 = |A|2 -∞+∞ dx exp{-x2/b2} = |A|2½.
|A|2 = 1/(bπ½).
(b)  Δx = <(<x2> - <x>2)>½.  <x> = 0 from symmetry.
<x2> = ∫-∞+∞ x2|Ψ(x)|2dx
= |A|2-∞+∞ x2exp{-x2/b2} dx = b2½-∞+∞ x'2exp{-x'2} dx' = b2/2.
Δx = b/2½.
(c)   Δp = (<p2> - <p>2))½.
pΨ(x) = (ħ/i)(∂/∂x)Ψ(x) = (p0 + iħx/b2)Ψ(x).
p2Ψ(x) = [(p0 + iħx/b2)2 + ħ2/b2]Ψ(x).
<p> = ∫-∞+∞ dx(p0 + iħx/b2)|Ψ(x)|2 = p0.
<p2> = ∫-∞+∞ dx[(p0 + iħx/b2)2 + ħ2/b2]|Ψ(x)|2
= p02 + ħ2/b2 -  ħ2/b4<x2> = p02 + ħ2/b2 - ħ2/(2b2) = p02 + ħ2/(2b2).
Δp = ħ/(2½b),  Δx Δp = (b/2½) ħ/(2½b) = ħ/2.
(d)  Δv/c = Δp/(mec) = ħ/(2mecΔx) = 1.05*10-34/[2*9.1*10-31*3*108*10-10] = 1.9*10-3.

#### Problem:

The state of a free particle is described by the following wave function
ψ(x) = 0 for x < −3a
ψ(x) = c for − 3a < x < a
ψ(x) = 0 for x > a
(a)  Determine c using the normalization condition.
(b)  Find the probability of finding the particle in the interval [0, a].
(c)  Compute <x> and the root-mean square deviation ∆x.
(d)  Calculate the momentum probability density.

Solution:

• Concepts:
Fundamental assumptions of QM
• Reasoning:
The expression for the mean value of an observable A in the normalized state |Ψ> is <A> = <Ψ|A|Ψ>.
If |Ψ> is not normalized then  <A> = <Ψ|A|Ψ>/<Ψ|Ψ>.
The root mean square deviation ΔA characterizes the dispersion of the measurement around <A>.
ΔA = (<(A - <A>2)>)½ = (<A2> - <A>2))½.
• Details of the calculation:
(a)  The normalization condition requires ∫-3aa|ψ(x)|2dx = 1, 4a c2 = 1, c = ½a-1.
(b)  Between x = -3a and x = a |ψ(x)|2 = 1/(4a).
The probability of finding the particle in the interval [0, a] is ∫0a|ψ(x)|2dx = ¼.
(c)  <x> = ∫-3aaψ(x)* x ψ(x) dx = ∫-3aa xdx/(4a) = -a.
<x2> = ∫-3aaψ(x)* x2 ψ(x) dx = ∫-3aa x2dx/(4a) = 7a2/3.
Δx = (<x2> - <x>2)½ (4a2/3)½ = 2a/(√3).
(d)  Denote the momentum probability density by|φ(p)|2.
φ(p) = (2π ħ)-½-3aaψ(x) exp(ipx/ ħ)dx = (-ic/p)(ħ/(2π))½[exp(-ipa/ħ) - exp(-3ipa/ħ)]
= (2c/p)(ħ/(2π))½exp(ipa/ħ) sin(2pa/ħ).
|φ(p)|2 = [ ħ/(2π a p2)] sin2(2pa/ħ).