Operators

1. Operators As we have seen, for a particle in a state of definite energy the probability distribution is independent of time. The expectation value of x is then given by In general, the expectation value of any function f(x) is given by

2. Operators If we know the momentum p of the particle as function of x, we can calculate the expectation value ‹p›. However, it is impossible in principle to find p as function of x since, according to uncertainty principle, both p and x can not be determined at the same time. To find ‹p› we need to know the distribution function for momentum. If we know Ψ(x), the distribution function can be found by Fourier analysis. It can be shown that ‹p› can be found from

3. Operators Similarly, ‹p2› can be found from Notice that in computing the expectation value the operator representing the physical quantity operates on Ψ(x), not on Ψ*(x). This is not important to the outcome when the operator is simply some function of x, but it is critical when the operator includes a differentiation, as in the case of momentum operator.

4. Expectation Values for p and p2 Find ‹p› and ‹p2› for the ground state wave function of the infinity square well.

5. In classical mechanics, the total energy written in terms of position and momentum variables is called the Hamiltonian function If we replace the momentum by the momentum operator pop and note that U = U(x), we obtain the Hamiltonian operator Hop: The time-independent Schrödinger equation can then be written:

6. The advantage of writing the Schrödinger equation in this formal way is that it allows for easy generalization to more complicated problems such as those with several particles moving in three dimensions. We simply write the total energy of the system in terms of position and momentum and replace the momentum variables by the appropriate operators to obtain the Hamiltonian operator for the system.

8. Minimum Energy of a Particle in a Box An important consequence of the uncertainty principle is that a particle confined to a finite space can not have zero kinetic energy. Let’s consider a one-dimensional box of lengthL. If we know that the particle is in the box,Δxis not larger thanL. This implies thatΔpis at leastħ/L. Let us take the standard deviation as a measureofΔp:

9. Minimum Energy of a Particle in a Box If the box is symmetric,pwill be zero since the particle moves to the left as often as to the right. Then and the average kinetic energy is:

10. Minimum Energy of a Particle in a Box The average kinetic energy of a particle in a box is: Thus, we see that the uncertainty principle indicate that the minimum energy of a particle in a box cannot be zero. This minimum energy is called zero-point energy.

11. The Hydrogen Atom The energy of an electron of momentum p a distance rfrom a proton is If we take for the order of magnitude of the position uncertaintyΔx = r, we have: (Δp2) = p2 ≥ ћ2/r2 The energy is then

12. The Hydrogen Atom There is a radiusrmat whichEis minimum. SettingdE/dr = 0yields rmandEm: rmcame out to be exactly the radius of the first Bohr orbit The ground state energy

13. The Hydrogen Atom The potential energy of the electron-proton system varies inversely with separation distance As in the case of gravitational potential energy, the potential energy of the electron-proton system is chosen to be zero if the electron is an infinity distance from the proton. Then for all finite distances, the potential energy is negative.

14. The Hydrogen Atom Like the energies of a particle in a box and of a harmonic oscillator, the energy levels in the hydrogen atom are described by a quantum number n. The allowed energies of the hydrogen atom are given by En = -13.6 eV/n2, n = 1,2,3,…… Energy-level diagram for the hydrogen atom. The energy of the ground state is-13.6 eV. Asnapproaches∞the energy approaches0.

15. Step Potential Consider a particle of energyEmoving in region in which the potential energy is the step function U(x) = 0, x<0 U(x) = V0, x>0 What happened when a particle moving from left to right encounters the step? The classical answer is simple: to the left of the step, the particle moves with a speed v = √2E/m

16. Step Potential Atx =0, an impulsive force act on the particle. If the initial energyEis less thanV0, the particle will be turned around and will then move to the left at its original speed; that is, the particle will be reflected by the step. IfEis greater than V0, the particle will continue to move to the right but with reduced speed given by v = √2(E – U0)/m

17. Step Potential We can picture this classical problem as a ball rolling along a level surface and coming to a steep hill of heighthgiven bymgh=V0. If the initial kinetic energy of the ball is less thanmgh, the ball will roll part way up the hill and then back down and to the left along the lower surface at it original speed. IfEis greater thanmgh, the ball will roll up the hill and proceed to the right at a lesser speed.

18. The quantum mechanical result: whenEis less thanV0,E<V0, the wave function does not go to zero at x=0 but rather decays exponentially. The wave penetrates slightly into the classically forbidden regionx>0, but it is eventually completely reflected.

19. Step Potential This problem is somewhat similar to that of total internal reflection in optics. ForE>V0, the quantum mechanical result differs from the classical result. Atx=0, the wavelength changes from λ1=h/p1 = h/√2mE to λ2=h/p2 = h/√2m(E-V0). When the wavelength changes suddenly, part of the wave is reflected and part of the wave is transmitted.

20. Reflection Coefficient Since a motion of an electron (or other particle) is governed by a wave equation, the electron sometimes will be transmitted and sometimes will be reflected. The probabilities of reflection and transmission can be calculated by solving the Schrödinger equation in each region of space and comparing the amplitudes of transmitted waves and reflected waves with that of the incident wave.

21. Reflection Coefficient This calculation and its result are similar to finding the fraction of light reflected from the air-glass interface. IfRis the probability of reflection, called the reflection coefficient, this calculation gives: wherek1is the wave number for the incident wave andk2is the wave number for the transmitted wave.

22. Transmission Coefficient The result is the same as the result in optics for the reflection of light at normal incidence from the boundary between two media having different indexes of refractionn. The probability of transmissionT, called the transmission coefficient, can be calculated from the reflection coefficient, since the probability of transmission plus the probability of reflection must equal 1: T + R = 1 In the quantum mechanics, a localized particle is represented by the wave packet, which has amaximum at the most probable position of the particle.

23. Time development of a one dimensional wave packet representing a particle incident on a step potential forE>V0. The position of a classical particle is indicated by the dot. Note that part of the packet is transmitted and part is reflected.

24. Reflection coefficientRand transmission coefficientTfor a potential stepV0high versus energyE(in unitsV0).

25. A particle of energyE0traveling in a region in which the potential energy is zero is incident on a potential barrier of heightV0=0.2E0. Find the probability thatthe particle will be reflected.

26. Lets consider a rectangular potential barrier of heightV0and withagiven by:U(x) = 0, x<0U(x) = V0, 0<x<aU(x) = 0, x>a

27. Barrier Potential We consider a particle of energyE, which is slightly less thanV0, that is incident on the barrier from the left. Classically, the particle would always be reflected. However, a wave incident from the left does not decrease immediately to zero at the barrier, but it will instead decay exponentially in the classically forbidden region0<x<a. Upon reaching the far wallof the barrier(x=a),the wave function must join smoothlyto a sinusoidal wave function to the right of barrier.

28. The potentials and the Schrödinger equations for the three regions are as follows: Region I (x<0) V = 0, Region II(0<x<a) V = V0, Region III (x>a) V = 0,

29. Barrier Potential If we have a beam of particle incident from left, all with the same energyE<V0, the general solution of the wave equation are, following the example for a potential step, wherek1 =√2mE/ħ andα = √2m(V0-E)/ħ This implies that there is some probability of the particle (which is represented by the wave function) being found on the far side of the barrier even though, classically, it should never pass through the barrier.

30. We assume that we have incident particles coming from the left moving along the +x direction. In this case the termAeik1xin region I represents the incident particles. The term Be-ik1xrepresents the reflected particles moving in the –x direction. In region III there are no particles initially moving along the -x direction. Thus G=0, and the only term in region III is Feik1x. We summarize these wave functions:

31. Barrier Potential For the case in which the quantity αa = √2ma2(V0 – E)/ħ2 is much greater than 1, the transmission coefficient is proportionaltoe-2αa, with α = √2m(V0 – E)/ħ2 The probability of penetration of the barrier thus decreases exponentially with the barrier thicknessaand with the square root of the relative barrier height(V0-E). This phenomenon is called barrier penetration or tunneling. The relative probability of its occurrence in any given situation is given by the transmission coefficient.

32. A wave packet representing a particle incident on two barriers of height just slightly greater than the energy of the particle. At each encounter, part of the packet is transmitted and part reflected, resulting in part of the packet being trapped between the barriers from same time.

33. A 30-eV electron is incident on a square barrier of height 40 eV. What is the probability that the electron will tunnel through the barrier if its width is (a) 1.0 nm? (b) 0.1nm?

34. The penetration of the barrier is not unique to quantum mechanics. When light is totally reflected from the glass-air interface, the light wave can penetrate the air barrier if a second peace of glass is brought within a few wavelengths of the first, even when the angle of incidence in the first prism is greater than the critical angle. This effect can be demonstrated with a laser beam and two 45° prisms.

35. α- Decay The theory of barrier penetration was used by George Gamov in 1928 to explain the enormous variation of the half-lives forαdecay of radioactive nuclei. Potential well shown on the diagram for anαparticle in a radioactive nucleus approximately describes a strong attractive force whenris less than the nuclear radiusR.Outside the nucleus the strong nuclear force is negligible, and the potential is given by the Coulomb’s law,U(r) = +k(2e)(Ze)/r, whereZeis the nuclear charge and2eis the charge of αparticle.

36. α- Decay An α-particle inside the nucleus oscillates back and forth, being reflected at the barrier at R. Because of its wave properties, when the α-particle hits the barrier there is a small chance that it will penetrate and appear outside the well at r = r0. The wave function is similar to that for a square barrier potential.

37. The probability that an α-particle will tunnel through the barrier is given by which is a very small number, i.e., the αparticle is usually reflected. The number of times per second N that the αparticle approaches the barrier is given by where v equals the particle’s speed inside the nucleus. The decay rate, or the probability per second that the nucleus will emit an α particle, which is also the reciprocal of the mean life time , is given by

38. The decay rate for emission ofαparticles from radioactive nuclei ofPo212. The solid curve is the prediction of equation The points are the experimental results.

39. Applications of Tunneling • Nanotechnology refers to the design and application of devices having dimensions ranging from 1 to 100 nm • Nanotechnology uses the idea of trapping particles in potential wells • One area of nanotechnology of interest to researchers is the quantum dot • A quantum dot is a small region that is grown in a silicon crystal that acts as a potential well • Nuclear fusion • Protons can tunnel through the barrier caused by their mutual electrostatic repulsion

40. Resonant Tunneling Device • Electrons travel in the gallium arsenide semiconductor • They strike the barrier of the quantum dot from the left • The electrons can tunnel through the barrier and produce a current in the device

41. Scanning Tunneling Microscope • An electrically conducting probe with a very sharp edge is brought near the surface to be studied • The empty space between the tip and the surface represents the “barrier” • The tip and the surface are two walls of the “potential well”

42. Scanning Tunneling Microscope • The STM allows highly detailed images of surfaces with resolutions comparable to the size of a single atom • At right is the surface of graphite “viewed” with the STM

43. Scanning Tunneling Microscope • The STM is very sensitive to the distance from the tip to the surface • This is the thickness of the barrier • STM has one very serious limitation • Its operation is dependent on the electrical conductivity of the sample and the tip • Most materials are not electrically conductive at their surfaces • The atomic force microscope (AFM) overcomes this limitation by tracking the sample surface maintaining a constant interatomic force between the atoms on the scanner tip and the sample’s surface atoms.

44. SUMMARY 1. Time-independent Schrödinger equation: 2.In the simple harmonic oscillator: the ground wave function is given: where A0is the normalization constant anda=mω0/2ħ. 3. In a finite square well of heightV0, there are only a finite number of allowed energies.

45. SUMMARY 4.Reflection and barrier penetration: When the potentials changes abruptly over a small distance, a particle may be reflected even thoughE>U(x). A particle may penetrate a region in whichE<U(x). Reflection and penetration of electron waves are similar for those for other kinds of waves.

46. The Schrödinger Equation in Three Dimensions The one-dimensional time-independent Schrödinger equation (1) is easily extended to three dimensions. In rectangular coordinates, it is (2) where the wave functionψand the potential energyUare generally functions of all three coordinates ,x, y, andz.

47. The Schrödinger Equation in Three Dimensions To illustrate some of the features of problems in three dimensions, we consider a particle in three-dimensional infinity square well given by U(x,y,z)=0for0<x<L,0<y<L, and0<z<L. Outside this cubical region,U(x,y,z)=∞.For this problem, the wave function must be zero at the edges of the well.

48. The Schrödinger Equation in Three Dimensions The standard method for solving this partial differential equation is guess the form of the solution using the probability. For a one-dimensional box along thexaxis, we have found the probability that the particle is in the regiondxat xto be A12sin2(k1x)dx(3) whereA1is the normalization constant, andk1=nπ/Lis the wave number. Similarly, for a box alongyaxis, the probability of a particle being in a regiondyatyis A22sin2(k2y)dy(4) The probability of two independent events occurring is the product of probabilities of each event occurring.

49. The Schrödinger Equation in Three Dimensions So, the probability of a particle being in regiondxat xand in regiondyatyis The probability of a particle being in the regiondx, dy, anddz isψ2(x,y,x)dxdydz, whereψ(x,y,z)is the solution of equation (2)

50. The Schrödinger Equation in Three Dimensions (2) The solution is of the form where the constantAis determined by normalization. Inserting this solution in the equation (2), we obtain for the energy: which is equivalent to: with px=ħk1and so on.