Lecture Notes on Quantum Mechanics-Part I IeThSnr ongthe Schrodinger Equation. Contents 1.wave ad I Motion 2 34 II.Statistical Interpretation of Wave Mechanics 556 8 891 IIL.Moment nd Ine 224 ty amplitude t of coordinate 556616778 D.Expectation value of dynamical quantity v.ehridmganatiac 202 VI.Review on basic concepts in quantum mechanics 25 L.INTRODUCTION:MATTER WAVE AND ITS MOTION The emergence and development of quantum mechanics began in early years of the previous century and accom-
Lecture Notes on Quantum Mechanics - Part I Yunbo Zhang Institute of Theoretical Physics, Shanxi University This is the first part of my lecture notes. I mainly introduce some basic concepts and fundamental axioms in quantum theory. One should know what we are going to do with Quantum Mechanics - solving the Schr¨odinger Equation. Contents I. Introduction: Matter Wave and Its Motion 1 A. de Broglie’s hypothesis 2 B. Stationary Schr¨odinger equation 3 C. Conclusion 4 II. Statistical Interpretation of Wave Mechanics 5 A. Pose of the problem 5 B. Wave packet - a possible way out? 6 C. Born’s statistical interpretation 7 D. Probability 8 1. Example of discrete variables 8 2. Example of continuous variables 9 E. Normalization 11 III. Momentum and Uncertainty Relation 12 A. Expectation value of dynamical quantities 12 B. Examples of uncertainty relation 14 IV. Principle of Superposition of states 15 A. Superposition of 2-states 15 B. Superposition of more than 2 states 16 C. Measurement of state and probability amplitude 16 1. Measurement of state 16 2. Measurement of coordinate 17 3. Measurement of momentum 17 D. Expectation value of dynamical quantity 18 V. Schr¨odinger Equation 19 A. The quest for a basic equation of quantum mechanics 19 B. Probability current and probability conservation 20 C. Stationary Schr¨odinger equation 22 VI. Review on basic concepts in quantum mechanics 25 I. INTRODUCTION: MATTER WAVE AND ITS MOTION The emergence and development of quantum mechanics began in early years of the previous century and accomplished at the end of the twentieth years of the same century. We will not trace the historical steps since it is a long story. Here we try to access the theory by a way that seems to be more ”natural” and more easily conceivable
2 A.de Broglie's hypothesis Inspiration:Parallelism between light and matter Wave:frequency ,w,wavelength A,wave vectork..... Particle:velocity v,momentum p,energy c of Yet,tl (corue ture .Matter particles should also posses another side of ture-the wave nature g=hv=hw 卫== A=h/p h/V2mE The state of(micro)-particle should be described by a wave function.Here are some examples of state functions: 1.Free particle of definite momentum and energy is described by a monochromatic traveling wave of definite wave vector and frequency a=cs(红x-2t+0） A'cos(kr-wt+o) =Am(后r-t+0) Replenish an imaginary part 物=A'si血（肛-方t+o we get the final form of wave function =V1+it=A'chevetpe-tu=Aetpre-tet which is the wave picture of free particle.In 3Dwe have 2.Hydrogen atom in the ground state will be shown later to be e-()”- This wave function shows that the motion of electron is in a "standing wave"state.This is the wave picture of above state
2 A. de Broglie’s hypothesis Inspiration: Parallelism between light and matter Wave: frequency ν, ω, wavelength λ, wave vector k · · · · · · Particle: velocity v, momentum p, energy ε · · · · · · • Light is traditionally considered to be a typical case of wave. Yet, it also shows (possesses) a corpuscle nature - light photon. For monochromatic light wave ε = hν = ~ω p = hν c = h λ = ~k, (k = 2π λ ) • Matter particles should also possess another side of nature - the wave nature ε = hν = ~ω p = h λ = ~k This is call de Broglie’s Hypothesis and is verified by all experiments. In the case of non-relativistic theory, the de Broglie wavelength for a free particle with mass m and energy ε is given by λ = h/p = h/√ 2mε The state of (micro)-particle should be described by a wave function. Here are some examples of state functions: 1. Free particle of definite momentum and energy is described by a monochromatic traveling wave of definite wave vector and frequency ψ1 = A ′ cos ( 2π λ x − 2πνt + φ0 ) = A ′ cos (kx − ωt + φ0) = A ′ cos ( 1 ~ px − 1 ~ εt + φ0 ) Replenish an imaginary part ψ2 = A ′ sin ( 1 ~ px − 1 ~ εt + φ0 ) , we get the final form of wave function ψ = ψ1 + iψ2 = A ′ e iφ0 e i ~ pxe − i ~ εt = Ae i ~ pxe − i ~ εt which is the wave picture of motion of a free particle. In 3D we have ψ = Ae i ~ ⃗p·⃗re − i ~ εt 2. Hydrogen atom in the ground state will be shown later to be ψ(r, t) = ( 1 πa3 0 )1/2 e − r a0 e − i ~ E1t This wave function shows that the motion of electron is in a ”standing wave” state. This is the wave picture of above state
3 B A FIG.1:Maupertuis'Principle B.Stationary Schradinger equation The parallelism between light and matter can go further Light: wave nature omitted geometric optics wave nature can not be omitted new mechanics namely quantum mechanics Quantum Mechanics二Wave Optics☐ Light wave geometric optics Fermat's principle wave propagati Matter wav particle dynam Principle of least action wave propagation I'resently unknown Now conide light wave propagation in a non-homogenousmedium B light path=n(问d Fermat's principle n=0 For a particle moving inapotential field V(),the principle of least action reads iV2mia=iV2me-v阿as=0 The corresponding light wave quation (Helmholtzequation)is (（-)保利-0
3 A B FIG. 1: Maupertuis’ Principle. B. Stationary Schr¨odinger equation The parallelism between light and matter can go further Light: wave nature omitted geometric optics wave nature can not be omitted wave optics Matter: wave nature omitted particle dynamics wave nature can not be omitted a new mechanics, namely quantum mechanics Particle Dynamics ⇐⇒ Geometric Optics ⇓ ⇓ Quantum Mechanics ? ⇐⇒ Wave Optics Here we make comparison between light propagation of monochromatic light wave and wave propagation of monochromatic matter wave Light wave geometric optics Fermat’s principle wave propagation Helmholtz equation Matter wave particle dynamics Principle of least action wave propagation Presently unknown Now consider light wave propagation in a non-homogeneous medium light path = ∫ B A n(⃗r)ds Fermat’s principle δ ∫ B A n(⃗r)ds = 0 For a particle moving in a potential field V (⃗r), the principle of least action reads δ ∫ B A √ 2mT ds = δ ∫ B A √ 2m(E − V (⃗r))ds = 0 The corresponding light wave equation (Helmholtz equation) is ( ∇2 − 1 c 2 ∂ 2 ∂t2 ) u (⃗r, t) = 0
which after the separation of variables reduces to w+-0 Here we note thatis a constant.Thus we arrived at a result of comparison as follows 3 y2m(E-V(F))ds=06 n()ds=0 Pruko☐名2+妥=司 威the e e广6cmea 720+A2m(E-V(川中=0 Substitute the known free particle sotion -er into the above equation.We find the unknown constant A equals to1/,therefore we have 2+0E-V=0 o the general cas.It inaform -+v= and bears the name"Stationary Schrodinger Equation" C.Conclusion quation. rom ti side of the motion of micro-particl.V2mvm(V(trented a refraction index of matter wave forhat ialdtricstV()What appe .Clasical mechanics:particles with total energy Ecan not arrive at places with E<V( r),t at the attenuated in its course of propagation
4 which after the separation of variables reduces to ∇2ψ + n 2ω 2 c 2 ψ = 0. Here we note that ω is a constant. Thus we arrived at a result of comparison as follows δ ∫ B A √ 2m(E − V (⃗r))ds = 0 ⇐⇒ δ ∫ B A n(⃗r)ds = 0 ⇓ ⇓ Presently unknown ? ⇐⇒ ∇2ψ + n 2ω 2 c 2 ψ = 0 ⇓ Presumed to be of the form ∇2ψ + An2ψ = 0 Here A is an unknown constant, and the expression √ 2mT plays the role of ”index of refraction” for the propagation of matter waves. The unknown equation now can be written as ∇2ψ + A [2m(E − V (⃗r))] ψ = 0 Substitute the known free particle solution ψ = e i ~ ⃗p·⃗r E = 1 2m ⃗p 2 V (⃗r) = 0 into the above equation. We find the unknown constant A equals to 1/~ 2 , therefore we have ∇2ψ + 2m ~ 2 [E − V (⃗r)] ψ = 0 for the general case. It is often written in a form − ~ 2 2m ∇2ψ (⃗r) + V (⃗r) ψ (⃗r) = Eψ (⃗r) and bears the name ”Stationary Schr¨odinger Equation” C. Conclusion By a way of comparison, we obtained the equation of motion for a particle with definite energy moving in an external potential field V (⃗r) - the Stationary Schr¨odinger Equation. From the procedure we stated above, here we stressed on the ”wave propagation” side of the motion of micro-particle. √ 2mT = √ 2m(E − V (⃗r)) is treated as ”refraction index” of matter waves. It may happen for many cases that in some spatial districts E is less than V (⃗r), i.e., E < V (⃗r). What will happen in such cases? • Classical mechanics: particles with total energy E can not arrive at places with E < V (⃗r). • Matter waves (Q.M.): matter wave can propagate into districts E < V (⃗r), but in that cases, the refraction index becomes imaginary. About Imaginary refraction index : For light propagation, imaginary refraction index means dissipation, light wave will be attenuated in its course of propagation. For matter waves, no meaning of dissipation, but matter wave will be attenuated in its course of propagation
Double Slit FIG.:Double II.STATISTICAL INTERPRETATION OF WAVE MECHANICS aves 3 made is hypothesis ectrons.r ons.etc. around stal Wave function is a complex function of its variables (红，t)=Ae-B到) v(r0.00-Viogze-t 1.Dynamical equation governing the motion of micro-particle is by itself a equation containing imaginary number This able the app be of space is ible fr A.Pose of the problem What kind of wave it is? .Optics:Electromagnetic wave wave propagating E(红，t)=oe(-2叫=%-网 Eo-amplitudefield strength Intensity E-energy density ·Acoustic wave U(,t)foloe(2)
5 FIG. 2: Double slit experiments. II. STATISTICAL INTERPRETATION OF WAVE MECHANICS People tried hard to confirm the wave nature of micro-particles, and electron waves were first demonstrated by measuring diffraction from crystals in an experiment by Davison and Germer in 1925. They scattered electrons off a Nickel crystal which is the first experiment to show matter waves 3 years after de Broglie made his hypothesis. Series of other experiments provided more evidences, such as the double slit experiments using different particle beams: photons, electrons, neutrons, etc. and X-ray (a type of electromagnetic radiation with wavelengths of around 10−10 meters). Diffraction off polycrystalline material gives concentric rings instead of spots when scattered off single crystal. Wave function is a complex function of its variables ψ(x, t) = Ae i ~ (px−Et) ψ (r, θ, ϕ, t) = 1 √ πa3 0 e − r a0 e − i ~ E1t 1. Dynamical equation governing the motion of micro-particle is by itself a equation containing imaginary number 2. The wave function describing the state of micro-particle must fit the general theory frame of quantum theory (operator formalism) - requirement of homogeneity of space. This means, the symmetry under a translation in space r → r + a, where a is a constant vector, is applicable in all isolated systems. Every region of space is equivalent to every other, or physical phenomena must be reproducible from one location to another. A. Pose of the problem What kind of wave it is? • Optics: Electromagnetic wave E(x, t) = ˆy0E0e i( 2π λ x−2πνt) = ˆy0E0 wave propagating z }| { e i(kx−ωt) E0 − amplitude → field strength Intensity E 2 0 → energy density • Acoustic wave U(x, t) = ˆy0U0e i( 2π λ x−2πνt)
Electro in the p A sing FIG.3:Electromagnetic Wave Propagation. o-amplitudemechanical displacement msity U哈→energy density 。Wave function 红，)=Ae(学-2）=Aetm-B Scalar wave.Amplitude-A-? Intensity-AP→ B.Wave packet-a possible way out? Two examples of localized wave packets .Superposition of waves with wave number between (o-Ak)and (+k)-square packet 因={A-AK<+as elsewhere =左 宏24恤a包 工 Amplitude Both wave number and spatial position have a spread-uncertainty relation △xπ/△k △x·△k≈T △r△p≈rh
6 FIG. 3: Electromagnetic Wave Propagation. U0 − amplitude → mechanical displacement Intensity U 2 0 → energy density • Wave function ψ(x, t) = Aei( 2π λ x−2πνt) = Ae i ~ (px−Et) Scalar wave, Amplitude - A →? Intensity - A 2 →? Early attempt: Intensity → material density, particle mass distributed in wave. But wave is endless in space, how can it fit the idea of a particle which is local. B. Wave packet - a possible way out? particle = wave packet - rain drop e ik0x− an endless train, How can it be connected with particle picture? Two examples of localized wave packets • Superposition of waves with wave number between (k0 − ∆k) and (k0 + ∆k) - square packet φ(k) = { A, k0 − ∆k < k < k0 + ∆k 0, elsewhere ψ(x) = 1 √ 2π ∫ φ(k)e ikxdk = 1 √ 2π ∫ k0+∆k k0−∆k Aeikxdk = 1 √ 2π 2A sin (∆kx) x | {z } Amplitude e ik0x Both wave number and spatial position have a spread - uncertainty relation ∆x ∼ π/∆k, ∆x · ∆k ≈ π ∆x · ∆p ≈ π~
M 1ANMM AAAAAN W A∧V A水AX1 IG.4:Wave packet formed by plane waves with different frequency Plane wave:elk /ave packet:ko and FIG.5:Spread of a wave packet .Superposition of waves with wave number in a Gaussian packet 2因=(e）e-a- Problem 1 The fou n It is the best Problem:Wave packet is But electros are localized in the atom (1).This viewpont e the e even in enatreaf cro-partic ea ave is cor C.Born's statistical interpretation function,which says that in ID asegives the probability of finding the particle at point at timetor more precisely Probability of findin between a al nd b.at tm
7 FIG. 4: Wave packet formed by plane waves with different frequency. FIG. 5: Spread of a wave packet. • Superposition of waves with wave number in a Gaussian packet φ(k) = ( 2α π )1/4 e −α(k−k0) 2 Problem 1 The Fourier transformation of φ(k) is also Gaussian. It is the best one can do to localize a particle in position and momentum spaces at the same time. Find the root mean square (RMS) deviation ∆x · ∆p =? Problem: Wave packet is unstable - The waves have different phase velocity and the wave components of a wave packet will disperse even in vacuum. But electrons are localized in the atom (1A˚). This viewpoint emphasize the wave nature of micro-particle while killing its particle nature. Another extreme viewpoint on wave particle duality is that wave is composed by large amount of particles (as waves in air). However experiments show clearly a single electron possesses wave nature. This viewpoint over-stressed the particle nature. C. Born’s statistical interpretation A particle by its nature is localized at a point, whereas the wave function is spread out in space. How can such an object represent the state of a particle? The answer is provided by Born’s Statistical Interpretation of the wave function, which says that in 1D case |ψ (x, t)| 2 gives the probability of finding the particle at point x, at time t - or, more precisely ∫ b a |ψ (x, t)| 2 dx = { Probability of finding the particle between a and b, at time t. }
◇ 1011121314151617161920212232425267 FIG.6:Histogram for the distribution etyfor the partice apperingn vome mentoprtiono the intnity Prob ility amplitud dxlw(,3F-(,t)p(,)7 Probability density D.Probability Bec use of the statistical interpretation,probability plays acentral role quantum mechanics.So we introduce ome notation and terminology. 14 people in a room.Let N(j)represent the number of people of age j. N(14)=1,N(15)=1,N(16)=3,N(22)=2,N(24)=2,V(25)=5 while N(17)for instance,is zero.The total number of people in the room is Questions about the distribution this particular age.In general the probability of getting age jis PG)-NU) PG)=1 0 2.What is the most probable age?Answer:25.In general the most probable j is the j for which P()is a maximum. getting a smaller result
8 FIG. 6: Histogram for the distribution. In 3D case, the relative probability for the particle appearing in a volume element d 3⃗r is proportional to the intensity of wave in that volume element dw ∝ |ψ (⃗r, t)| 2 d 3 ⃗r = ψ ∗ (⃗r, t) Probability amplitude z }| { ψ (⃗r, t) | {z } Probability density d 3 ⃗r D. Probability Because of the statistical interpretation, probability plays a central role quantum mechanics. So we introduce some notation and terminology. 1. Example of discrete variables 14 people in a room. Let N(j) represent the number of people of age j. N(14) = 1, N(15) = 1, N(16) = 3, N(22) = 2, N(24) = 2, N(25) = 5 (1) while N(17), for instance, is zero. The total number of people in the room is N = ∑∞ j=0 N(j) Questions about the distribution 1. If you selected one individual at random from this group, what is the probability that this person’s age would be 15? Answer : One chance in 14, since there are 14 possible choices, all equally likely, of whom only one has this particular age. In general the probability of getting age j is P(j) = N(j) N Notice that the probability of getting either 14 or 15 is the sum of the individual probabilities (in this case, 1/7). In particular, the sum of all the probabilities is 1 - you’re certain to get some age: ∑∞ j=0 P(j) = 1 2. What is the most probable age? Answer : 25. In general the most probable j is the j for which P(j) is a maximum. 3. What is the median age? Answer : 23, for 7 people are younger than 23, and 7 are older. In general, the median is that value of j such that the probability of getting a larger result is the same as the probability of getting a smaller result
FIG.7:Two histogram with different. 4.What is the average(or mean)age?Answer: 国++300+2@+2e+5四-警-21 14 In general the average value ofj(which we shall write thus:())is 6)=2N0-2PU 1=0 a nechanics it is called the expectation value.Nevertheless the yalue is not sgay9器seew 的=2PU) In general,the average value of some function of j is given by U6》-立f6PU j= Beware(2〉≠)2 o三/(a)2〉=VG)-02 △j=j- For the two distributions in the above figure,we have 1=V25.2-25-V0.2 2=V31.0-25=v6 Erample of riables It is simple enough to generalize to continuous distribution.Technically we need "infinitesimal intervals".Thus =p(r)dr
9 FIG. 7: Two histogram with different σ. 4. What is the average (or mean) age? Answer: (14) + (15) + 3 (16) + 2 (22) + 2 (24) + 5 (25) 14 = 294 14 = 21 In general the average value of j (which we shall write thus: ⟨j⟩) is ⟨j⟩ = ∑jN(j) N = ∑∞ j=0 jP(j) In quantum mechanics it is called the expectation value. Nevertheless the value is not necessarily the one you can expect if you made a single measurement. In the above example, you will never get 21. 5. What is the average of the squares of the ages? Answer: You could get 142 = 196, with probability 1/14, or 152 = 225, with probability 1/14, or 162 = 256, with probability 3/14, and so on. The average is ⟨ j 2 ⟩ = ∑∞ j=0 j 2P(j) In general, the average value of some function of j is given by ⟨f(j)⟩ = ∑∞ j=0 f(j)P(j) Beware ⟨ j 2 ⟩ ̸= ⟨j⟩ 2 . Now, there is a conspicuous difference between the following two histograms, even though they have the same median, the same average, the same most probable value, and the same number of elements: The first is sharply peaked about the average value, whereas the second is broad and flat. (The first might represent the age profile for students in a big-city classroom, and the second, perhaps, a rural one-room schoolhouse.) We need a numerical measure of the amount of ”spread” in a distribution, with respect to the average. The most effective way to do this is compute a quantity known as the standard deviation of the distribution σ ≡ √⟨ (∆j) 2 ⟩ = √ ⟨j 2⟩ − ⟨j⟩ 2 ∆j = j − ⟨j⟩ For the two distributions in the above figure, we have σ1 = √ 25.2 − 25 = √ 0.2 σ2 = √ 31.0 − 25 = √ 6 2. Example of continuous variables It is simple enough to generalize to continuous distribution. Technically we need ”infinitesimal intervals”. Thus { Probability that an individual (chosen at random) lies between x and (x + dx) } = ρ(x)dx
10 h FIG.:The probability density in the Examplep() 2点水Temh:bt and6ae Pu=de)de and the ruleswe deduced for discrete distributions translate in the obvious way: (回=xpe)da Uen=CIee达 a2=(△x〉-x)-2 it falls:i distance ut beeth Inongin he distancetime e time near the top,so the average x0=)9t2 片e共 要-√景=后“ Evidently the probability density is 网-2症0≤rs) 1 。2后-a(n0-1
10 FIG. 8: The probability density in the Example ρ(x). ρ(x) is the probability of getting x, or probability density. The probability that x lies between a and b (a finite interval) is given by the integral of ρ(x) Pab = ∫ b a ρ(x)dx and the rules we deduced for discrete distributions translate in the obvious way: 1 = ∫ +∞ −∞ ρ(x)dx ⟨x⟩ = ∫ +∞ −∞ xρ(x)dx ⟨f(x)⟩ = ∫ +∞ −∞ f(x)ρ(x)dx σ 2 ≡ ⟨ (∆x) 2 ⟩ = ⟨ x 2 ⟩ − ⟨x⟩ 2 Example: Suppose I drop a rock off a cliff of height h. As it falls, I snap a million photographs, at random intervals. On each picture I measure the distance the rock has fallen. Question: What is the average of all these distance? That is to say, what is the time average of the distance traveled? Solution: The rock starts out at rest, and picks up speed as it falls; it spends more time near the top, so the average distance must be less than h/2. Ignoring air resistance, the distance x at time t is x(t) = 1 2 gt2 The velocity is dx/dt = gt, and the total flight time is T = √ 2h/g. The probability that the camera flashes in the interval dt is dt/T, so the probability that a given photograph shows a distance in the corresponding range dx is dt T = dx gt √ g 2h = 1 2 √ hx dx Evidently the probability density is ρ(x) = 1 2 √ hx ,(0 ≤ x ≤ h) (outside this range, of course, the probability density is zero.) We can check the normalization of this result ∫ h 0 1 2 √ hx dx = 1 2 √ h ( 2x 1/2 )h 0 = 1
