XI. INTRODUCTION TO QUANTUM MECHANICS. C. Cohen-Tannoudji et al., Quantum Mechanics I, Wiley. Outline: Electromagnetic waves and photons

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1 XI. INTRODUCTION TO QUANTUM MECHANICS C. Cohen-Tannoudji et al., Quantum Mechanics I, Wiley. Outline: Electromagnetic waves and photons Material particles and matter waves Quantum description of a particle: wave packets Particle in a time-independent scalar potential

2 Electromagnetic waves and photons Light quanta and the Planck-Einstein relations Planck (1900): the hypothesis of the quantization of energy: For an electromagnetic wave of frequency ν, the only possible energies are integral multiples of the energy quantum hν (or ω), where h is a new fundamental constant. Einstein (1905): the particle theory of light: Light consists of a beam of photons, each possessing an energy hν. (Experimental verification: Compton (1924)).

3 The interaction of an electromagnetic wave with matter occurs by means of elementary indivisible processes, in which the radiation appears to be composed of particles, the photons. Planck-Einstein relations: E = hν = ω p = k = h 2π where h J.s is the Planck constant, ω = 2πν, and vector. k = 2π/λ is the wave

4 Wave - particle duality Young s slit experiment If both slits, F 1 and F 2 are open, the intensity of the light emitted from a monochromatic light source S on the screen I(x) shows interference fringes, i.e. I(x) I 1 (x) + I 2 (x)

5 Wave theory provides natural interpretation of this phenomenon: consider the electric fields E 1 (x) and E 2 (x) (in complex notation) produced at x by slits F 1 and F 2 respectively. The total field when both slits are open is E(x) = E 1 (x) + E 2 (x) Using the complex notation, the intensity I(x) is then I(x) E(x) 2 = E 1 (x) + E 2 (x) 2 = E 1 (x) 2 + E 2 (x) E 1 (x) E 2 (x) cos(θ 1 θ 2 ) E 1 (x) 2 + E 2 (x) 2 = I 1 (x) + I 2 (x) The wave theory predicts that diminishing the intensity of the source S will cause the interference fringes to diminish in intensity but not to vanish.

6 What happens when S emits photons one by one? Neither the predictions of the wave theory nor of the particle theory are verified!: (i) for a long exposure time (i.e. a large number of photons is captured on the screen E) the fringes have not disappeared, and thus the purely corpuscular theory must be rejected; (ii) for a short exposure time (just a few photons captured), each photon produces a localized impact on E and not a weak interference pattern the purely wave interpretation must also be rejected.

7 As more and more photons strike the photographic plate, the individual impacts, seemingly distributed in a random manner, begin to lead to the distribution of the impacts which is continuous and which exhibits interference fringes. Each photon has passed both slits simultaneously

8 Quantum unification of the two aspects of light We need to fundamentally reconsider the concepts of classical physics to unite two aspects, wave or particle, of light. For example, if we put a photodetector behind F 2 only a half of the photons pass through F 2 and the other half through F 1, giving just a single slit signal on E: when one performs a measurement on a microscopic system, one disturbs it in a fundamental fashion; it is impossible to observe the interference pattern and to know at the same time through which slit each photon has passed. Therefore, we are led to question the concept of a particle s trajectory which is fundamental one of classical physics.

9 Moreover, as the photons arrive one by one, their impacts on the screen build up the interference pattern for a particular photon, we are not certain in advance where it will strike the screen; as these photons are emitted under the same conditions, it implies that the classical idea that the initial conditions completely determine the subsequent motion of a particle is invalid in quantum mechanics.

10 The concept of wave-particle duality: (i) the particle and wave aspects of light are inseparable: Light behaves simultaneously like a wave and like a flux of particles, the wave enabling us to calculate the probability of the manifestation of a particle; (ii) predictions about the behaviour of a photon can only be probabilistic; (iii) the information about a photon at time t is given by the wave E( r, t) which is a solution of Maxwell s equation, and which characterizes the state of the photons at time t. E( r, t) is interpreted as the probability amplitude of a photon appearing, at time t, at the point r = (x, y, z). The corresponding probability is proportional to E( r, t) 2.

11 Comments: (i) If E 1 and E 2 are two solutions of Maxwell s equation (which are linear and homogeneous) then E = λ 1 E 1 + λ 2 E 2 where λ 1, λ 2 are constants, is also a solution. This implies superposition principle and its consequences, e.g. interference, diffractions etc. (ii) The theory merely allows to calculate the probability of the occurrence of a given event. Experimental verifications are to be founded on the repetition of a large number of identical experiments (a large number of identically prepared particles, photons).

12 (iii) Optical analogy: we can regard the quantum state of a material particle characterized by a wavefunction ψ( r, t) in a similar way as we regard the photon state E( r, t). The fact that ψ( r, t) is complex is essential in quantum mechanics.

13 Principle of spectral decomposition Consider a plane monochromatic light wave E( r, t) = E 0 e P e i(kz ωt) polarized in the direction e P. The corresponding intesity is I E 0 2. After the passage of the wave through the analyzer, the plane wave is polarized along 0x: E ( r, t) = E 0 e xe i(kz ωt) and the intensity is given by the Malus law as I = I cos 2 θ

14 What will happen on the quantum level, when I is weak and photons reach the analyzer one by one? (1) photon either crosses the analyzer or it is entirely absorbed by it; (2) we cannot in general predict with certainty whether a given photon will pass or will be absorbed; (3) for a large number N of equally prepared photons, the result will correspond to the classical law, i.e. N cos 2 θ photons will pass.

15 These observations have several important implications: (i) the measurement device can give only certain privileged results, called eigen results. There is quantization of the results of the measurement (in contrast to the classical case). (ii) to each of the eigen results corresponds an eigenstate. For example, e P = e x (in the case that the photon pass the analyzer with certainty), or e P = e y (if it is absorbed with certainty). If the particle is, before the measurement, in one of the eigenstates, the result of this measurement is certain: it can only be the associated eigen result.

16 (iii) when the state before the measurement is arbitrary, only the probabilities of obtaining the different eigen results can be predicted. To find these one decomposes the state of the particles into a linear combination of the various eigenstates this is the spectral decomposition: e P = e x cos θ + e y sin θ where cos θ and sin θ are the probability amplitudes. The corresponding probabilities, which are cos 2 θ and sin 2 θ, statisfy cos 2 θ + sin 2 θ = 1. (iv) The measurement disturbs the microscopic system in a fundamental fashion: after passing through the analyzer, the light is completely polarized along e x.

17 Material particles and matter waves The de Broglie relations Atomic emission and absorption spectra are composed of narrow lines, that is, a given atom emits and absorbs only photons with well defined frequencies (energies) the energy of the atom is quantized (Bohr-Sommerfeld (empirical) quantization of electron orbits) hν i j = E i E j where l.h.s. corresponds to the energy of the photon, and r.h.s. to that of the atom.

18 de Broglie hypothesis: material particles, just like photons, can have a wavelike aspect. This permits derivation of the Bohr-Sommerfeld quantization rules: the various permitted energy levels appear as analogues of the normal modes of a vibrating string Davisson and Germer (1927) confirmed the de Broglie hypothesis by showing that interference patterns could be obtained with material particles like electrons.

19 Consider a material particle of energy E and momentum p and associate with it a wave of angular frequency ω = 2πν and wave vector k. Then the relations E = hν = ω p = k imply the de Broglie relation for the wavelength associated with a material particle λ = 2π k = h p

20 Examples: (i) a dust particle: diameter d = 1 µm, m = kg, speed v = 10 3 ms 1 : λ = m = Å (ii) a thermal neutron: m n kg, the speed is calculated as: 1 2 m nv 2 = p2 2m n 3 2 kt where k = JK 1 is the Boltzmann constant. De Broglie wavelength is then λ = h p = h 3mn kt which for T = 300 K gives λ 1.4 Å, which is about the interatomic distance in a crystal.

21 (iii) an electron, m e = kg, accelerated through a potential difference V to the kinetic energy E = qv = J Using the expression for the kinetic energy we obtain This gives numerically E = p2 2m e λ = h p = h 2me E λ = V m 12.3 V Å

22 Wavefunctions. Schrödinger equation We now apply the ideas introduced for the case of the photon to all material particles: (i) The quantum state of a particle is characterized by a wavefunction ψ( r, t), which contains all the information that is possible to obtain about the particle; (ii) ψ( r, t) is interpreted as a probability amplitude of the particle s presence at the point r at time t; The probability dp( r, t) of the particle being at time t in a volume element d 3 r = dx dy dz situated at the point r is interpreted as the corresponding probability density where C is a normalization constant. dp( r, t) = C ψ( r, t) 2 d 3 r

23 (iii) The principle of spectral decomposition applies to the measurement of an arbitrary physical quantity: - the result found must belong to a set of eigen results {a}; - with each eigenvalue a is associated an eigenstate, i.e. an eigenfunction ψ a ( r) such that if ψ( r, t 0 ) = ψ a ( r), the measurement (at t 0 ) will always yield a; - for any ψ( r, t), the probability of finding the eigenvalue a for a measurement at time t 0 is found by decomposing ψ( r, t 0 ) in terms of the functions ψ a ( r): ψ( r, t 0 ) = c a ψ a ( r) a

24 Then P a = c a 2 a c a 2 where the denominator ensures that the total probability a P a = 1 (normalization). - if the measurement indeed yields a, the wavefunction immediately after the measurement is: ψ ( r, t 0 ) = ψ a ( r) This is known as wavefunction collapse.

25 (iv) The evolution of the wavefunction ψ( r, t) of a particle of mass m subjected to a potential V( r, t) is governed by the Schrödinger equation i 2 ψ( r, t) = ψ( r, t) + V( r, t)ψ( r, t) t 2m where = 2 / x / y / z 2 is the Laplace operator. The fact that the Schrödinger equation is linear and homogeneous in ψ implies that there exists a superposition principle which, combined with the interpretation of ψ as a probability amplitude, is the source of wavelike effects, that is, it leads to wave-particle duality.

26 Comments: (i) For a system composed of only one particle, the total probability of finding the particle anywhere in space, at time t, equals to 1: dp( r, t) = 1 Since dp( r, t) = C ψ( r, t) 2 d 3 r, this implies that the wavefunction must be squareintegrable: ψ( r, t) 2 d 3 r < The normalization constant C is then given by the relation 1 C = ψ( r, t) 2 d 3 r

27 We say that a wavefunction is normalized if C = 1: ψ( r, t) 2 d 3 r = 1 (ii) ψ( r, t) is interpreted as a probability amplitude of the particle s presence at the point given by the spatial coordinates r at time t: the wavefunction ψ( r, t) characterizes a quantum state in the coordinate representation. (iii) In general, it may be useful to represent a quantum state by other than coordinate representation. For example, ψ( p) is interpreted as a probability amplitude that the particle has the momentum p (at some fixed t); and the wavefunction ψ( p) characterizes a quantum state in the momentum representation.

28 (iv) The relation between the wavefunctions in the coordinate and momentum representations is given by the Fourier transform. In one-dimensional systems this is given as ψ(p) = The inverse formula is then Remark: ψ(x) = 1 + 2π 1 + 2π dxe ipx/ ψ(x) dpe ipx/ ψ(p) while ψ(x) is an eigenstate of the coordinate operator, ψ( p) is an eigenstate of the momentum operator.

29 Quantum description of a particle. Wave packets Free particle For V( r, t) = 0 (or constant), the Schrödinger equation becomes i 2 ψ( r, t) = ψ( r, t) t 2m This equation is satisfied by solutions of the plane wave form where A is a constant and ψ( r, t) = Ae i( k r ωt) ω = E = p2 /2m = k2 2m

30 We see that ψ( r, t) 2 = A 2 so the plane wave of this type represents a particle whose probability of presence is uniform throughout all space. The priciple of superposition implies that every linear combination of plane waves satisfying ω = k 2 /2m will also be a solution of the Schrödinger equation for a free particle. Such a superposition can be written as 1 ψ( r, t) = (2π) 3/2 g( k)e i( k r ωt) d 3 k where d 3 k = dk x dk y dk z represents the infinitesimal volume element in k-space; g( k) can be complex and must be sufficiently regular. Moreover any square-integrable function can be written in the form above.

31 For the sake of simplicity, we will study one-dimensional wave packets obtained from a superposition of plane waves ψ(x, t) = 1 2π + g(k)e i(kx ω(k)t) dk We will be interested in the form of the wavepacket at a given time (t = 0): ψ(x, 0) = 1 g(k)e ikx dk 2π Notice that g(k) is simply the Fourier transform g(k) = 1 ψ(x, 0)e ikx 2π (Remark: this is actually valid for any potential.)

32 Form of the wavepacket at a given time ψ(x, 0) = 1 2π g(k)e ikx dk Imagine that g(k) has the shape such that (i) it has a pronounced peak centered at k = k 0, and (ii) it has a width (e.g. at half of its maximum) of k.

33 We will start with a simple case of a superposition of only three plane waves (rather than infinitely many). The wave vectors of these plane waves are k 0, k 2 + k 0, k 2 + k 0 and their amplitudes are proportional respectively to 1, 1/2, 1/2. ψ(x, 0) = g(k [ 0) 2π e ik0x ei( k 0 k 2 ) x ei ( k0 + k ) ] x 2 = g(k [ ( )] 0) e ik 0x k 1 + cos 2π 2 x

34 ψ(x, 0) = g(k [ 0) 2π e ik0x ei( k 0 k 2 ) x ei ( k0 + k ) ] x 2 = g(k [ ( )] 0) e ik 0x k 1 + cos 2π 2 x

35 ψ(x) has maximum at x = 0. At this point all three waves are in phase and interfere (add) constructively. As we move away from x = 0, the waves become more and more out of phase and ψ(x) decreases. The interference is completely destructive when the phase shift between e ik 0x and e i( k 0 ± k ) 2 x is equal to ±π.

36 ψ(x) goes to zero when x = ± x 2 where x. k = 4π (Note that cos ( k 2 x ) = cos ( k 2 x 2 ) = cos π = 1.) so the smaller the width k of g(k), the larger the width x of ψ(x) (the distance between two zeroes of ψ(x). For a continuous superposition of an infinite number of plane waves, ψ(x) can have only one maximum.

37 Example: Consider a one-dimensional wavepacket given in the momentum representation by a Gaussian function: whose width is p = σ. ψ(p) = e p2 2σ 2 Calculate ψ(x) by performing the Fourier transform and show that the Gaussian wavepacket satisfies the minimal bound on the Heisenberg uncertainty principle x p.

38 ψ(x) = = = = 1 + 2π 1 + 2π 1 + 2π 1 + 2π e p2 2σ 2 e +ipx dp ( e ( e p 2 2σ 2 ipx x2 σ e 1 2σ 2 ( ) e x2 σ dp p 2σ ixσ 2 ) 2 e x2 σ dp p ixσ2 ) 2 e x2 σ dp

39 substituting p = p ixσ2, dp = dp, and integrating using the formula + e a2 p 2 dp = π/a, we obtain = 1 e x2 σ π e p 2 2σ 2 dp = σ 2π 2π e x2 σ = σ e x2 σ = σ e x2 2( x) 2 This is the Guassian wavepacket in the coordinate representation with the width that satisfies the lower bound Heisenberg uncertainty relation x = σ = p x p =

40 Considering the general wavepacket ψ(x, 0) = 1 2π g(k)e ikx dk we see that its form also results from an interference phenomenon: ψ(x, 0) has maximum when the different plane waves interfere constructively. Let α(k) be the argument of g(k): g(k) = g(k) e iα(k) Assume that α(k) [ varies smoothly within the interval k 0 k 2, k 0 + k ] 2, where g(k) is appreciable. For sufficiently small k we can expand α(k) as [ ] dα α(k) α(k 0 ) + (k k 0 ) dk k=k 0

41 This allows us to rewrite the wavefunction ψ(x, 0) ei[k 0x+α(k 0 )] 2π + g(k) e i(k k 0)(x x 0 ) dk with How ψ(x, 0) varies with x? x 0 = [ ] dα dk k=k 0 (a) when x x 0 is large, the function of k which is to be integrated oscillates a very larger number of times within the interval k the integral becomes negligible as the contributions from successive oscillations cancel out.

42 (b) if x x 0, the function to be integrated over k oscillates hardly at all and ψ(x, 0) is maximum. The position x M (0) of the center of the wavepacket is therefore [ ] dα x M (0) = x 0 = dk Stationary phase condition: k=k 0 the derivative with respect to k of the phase is zero at k = k 0, that is at k = k 0. d dα (kx + α(k)) = x + dk dk = 0

43 When x moves away from x 0, ψ(x, 0) decreases. This decrease becomes appreciable if e i(k k 0)(x x 0 ) oscillates approximately once when k traverses the domain k, i.e. k(x x 0 ) 1 If x is an approximate width of the wavepacket, we have the lower bound of the expression k x 1 This is a classical relation between the widths of two functions which are Fourier transforms of each other.

44 Heisenberg uncertainty relation Recall that a plane wave e i(k 0x ω 0 t) corresponds to a constant probability density for the presence of the particle along the x axis for all walues of t, that it: (i) x is infinite; (ii) only one angular frequency ω 0 and one wave vector k 0 are involved and thus according to the de Broglie relations, the energy E = ω 0 and momentum p = k of the particle are well defined. (In this case g(k) = δ(k k 0 ), a delta function.)

45 Alternatively, this property can be interpreted using the principle of spectral decomposition: a particle described by ψ(x, 0) = Ae ikx has well defined momentum = a measurement of the momentum at t = 0 will definitely yield p = k, i.e. e ikx characterizes the eigenstate corresponding to p = k. In the formula ψ(x, 0) = 1 2π g(k)e ikx dk, ψ(x, 0) appears as a linear superposition of the momentum eigenfunctions in which the coefficient of e ikx is g(k) g(k) 2 is proportional to the probability of finding p = k, if at t = 0 one measures the momentum of a particle whose state is described by ψ(x, t).

46 As the possible values of p form a continuous set (like those of x), g(k) 2 is proportional to probability density: d P(k) of obtaining a value between k and (k + dk) is (to within a constant factor) g(k) 2 dk. More precisely, if we write ψ(x, 0) = 1 2π ψ(p) e ipx/ dp ψ(p) and ψ(x, 0) satisfy the Bessel-Parseval relation + ψ(x, 0) 2 dx = + ψ(p) 2 dp

47 If the common value of these integrals is C then the probability of the particle being found at t = 0 between x and x + dx is dp(x) = 1 C ψ(x, 0) 2 dx and the probability that the measurement of momentum yields a results between p and p + dp is d P(p) = 1 C ψ(p) 2 dp We can write the inequality k x 1, using p = k where p is the width of ψ(p), as The Heisenberg uncertainty relation: x p It is impossible to define at any t both the position and the momentum of a particle to an arbitrary degree of accuracy.

48 Time evolution of a free wavepacket Consider a free particle whose state is described by the one-dimensional wavepacket ψ(x, t) = 1 2π + g(k)e i(kx ω(k)t) dk A given plane wave e i(kx ωt) propagates along the x axis with the phase velocity v φ (k) = ω k since it depends on x and t only via (x ω k t). In the case of an electromagnetic wave propagating in a vacuum, all the waves of a wavepacket propagate at the same velocity as v φ = c which is independent of k.

49 However in a dispersive medium v φ (k) = c n(k) where n(k) is the index of the medium which depends on the wavelength. The case considered here corresponds to a dispersive medium as v φ (k) = k 2m E = ω = 2 k 2 2m = p2 2m and different waves will have unequal phase velocities. Therefore the velocity of the maximum x M of the wavepacket is not an average phase velocity ω 0 k 0 = k 0 2m.

50 For the sake of simplicity, let us consider the superposition of three waves again ψ(x, t) = g(k { [( 0) e i(k 0x ω 0 t) 1 k0 k ) ( + 2 x ω0 ω ) ] 2 t + e i[( k 0 + k ) ( 2 x ω0 + ω ) ]} 2 t 2π = g(k 0) 2π e i(k 0x ω 0 t) 2 ei [ 1 + cos ( k 2 x ω )] 2 t The maximum of ψ(x, t) which was at x = 0 at t = 0 is now at x M (t) = ω ( k t ω ) 0 t k 0

51 (a) Since the maxima (2) coincide at x = 0, the constructive interference leads to the maximum of the wavepacket. (b) Since the phase velocity increases with k, the maximum (3) of the wave (k 0 + k/2) will gradually catch up with that of the wave (k 0 ), which will in turn catch up with the wave (k 0 k/2): x M (t) = ω k t

52 We can also determine the shift of the wavepacket using the stationary phase method: in order to go from ψ(x, 0) to ψ(x, t), we need to change g(k) g(k)e iω(k)t We replace the argument α(k) of g(k) by α(k) ω(k)t and x M (0) = x 0 = [dα/dk] k=k0 then leads to [ ] dω x M (t) = x 0 = t dk k=k 0 [ dα dk ] k=k 0

53 The velocity of the maximum of the wavepacket is [ ] dω v G (k 0 ) = dk k=k 0 and is called the group velocity of the wavepacket. With the dispersion relation ω = k2 2m, we get v G (k 0 ) = k 0 m = 2v φ(k 0 ) In the cases where x and p can be made negligible, the maximum of the wavepacket moves like a particle which obeys the laws of classical mechanics (with velocity v = p 0 /m).

54 Particle in time-independent scalar potential. Separation of variables. Stationary states. i 2 ψ( r, t) = ψ( r, t) + V( r, t)ψ( r, t) t 2m Existence of stationary states: We are looking for solutions of this equation of the form ψ( r, t) = φ( r)χ(t) i φ( r) d dt χ(t) = χ(t) 2 2m φ( r) + χ(t)v( r, t)φ( r) i dχ(t) 1 = 2 χ(t) dt φ( r) 2m φ( r) + V( r, t)

55 In the last equation the l.h.s. is a function of time and r.h.s. is a function of r. The equality is possible if both sides equal to a constant. Let us say and in the same way i dχ(t) χ(t) dt = ω χ(t) = Ae iωt 2 φ( r) + V( r, t)φ( r) = ωφ( r) 2m Setting A = 1 we get ψ( r, t) = φ( r)e iωt the solution of the Schrödinger equation above where time and space variables have been separated.

56 The wavefunction of this form ψ( r, t) = φ( r)e iωt is called a stationary solution of the Schrödinger equation: it leads to a time-independent probability density: ψ( r, t) 2 = φ( r) 2 In a stationary solution, only one ω appears, so a stationary state is a state with well defined energy E = ω (energy eigenstate).

57 Provided the potential energy is time independent (i) in classical mechanics, the total energy is a constant of motion; (ii) in quantum mechanics, there exist well-defined energy states: 2 + V( r, t) φ( r) = Eφ( r) 2m or Hφ( r) = Eφ( r) This is the eigenvalue equation for the Hamiltonian, the differential operator representing the total energy. E is the eigenvalue (total energy) and φ( r) is the corresponding eigenvector (the eigenvalue equation reflects energy quantization). The Hamiltonian is a linear operator, that is H[λ 1 φ 1 + λ 2 φ 2 ] = λ 1 Hφ 1 + λ 2 Hφ 2.

58 Superposition of stationary states We label various energy values by an index n: Hφ n ( r) = E n φ n ( r) where the stationary states are ψ n ( r, t) = φ n ( r)e ie nt/ The Schrödinger equation, as a linear equation, has a whole series of solutions of the form ψ( r, t) = c n φ n ( r)e ie nt/ where c n are complex numbers. In particular ψ( r, 0) = c n φ n ( r) n n

59 Any function ψ( r, 0) can always be decomposed in terms of eigenfunctions of H. The coefficients c n are thus always determined by ψ( r, 0). The states time-dependence arises from the terms e ie nt/ where En is the eigenvalue associated with the eigenvector (eigenstate) φ n ( r).

60 Stationary states of a particle in one-dimensional square potentials Behavior of a stationary wavefunction φ(x): d 2 2m dx2φ(x) + (E V) φ(x) = 0 2 where V(x) = V in some region of space.

61 Several cases: (i) E > V: E V = 2 k 2 2m The solution: φ(x) = Ae ikx + A e ikx where A, A C. (ii) E < V (classically forbiden): we introduce a positive constant ρ: V E = 2 ρ 2 2m The solution: where B, B C. φ(x) = Be ρx + B e ρx (iii) E = V: φ(x) is a linear function of x.

62 Behavior of φ(x) at a potential energy discontinuity φ(x) and dφ(x)/dx are continuous (only d 2 φ(x)/dx 2 is discontinuous at the potential energy discontinuity) matching conditions: φ(x) and dφ(x)/dx for the regions with different values of V must match at the values of x where the potential energy discontinuity is present.

63 Simple cases: Potential step Several cases: (i) E > V 0 : partial reflection Set 2mE 2m(E V0 ) k 1 = 2 k 2 = 2 Solutions: region I (x < 0) and region II (x > 0), respectively: φ I (x) = A 1 e ik 1x + A 1 e ik 1x φ II (x) = A 2 e ik 2x + A 2 e ik 2x We will choose to limit ourselves to the case of an incident particle coming from x =, i.e. we put A 2 = 0.

64 Matching conditions: [ ] dφi (x) dx x=0 φ I (0) = φ II (0) A 1 + A 1 = A 2 [ ] dφii (x) = A dx 1 k 1 A 1 k 1 = A 2 k 2 x=0 A 1 A 1 = k 1 k 2 k 1 + k 2 & A 2 A 1 = 2k 1 k 1 + k 2 The incident particle and reflected particle, p = k 1 : φ I (x) = A 1 e ik 1x + A 1 e ik 1x (Note that there is no reflected particle in classical mechanics for E > V.) The transmitted particle: φ II (x) = A 2 e ik 2x

65 The reflection coefficient The transmission coefficient where For E >> V 0, T 1. R = A 1 A 1 T = k 2 k 1 2 = 1 4k 1k 2 (k 1 + k 2 ) 2 A 2 2 A 1 R + T = 1 = 4k 1k 2 (k 1 + k 2 ) 2

66 (ii) E < V: total reflection 2m(V0 E) ρ 2 = 2 φ II (x) = B 2 e ρ2x + B 2 e ρ 2x For the solution to remain bounded for x, it is necessary that B 2 = 0 φ II (x) = B 2 e ρ2x where the r.h.s. of the equation corresponds to an evanescent wave. The matching conditions at x = 0 give A 1 = k 1 iρ 2 A 1 k 1 + iρ 2 B 2 2k = 1 A 1 k 1 + iρ 2

67 The reflection coefficient R = A 1 A 1 2 = k 1 iρ 2 k 1 + iρ 2 2 = 1 As in classical mechanics, the particle is always reflected. However, the particle has a non-zero probability of presence in the region of space where it is classically forbiden (e ρ 2x ). Also A 1 /A 1 is a complex number, which implies a phase shift upon reflection.

68 Potential bariers (i) E > V 0 : resonances φ I (x) = A 1 e ik 1x + A 1 e ik 1x φ II (x) = A 2 e ik 2x + A 2 e ik 2x φ III (x) = A 3 e ik 3x + A 3 e ik 3x

69 Considering the incident particle coming from x =, we take A 3 the matching conditions yields A 1 = cos(k 2l) i k2 1 + k2 2 sin(k 2k 1 k 2 l) eik1l A 3 2 = 0, and applying The reflection coefficient R = A 1 = ik2 1 k2 2 2k 1 k 2 sin(k 2 l)e ik 1l A 3 A 1 A 1 2 = (k 2 1 k2 2 )2 sin 2 (k 2 l) 4k 2 1 k2 2 + (k2 1 k2 2 )2 sin 2 (k 2 l)

70 The transmission coefficient T = A 3 2 4k 1 2 A = k k 1 2k2 2 + (k2 1 k2 2 )2 sin 2 (k 2 l) = 4E(E V 0 ) 4E(E V 0 ) + V 0 2 [ sin2 2m(E V0 )l/ ] [ V T oscillates periodically between its minimal value 1 + 4E(E V 0 )] and its maximal value T = 1. The resonances, obtained when T = 1, correspond to k 2 l = nπ, that is to the values of l which are integral multiples of the half-wavelength of the particle in the region II (standing waves in II).

71 (ii) E < V 0 : tunnel effect φ II (x) = A 2 e ρ 2x + A 2 e ρ 2x The matching conditions at x = 0 and x = l allow us to calculate the transmission coefficient (k 2 is replaced by iρ 2 ) T = A 3 2 4E(V A = 0 E) 1 4E(V 0 E) + V 0 2 [ sinh2 2m(V0 E)l/ ] (R = 1 T) When ρ 2 l >> 1 T 16E(V 0 E) V 0 2 e 2ρ 2l The particle has a nonzero probability of crossing the potential barrier (due to evanescent wave). This is called tunnel effect.

72 A particle in an infinite square well potential V(x) = 0 in the region 0 < x < a V(x) = everywhere else k = 2mE 2 φ(x) must be zero outside of the region 0 < x < a and continuous at x = 0 and x = a since φ(0) = 0, A = A and φ(x) = Ae ikx + A e ikx φ(x) = 2iA sin(kx)

73 Moreover, φ(a) = 0, so sin(ka) = 0 k = nπ E n = n2 π 2 2 a 2ma 2 The energy E n is quantized (with n = 1, 2, 3 etc.) and corresponds to the bound states. (Recall: normal modes of a string with fixed boundary conditions). The energy eigenfunctions have to be normalized, that is we demand a a φ(x) 2 dx = 4A 2 sin 2 (kx)dx = 2A 2 1 a = 1 A = 2a 0 0 where we used the formula sin 2 (x )dx = 4 1 sin(2x ) x. The solution is 2 φ(x) = a sin(kx) (The imaginary unit i = e iπ/2 represents physically irrelevant global phase.)