Problem:Wave packet is unstable-The waves have different phase velocity and the wave components of a wave packet will disperse even in vacuum.But electrons are localized in the atom(1A).This viewpoint emphasize the wave nature of micro-particle while killing its particle nature. Another extreme viewpoint on wave particle duality is that wave is composed by large amount of particles (as waves in air).However experiments show clearly a single electron possesses wave nature.This viewpoint over-stressed the particle nature C.Born's statistical interpretation A particle by its nature is localized at a point,whereas the wave function is spread out in space.How can such an object represent the state of a particle?The answer is provided by Born's Statistical Interpretation of the wave function,which says that in 1D case(,t) gives the probability of finding the particle at point,at time tor,more precisely fleG.OPa- Probability of finding the particle between a and b,at time t. In 3D case,the relative probability for the particle appearing in a volume element dis proportional to the intensity of wave in that volume element Probability amplitude d加x(,t=(,)D Probabllity density D.Probability Because of the statistical interpretation,probability plays a central role quantum me- chanics.So we introduce some notation and terminology. 1.Erample of discrete variables 14 people in a room.Let N(j)represent the number of people of age j. N(14=1,N(15)=1.N(16)=3,N(22)=2,N(24)=2.N(25)=5 (1) 11
Problem: Wave packet is unstable - The waves have different phase velocity and the wave components of a wave packet will disperse even in vacuum. But electrons are localized in the atom (1A˚). This viewpoint emphasize the wave nature of micro-particle while killing its particle nature. Another extreme viewpoint on wave particle duality is that wave is composed by large amount of particles (as waves in air). However experiments show clearly a single electron possesses wave nature. This viewpoint over-stressed the particle nature. C. Born’s statistical interpretation A particle by its nature is localized at a point, whereas the wave function is spread out in space. How can such an object represent the state of a particle? The answer is provided by Born’s Statistical Interpretation of the wave function, which says that in 1D case |ψ (x, t)| 2 gives the probability of finding the particle at point x, at time t - or, more precisely Z b a |ψ (x, t)| 2 dx = Probability of finding the particle between a and b, at time t. In 3D case, the relative probability for the particle appearing in a volume element d 3~r is proportional to the intensity of wave in that volume element dw ∝ |ψ (~r, t)| 2 d 3 ~r = ψ ∗ (~r, t) Probability amplitude z }| { ψ (~r, t) | {z } Probability density d 3 ~r D. Probability Because of the statistical interpretation, probability plays a central role quantum mechanics. So we introduce some notation and terminology. 1. Example of discrete variables 14 people in a room. Let N(j) represent the number of people of age j. N(14) = 1, N(15) = 1, N(16) = 3, N(22) = 2, N(24) = 2, N(25) = 5 (1) 11
1011121314151617181920212223242526 FIG.6:Histogram for the distribution while N(17),for instance,is zero.The total number of people in the room is N-∑NO) j=0 Questions about the distribution 1.If you selected one individual at random from this group,what is the probability that this person's age would be 15?Answer:One chance in 14,since there are 14 possible choices,all equally likely,of whom only one has this particular age.In general the probability of getting age j is PU)= Notice that the probability of getting either14 or 15 is the sum of the individual probabilities (in this case,1/7).In particular,the sum of all the probabilities is 1- you're certain to get some age: P0)=1 j=0 2.What is the most probable age?Answer:25.In general the most probable j is the j for which P(j)is a maximum. 3.What is the median age?Answer:23,for 7 people are younger than 23,and 7 are older.In general,the median is that value of j such that the probability of getting a larger result is the same as the probability of getting a smaller result. 4.What is the average (or mean)age?Answer: (14+(15+3(16)+222+2(24+52-294=21 14 14 12
FIG. 6: Histogram for the distribution. while N(17), for instance, is zero. The total number of people in the room is N = X∞ j=0 N(j) Questions about the distribution 1. If you selected one individual at random from this group, what is the probability that this person’s age would be 15? Answer : One chance in 14, since there are 14 possible choices, all equally likely, of whom only one has this particular age. In general the probability of getting age j is P(j) = N(j) N Notice that the probability of getting either 14 or 15 is the sum of the individual probabilities (in this case, 1/7). In particular, the sum of all the probabilities is 1 - you’re certain to get some age: X∞ j=0 P(j) = 1 2. What is the most probable age? Answer : 25. In general the most probable j is the j for which P(j) is a maximum. 3. What is the median age? Answer : 23, for 7 people are younger than 23, and 7 are older. In general, the median is that value of j such that the probability of getting a larger result is the same as the probability of getting a smaller result. 4. What is the average (or mean) age? Answer: (14) + (15) + 3 (16) + 2 (22) + 2 (24) + 5 (25) 14 = 294 14 = 21 12
In general the average value of j(which we shall write thus:(j))is 0=2N0-2P60) 0 In quantum mechanics it is called the expectation value.Nevertheless the value is not necessarily the one you can expect if you made a single measurement.In the above example,you will never get 21 5.What is the average of the squares of the ages?Answer:You could get 142- 196,with probability 1/14,or 152=225,with probability 1/14,or 162=256,with probability 3/14,and so on.The average is G)=-∑P0) j= In general,the average value of some function of j is given by G=∑fG)PO) Beware(2)≠) Now,there is a conspicuous difference between the following two histograms,even though they have the same median,the same average,the same most probable value,and the same number of elements:The first is sharply peaked about the average value,whereas the second is broad and flat.(The first might represent the age profile for students in a big-city classroom,and the second,perhaps,a rural one-room schoolhouse.)We need a numerical measure of the amount of"spread"in a distribution,with respect to the average.The most effective way to do this is compute a quantity known as the standard deviation of the distribution σ三V(A)2〉=VG)-》 △=j-》 For the two distributions in the above figure,we have 01=V25.2-25=V0.2 02=V31.0-25=V6 13
In general the average value of j (which we shall write thus: hji) is hji = PjN(j) N = X∞ j=0 jP(j) In quantum mechanics it is called the expectation value. Nevertheless the value is not necessarily the one you can expect if you made a single measurement. In the above example, you will never get 21. 5. What is the average of the squares of the ages? Answer: You could get 142 = 196, with probability 1/14, or 152 = 225, with probability 1/14, or 162 = 256, with probability 3/14, and so on. The average is j 2 ® = X∞ j=0 j 2P(j) In general, the average value of some function of j is given by hf(j)i = X∞ j=0 f(j)P(j) Beware hj 2 i 6= hji 2 . Now, there is a conspicuous difference between the following two histograms, even though they have the same median, the same average, the same most probable value, and the same number of elements: The first is sharply peaked about the average value, whereas the second is broad and flat. (The first might represent the age profile for students in a big-city classroom, and the second, perhaps, a rural one-room schoolhouse.) We need a numerical measure of the amount of ”spread” in a distribution, with respect to the average. The most effective way to do this is compute a quantity known as the standard deviation of the distribution σ ≡ q (∆j) 2 ® = q hj 2i − hji 2 ∆j = j − hji For the two distributions in the above figure, we have σ1 = √ 25.2 − 25 = √ 0.2 σ2 = √ 31.0 − 25 = √ 6 13
六由。 。。 FIG.7:Two histogram with different o. 2.Erample of continuous variables It is simple enough to generalize to continuous distribution.Technically we need"infini- tesimal intervals".Thus Probability that an individual (chosen p(x)da at random)lies between r and (+dr) p(r)is the probability of getting r,or probability density.The probability that lies between a and b(a finite interval)is given by the integral of p() and the rules we deduced for discrete distributions translate in the obvious way: 1 = xp(r)dr n-ronoyke o2=(A)〉=(x2)-(2 Example:Suppose I drop a rock off a cliff of height h.As it falls,I snap a million photographs,at random intervals.On each picture I measure the distance the rock has fallen.Question:What is the average of all these distance?That is to say,what is the time average of the distance traveled? Solution:The rock starts out at rest,and picks up speed as it falls;it spends more time near the top,so the average distance must be less than h/2.Ignoring air resistance,the distancer at timet is () 14
FIG. 7: Two histogram with different σ. 2. Example of continuous variables It is simple enough to generalize to continuous distribution. Technically we need ”infinitesimal intervals”. Thus Probability that an individual (chosen at random) lies between x and (x + dx) = ρ(x)dx ρ(x) is the probability of getting x, or probability density. The probability that x lies between a and b (a finite interval) is given by the integral of ρ(x) Pab = Z b a ρ(x)dx and the rules we deduced for discrete distributions translate in the obvious way: 1 = Z +∞ −∞ ρ(x)dx hxi = Z +∞ −∞ xρ(x)dx hf(x)i = Z +∞ −∞ f(x)ρ(x)dx σ 2 ≡ (∆x) 2 ® = x 2 ® − hxi 2 Example: Suppose I drop a rock off a cliff of height h. As it falls, I snap a million photographs, at random intervals. On each picture I measure the distance the rock has fallen. Question: What is the average of all these distance? That is to say, what is the time average of the distance traveled? Solution: The rock starts out at rest, and picks up speed as it falls; it spends more time near the top, so the average distance must be less than h/2. Ignoring air resistance, the distance x at time t is x(t) = 1 2 gt2 14
FIG.8:The probability density in the Example p(). The velocity is dr/dt=gt,and the total flight time is T=v2h/g.The probability that the camera flashes in the interval dt is dt/T,so the probability that a given photograph shows a distance in the corresponding range dr is dt dzg 1 T9元V2苏2VW Evidently the probability density is 1 回=2W后0≤r≤ (outside this range,of course,the probability density is zero.) We can check the normalization of this result Ch 1 1 人2wm=2w元2r6=1 The average distance is ==a(传心-咖 which is somewhat less than h/2,as anticipated.Figure shows the graph of p().Notice that a probability density can be infinite,though probability itself(the integral of p)must of course be finite (indeed,less than or equal to 1). Problem 2 Griffiths,page 12,Problem 1.2,1.3 15
FIG. 8: The probability density in the Example ρ(x). The velocity is dx/dt = gt, and the total flight time is T = p 2h/g. The probability that the camera flashes in the interval dt is dt/T, so the probability that a given photograph shows a distance in the corresponding range dx is dt T = dx gt r g 2h = 1 2 √ hx dx Evidently the probability density is ρ(x) = 1 2 √ hx ,(0 ≤ x ≤ h) (outside this range, of course, the probability density is zero.) We can check the normalization of this result Z h 0 1 2 √ hx dx = 1 2 √ h ¡ 2x 1/2 ¢h 0 = 1 The average distance is hxi = Z h 0 x 1 2 √ hx dx = 1 2 √ h µ 2 3 x 3/2 ¶h 0 = h/3 which is somewhat less than h/2, as anticipated. Figure shows the graph of ρ(x). Notice that a probability density can be infinite, though probability itself (the integral of ρ) must of course be finite (indeed, less than or equal to 1). Problem 2 Griffiths, page 12, Problem 1.2, 1.3 15