量子力学英文格里菲斯Chapter2PPT课件.ppt

1、1Outline2概况1您的内容打在这里，或者通过复制您的文本后。概况2您的内容打在这里，或者通过复制您的文本后。概况3您的内容打在这里，或者通过复制您的文本后。+整体概况3In Chapter 1, we have studied a lot about the wave function and how you use it to calculate various quantities of interest.Question: How do you get (x,t) in the first place ? How do you go about solving the Schrdi

2、nger equation ?4 In this Chapter，we assume that the potential (or potential energy function), V(x,t)=V(x), of the system, is independent of time t !5Now the left side is a function of t alone, and the right side is a function of x alone.6 The only way this can be possibly be true is if both sides ar

3、e in fact constant, we shall call the separation constant E. Then Separation of variables has turned a partial differential equation into two ordinary differential equations (Eq. 2.3 and 2.4). 7 The first equation 2.3 is easy to solve, the general solution of Eq.2.3 is The second equation 2.4 is cal

4、led the time -independent Schrdinger equation, we can go on further with it until the potential V(x) is specified. The rest of this chapter will be devoted to solving the time-independent Schrdinger equation 2.4, for a variety of simple potentials. But before we get to that we would like to consider

5、 further the question: 8Whats so great about separable solution ? 可分离的解（即 (x,t)=(x) f(t) ）为何如此重要？ After all, most solutions to the (time-dependent) Schrdinger equation do not take the form (x)f(t). We will offer three answers two of them physical and one mathematical: 9NOTE: for normalizable solutio

6、ns, E must be real (see Problem 2.la). 10Nothing ever happens in the Stationary State (x,t) ! 1112 Please distinguish the operator with “hat” () to its dynamical variable in Eq.2.12. 13Conclusion: A separable solution has the property that every measurement of the total energy is certain to return t

7、he value E. (Thats why we chose that letter E for the separation constant.) 3. The general solution is a linear combination of separable solutions14 Now the (time-dependent) Schrdinger equation (Eq. 1.1) has the property that any linear combination5 of solutions is itself a solution.15 It so happens

8、 that every solution to the (time -dependent) Schrdinger equation can be written in this form it is simply a matter of finding the right constants (c1, c2, c3, c4, ）so as to fit the initial conditions for the problem at hand. Once we have found the separable solutions, then, we can immediately const

9、ruct a much more general solution, of the form 16 Youll see in the following sections how all this works out in practice, and in Chapter 3 well put it into more elegant language, but the main point is this: Once youve solved the time-independent Schrdinger equation, youre essentially done; getting f

10、rom there to the general solution of the time-dependent Schrdinger equation is simple and straightforward. 17Brief summary of section 2.1V(x,t)=V(x)boundary conditions1819 The first example to solve the Schrdinger equation is the infinite square well20A particle in this potential is completely free,

11、 except at the two ends (x = 0 and x = a), where an infinite force prevents it from escaping. Outside the well, (x,t)=0 (the probability of finding the particle there is zero). Inside the well, where V=0, the time-independent Schrdinger equation (Equation 2.4) reads 21Equation 2.17 is the (classical

12、) simple harmonic oscillator equation; the general solution is Typically, these constants are fixed by the boundary conditions of the problem. What are the appropriate boundary conditions for (x)? 22 For the infinite square well, both (x) and d (x)/dx are continuous at the two ends (x = 0 and x = a)

13、 !NOTE: only the first condition of these is applied since the potential goes to infinity here! Continuity of (x) requires that 2324 Curiously, the boundary condition at x = a does not determine the constant A, but rather the constant k, and hence the possible values of E can be obtained from Eq.2.1

14、7 and 2.22: In sharp contrast to the classical case, a quantum particle in the infinite square well cannot have just any old energy only these special allowed values. 25 As promised, the time-independent Schrdinger equation has delivered an infinite set of solutions, one for each integer n. The firs

15、t few of these are plotted in Figure 2.2: 26 they look just like the standing waves on a string of length a. 1, which carries the lowest energy, is called the ground state; the others, whose energies increase in proportion to n2, are called excited states. 27 The wave functions n(x) have some intere

16、sting and important properties: 28dxxanmxanmaa0coscos1axanmnmxanmnm0sin)(1sin)(10)()sin()()sin(1nmnmnmnm29 Note that: this argument does not work if mn (can you spot the point at which it fails?); in that case normalization tells us that the integral is 1. In fact, we can combine orthogonality and n

17、ormalization into a single statement :We say that the s are orthonormal.304. They are complete In the sense that any other function, f ( x ), can be expressed as a linear combination of them n(x) : “Any” function can be cxpanded in this way is sometimes called Dirichlets (狄利克雷) theorem. The expansio

18、n coefficients (cn) can be evaluated for a given f (x) - by a method called Fouriers trick(技巧), which beautifully exploits the orthonormality of n(x) : Multiply both sides of Equation 2.28 by m*(x) , and integrate. 31.)()()()(1*1*mnmnnnmnnmccdxxxcdxxfx Thus the mth coefficient in the expansion of f

19、(x) is given by dxxfxcmm)(*32 These four properties are extremely powerful, and they are not peculiar(特有的) to the infinite square well. The first is true whenever the potential itself is an even function; The second is universal, regardless of the shape of the potential. Orthogonality is also quite

20、general - we1 show you the Proof in Chapter 3. Completeness holds for all the potentials you are likely to encounter, but the Proofs tend to be nasty and laborious; Im afraid most physicists simply assume completeness and hope for the best. 33 The stationary states (Equation 2.6) for the infinite sq

21、uare well are evidently by using of Eq. 2.23 and 2.24 :tiExtxnnnexp)(),(tmanixana2222expsin2 The most general solution to the (time-dependent) Schrdinger equation is a linear combination of stationary states Eq.2.31: 1),(),(nnntxctx12222expsin2nntmanixanac34 In general, when t = 0, according to Equa

22、tion 2.32, we can fit any prescribed initial wave function, (x,0), by appropriate choice of the coefficients cn:)()0 ,(1xcxnnn The completeness of the (x,0)s (confirmed in this case by Dirichlets theorem) guarantees that we can always express (x,0) in this way, and their orthonormality licenses the

23、use of Fouriers trick to determine the actual coefficients. For example, infinite square well, we have 35Given the initial wave function, (x,0)We first compute the expansion coefficients cn，by using of Equation 2.33 Then plug these into Equation 2.32 to obtain (x,t) Armed with the wave function, we

24、are in a position to compute any dynamical quantities of interest, using the procedures in Chapter 1. And this same ritual applies to any potentialthe only things that change are the functional form of the s and the equation for the allowed energies. 36Homework: Example 2.1, Example 2.2 Problem 2.7,

25、 Problem 2.3737 The paradigm for a classical harmonic oscillator is a mass m attached to a spring (弹力) of force constant k. The motion is governed by Hookes law: 38Of course, theres no such thing as a perfect simple harmonic oscillator if stretch it too far the spring is going to break, and typicall

26、y Hookes law fails long before that point is reached. But practically any potential is approximately parabolic(抛物线的), in the neighborhood of a local minimum (Figure 2.3). 3940Thats why the simple harmonic oscillator is so important: Virtually(实际上; 事实上) any oscillatory motion is approximately simple

27、harmonic, as long as the amplitude is small.41 In the literature you will find two entirely different approaches to this problem. The Quantum problem is to solve the Schrdinger equation for the potential As we have seen, it suffices to solve the time independent Schrdinger equation: 42 The second is

28、 a diabolically(魔鬼似地) clever(聪明的) algebraic(代数的) technique, using so-called ladder(阶梯) operators. We will study the algebraic method firstly, because it is quicker and simpler (and more fun). The first is a straightforward “brute force”(强力) solution to the differential equation, using the method of

29、power series expansion; it has the virtue(优点) that the same strategy(策略) can be applied to many other potentials (in fact, we will use it in Chapter 4 to treat the Coulomb potential). 43 The idea is to factor the term in square brackets. If these were numbers easy: To begin with, lets rewrite Equati

30、on 2.39 in a more suggestive form: Here, however, its not quite so simple, because u and v are operators, and operators do not, in general, commute (i.e. uvvu ). Still, this does invite us to take a look at the expressions, from Eq.2.40, we assume 44 Warning: Operators can be slippery to work with i

31、n the abstract, and you are bound to make mistakes unless you give them a “test function”, f (x), to act on. At the end you can throw away the test function, and youll be left with an equation involving the operators alone. “product” 积 45In the present case, we have 46 By using of Eq.2.42, the Schrd

32、inger equation 2.40 becomes Notice that the ordering of the factor a+ and a is important here !47The same argument, with a+ on the left, yields Thus, Eq.2.42 Eq.2.44, we have 48Now, here comes the crucial (关键的) step: 49 Here, then, is a wonderful machine for grinding out new here solutions, with hig

33、her and lower energies if we can just find one solution, to get stared! 50We called a ladder operators, because they allow us to climb up and down in energy; a+ is called the raising operator, and a the lowering operator. The “ladder” of states is illustrated in Figure 2.4. 51But wait ! What if we a

34、pply the lowering operator a repeatedly ? Eventually were going to reach a state with energy less than zero , which (according to the general theorem in Problem 2.2) does not exist ! At some point the machine must fail. How can that happen？We know that a is a new solution to the Schrdinger equation,

35、 but there is no guarantee that it will be nonmalleable it might be zero, or its square integral might be infinite. Problem 2.11 rules out the latter possibility. 52Conclusion: There must occur a “lowest rung(阶梯)” ( lets call it 0）such that(see Eq.2.41)53 To determine the energy of this state, we pl

36、ug it into the Schrdinger equation (in the form of Equation 2.46) 54With our foot now securely planted on the bottom rung (the ground state of the quantum oscillator), we simply apply the raising operator to generate the excited state:This method does not immediately determine the normalization fact

37、or An, which will be worked out by yourself in Problem 2.12.55 We wouldnt want to calculate 50 in this way, but never mind: We have found all the allowed energies, and in principle we have determined the stationary states the rest is just computation. 56Homework: Example 2.557 We return now to the S

38、chrdinger equation for the harmonic oscillator (Equation 2.39): We introduce the dimensionless variable 58 Our problem is to solve Equation 2.56, and in the process obtain the “allowed” values of K (and hence of E from Eq.2.57). 59 Note that the B term in Eq.2.59 is clearly not normalizable (it blow

39、s up as |x|); the physically acceptable solutions, then, have the asymptotic form This suggests that we “peel off” the exponential part 60so the Schrdinger equation (Eq.2.56) becoms 6162 This recursion(递推) formula is entirely equivalent to the Schrdinger equation itself. From Eq.2.65: given a0 , Eq.

40、2.65 enables us (in principle) to generate a2, a4, a6, given a1 , Eq.2.65 generates a3, a5, a7, . Let us write 63 Thus equation 2.65 determines h() in terms of two arbitrary (a0 and a1) which is just what we would expect, for a second-order differential equation. However, not all the solutions so ob

41、tained are normalizable. For at very large j, the recursion formula 2.65 becomes (approximately) 64 Now, if h goes like exp(2), then (remember ? thats what were trying to calculate) goes like exp(2/2) (Equation 2.61), which is precisely the asymptotic behavior we dont want. 65 There is only one way

42、to wiggle out of this: For normalizable solutions the power series must terminate. There must occur some “highest” j (call it n) such that the recursion formula spits out an+2 = 0 . Conclusion: the series hodd will be truncated at some highest n; the series heven must be zero from the start. For phy

43、sically acceptable solutions, then, we must have for some positive integer n, which is to say (referring to Equation 2.57) that the energy must be of the form 66 Thus, we recover, by a completely different method, the fundamental quantization condition we found algebraically in Equation 2.50.(Eq.2.6

44、8 is obtained by putting K=2n+1 into Eq.65)6768 In general, hn() will be a polynomial of degree n in , involving even powers only: if n is an even integer, and odd powers only, if n is an odd integer. Apart from the overall factor (a0 or a1) they are the so-called Hermite polynomials, Hn(). The firs

45、t few of them are listed in Table 2.1. 69 By tradition, the arbltrary multiplicative factor is chosen so that the coefficient of the highest power of is 2n. With this convention, the normalized stationary states for the harmonic oscillator are They are identical (of course) to the ones we obtained a

46、lgebraically in Equation 2.50. In Figure 2.5a we can plot n(x) for the first few ns.7071 The quantum oscillator is strikingly(醒目地) different from its classical counterpart not only are the energies quantized, but the position distributions have some bizarre (怪诞的) features. For instance: The probabil

47、ity of finding the particle outside the classically allowed range (that is, with x greater than the classical amplitude for the energy in question) is not zero (see Problem 2.15). In all odd states the probability of finding the particle at the center of the potential well is zero. Only at relativel

48、y large n do we begin to see some resemblance(相似) to the classical case. 72In Figure 2.5b, we have superimosed the classical position distribution on the quantum one (for n=100); if you smoothed out the bumps in the latter, the two would fit pretty well.Fig.2.5 (b) Graph of |100|2, with the classica

49、l distribution superimposed. 7374 We turn next to what should have been the simplest case of all: the free particle V(x)=0 everywhere. As youll see in a moment, the free particle free particle is in fact a surprisingly(令人惊讶地) subtle(难以捉摸的) and tricky(难处理的) example. The time-independent Schrdinger eq

50、uation reads 75 So far, its the same as inside the infinite square well (Equation 2.17), where the potential is also zero. This time, however, we prefer to write the general solution of the free particle in exponential form (instead of sines and cosines) for reasons that will appear in due course: U