8.04 Quantum Physics Lecture IV. ψ(x) = dkφ (k)e ikx 2π

Size: px

Start display at page:

Download "8.04 Quantum Physics Lecture IV. ψ(x) = dkφ (k)e ikx 2π"

Mark Harvey
5 years ago
Views:

1 Last time Heisenberg uncertainty ΔxΔp x h as diffraction phenomenon Fourier decomposition ψ(x) = dkφ (k)e ikx π ipx/ h = dpφ(p)e (4-) πh φ(p) = φ (k) (4-) h Today how to calculate φ(k) interpretation of ψ(x) and φ(k): probability amplitude & probability density measurement The larger φ (k), or φ(p), the more the wavefunction ψ(x) resembles the plane wave, e ikx, that has definite momentum p = hk, and the more the particle described by the wavefunction ψ(x) is likely to be found to have momentum hk, if the momentum of the particle is measured. Conversely, if the particle s momentum is exactly p = p 0 = hk 0, then the particle s wavefunction must be ψ(x) = e ik 0x, and the particle is equally likely to be found anywhere in space, Δx. To localize a particle in space, we need to add other Fourier components close to k 0. constructive interference k 0 k k 0 k 0 + k destructive interference Figure I: Addition of Fourier components close to k 0 yields a wavepacket localized in space. IV-

2 To confine a wavefunction to a small region Δx of space, one needs many Fourier components, i.e., many plane waves of different momentum k [k 0, k 0 + Δk]. We will prove mathematically that, with suitably defined uncertainties Δx and Δk: ΔxΔk h ΔxΔp follows strictly from the Fourier = decomposition. Using Δp = hδk (4-3) we then arrive at = Heisenberg uncertainty relation (4-4) From this viewpoint, the Heisenberg uncertainty relation follows from the decomposition of a wave into plane waves e ikx, i.e., waves with definite wavevector k, and from the relation between the wavevector k and the momentum p = hk. The smaller the region Δx, the more Fourier components p = hk are necessary to produce destructive interference everywhere outside Δx. Corollary: motion of particles. From the motion of particles and plane waves from Fig. I, it follows that if you change the relative phase of the different Fourier components, the constructive interference will occur somewhere else in space. This makes sense. We should be able to place a particle at a different location in space while maintaining the momentum distribution. It follows that if the relative phase between the plane waves changes continuously, the location of constructive interference (i.e., the location of the particle) will move in space. In terms of wave mechanics, the motion of the particle is simply due to a change of phase between the Fourier components (i.e., the plane waves). Then, to reproduce CM, the phase of the Fourier components must rotate in time at a frequency that depends on the momentum p. Corollary: time-energy uncertainty. If instead of Fourier transforming the posi t 0 ) we fix the position x = x 0 and study tion coordinate at fixed time Ψ(x, t 0 ) Φ(k, the time variation of the wavefunction Ψ(x 0, t), then we can Fourier decompose the wavefunction into frequency components. ψ(t) = dωf(ω)e iωt (4-5) π Convention. Positive frequency ω corresponds to negative phase evolution. Applying the same mathematical and logical arguments as before, we arrive at: ΔωΔt = time-frequency uncertainty (4-6) h ΔEΔt = energy-time Heisenberg uncertainty relation (E = hω) (4-7) IV-

3 J O J Figure II: Wave function vs. time. Probability amplitude and probability density For light or other waves, the energy per unit volume (or per unit length) is proportional to the square of the electric field. Since the number of photons per unit volume is proportional to E, we postulate in analogy: ( ) Probability to find a particle = ψ(x) dx. (4-8) between x and x + dx ψ(x) is called the probability density (probability per unit length). The waveform O Figure III: The magnitude squared ψ(x) of the wavefunction is the probability density to find the particle in a region of space. The probability to find the particle in the spatial interval [x, x + dx] is given by ψ(x) dx. ψ(x) is also called the probability amplitude (more exactly: probability density amplitude). In contrast to EM fields, Ψ is truly a complex quantity. The requirement that the particle be found somewhere in space leads to the normalization condition: dx ψ(x) =. (4-9) In the homework, you will prove Parseval s theorem: Proof. If φ (k), φ(p), is the Fourier transform of a wavefunction ψ(x), i.e., if ψ(x) = dkφ (k)e ikx = dpφ(p)e ipx/ h π πh IV-3

4 F F F then, dx ψ(x) = dk φ(k) = dp φ(p). It follows that if the wavefunction ψ(x) is normalized, so is φ(p). We already argued that if φ(p) is peaked around some value p 0, the motion of the particle will be similar to that of a classical particle with momentum p 0, (plane wave e ip 0x/h ). Taking into account Parseval s theorem, it is reasonable to interpret φ(p) as the probabilty amplitude for momentum, i.e., ( ) Probability to find a particle momentum = φ(p) dp. (4-0) between p and p + dp Similarly, φ (k) is the probability density for the wavevector k. ote that Δx (or B F F Figure IV: The probability density in momentum space is given by φ(p). Δp) is not the uncertainty of the measurement device, but associated with the particle itself. If our measurement apparatus can resolve with resolution Δx app Δx, and we repeat the experiment with an identically prepared particle many times, we will observe a histogram.in the limit of a very large number of measurements, the histogram reproduces the probability density ψ(x) or φ(p). B K J? A I, = F F, B K J? A I, F = F F, F E F E Figure V: Probability density reconstruction in position space. Figure VI:... and in momentum space. ote. After measuring a particular value x with apparatus uncertainty Δx app, the particle is no longer described by the original wavefunction ψ(x), but by a new wavefunction ψ (x) that is consistent with the outcome of the measurement result ( collapse IV-4

5 & O of the wavefunction ). In particular, if Δx app Δx, the spread of the new wavefunction ψ (x) in momentum will be much larger than before, consistent with the Heisenberg uncertainty (Δ x Δx ψ = Δx app ). h Δ xδ p = Δx app Δ p, (4-) or h h Δ p = Δp. Δxapp Δx (4-) If you now choose to measure momentum with resolution Δp app Δ p, the uncertainty in position will again increase and so on. ψ(x) = dkφ (k)e ikx = dpφ(p)e ipx/ h π πh (4-3) How to determine the momentum distribution φ(p) given the wavefunction ψ(x) in position? The expansion coefficients φ(k) are given by the inverse of the Fourier decomposition: φ(k) = dxψ(x)e ikx π Proof. dkφ(k)e ikx = dk dx ψ(x )e ikx e ikx π π = dx ψ(x ) π dke ik(x x ) What is the value of dke ik(x x )? Qualitatively, if x = x, the integrand oscillates in the complex plane many times a k ±, so the integral is zero. If x = x, the integral is dk and diverges. So the function I(y) = dkeiky looks something like: O O O O A = O Figure VII: Sketch of the function represented by the integral I(y). Figure VIII: Convolution of I(y) and a Gaussian function to determine the area under I(y). IV-5

6 How bad is the divergence? Let us calculate the area under the curve α real and positive. dyi(y)e αy = dy dke iky e αy = dye αy +iky To calculate the integral, we note without proof that: π dye α(y β) = for any complex α, β with Re(α) 0. (4-5) α ( ) ik k = α y (4-6) α 4α ik k dyi(y)e αy = dk dye α(y α ) 4α independent of the value α! Since our probing function e αy, we conα 0 clude that the area under our function I(y) is finite and equal to π. We define a generalized function (mathematically, a distribution) by { 0 for x = 0, δ(x) = (4-8) for x = 0. dk (4-4) To bring the above integral, Eq. (4-4), into the desired form, we expand the exponent: ( ( ) ) ik ik k αy + iky = α y α y + α 4α π k = dke 4α α π = π 4α α = π (4-7) with the property, dxδ(x) =. (4-9) This is the Dirac delta function. We can think of it as the limiting case of a function of finite width (e.g., a Gaussian or square function) that is made narrower and narrower, while keeping the area under it constant. We have dkeiky = πδ(y). IV-6

7 M & ) H A ) H A = Figure IX: The delta function can also be expressed as δ(x) = x lim w 0 πw e w. Figure X: Dirac delta function. Properties of the delta function What is dxf(x)δ(x x 0 )? For a function f(x) that is sufficiently smooth (regular) at x = 0 and using δ(x x 0 ) = 0 for x = x 0, we have Therefore, we have: x0 +ɛ dxf(x)δ(x x 0 ) = dxf(x)δ(x x 0 ) x 0 ɛ x0 +ɛ = f(x 0 ) dxδ(x x 0 ) x 0 ɛ = f(x 0 ) dxδ(x) = f(x 0 ). (4-0) dxf(x)δ(x x 0 ) = f(x 0 ). (4-) Convolution of a function f(x) with δ(x x 0 ) projects out the value of the function A A Figure XI: Convolving a sufficiently smooth function with a Dirac delta function projects out the function value at one point. IV-7

8 G Figure XII: The δ-function is also the derivative of the Heavyside step function. at x 0. Without proof, we note that δ(x) = d Θ(x). Derivative of the delta function dx can be defined. Integration by parts yields dxf(x)δ (x x 0 ) = f (x x 0 ). (4-) Convolution with δ projects out the negative derivative at x 0. IV-8

SOLUTIONS for Homework #2. 1. The given wave function can be normalized to the total probability equal to 1, ψ(x) = Ne λ x.

SOLUTIONS for Homework #2. 1. The given wave function can be normalized to the total probability equal to 1, ψ(x) = Ne λ x. SOLUTIONS for Homework #. The given wave function can be normalized to the total probability equal to, ψ(x) = Ne λ x. () To get we choose dx ψ(x) = N dx e λx =, () 0 N = λ. (3) This state corresponds to