Uncertainty in Free Particle Waves

Uncertainty in Free Particle Waves Now let us return to the free particle waves and our loose understanding of the relation between knowledge of the position, and knowledge of the momentum, i.e., that the narrower we make the wave packet, the more k-values we must include and therefore the less we know about the wavelength, and thus (through the de Broglie relations) the momentum of the packet. We first will use a somewhat different formalism for the wave packet: ∞ dk i()kx −ωt Ψ()x,t = ∫ φ()k e − ∞ 2π where we have now used the complex form for the plane wave function. This shouldn’t cause us any problems as we have now defined the probability density as Ψ*Ψ which should give us real positive definite values. We choose a momentum distribution function that we are familiar with, and have evaluated the variance in k, σk. We will use the Gaussian momentum distribution: 2 − ()k − k0 −1 4 4σ 2 φ k = 2πσ 2 e k () ()k * so that φ φ is centered about k0 and has a width σk. In order to understand quantitatively the relationship between the spread and position and the spread in momentum, let’s now calculate the position variance σx. To do that we will insert φ into the equation for Ψ and evaluate at t = 0: −()k − k 2 ∞ 0 dk −1 4 4σ 2 Ψ()x 0 = 2πσ 2 e k eikx , ∫ ()k − ∞ 2π ’ ’ let k = k – k0 so that k = k + k0, then, −k′2 ∞ 2 dk′ 2 −1 4 4σ i()k′ + k x Ψ()x 0 = 2πσ e k e 0 , ∫ ()k − ∞ 2π − k′2 ∞ 2 2 −1 4 ik x dk′ 4σ ik′x = 2πσ e 0 e k e ()k ∫ − ∞ 2π − k′2 ∞ ′ 2 + ik x 2 −1 4 ik x dk′ 4σ = 2πσ e 0 e k ()k ∫ − ∞ 2π Now, we manipulate the term in the exponent to make integration easier: − k′2 − 1 + ik′x = ()k′2 − i4σ 2 xk′ 4σ 2 4σ 2 k k k − 1 2 = ()k′ − i2σ 2 x − σ 2 x2 4σ 2 k k k So that our wavefunction now becomes: 2 − k′ − i2σ 2 x ( k ) 2 2 ∞ 2 −1 4 ik x −σ x dk′ 4σ 2 Ψ x 0 = 2πσ e 0 e k e k (), ()k ∫ − ∞ 2π 2 2 2 ⎡ 4σ π ⎤ 2 −1 4 ik x −σ x k = ()2πσ e 0 e k ⎢ ⎥ k ⎢ 2π ⎥ ⎣ ⎦ ______________________________________________________________________ NOTE: Had we been given the distribution in x originally, we could have found the distribution in k in the same way, namely that: ∞ dk i()kx −ωt φ()k = ∫ φ()x e − ∞ 2π This is the Fourier transform of the wave functions between momentum and position spaces and we will find this useful later. ______________________________________________________________________ Ignoring all the constant terms, it is clear to see that Ψ*Ψ will yield a Gaussian distribution in x with a standard deviation of 2/σk. Let’s then look at the relation between the two spreads: 1 σ = ⇒ x 2σ k 1 σ σ = x k 2 But, from the relation between the wave number k, and p, p=ћk, we finally have: σ = σ ⇒ px h k σ σ = h x px 2 It turns out that had we begun with any other momentum distribution, this result would have been larger, hence, we now have from basic wave diagnosis a minimum value of the combined uncertainty in position and momentum: σ σ ≥ h x px 2 This is one form of Heisenberg’s Uncertainty Principle, a cornerstone of quantum theory. Of course, this one-dimensional form can be extended to the other two dimensions: h h h σ xσ p ≥ , σ yσ p ≥ , σ zσ p ≥ x 2 y 2 z 2 Following the same reasoning, One can also show that energy and time obey the same relationship, σ σ ≥ h . E t 2 .

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