Quantum Physics-Expectation Values and Compton Scattering

� � � � 8.04 Quantum Physics Lecture VLast time • |ψ(x)|2 and |φ(k)|2 as probability densities in position and momentum space, respectively... • Parseval’s theorem: ∞ dx|ψ(x)|2 = ∞ dk|φ(k)|2 −∞ −∞ Normalization: 1= ∞ dxψ(x)2 = ∞ dkφ(k)2 • −∞ ||−∞ ||Measurement: • Measurement of x with resolution Δxapp < Δx yields random result within Δx (with probability given by |ψ(x)|2Δxapp) and changes the momentum distributtion back action of the measurement. • You can know the probability distributions |ψ(x)|2 and |φ˜(p)|2 exactly and simultaneously without violating Heisenberg uncertainty. Ina particulla realization of the experiment, you cannot predict the outcomes of a position measurement followed by a momentum measurement to better than ΔxΔp ≥ h¯.2 • Since ψ(x) uniquely determines φ(k), the wavefunction ψ(x) in position space encodes both the spatial and the momentum distribution of the particlle Different ψ1(x), ψ2(x) with the same probability density |ψ1(x)|2 = |ψ2(x)|2 describe particles with the same position distribution but different momentum distributions. φ(k) = 1 √2π � ∞ −∞ dxψ(x)e−ikx ψ(x) = dkφ(k)e ikx 1 √2π ∞� −∞ = Fourier transform (5-1)⇒ = inverse Fourier transform (5-2)⇒ Dirac delta function: • 0xArea=1wFigure I: Dirac delta function.Massachusetts Institute of Technology V-1 � � 8.04 Quantum Physics Lecture Vx1 2 δ(x) = lim e− (5-3) w0 √2πw 2w2 �→0 for x = x0,δ(x − x0)= � (5-4) � ∞ for x = x0. ∞ dxδ(x − x0) = 1 (5-5) � −∞∞ dxδ(x − x0)f(x)= f(x0) (5-6) −∞ � ∞ dxe±ikx =2πδ(k) (5-7) �−∞∞ dke±ikx =2πδ(x) (5-8) −∞ Today • Finish proof: inverse Fourier transform • Exact definition of expectation values, uncertainties • Proof (see notes): ΔxΔk ≥ 1 ,ΔxΔp ≥ h¯,ΔωΔt ≥ 1 ,ΔEΔt ≥ ¯h 222 2 • Compton scatteringPhotoelectric effect• Back to our integral I(y)= ∞ dkeiky. We have already determined the area under −∞I(y) as being equal to 2π. Consequently, we have another useful identity: ∞ dkeiky =2πδ(y) (5-9) �−∞∞ dxeikx =2πδ(k) (5-10) −∞ We can now finish our proof of the inverse Fourier transformation.Massachusetts Institute of Technology V-2 � � � 8.04 Quantum Physics Lecture VInverse Fourier Transform.˜1 ∞ dxψ(x)e−ikx φ(k)= √2π�� −∞ � 1 ∞ 1 ∞∞ √2π dkφ˜(k)e ikx =2π dx�ψ(x�) dkeik(x−x�) −∞ �−∞ −∞ 1 ∞ = dx�ψ(x�)2πδ(x − x�)2π −∞ = ψ(x) φ(k) = 1 √2π � ∞ −∞ dxψ(x)e−ikx ψ(x) = 1 √2π � ∞ −∞ dkφ(k)e ikx = Fourier transform (5-11)⇒ inverse Fourier transform = (5-12)⇒ (Fourier decomposition) Note that since φ(k), and hence the momentum distribution |φ˜(p)|2 is completely determmine by ψ(x), the wavefunction carries momentum information as well as spatial information about the particle. Expectation values and a precise definition of uncerttaint Given that |ψ(x)|2dx is the probability to find the particle within [x,x + dx], we can calculate the expectation value of the particle’s position as ∞ 2�x� = −∞ dxx|ψ(x)|= average position. (5-13)⇒ Similarly, we can define � ∞n2�x � = dxxn|ψ(x)|(5-14) −∞ and for any function of position f(x), ∞ 2�f(x)� = dxf(x)|ψ(x)|. (5-15) −∞ We define the uncertainty of the particle’s position Δx via the following relation: (Δx)2 = �(x −�x�)2� ≥ 0= ⇒ precise definition ofΔx (5-16) Massachusetts Institute of Technology V-3 �������������8.04 Quantum Physics Lecture VWe can expand the RHS: �(x −�x�)2� = �x 2 − 2x�x� + �x�2� = �x 2�− 2�x��x� + �x�2 (Δx)2 = �x 2�−�x�2 (5-17) Similarly, since |φ˜(p)|2 is the probability density for momentum, ∞ ∞ φ˜(p)2 2�p� =dpp−∞∞ dk(¯hk)φ(k)(5-18)||||=�∞−∞ =φ˜(p)2 2�f(p)� = dpf(p)−∞ dkf(¯hk)φ(k)(5-19)||||−∞ (Δp)2 = �(p − �p�)2� = �p 2� − �p�2 Using these definitions we can prove the following theorem: (5-20) Theorem 5.1. For any function ψ(x), we have the inequality ΔxΔk ≥ 1 2 . (5-21) Multiplication by ¯h then yields the Heisenberg uncertainty relation ΔxΔp ≥ ¯h 2 . (5-22) The equality occurs only for Gaussian functions. Proof. Consider the positive quantity xψ(x)+ λdψdx2 ≥ 0, (5-23)∞ I(λ)= dx−∞ where λ is a real number. We can assume without loss of generality that the origin of the coordinate system is chosen at the expectation value of the particle’s position, such that �x� = 0. When we factor out the integrand in I(λ), there are three terms. The first is∞ dxx2|ψ(x)|2 == (Δx)2 (5-24) −∞ Massachusetts Institute of Technology V-4 � � 8.04 Quantum Physics Lecture Vsince �x� = 0. The second term can be integrated by parts. � �����∞ dψ ∗ dψ ∞ d dxλx ψ + ψ∗ = dxλx (ψ∗ψ)dxdx dx−∞ −∞ � ∞x=∞ 2 = λxψ∗(x)ψ(x)|x=−∞ − dxλ|ψ(x)|� � ��−∞ � � =0 if ψ(x) van-=λ since ψ(x) is ishes sufficiently normalized fast as x→ ±∞ = λ (5-25) The last term can be represented in terms of the Fourier Transform. λ2 � ∞ dx ��� dψ ���2 = λ2 � ∞ dx � d 1 � ∞ dkφ˜(k)e ikx � � dx � dx √2π−∞ � −∞ � −∞ �d 1 ∞∗· dx √2π dk�φ˜(k�)e ik�x � −∞ ��� ��� = λ2 ∞ dx ∞ dkik φ˜(k)e ikx ∞ dk�ik�φ˜(k�)e ik�x ∗ 2π �−∞ � −∞ � −∞ = λ2 ∞ dk ∞ dk�kk�φ˜(k)φ˜∗(k�) ∞ dxei(k−k�)x 2π −∞ −∞ �−∞ � � �� 2πδ(k−k�) = λ2 ∞ dk ∞ dk�kk�φ˜(k)φ˜∗(k�)δ(k − k�) �−∞ −∞∞ = λ2 dkk2|φ(k)|2 −∞ = λ2�k2� = λ2(Δk)2 . (5-26) Where in the last step we have again assumed without loss of generality a coordinate system where �p� = 0. Consequently, we have I(λ) = (Δx)2 − λ + λ2(Δk)2, and I(λ) ≥ 0 for all λ λ 11 = (Δk)2 λ2 − (Δk)2 + 4(Δk)2 − 4(Δk)2 + (Δx)2 ��211 = (Δk)2 λ − 2(Δk)2 + (Δx)2 − 4(Δk)2 . (5-27) I(λ) ≥ 0 thus requires (Δx)2 ≥ 1 or ΔxΔk ≥ 1 .4(Δk)2 2 Massachusetts Institute of Technology V-5 8.04 Quantum Physics Lecture VYou will show in a homework problem that the equality holds for Gaussian wavepackets. We note without proof that the equality holds only for Gaussian wavepackets. The same calculation in the time-frequency domain yields 1 ΔωΔt ≥ (5-28)2 or, multiplying by ¯h, E =¯hω, h¯ΔEΔt ≥ (5-29)2 ⇒ energy-time uncertainty. The Heisenberg uncertainty arises because of our inability to measure a frequency (wavelength) accurately in a finite time interval (position interval). Compton Scattering: Electrons scattering x-rays When visible light is scattered by matter (electrons), the scattered light has approximattel the same frequency as the incident light. For x-rays, Compton (1922) observed e-qww'Figure II: Compton Scattering that the wavelength of the scattered light increases, and that the increase is longer for larger scattering angle θ. For backscattering (θ = π) the wavelength shift Δλ = λ� −λ is equal to Δλ =4.85 × 10−12 m, independent of the wavelength of the x-rays. Comptto interpreted the effect as a (relativistic) scattering process between two particles, an electron and an x-ray photon, obeying energy and momentum conservation. Electrro initially at rest, treat it as free particle (initial energy of electron bound in atom ∼ 10 eV, x-ray photon energy 10 keV). Massachusetts Institute of Technology V-6 8.04 Quantum Physics Lecture Vlinq=0linq=p/4(a) θ = 0 (b) θ = π/4 linq= p/2lin=0.71Aq=p0.0485A(c) θ = π/2 (d) θ = π Figure III: . . . e-qw, kw', k'E', p'ω = c|k| = 2πc λ p =0 ω� = c|k�| = 2λπc � Figure IV: Compton ScatteringMassachusetts Institute of Technology V-7 8.04 Quantum Physics Lecture VElectron: E = m0c 2 E�2 = p�2 c 2 + m 20c 4 Momentum conservation: h¯k = p� +¯hk� Energy conservation: ¯hωhω = E� +¯In your homework, you will show that these assumptions result in a wavelength shift Δλ = λ� − λ = λc(1 − cos θ)=⇒ Compton shift (5-30) where λc = h =0.024˚aA is constant called the Compton wavelength of the mec electron. According to λdB = hp , it is the deBroglie wavelength associated with the (fictional) momentum p = mec. This formula explains the shifted inelastic scattering peak, the unshifted elastic peak observed at λ� = λ can be explained by the scattering of strongly bound inner-shell electrons. In this case, the whole atom (with 104 times larger mass) absorbs the momentum, resulting in a much smaller recoil energy shift that was unobservable to Compton. The energy loss of the x-ray photon is due to the KE imparted to the electron as a consequence of momentum conservation in the scattering process of an individual photon. If h 0 (i.e., E = hν and p = hk¯→ hassociated with individual photons become very small, λc = 0, and there is mec →no shift): the Compton shift is a QM effect associated with the “grainy” quantized nature of light. Compton scattering shows that there are individual photons carrying ¯energy E = hω and momentum p ¯= c and that the interaction between ¯= hk hω , electrons and light can be viewed as a scattering process obeying momentum and energy conservation. Massachusetts Institute of Technology V-8