� � 8.04 Quantum Physics Lecture IVLast time ¯• Heisenberg uncertainty ΔxΔpx ≥ h as diffraction phenomenon 2 • Fourier decomposition ψ(x)= 1 ∞ dkφ˜(k)e ikx √2π −∞ 1 ∞ ipx/¯h = dpφ(p)e(4-1)√2πh¯−∞ 1 φ(p)= φ˜(k) (4-2)√h¯Today • how to calculate φ(k) • interpretation of ψ(x) and φ(k):probability amplitude & probability densitymeasurement • The larger φ˜(k), or φ(p), the more the wavefunction ψ(x) resembles the plane wave, eikx, 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 = p0 =¯hk0, then the particle’s wavefunction must be ψ(x)= eik0x, 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 k0. constructive interferencedestructive interferencek0−D kk0k0+D kFigure I: Addition of Fourier components close to k0 yields a wavepacket localized in space. Massachusetts Institute of Technology IV-1 � 8.04 Quantum Physics Lecture IVTo confine a wavefunction to a small region Δx of space, one needs many Fourier components, i.e., many plane waves of different momentum k ∈ [k0,k0 +Δk]. We will prove mathematically that, with suitably defined uncertainties Δx and Δk: follows strictly from the Fourier 1 = decomposition. Using Δp =¯hΔk (4-3)ΔxΔk ≥ ⇒2 we then arrive at h¯= Heisenberg uncertainty relation (4-4)ΔxΔp ≥ ⇒2 From this viewpoint, the Heisenberg uncertainty relation follows from the decomposittio of a wave into plane waves eikx, 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 interferennc (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˜tion coordinate at fixed time Ψ(x,t0) Φ(k,t0) we fix the position x = x0 and study →the time variation of the wavefunction Ψ(x0,t), then we can Fourier decompose the wavefunction into frequency components. 1 ∞ ψ(t)= √2π dωf(ω)e−iωt (4-5) −∞ Convention. Positive frequency ω corresponds to negative phase evolution. Applying the same mathematical and logical arguments as before, we arrive at: 1 ΔωΔt ≥ 2 = ⇒ time-frequency uncertainty (4-6) ¯h energy-time Heisenberg unΔEEΔ ≥ 2 = ⇒ certainty relation (E = ¯hω) (4-7) Massachusetts Institute of Technology IV-2 |� |� � � � 8.04 Quantum Physics Lecture IVty(x0,t)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 proportioona to the square of the electric field. Since the number of photons per unit volume is proportional to E2, we postulate in analogy: Probability to find a particle between x and x + dx = |ψ(x)|2dx. (4-8) |ψ(x)|2 is called the probability density (probability per unit length). The waveform x|y(x)|2Figure III: The magnitude squared |ψ(x)|2 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)|2dx. ψ(x) is also called the probability amplitude (more exactly: probability density amplitudde) 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)|2 =1. (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)= 1 ∞ dkφ˜(k)e ikx =1 ∞ dpφ(p)eipx/¯h√2π −∞ √2πh¯−∞ Massachusetts Institute of Technology IV-3 � � 8.04 Quantum Physics Lecture IVthen, � ��∞ 2 ∞ ˜2 ∞ 2dx|ψ(x)|= dk|φ(k)|= dp|φ(p)|. −∞ −∞−∞ It follows that if the wavefunction |ψ(x)|2 is normalized, so is |φ(p)|2 . We already argued that if φ(p) is peaked around some value p0, the motion of the particle will be similar to that of a classical particle with momentum p0, (plane wave eip0x/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 between p and p + dp = |φ(p)|2dp. (4-10) Similarly, |φ˜(k)|2 is the probability density for the wavevector k. Note that Δx (or p|f(p)|2pp+dpFigure IV: The probability density in momentum space is given by |φ(p)|2 . Δp) is not the uncertainty of the measurement device, but associated with the particle itself. If our measurement apparatus can resolve with resolution Δxapp � 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)|2 or |φ(p)|2 . x# of outcomesDxappDxxiFigure V: Probability density recon-struction in position space. Figure VI: . . . and in momentum space. Note. After measuring a particular value x with apparatus uncertainty Δxapp, the particle is no longer described by the original wavefunction ψ(x), but by a new wave-function ψ˜(x) that is consistent with the outcome of the measurement result (“collapse p# of outcomesDppiDpapp Massachusetts Institute of Technology IV-4 � � � � � � � � � 8.04 Quantum Physics Lecture IVof the wavefunction”). In particular, if Δxapp � Δx, the spread of the new wave-function ψ˜(x) in momentum will be much larger than before, consistent with the Heisenberg uncertainty (Δ˜x ≡ Δxψ˜=Δxapp). h¯Δ˜xΔ˜p =ΔxappΔ˜p � , (4-11)2 or h¯h¯Δ˜=Δp. (4-12)p � 2Δxapp � 2Δx If you now choose to measure momentum with resolution Δpapp � Δ˜p, the uncertaiint in position will again increase and so on. ψ(x)= 1 ∞ dkφ˜(k)e ikx =1 ∞ dpφ(p)eipx/¯h (4-13)√2π −∞ √2πh¯−∞ How to determine the momentum distribution |φ(p)|2 given the wavefunction ψ(x) in position? The expansion coefficients φ(k) are given by the inverse of the Fourier decomposition: 1 ∞ φ(k)= √2π dxψ(x)e−ikx −∞ Proof. 1 ∞ 1 ∞∞ √2π dkφ(k)e ikx =2π dk dx�ψ(x�)e−ikx� e ikx −∞ �−∞ −∞ � =1 ∞ dx�ψ(x�) ∞ dkeik(x−x�) 2π −∞ −∞ What is the value of ∞ dkeik(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 1 and diverges. So the “function” I(y)= ∞ dkeiky looks −∞ · −∞something like: I(y=0)=08I(y)I(y=0)=0I(0)=0y0e-ay2yFigure VII: Sketch of the “function” represented by the integral I(y). Figure VIII: Convolution of I(y) and a Gaussian function to determiin “the area under I(y)”. Massachusetts Institute of Technology IV-5 � � � � � � � � � � 8.04 Quantum Physics Lecture IVHow bad is the divergence? Let us calculate the area under the curve α real and→positive. ∞ dyI(y)e−αy2 = ∞ dy ∞ dkeikye−αy2 −∞ �−∞ �−∞ = ∞ dk ∞ dye−αy2+iky (4-14) −∞ −∞ To calculate the integral, we note without proof that: ∞ π dye−α(y−β)2 = for any complex α, β with Re(α) ≥ 0. (4-15)α−∞ To bring the above integral, Eq. (4-14), into the desired form, we expand the exponeent � ��2 � ikik k2 −αy2 + iky = −αy 2 − αy +2α − 4α ��2ik k2 = −αy − 2α − 4α (4-16) 2∞ ∞∞ ik k2 dyI(y)e−αy2 = dk dye−α(y− 2α )− 4α −∞ �−∞ � −∞ π ∞ k2 = dke− 4α α � −∞ = π √π 4α α · =2π (4-17) ⇒ independent of the value α! Since our probing function e−αy2 −−→0 1, we conέ→clude that the area under our function I(y) is finite and equal to 2π. We define a “generalized function” (mathematically, a distribution) by 0 for x =0,δ(x)= �(4-18) ∞ for x =0. with the property, � ∞ dxδ(x)=1. (4-19) −∞ 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 =2πδ(y).−∞ Massachusetts Institute of Technology IV-6 � � � 8.04 Quantum Physics Lecture IV0xArea=1w8d(x)0xArea=1Figure IX: The delta function Figure X: Dirac delta function. can also be expressed as δ(x)= 2 x limw→0 √21 πw e− 2w2 . Properties of the delta function What is ∞ dxf(x)δ(x−x0)? For a function f(x) that is sufficiently smooth (regular) −∞at x = 0 and using δ(x − x0)=0 for x = x0, we have �� x0+�∞ dxf(x)δ(x − x0)= dxf(x)δ(x − x0) −∞ x0−� � x0+� = f(x0) dxδ(x − x0) �x0−� ∞ = f(x0) dxδ(x)= f(x0). (4-20) −∞ Therefore, we have: ∞ dxf(x)δ(x − x0)= f(x0) . (4-21) −∞ Convolution of a function f(x) with δ(x−x0) “projects out” the value of the function x0xx0-ex0+ef(x0)d(x)Figure XI: Convolving a sufficiently smooth function with a Dirac delta function projects out the function value at one point. Massachusetts Institute of Technology IV-7 � 8.04 Quantum Physics Lecture IV0xq(x)1Figure XII: The δ-function is also the derivative of the Heavyside step function. at x0. 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 − x0)= −f�(x − x0). (4-22) Convolution with δ� projects out the negative derivative at x0. Massachusetts Institute of Technology IV-8