MIT 8.03SC Fall 2016 Lecture 15: Uncertainty Principle, 2D Waves

8.03 Lecture 15 Review: Fourier Transform and Narrow band signal transmission. f (t) = fs (t) cos ω0 t In the small bandwidth approximation we have carrier  z }| { ψ(x, t) ≈ Re fs (t − x/vg ) e−i(t−x/vg )ω0 | {z envelope } Where the envelope is traveling at group velocity vg = dω dk and the carrier is traveling at phase ω0 ω0 k0 AM radio: typical ω0 : 0.3 − 30M Hz and ∆ω : 5kHz. We have ∆ω  ω0 Makes sense! velocity vp = Example: f (t) = eΓ|t| What will be the corresponding C(ω)? From the previous lecture: 1 ∞ dt e−Γ|t| eiωt 2π −∞  Z 0 Z ∞ 1 −Γt iωt +Γt iωt dt e e dt e e + 2π −∞ 0   1 1 i + 2π Γ + iω Γ − iω 1 2Γ 2 2π Γ + ω 2 Γ 2 π(Γ + ω 2 ) Z C(ω) = = = = = If we plot C(ω) as a function of ω Large Γ gives narrow pulse (f (t) narrow) but one will get wide C(ω) Small Gamma gives wide pulse (f (t) wide) but will get narrow C(ω) In your pset, you will work on another function form: the Gaussian wave. We can demonstrate this using a precie mathematical definition of the spread of the signal. 1. We define the intensity of the signal to be proportional to |f (t)|2 2. Average value of any function weighted with the signal’s intensity Z ∞ hg(t)i = dt g(t)|f (t)|2 −∞ Z ∞ dt |f (t)|2 −∞ 3. Spread of time: Z ∞ ∆t2 ≡ h[t − hti]2 i = dt (t − hti)f (t) −∞Z ∞ 2 dt |f (t)|2 −∞ mean square deviation from the average time. Similarly spread of the frequency spectrum ∆ω 2 ≡ h[ω − hωi]2 i We will prove ∆ω · ∆t ≥ 1/2 4. We also realize that Z ∞ −∞ ∂ ∞ dω C(ω)e−iωt ∂t −∞ ∂ = i f (t) ∂t dω ωC(ω)e−iωt = i Z Z ∞ dt f (t)∗ i −∞ Z ⇒ hωi = ∂ f (t) ∂t ∞ dt f (t) 2 −∞ Z ∞  −∞ ∆ω 2 ≡ h[ω − hωi]2 i = ∂ − hωi f (t) ∂t  i dt Z ∞ dt f (t) 2 2 −∞ Take home message: ω ←→ i ∂ ∂t 5. We will use a trick which leads to Heisenberg Uncertainty principle in Quantum Mechanics: Consider a function γ(t) ∂ γ(κ, t) ≡ [t − hti] − iκ i − hωi ∂t = (T − iκΩ)f (t)    f (t) Where κ is a free parameter. Consider: Z ∞ 2 γ(κ, t) dt R(κ) = Z−∞∞ 2 f (t) dt −∞ Both the numerator and denominator are positive−→ R > 0 γ(κ, t) 2 = (T − iκΩ)f (t) · (T + iκΩ∗ )f ∗ 2 = Tf + Ωf 2 + iκ[T f Ω∗ f ∗ − Ωf T f ∗ ] Take the last term and simplify: ∂ > f ∗ − i ∂ − hωi  > fTf∗  iκ T f −i −  hωi ∂t ∂t    ∂f ∗ ∂f ∗ + f = κT f ∂t ∂t ∂ = κT (f f ∗ ) ∂t Now integrate over all time:    Z ∞ dt κT = −∞ = κ Tff∗ | = −κ −∞ Z =0 ∞ −∞  ∂ (f f ∗ ) ∂t ∞ {z  −κ −∞ } dt |f | Z ∞ |f |2 ∂T dt ∂t |{z} =1 2  Where in the second step we have assumed f is localized such that f (t = ±∞) = 0 (1): Z ∞ ∆t2 = 2 [t − hti]2 f (t) dt −∞Z ∞ 2 f (t) dt −∞ (2): κ2 2 2 κ ∆ω = Z ∞  ∂ i −∞ ∂t  2 − hωi f (t) dt Z ∞ 2 f (t) dt −∞ (3.) using the result from above Z ∞ κ −κ = Z |f (t)|2 dt −∞ ∞ 2 f (t) dt −∞ 2 ⇒ R = ∆t + κ2 ∆ω 2 − κ > 0 dR 1 We know = 0. R(κ) minimize at κ = κmin = dκ 2∆ω 2 1 ≥0 ⇒ R(κmin ) = ∆t2 − 4∆ω 2 Rearranging, we get the Uncertainty principle!!! ∆t∆ω ≥ 1 2 AM station broadcast: Bandwidth: ∆ω = 2π∆ν ≈ 3 × 104 s−1 ⇒ ∆t is a few ×10−5 seconds. This means we cannot tell signals apart when ∆t < O(10−5 s) Quantum Mechanics: We can rewrite Uncertainty as: v∆t · ∆ω/v ≥ 1/2 ⇒ ∆x · ∆k ≥ 1/2 In quantum mechanics, p = ~k where Planck’s Constant ~ ≡ h/2π. ~ ≈ 6.6×10−10 eV·s or 1×10−34 J·s. We can rewrite our uncertainty principle as: ~ 2 This is the mathematical statement of the fact that the position and momentum of a particle cannot be specified simultaneously. ⇒ ∆x∆p ≥ Let’s move to two and three dimensional waves. In this example, we have a beaded mesh. Index: (jx , jy ) where each one runs from 1 to 4. First we consider an infinite system by removing the wall and use space translation symmetry: X direction: Eigenstate of M −1 k is eikx x Y direction: Eigenstate of M −1 k is eiky y ψ(x, y) = Aeikx x eiky y ~ = Aeik·~r Where ~k ≡ kx x̂ + ky ŷ and ~r ≡ xx̂ + y ŷ Time dependent displacement: h i ~ ψ(x, y, t) = Re Aeik·~r e−iωt 4TH kx aH 4TV ky aV sin2 + sin2 maH 2 maV 2 Where kx an ky are arbitrary for the moment. In general the two dimensional problem can be infinitely hard!! (But in this special case it is solvable). Before introducing the boundary (infinitely long system) ω2 = 1. Normal mode in 1D: always two normal modes: e±ikx 2. But in 2D, in general, a fixed ω gives an infinite number of solutions (if we lower kx , we can always increase ky to compensate!) Adding walls back in: Boundary conditions give: nx π LH ny π ky = LV kx = LH = 5aH LV = 5aV nx πx ny πy = Anx ,ny sin sin LH LV  ψ(x, y)nx ,ny General solution: linear combination of ψnx ,ny