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.
(m_e = 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|²bπ^½.
  |A|² = 1/(bπ^½).
  (b)  Δx = <(<x²> - <x>²)>^½.  <x> = 0 from symmetry.
  <x²> = ∫_-∞^+∞ x²|Ψ(x)|²dx
  = |A|²∫_-∞^+∞ x²exp{-x²/b²} dx = b²/π^½∫_-∞^+∞ x'²exp{-x'²} dx' = b²/2.
  Δx = b/2^½.
  (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/(2^½b),  Δx Δp = (b/2^½) ħ/(2^½b) = ħ/2.
  (d)  Δv/c = Δp/(m_ec) = ħ/(2m_ecΔx) = 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 ∫_-3a^a|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 ∫₀^a|ψ(x)|²dx = ¼.
  (c)  <x> = ∫_-3a^aψ(x)^* x ψ(x) dx = ∫_-3a^a xdx/(4a) = -a.
  <x²> = ∫_-3a^aψ(x)^* x² ψ(x) dx = ∫_-3a^a 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∫_-3a^aψ(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).