in Free

Now let us return to the waves and our loose understanding of the relation between of the , and knowledge of the , i.e., that the narrower we make the packet, the more k-values we must include and therefore the less we know about the , and thus (through the de Broglie relations) the momentum of the packet.

We first will use a somewhat different formalism for the : ∞ dk i()kx −ωt Ψ()x,t = ∫ φ()k e − ∞ 2π where we have now used the complex form for the plane wave . This shouldn’t cause us any problems as we have now defined the 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 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 ⎥ ⎣ ⎦

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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 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 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 , a cornerstone of 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 and time obey the same relationship, σ σ ≥ h . E t 2