1. Young/Freeman University Physics 11e

2. Ch 40 Quantum Mechanics © 2005 Pearson Education

3. Wave functions with a precise energy For a given wave function, we cannot predict where a particle will be found. We are faced with probabilities. For a given wave function, we cannot predict where a particle will be found. We are faced with probabilities. For the energy it is different. Special wave functions can be calculated such that the energy is well defined. These wave functions are found by solving the time independent Schr ö dinger equation. For the energy it is different. Special wave functions can be calculated such that the energy is well defined. These wave functions are found by solving the time independent Schr ö dinger equation. Several examples will be described. Several examples will be described.

4. Big surprise: ‘ energy quantization ’ Classical localised systems have quantised energy values. This fact explains emission and absorption events by atoms and molecules.

5. A classical localized particle If the potential energy has the shape shown in the figure a particle with energy E will be confined to a region [x 1,x 2 ]. We say that the particle is localized.

6. Classical physics  Quantum physics When a particle with such a potential energy is treated by quantum mechanics we find that all energy values are not allowed, energy is quantized. When a particle with such a potential energy is treated by quantum mechanics we find that all energy values are not allowed, energy is quantized. We will consider three examples of this statement 1) a particle in a box, 2) a finite box 3) a harmonic oscillator We will consider three examples of this statement 1) a particle in a box, 2) a finite box 3) a harmonic oscillator

7. Energy E calculation in QM. Energy values and related wave functions of a quantum system are calculated by solving the time independent Schrodinger equation. Energy values and related wave functions of a quantum system are calculated by solving the time independent Schrodinger equation. The problem is to find a function  (x) and E which satisfy this equation and where  (x) is normalizable. The problem is to find a function  (x) and E which satisfy this equation and where  (x) is normalizable.

8. Moreover … Moreover the function  (x) and its derivative  ‘ (x) should be continuous everywhere. Moreover the function  (x) and its derivative  ‘ (x) should be continuous everywhere. This is required because a derivative  ‘ (x) does not exist when the function  (x) is not continuous. Also a second derivative  ” (x) exists only when the derivative  ‘ (x) is continuous. This is required because a derivative  ‘ (x) does not exist when the function  (x) is not continuous. Also a second derivative  ” (x) exists only when the derivative  ‘ (x) is continuous. ( the second derivative  ” (x) occurs in the SE.) ( the second derivative  ” (x) occurs in the SE.)

9. 40.1 Particle in a box We have a very simple situation for a localized particle when we take U(x)=0 inside an interval [0,L] and U(x) is infinitely high outside that interval. The latter implies that a particle cannot come there or  (x) =0 for x L.

10. General solution of the SE inside the box We check by substitution that the general solution is of the form We check by substitution that the general solution is of the form   (x) = Asin(kx) +  cos(kx)  with A and B are (complex) constants and k is related to the energy by 2mE =( ћ k) 2.

11. Energy values Because of continuity:  (x) =0 at x=0 and Because of continuity:  (x) =0 at x=0 and  (x) =0 at x=L. It follows from the first requirement that B=0 and from the second that kL = n  with n an integer n=1,2,3... It follows from the first requirement that B=0 and from the second that kL = n  with n an integer n=1,2,3... Therefore we find the quantized energy values Therefore we find the quantized energy values E n =n 2 E 1 with E 1 = h 2 /(8mL 2 ) n=1,2,3, … E n =n 2 E 1 with E 1 = h 2 /(8mL 2 ) n=1,2,3, …

12. Final results For a particle in a box there are wave functions  n (x) with a well defined energy E n © 2005 Pearson Education The solutions are normalized when C = sqrt(2/L)

13. © 2005 Pearson Education energy levels, particle in a box

14. Probability and normalization Normalization condition Particle in a box

15. The measurement in QM QM says: every measurement influences the particle and thus the wave function. QM says: every measurement influences the particle and thus the wave function. Rule: Rule: ‘‘ Immediately after the measurement the wave function is restricted to what has been measured. A new wave function normalization is required ”.

16. A position measurement. Say we measure the particle described by  (x) to be in the interval [a,b]. Immediately after the measurement the new wave function is the previous wave function limited to the interval [a,b].

17. An energy measurement Suppose before the energy measurement the wave function is Suppose before the energy measurement the wave function is  (x) = 1/sqrt(5)( 1 (x) +2i 2 (x))  (x) = 1/sqrt(5)(  1 (x) +2i  2 (x)) If we measure the energy we can only find a quantized value. In this case the wave function is a sum of 1 (x) and 2 (x). Following QM we may find E 1 with probability 1/5 and E 2 with probability 4/5. If we measure the energy we can only find a quantized value. In this case the wave function is a sum of  1 (x) and  2 (x). Following QM we may find E 1 with probability 1/5 and E 2 with probability 4/5.

18. What do you think? Say we measure the energy and find E 1. What is the normalized wave function after the measurement? (hint: consider what happened after a position measurement) Say we measure the energy and find E 1. What is the normalized wave function after the measurement? (hint: consider what happened after a position measurement)  (x) = ?

19. Understanding emission and absorption processes. Absorption. When a particle is in the state with n=1 it can absorp a photon when this has a correct energy. Similarly a particle in an excited state (n>1) can fall back to a lower state by emitting a photon The same photon frequencies are observed.

20. 40.2 Potential Wells Another simple example is a box with a finite high potential wall U o. Now we must distinguish between energies lower than U o and energies higher than U o. The result: energies below U o are quantized, those higher than U o are not.

21. Solutions for the square well potential For E<U o the wave function extends outside the well but far away goes to zero and the energy is quantized. The number of such quantized energies is limited and depends on U o. These functions look very much like the solutions for a infinite box potential

22. Probabilities

23. 40.4 An atomic spring Another simple system is a particle with a potential energy like a spring. Then U(x) = ½ kx 2 with k the spring constant. The solution of the time independent Schrodinger equation in this case requires higher mathematics

24. Results The shape of the wave functions is similar to a particle in a square well potential.  The energy levels are equally spaced and of the form with n=0,1,2,3…  2 = k/m m =particle mass

25. Non localised systems When the particle is not localised in the classical sense the energies are not quantized. When the particle is not localised in the classical sense the energies are not quantized. We discuss two examples. We discuss two examples. 1) a free particle 2) a particle with a potential energy in the shape of a potential barrier. This represents a situation where a particle is scattered by an obstacle. 1) a free particle 2) a particle with a potential energy in the shape of a potential barrier. This represents a situation where a particle is scattered by an obstacle.

26. A free particle: U(x) =0 everywhere The general solution of the SE is of the form The general solution of the SE is of the form  (x)= A cos(kx) + Bsin(kx) where E = ( ћ k) 2 /2m E = ( ћ k) 2 /2m There is no restriction on k and so every energy E is possible: no quantization. There is no restriction on k and so every energy E is possible: no quantization. Note that  (x) is periodic:  (x+ )=  (x) for =2  /k. Note that  (x) is periodic:  (x+ )=  (x) for =2  /k. As the energy is also written as E=p 2 /2m we find p= ћ k = h/ which is the relation of the Broglie we mentioned earlier. This was to be expected. As the energy is also written as E=p 2 /2m we find p= ћ k = h/ which is the relation of the Broglie we mentioned earlier. This was to be expected.

27. Quantum scattering Classically a particle can approach an obstacle and for E<Uo will bounce back. In QM this is not so. To see what happens we look for a solution of the SE with an energy E<Uo.

28. The tunnel effect for E<Uo For x L we have the situation of a free particle with a solution cos(kx) or sin(kx). Unlike the classical situation we see that there is a chance that the particle is found at x>L. This remarkable behavior is called the tunnel effect. It has several applications.

29. Application: a scanning tunneling microscope (Binnig and Rohrer) A fine tip moves slightly above a surface. A potential difference is set up between surface and tip: tip negative and sample positive. Because of the tunnel effect electrons tunnel through the gap between tip and surface

30. Scanning tunneling microscope When the tip comes closer to the surface the current rises Keeping the current constant requires adaptation of the tip height. So surface geometry is recorded. The tip scans the surface and represents the elevations in an image.

31. A view of a crystal surface Crystal surface is positively charged with respect to tip. Because of the tunnel effect electrons can jump over the barrier from tip to the crystal Individual atoms can be observed on the surface

32. Tunneling application: Emission of an  particle by a nucleus. In a nucleus of an atom two protons and 2 neutrons form a very stable group called an  particle. The potential of the  particle is shown in the figure. We observe that sometimes an  particles leaks out of the nucleus by tunneling. We speak about  decay.

33. 40.5 Three-Dimensional Problems three-dimensional Schrödinger equation © 2005 Pearson Education It is very difficult to find solutions  (x) and energy values E which satisfy this equation. If more particles are involved it is even much more difficult. Normally one resorts to computer calculations.

34. END © 2005 Pearson Education THE END

35. Differential equations The Schrodinger equation is a differential equation of second order. We mentioned already that a differential equation has a general solution. The Schrodinger equation is a differential equation of second order. We mentioned already that a differential equation has a general solution. For a better understanding some basic information is given in the next slides. For a better understanding some basic information is given in the next slides.

36. A first order differential equation Consider an unknown function of a variable x, we note y(x). Consider an unknown function of a variable x, we note y(x). Say we are given that y(x) satisfies y ’ (x) =2. This is called a first order differential equation as only a first derivative enters. Say we are given that y(x) satisfies y ’ (x) =2. This is called a first order differential equation as only a first derivative enters. Then how is the function y(x)? Then how is the function y(x)? We see that there are many functions satisfying this requirement: e.g. y(x) =2x, y(x) =2x+5, y(x) = 2x-1 etc. We see that there are many functions satisfying this requirement: e.g. y(x) =2x, y(x) =2x+5, y(x) = 2x-1 etc. In general a solution has the form y(x) = 2x + c with c some constant. In general a solution has the form y(x) = 2x + c with c some constant.

37. A first order differential equation The solution of a differential equation is thus not uniquely defined. The solution of a differential equation is thus not uniquely defined. For a unuique solution a further condirion is required like demanding that y(x=0)=7. There there is only one solution y(x) = 2x+7. For a unuique solution a further condirion is required like demanding that y(x=0)=7. There there is only one solution y(x) = 2x+7.

38. A second order differential equation Say we are given that y(x) satisfies y ” (x) =2. This is called a second order differential equation as a second derivative enters the equation. Say we are given that y(x) satisfies y ” (x) =2. This is called a second order differential equation as a second derivative enters the equation. What about this function y(x)? What about this function y(x)? Also here there are many functions satisfying this requirement: e.g. y(x) =x 2, y(x) = x 2 +x+8,, y(x) = x 2 -5 etc. Also here there are many functions satisfying this requirement: e.g. y(x) =x 2, y(x) = x 2 +x+8,, y(x) = x 2 -5 etc. In general a solution has the form y(x) = x 2 +ax+b with a and b some constant. In general a solution has the form y(x) = x 2 +ax+b with a and b some constant.

39. A unique solution? To have a unique solution again further conditions have to be imposed. Now two constants should be fixed, so we need two conditions. To have a unique solution again further conditions have to be imposed. Now two constants should be fixed, so we need two conditions. We might ask a solution satisfying {y(x=0)=3 and y ’ (x=0)=1} or {y(x=0)=3 and y(x=1)=2}. We might ask a solution satisfying {y(x=0)=3 and y ’ (x=0)=1} or {y(x=0)=3 and y(x=1)=2}.

40. The Schrodinger equation The SE is a second order differential equation. The SE is a second order differential equation. Two conditions for uniqueness are that the solution goes to zero at plus and minus infinity. Two conditions for uniqueness are that the solution goes to zero at plus and minus infinity. 1)  (x) → 0 as x → +∞ 1)  (x) → 0 as x → +∞ 2)  (x) → 0 as x → - ∞ 2)  (x) → 0 as x → - ∞

41. Questions (you may discuss these questions with your friends). 1. Consider example 40.1 in this ppt. Suppose a photon is absorped so that the system jumps from the state n=1 to n=2. Is this photon transition in the visible part of the spectrum? 1. Consider example 40.1 in this ppt. Suppose a photon is absorped so that the system jumps from the state n=1 to n=2. Is this photon transition in the visible part of the spectrum? 2. A two-atomic molecule may vibrate. In a simple model we may describe it as a vibrating atomic spring. For HCl the observed force constant is 482 N m-1. Considering the transition from n=1 to n=2, is the frequency of this transition in the visible part of the spectrum? What is the transition energy in eV. Is this energy larger or smaller than typical atomic energies. (hint: an atom is roughly like a ball of diameter L= 10 -10 m) 2. A two-atomic molecule may vibrate. In a simple model we may describe it as a vibrating atomic spring. For HCl the observed force constant is 482 N m-1. Considering the transition from n=1 to n=2, is the frequency of this transition in the visible part of the spectrum? What is the transition energy in eV. Is this energy larger or smaller than typical atomic energies. (hint: an atom is roughly like a ball of diameter L= 10 -10 m)