Fundamental of Quantum Mechanics - Pham Tan Thi

Nội dung text: Fundamental of Quantum Mechanics - Pham Tan Thi

1. Fundamental of Quantum Mechanics Pham Tan Thi, Ph.D. Department of Biomedical Engineering Faculty of Applied Sciences Ho Chi Minh University of Technology
2. Quantum Mechanics
3. Diffraction of Light/Electron by One Slit
4. Double-slits Experiments for Electrons (Wave-like character of electrons)
5. Wave-Particle Duality Energy in Energy out A beam of light can be thought as a flux of particle an electromagnetic wave (Newton/Planck/Einstein) (Huygens/Maxwell/Hertz) Wavelength λ Electric Field Zero mass, speed: c = 3 x 108 m/s Energy carried by each particle: Direction of propagation h = 6.6262 x 10-34 J.s (Planck) c = λν ν (frequency) = 1/T (period) Dispersion relation (Free space) Particle (Photon): Wave (Electromagnetism): - Photoelectric Effect - Interference - Compton Effect - Diffraction
6. de Broglie’s Hypothesis “Light can behave like a particle and Matter/Electron can behaves like a wave Energy of Photon by Special Relativity: E = pc Energy of Photon by Planck’s theory: E = h⌫ de Broglie’s wavelength: h = p Wave if λ > Scale It turns out that everything’s kind of mixed together at the fundamental microscopic scale.
7. For a Non-relativistic Free Particle Momentum is p = mv, here is v is the speed of the particle Total energy of a free particle, E, is kinetic energy: h h h B = = = p mv p2mE p2 mv2 E = K = = 2m 2 Bullet: m = 0.1 kg; v = 1000 m/s 36 ➔ 6.63 10 m B ⇡ ⇥
8. What is the de Broglie’s Wavelength of an Electron? What is the de Broglie’s wavelength of an electron moving at 2.2 x 106 m/s
9. What is the de Broglie’s Wavelength of an Electron? What is the de Broglie’s wavelength of an electron moving at 2.2 x 106 m/s Now this is really really fast: 2.2 million meters per second. But it’s not relativistic. It’s is still slow compared to the speed of light so we can still do everything fairly classically. 31 6 24 p = mv =9.1 10 2.2 10 [kg m/s] = 2 10 N.s ⇥ ⇥ ⇥ ⇥ 34 24 10 = h/p =6.626 10 /2 10 =3.33 10 m ⇥ ⇥ ⇥ This is important number. This speed is a kind of average speed of an electron in ground state of hydrogen.
10. AcceleratedWhat is the deCharges Broglie’s (electrons, Wavelength protons) of a Person?Produce de Broglie’s theory can be extended to show that all matter exhibits the same wave-particle duality as light. This means everything in the universe can act like a wave: h = and h = 6.626 x 10-34 m2kg s-1 mv This shows that an object’s wavelength gets smaller when the more massive it is, and the faster it is moving. If a person has a mass of 75 kg, and is jogging at 8 km/h (which is about 2.2 m/s), then
11. AcceleratedWhat is the deCharges Broglie’s (electrons, Wavelength protons) of a Person?Produce de Broglie’s theory can be extended to show that all matter exhibits the same wave-particle duality as light. This means everything in the universe can act like a wave: h = and h = 6.626 x 10-34 m2kg s-1 mv This shows that an object’s wavelength gets smaller when the more massive it is, and the faster it is moving. If a person has a mass of 75 kg, and is jogging at 8 km/h (which is about 2.2 m/s), then 31 6.626 10 36 = ⇥ =4.016 10 m 7.5 2.2 ⇥ ⇥ This is about 700 billion billion times smaller than the classical electron radius, which is about 2.8 x 10-15 m. Diffraction works best if the slit is about the same size as the wavelength, and so this explains why we do not notice wave-like behavior in human.
12. Energy in Energy out Terminology Particle: Wave: A beam of light can be thought as Our traditional Our traditional understanding of a particle understanding of a wave a flux of particle an electromagnetic wave (Newton/Planck/Einstein) (Huygens/Maxwell/Hertz) Wavelength λ Electric Field Direction of propagation “Localized” - “De-localized” - definite position, spread out in space momentum, and time *Disturbance in the medium
13. 38.4 Wave–Particle Duality, Probability, and Uncertainty 1277 h (photon energy in terms E = hƒ = 2pƒ =Uv (38.19b) 2p of angular frequency) Using Eqs. (38.19) in Eq. (38.18), we can rewrite our photon wave equation as (wave function for a Ey x, t = Asin px x - Et U photon with x-momentum (38.20) px and energy E) 1 2 31 2> 4 Since this wave function has a definite value of x-momentum px, there is no uncertainty in the value of this quantity: Âpx = 0 . The Heisenberg uncertainty principle, Eq. (38.17), says that ÂxÂpx ÚU2. If Âpx is zero, then Âx must be infinite. Indeed, the wave described by Eq. (38.20) extends along the entire x-axis and has the same amplitude everywhere. The> price we pay for knowing the pho- ton’s momentum precisely is that we have no idea where the photon is! In practical situations we always have some idea where a photon is. To describe this situation, we need a wave function that is more localized in space. We can create one by superimposing two or more sinusoidal functions. To keep things simple, we’ll consider only waves propagating in the positive x-direction. For example, let’s add together two sinusoidal wave functions like those in Eqs. (38.18) and (38.20), but with slightly different wavelengths and frequencies and hence slightly different values px1 and px2 of x-momentum and slightly different values E1 and E2 of energy. The total wave function is Ey x, t = A1sin p1x x - E1t U + A2sin p2x x - E2 t U (38.21) Consider1 what2 this wave31 function looks2> 4 like at a31 particular instant2> 4 of time, say t = 0. At this instant Eq. (38.21) becomes Ey x, t = 0 = A1sin p1x x U + A2sin p2x x U (38.22) Figure 38.19a 1is a graph2 of the individual1 > 2 wave functions1 > at2 t = 0 for the case A2 =-A1, and Fig. 38.19b graphs the combined wave function Ey x, t = 0 given by Eq. (38.22). We saw something very similar to Fig. 38.19b in our dis- cussion of beats in Section 16.7: When we superimposed two sinusoidal1 waves2 with slightly different frequencies (see Fig. 16.24), the resulting wave exhibited amplitude variations not present in the original waves. In the same way, a photon represented by the wave function in Eq. (38.21) is most likely to be found in the regions where the wave function’s amplitude is greatest. That is, the photon is localized. However, the photon’s momentum no longer has a definite value because we began with two different x-momentum values, px1 and px2. This agrees with the Heisenberg uncertainty principle: By decreasing the uncertainty in the photon’s position, we have increased the uncertainty in its momentum. 38.19 (a) Two sinusoidal waves with slightly different wave numbers k and hence slightly different values of momentum px =Uk shown at one instant of time. (b) The superposition of these waves has a momentum equal to the average of the two individual values of momentum. The amplitude varies, giving the total wave a lumpy character not possessed by either individualSuperposition wave. Principle Ey(x) (a) 0 x Ey(x) (b) 0 x
14. Heisenberg Uncertainty Principle Heisenberg uncertainty states that only one of the “position” or “momentum” can be measured accurately at a single moment within the instrument limit. or It is impossible to measure both the position and momentum simultaneously with unlimited accuracy. x uncertainty in position ! p uncertainty in momentum x ! then ~ h xpx ~ = 2 2⇡
15. Heisenberg Uncertainty Principle If Δx is measured accurately, i.e., x 0 ⇒ px ! !1 The principle applies to all canonically conjugate pairs of quantities in which measurement of one quantity affects the capacity to measure the other. For examples, Energy E and time t ~ Et 2 and Angular momentum L and angular position θ ~ L✓ 2
16. Determination of the Position of a Particle by a Microscope Suppose we want to determine accurately the position and momentum of an electron along x-axis using an ideal microscope free from all mechanical and optical defects. The limit of resolution of the microscope is x = 2sini here i is semi-vertex angle of the cone of rays entering the objective lens of the microscope. Δx is the order of uncertainty in the x- component of the position of the electron.
17. Determination of the Position of a Particle by a Microscope We can’t measure the momentum of the electron prior to illumination. So there is uncertainty in the measurement of momentum of the electron. The scattered photon can enter the microscope anywhere between the angular range +i/-i. The momentum of the scattered photon is (according to de Broglie) h p = Its x-component can be given as 2h p = sini x
18. Determination of the Position of a Particle by a Microscope The product of the uncertainties in the x-components of position and momentum for the electron is 2h xp = sini x 2sini ⇥ ✓ ◆ ~ xp = h> x 2 This is in agreement with the uncertainty relation.
19. Applications of Heisenberg Uncertainty Principle (i) Non-existence of electron in nucleus Order of diameter of an atom ~5 x 10-15 m 15 If electron exists in the nucleus then (x) =5 10 m max ⇥ ~ xp x 2 ~ (x) (p ) = max x min 2 ~ 20 1 (px)min = =1.1 10 kg.m s 2(x)max ⇥ then E = pc = 20 MeV Thus kinetic energy of an electron must be greater than 20 MeV to be a part of nucleus. Experiments show that the electrons emitted by certain unstable nuclei don’t have energy greater than 3-4 MeV. Thus we can conclude that the electrons cannot be present within nuclei.
20. Concept of Bohr Orbit violates Uncertainty Principle ~ Recall Heisenberg Uncertainty xp x 2 p2 Kinetic energy is defined as E = 2m pp mvp x E = E = = p m m t Et = xp ~ then Et 2
21. Concept of Bohr Orbit violates Uncertainty Principle According to the concept of Bohr orbit, energy of an electron in an orbit is constant, i.e., ∆E = 0 ~ Et 2 ➔ t !1 All energy states of the atom must have an infinite life-time. But the excited states of the atom have life-time ~10-8 sec. The finite life-time ∆t gives a finite width (uncertainty) to the energy levels.
22. Recall Two-slit Interference Experiment
23. Taylor’s experiment (1908): double slit experiment with very dim light: interference pattern emerged after waiting for few weeks. Interference cannot be due to interaction between photons, i.e. cannot be outcome of destructive or constructive combination of photons. Interference pattern is due to some inherent property of each photon - it “interferes with itself” while passing from source to screen. Photons don’t “slit” —> light detectors always show signals of same intensity Slits open alternately: get two overlapping single-slit diffraction patterns - no two-slit interference Add detector to determine through which slit photon goes: —> no two interference Interference pattern only appears when experiments provides no means of determining through which slit photon passes.
24. Quantum Mechanics Interpretation • Patterns on screen are result of distribution of photons. • No way of anticipating where particular photon will strike. • Impossible to tell which path photon took — cannot assign specific trajectory to photon • cannot suppose that half went through one slit and half through other • can only predict how photons will be distributed on screen (or over detector(s)) • interference and diffraction are statistical phenomena associated with probability that, in given experimental setup, a photon will strike a certain point
25. Double-slit Exp. — Wave vs Quantum Wave theory: Quantum theory: Patterns of fringes: Patterns of fringes: Intensity bands due to variations Intensity bands due to variations in square of amplitude, A2, of in probability, P, of a photon resultant wave on each point on striking points on screen. screen. Role of the slits: Role of the slits: to provide two coherent sources to present two potential routes by of the secondary waves that which photon can pass from interfere on the screen. source to screen.
26. Wave Function • The quantity with which Quantum Mechanics concerned is the wave function of a body. • Wave function ψ is a quantity associated with a moving particle. It is a complex quantity. • |ψ|2 is proportional to the probability of finding a particle at a particular point in a particular time. It is the probability density. 2 = ⇤ | | ψ is the probability amplitude. Thus if = A + iB then ⇤ = A iB 2 2 2 2 = ⇤ = A i B = A + B | |
27. Properties of Wave Function 1. It must be finite everywhere. If ψ is infinite for a particular point, it means an infinitely large probability of finding the particles at that point. This would violate the uncertainty principle. 2. It must be single valued. If ψ has more than one value at any point, it mean more than one value of probability of finding the particle at that point which is obviously ridiculous. 3. It must be continuous and have a continuous first derivative everywhere. ~2 @2 (x) + U(x) (x)=E (x) 2m @x2
28. Derive the Schrodinger Equation i(kx !t) i✓ Wave function: = e e = cos✓ + isin✓ i✓ e = cos✓ isin✓ E = K.E. + P.E. i✓ i✓ 1 e + e E = mv2 + U = cos✓ 2 2 2 i✓ i✓ p e e E = + U =sin✓ 2m 2i Differential equation (1) with respect to position: i(kx !t) = e @ (kx !t) = ike @x @ (kx !t) = ike = ik @x
29. Derive the Schrodinger Equation (Cont.) Second differential equation (1) with respect to position: 2 @ 2 2 (kx !t) 2 2 = i k e = i k @t2 p 2⇡ h h Substitute to (1.2): k = where k = , p = , ~ = ~ 2⇡ 2 2 @ 2 2 2 2 @ = i k p = ~ @x2 @x2 p2 p2 E = + U E = + U 2m 2m ~2 @2 ~2 + U = E General Form, 2 + U = E 2m @x2 2mr
30. 40.2 Particle in a Box 1339 In Section 40.1 we solved this problem for the case U x = 0, corresponding 40.8 The Newtonian view of a particle to a free particle. The allowed wave functions and corresponding energies are in a box. 1 2 2 2 A particle with mass m moves along a ikx U k c x = Ae E = (free particle (40.24) straight line at constant speed, bouncing 2m between two rigid walls a distance L apart. The wave number1 k2 is equal to 2p l, where l is the wavelength.2 We found that k can have any real value, so the energy E of a free particle can have any value m from zero to infinity. Furthermore,> the particle can be found with equal probabil- v ity at any value of x from - q to + q. Now let’s look at a simple model inParticle which a particle in isa bound boxso withthat it cannot “Infinitely40.2 HardParticle Walls”in a Box 1339 escape to infinity, but rather is confined to a restricted region of space. Our sys- L In Section 40.1 we solved this problem for the case U x 0, corresponding 40.8 The Newtonian view of a particle tem consists of a particle confined between two rigid walls separated= by a dis- Boundary conditions and x tance L (Fig. 40.8).to a free Theparticle. motion The allowed is purely wave+∞ one functions dimensional, andx ≤ corresponding 0 with the energies particle are innormalization a box. determine ψ(x) 1 2 2 2 A particle0 with mass m moves along a L moving along the x-axis only and theikx walls Uatk x = 0 and x = L. The potential c U(x)x = Ae = E 0= (free 0 the particle can be found with equal probabil- free to move within a long, straight molecule or along a very thin wire. v ity at any value of x fromSince - q the to + qwalls. are impenetrable, Now let’s look at athere simple modelis a zeroin which probability a particle is bound ofso that it cannot Wave Functionsescape for to infinity, a Particle but rather in is confineda Box to a restricted region of space. Our sys- L tem consists of a particlefinding confined particle between outsidetwo rigid walls the separated box. by a dis- To solve the Schrửdinger equation for this system, we begin with some restric- 40.9 The potential-energy functionx for a tions on the particle’stance L stationary-state (Fig. 40.8). The wave motion function is purely c onex . dimensional, Because the with particle the particle is moving along the x-axis only and the walls at x 0 and x L. The potential particle in0 a box. L confined to the region 0 x L, weZero expect probability the probability= means: distribution= func- energy corresponding to the rigid walls is infinite, so the particle cannot escape; The potential energy U is zero in the interval 2 2 2 1 2 tion ƒ ° x, t ƒ between= ƒc x theƒ walls, and the the potential wave|ψ(x)| function energy= 0, isfor czero xx (Fig.≤ to 0 40.9). beand zero This x ≥ outside situation L thatis often 0 , x , L and is infinite everywhere outside region. This agreesdescribed with as a the “particle Schrửdinger in a box.” equation:This model Ifmight the represent term U anx electronc x that in is this interval. Eq. (40.23)1 is2 tofree be tofinite,1 move2 then within c a xlong,The must straight wave be zero molecule function where1 or2 alongU mustx a isvery infinite. also thin wire. be zero ` ` Furthermore, c x must be a continuous function to be a mathematically1 2 1 well-2 Wave Functions forat athe Particle walls in (x a = Box 0 and x = L), since behaved solution to the Schrửdinger1 2 equation. This implies1 2 that c x must be To solve the Schrửdingerthe equation wave for function this system, must we begin be with some restric- zero at the region’s1 boundary,2 x = 0 and x = L. These two conditions serve as 40.9 The potential-energy function for a tions on the particle’s stationary-statecontinuous. wave function c x . Because the particle is particle in a box. boundary conditions for the problem. They should look familiar, because1 2 they U(x) U(x) confined to the region 0 x L, we expect the probability distribution func- 2 2 The potential energy U is zero in the interval are the same conditionstion ° x, t that= wec usedx and to the find wave the function normal c modesx1 2 to be of zero a vibrating outside that U 5 0 ƒ ƒ ƒ Mathematically,ƒ ψ(0) = 0 and ψ(L) = 0 0 , x , L and is infinite everywhere outsidex string in Sectionregion. 15.8 (Fig. This agrees40.10); with you the should Schrửdinger review equation: that discussion. If the term U x c x in this interval. An additionalEq. condition (40.23)1 is2 is to that be finite, 1to2 calculate then c x the must second be zero derivative where1 2 U xd2 isc infinite.x dx 2 in 0` ` L 1 2 1 2 Eq. (40.23), the firsFurthermore,t derivative c dxc mustx dx be musta con tinuousalso befunction continuous to be a except mathematically at points well- behaved solution to the Schrửdinger1 2 equation. This implies1 2 that c x must be where the potential energy becomes1 2 infinite (as it does at the walls of1 the2> box). zero at the region’s boundary,1 2> x = 0 and x = L. These two conditions serve as This is analogousboundary to the condirequirementtions for thethat problem. a vibrating They shouldstring, looklike familiar,those shown because1 2 in they U(x) U(x) Fig. 40.10, can’tare have the any same kinks conditions in it that(which we usedwould to findcorrespond the normal to modesa discontinuity of a vibrating U 5 0 in the first derivativestring ofin Sectionthe wave 15.8 function) (Fig. 40.10); except you should at the review ends ofthat the discussion. string. x An additional condition is that to calculate the second derivative d2c x dx 2 in 0 L We now solve for the wave functions in the region 0 x L subject to the Eq. (40.23), the first derivative dc x dx must also be continuous except at points above conditions. In this region U x = 0, so the wave function in this region1 2> where the potential energy becomes infinite (as it does at the walls of the box).40.10 Normal modes of vibration for a must satisfy 1 2> This is analogous to the requirement that a vibrating string, like those shown in string with length L, held at both ends. 2 2 1 2 Fig. 40.10,U d can’tc x have any kinks in it (which would correspond to a discontinuity in -the first derivative= ofE thec wavex function)(particle except in aat box) the ends of the string.(40.25) Each end is a node, and there are n2 1 2m 2 We now dxsolve1 2 for the wave functions in the region 0 x L subject to the additional nodes between the ends. Equation (40.25)above is the conditions. same Schrửdinger In this region1 2 equation U x = 0,as so for the a wave free functionparticle, in sothis it regionis must satisfy 40.10 Normal modes of vibration for a tempting to conclude that the wave functions and energies are given by Eq. (40.24). string with length L, held at both ends. 2 2 1 2 n5 3 ikx U d c x It is true that c x = Ae satisfies- the Schrửdinger= Ec x equation(particle inwith a box) U x = 0,(40.25) is Each end is a node, and there are n2 1 2m 2 ikx continuous, and has a continuous firstdx 1derivative2 dc x dx = ikAe . However, additional nodes between the ends. this wave function1Equation2 does no(40.25)t satisfy is the the same boundarySchrửdinger 1conditions2 equation that as for c ax free must1 particle,2 be zero so it is tempting to conclude that the wave functions 1and2> energies are given by Eq. (40.24). n5 2 at x = 0 and x = L: At x = 0 the ikx wave function in Eq. (40.24) is equal to n5 3 0 It is true that c x = Ae ikL satisfies the Schrửdinger equation with U x = 0, is Ae A, and at x L it is equal to Ae . (These would be equal 1to 2zero ifikx A 0, = continuous,= and has a continuous first derivative dc x dx = ikAe . However,= but then the wavethis function wave function would1 2 does be nozerot satisfy and thethere boundary would conditions be no particle that c xat mustall!)1 2 be zero n5n25 1 The way out atof xthis= 0dilemma and x = isL: to At recall x = 0Example the wave 40.2 function (Section 1 in2> Eq. 40.1), (40.24) in iswhich equal to 0 ikL we found that aAe more= A , generaland at x = stationary-stateL it is equal to Ae solution. (These towould the be time-independent equal 1to 2zero if A = 0, L Schrửdinger equationbut then with the wave U x function0 iswould be zero and there would be no particle at all!) The way out of this= dilemma is to recall Example 40.2 (Section 40.1), in which n5 1 The length is an integral number we found that c a morex generalA eikx stationary-stateA e-ikx solution to the time-independent(40.26) 1 2 = 1 + 2 of half-wavelengths:L L 5 nln/2. Schrửdinger equation with U x = 0 is The length is an integral number 1 2 c x A eikx A e-ikx (40.26) 1 2 = 1 + 2 of half-wavelengths: L 5 nln/2. 1 2
31. Schrodinger Equation for Particle in a Box ~2 @2 (x) + U(x) (x)=E (x) 2m @x2 +∞ x ≤ 0 For a particle in a box with U(x) = 0 0 < x < L infinitely hard wall, -∞ x ≥ L ~2 @2 (x) = E (x) 2m @x2 We can re-write it as @2 (x) 2mE = (x) @x2 ~2 @2 (x) 2mE = k2 (x) k2 = @x2 ~2
32. @2 (x) = k2 (x) @x2 The most general solution to this differential equation is: ψ(x) = Asin(kx) + Bcos(kx) A and B are constants which are determined from the properties of ψ(x) as well as boundary and normalization conditions 1. Sin(x) and Cos(x) are finite and single-valued functions 2. Boundary Condition: ψ(0) = ψ(L) = 0 • ψ(x) = Asin(k0) + Bcos(k0) = 0 => B = 0 => ψ(x) = Asin(kx) • ψ(L) = Asin(kL) = 0 => sin(kL) = 0 => kL = nπ, n = ±1, ±2,
33. ⇡ k = n (n = 1, 2, 3, ) n L Particle in a Box: Energy Levels: 2 ⇡ 2 ~2k2 ~ n ⇡2~2 h2 E = n = L = n2 = n2 n 2m 2m 2mL2 8mL2 ✓ ◆ ✓ ◆ The allowed wave functions are given by n⇡ (x)=Asin x n L ⇣ ⌘ The normalized wave function: 2 n⇡ (x)= sin x n L L r ⇣ ⌘
34. Wave function: Energy Levels: 2 2 n⇡ ⇡ 2 2 (x)= sin x En = 2 n = n Eo n L L 2mL r ⇣ ⌘ ⇡2~2 Ground state (n =1) energy, E1 = Eo E = o 2mL2
35. Example 1: Electron in a 10 nm wide well with infinite barriers. Calculate Eo for L = 10 nm. Example 2: Assume that a photon is absorbed and the electron is transferred from the ground state (n =1) to the second excited state (n = 3). What was the wavelength of the photon?
36. Example 1: Electron in a 10 nm wide well with infinite barriers. Calculate Eo for L = 10 nm. 2 2 2 ⇡ ~ E = n E E = E = n o 1 o 2mL2 2 34 2 3.14 (1.05 10 ) E = ⇥ o 2 9.1 10 31 (10 10 9)2 ⇥ ⇥ ⇥ ⇥ 22 E 6 10 J=0.00375 eV = 3.75 meV o ⇡ ⇥ Example 2: Assume that a photon is absorbed and the electron is transferred from the ground state (n =1) to the second excited state (n = 3). What was the wavelength of the photon? Eground = E1 = Eo =0.00375 eV E = E 32 =9 0.00375 eV 0.0338 eV 3 o ⇥ ⇥ ⇡ (h⌫)=E E =0.0338 0.00375 0.030 eV 3 1 ⇡ 1240 = 413333 nm 41àm 0.03 ⇡ ⇡
37. Particle in a box with “Finite Walls” ~2 @2 + U = E U now has a finite value 2m @x2
38. Tunneling Effects
39. Tunneling Effects