CHAPTER SEVEN AN INTRODUCTION TO PROBABILITY CALCULATIONS FOR SMALL SYSTEMS

Transcription

1 CHAPTER SEVEN AN INTRODUCTION TO PROBABILITY CALCULATIONS FOR SMALL SYSTEMS Introduction In the last chapter we discussed how the different states of a system are specified, both quantum mechanically and classically. Once the different states are uniquely specified, we typically examine the number of unique microstates which all have the same macroscopic characteristics of interest (say the same pressure T ÑÞ If we assume that all of the microstates of the system are equally likely, then we would expect the most likely configuration of the system to be that macroscopic state which corresponds to the largest number of microstates. In order to determine which of the macroscopic states of a system are indeed the most likely, and to examine the fluctuations in the macroscopic measurements of these states, we must examine the concepts of probability and statistics. The Definition of Probability In order to define the concept of probability, we examine the results of tossing a coin. The probability, c( h), of obtaining heads in this experiment is given by H c( h) Ð2Ñ (7.1) H where H is the total number of possible outcomes of the experiment, and where H(h) is the number of different ways of obtaining the desired result! Thus the probability of obtaining heads in a single coin toss is (one possible way of getting heads out of two possible outcomes heads or tails). Of course, this result depends upon the a priori assumption that each of the possible outcomes is equally likely! (This would not be true, for example, for a weighted coin or rounded die.) Although the probability of getting heads in a single coin toss is, if I take a single coin out of my pocket and flip it four (4) times I may actually get (where stands for heads, and stands for tails). I find that I get heads only of the time rather than. Trying this same experiment again, but flipping the % $ coin eight (8) times, I might get +. I find that now I get heads % of the time. I dont get in either case! However, if I repeat this experiment (tossing the coin) enough times, I find that the number of times I get heads begins to converge to (although I may need to flip the coin a very large number of times to see this). Thus, when we speak of probabilities, we really think of repeating the same experiment a large number of times ( ). This can be accomplished, in principle, by performing the same identical experiment many times over and over again with the same system (coins, die, boxes of atoms, etc.) or it can be accomplished by performing the same experiment in a large number of identical systems all at the same time. We call the large number of identical experiments an ensemble of experiments. The probability, then, is just the relative number of times that a given result occurs in the ensemble. A single experiment is just one member of a larger ensemble. If we repeat the same experiment several times we are just sampling the ensemble. Thus, although we expect 50% of an ensemble of tossed coins to exhibit heads and 50% to exhibit tails, any random selection from the ensemble will give different results. But, if we examine a large number of the members of the ensemble, we will begin to get results close to 50%. An additional point to consider is the case where there is some constraint which limits the number of accessible states. As an example lets look at the case of the semi-classical, one-dimensional simple harmonic oscillator with constant energy. We pointed out in the last chapter that the region of phase space accessible to this system was the area between the ellipse corresponding to the energy I and the ellipse corresponding to energy I $ I. [We designate this system as one which has energy I with an uncertainty $ I.] The volume of phase space lying between I and I $ I, contains H( E) different cells in phase space (each with a volume 2o). The probability of finding this system with constant energy I in a particular region is given by H H c ( ) ( E;xk) E( xk) xk H( E) H (7.2) E x k

2 Chapter Seven: Probability for Small Systems 2 where we explicitly indicate a subset, H( E àb 5 ) of all the states in which the system might be found with constant energy I and also within the region B 5. Example 1: Consider throwing die. The probability of the die landing with one dot up is given by number of ways to get c( ) ( ) 1 (7.3) number of possible outcomes 6 Example 2: Consider drawing a single card from a full deck of 52 cards. The probability of drawing an ace (of any kind) is given by number of aces in deck c ( ace) % (7.4) number of different cards in deck & The probability of drawing a particular ace, say the ace of clubs, would be number of ace of clubs in the deck c Ðace of clubsñ number of differenct cards in deck & (7.5) Now, if, after drawing the first card, we had obtained the ace of clubs, the probability of obtaining a second ace (of any kind) from this same deck of cards would be number of aces remaining in deck 3 c $ ( ace) (7.6) number of cards remaining in deck 51 The probability of drawing the ace of diamonds from this deck on the second draw would be number of ace of diamonds c % Ðace of diamonds Ñ (7.7) number of cards in deck & Notice that the probability of drawing an ace (of any kind) and then drawing a particular ace (say the ace of clubs) depends upon whether or not you drew the ace of clubs on the first round or not. Thus these two probabilities are not independent. The Probability Sample Space. Another way of defining probability is based upon what we will call a sample space. A sample space is just a list of all possible, mutually exclusive, outcomes of an experiment and their associated probabilities. We call each individual outcome a point in the sample space. A uniform sample space is one in which all the points are equally likely and mutually exclusive. An example of a uniform sample space is the case where a single coin is flipped. Each of the two outcomes is equally likely. Most sample spaces, however, are not uniform. As an example of a non-uniform sample space, consider the case where two die are rolledþ The possible outcomes of this toss are listed below: ß ß ß$ ß% ß& ß ß ß ß$ ß% ß& ß $ß $ß $ß$ $ß% $ß& $ß %ß %ß %ß$ %ß% %ß& %ß &ß &ß &ß$ &ß% &ß& &ß ß ß ß$ ß% ß& ß

3 Chapter Seven: Probability for Small Systems 3 If we assume that each outcome is equally likely, then we can determine the probability of obtaining various results simply by adding the number of ways each result can be obtained. For example, the number of ways of obtaining a sum of 5 is the number of sample points which add to five (4,1 3,2 2,3 1,4). The probability of getting a sum of 5 is, therefore, Likewise, the probability of obtaining a sum of 10 is cð&ñ cð!ñ % $ * $ $ If we are interested in the sum of the two dice we can generate a different sample space, in which the possible outcomes of this experment correspond to the sample points and the relative number of different ways that this result can be obtained would complete this particular sample space: Sample Space for the Sum of Two Die Sample Point: Associated Probability: Probabilities of Multiple Events We now consider multiple events and their associated probabilities. We will designate the probability of one event (call it E) by c( A ), and another event (say F) by c( B ). We can visualize the possible probabilities by using a representation of sample space. Consider the grid of points below. A AB B The grid represents points in the sample space. We can enclose those points in which event E occurs, and those points in which event F occurs. If there is no overlap, the events are independent. If, however, the two regions overlap then we can talk about the probability that both E and F occur, c( A and B) c( EF) Þ Several other possibilities also arise, such as the probability that either A or B occur, c( A or B); the probability that either A or B occur or both, c( A B ); the probability that F has occurred if we know that E has already occurred q q ca( B ); the probability that neither A nor B occur, c( A and B); etc. There are a few important relationships among these probabilities which we will now introduce without proof. For further explanations you might refer to Boas. Many of these can be easily reasoned out based upon the representative sample space above. In the equations below, R represents the number of sample points in the sample space, N( A) the number of sample points in which event E occurs, etc. c( ) (7.8) A RÐEÑ R

4 Chapter Seven: Probability for Small Systems 4 B c( AB) RÐE Ñ (7.9) R B c A ( B) RÐE Ñ (7.10) R( A) c( AB) 0 if A and B are mutually exclusive (7.11) c( A B) c( A) c( B) c( AB) (7.12) c( A) c ( B) c( AB) c( B) c ( A) c( BA) (7.13) A B ca( B) c( B ) if A and B are independent (7.14) Statistically Independent Events Of particular interest are those cases where the occurance of one outcome ( E or F) does not influence the occurance of the other. These are said to be independent occurances. For this case, the probability of obtaining E and then F is given by c( A and B) c( AB) c( A) c( B) (7.15) Example: Consider throwing two die. The probability of obtaining a six on the first role is, and the probabilility of obtaining a six on any subsequent roll (or on the roll of another die) is also. The two rolls (or die) are independent of one another. The probability of obtaining two sixes on two subsequent rolls, since these rolls are independent, is therefore given by $ c( and ) c( ) c( ). Lets show that this is indeed correct. By definition the probability of an event is the number of times the event can occur divided by the total number of all possible events for the specific experiment. How many possible combinations of numbers are there with two dice? For each face upward on die E there are 6 possible orientations of die F. Thus, there are different possible combinations of the dice. How many different combinations will give us 12 (6 on both die)? There is only one! To see this examine the table on the next page. Notice that there is only one way to obtain snake eyes and box cars, but that there are several ways to obtain other total values, for example 7 or 11! For example, there are two ways to obtain 11; with a 6 and 5, and a 5 and 6, where we differentiate between die E and die F. Thus, the most probable number to arise in throwing two dice is 7! Notice the difference between specifying the system in this way and by using the sample space designation used earlier. Sometimes the probability that an event occurs depends upon whether or not another event has occured. We designate by ( ) the probability that event occurs given the fact that event has already occured. c A F F E Example 1: What is the probability of picking two aces from a 52 card deck? The probability of picking one ace is given by number of aces c(first ace) (7.16) number of cards in deck % &

5 Chapter Seven: Probability for Small Systems 5 The probability of picking a second ace, after the first one is removed, is given by number of remaining aces c(second ace) (7.17) number of remaining cards in deck $ & From our earlier arguments, the probability of picking two aces in sequence must, therefore, be % $ c(two aces) c(first ace) c(second ace) (7.18) & & Example 2: A meeting is held in which 40 doctors and 10 psyhcologists attend. A committee of 3 is selected at random. What is the likelyhood that at least 1 psychologist will be on the committee? We attack this problem much like the one of aces drawn from a deck. We choose one member of the committee at a time. Lets consider the likelyhood that no psychologist is chosen for the committee. If the probability that no psychologist is on the committee is ;, then the probability that at least one on the committee is a psychologist is : ;Þ For the first member, the probability that no psychologist is chosen will be 40/50. For the second selection, if no psychologist was selected for the first position, the probability will be 39/49. For the last position, we get 38/48. So the probability of selecting all doctors after three random selections is %! $* $) c(all doctors) (7.19) &! %* %) The probability that at least one of the committee members will be a psychologist is, therefore, 0.496! Events with Two Possible Outcomes. When we toss a coin we assume that the likelyhood of obtaining heads is equal to the likelyhood of obtaining tails (i.e.,. This would be true provided the coin is balanced Ñ properly. But what if the coin is bent? In this case the probabilities of obtaining the two possible results may not be equal. If we designate the probability of obtaining heads by c( h ), then we can designate the probability of not q q getting heads by the expression c( 2 ), where we let 2 stand for not-h. It should be obvious that we will obtain either heads or not-heads for any given experiment, so the probability of getting heads plus the probability of not getting heads is a certainty, or q c( h ) c( h ) (7.20) or q c( 2 ) c( h ) (7.21) This idea can be applied to any situation where we are interested only in the probability of one type of event as compared to any other event, no matter how many different outcomes may actually be possible. To simplify notation, we will designate the probabiltiy of obtaining the desired outcome by : and the probability of any other outcome by ;. This means that the equations above would be written p q (7.22) ; : (7.23) Example 1: The probability of obtaining two dots up when a die is thrown is & probability of obtaining any other result is., while the Example 2: For a spin particle, the probability of its z-component being is and the probability of it being is also. Permutations and Combinations

6 Chapter Seven: Probability for Small Systems 6 Assume that we have a true die so that the probability of getting any one face up is. The probability & of getting an ace (one up) is just while the probability of getting anything else is Now assume that we have Þ five such dice and want to know the probability of throwing 2 aces. When we roll the first die, the probability of & throwing an ace is and the probability of anything else is. Since the probabilities of the individual die are independent (one does not influence the other) the probability that the second die will be an ace is also. Thus, the probability of getting two aces on the first two throws, but not getting an ace on the next three must be & & & cðeerrrñ!þ! Likewise, the probability of getting and ace on the first throw and on the third throw would be & & & cðererrñ!þ! and is the same result. Obviously, there are a number of different ways that we can obtain just two aces. We typcially want to know the probability of obtaining two aces out of five throws independent of the order in which the ace occurs, or of the identity of an individual die. But just how many ways is that? To answer this we need to talk about permutations and combinations. To understand the concept of permutations we will consider the example of a State Dinner for which there are R persons to be seated at the table (where there are R chairs we dont want to leave anyone out). We might wonder how many different ways we could arrange the nametags for the individual plates (i.e., how many permutations TÐR ± RÑ are there of R persons among R chairs). Lets assume that all the nametags were placed in a large box and that we would draw them out one at a time and place them by the plates. When we draw the first name, there are R nametags to choose from thus there are R possibilities. For the second plate we have R 1 possible choices. Thus, in selecting the nametags for two different plates we would have R( R 1) possible combinations. For the next plate there are R 2 possible choices, etc., which means that by the time we have chosen R nametags and placed them by R plates, there are TÐR ± RÑ R( R 1)( R 2) â(1) R! (7.24) different ways that the nametags could have been selected. This means that there are R! different seating arrangements if we do not restrict the seating order (for example if we do not care who sits at the head table). In our example of throwing five dice, we might consider numbering each individual die from one to five so that they are distinguishable from one another (just like each die having its own name). The dice are then placed into a sack and are pulled out of the sack one at a time and rolled. According to what we have stated above, there are &x different ways we can reach into our bag, pull out a die and roll the die, and for each possible combination we have a probability of getting an ace. Now suppose we have only 8 ( 8 < R ) plates for our state dinner. (We will ignore the fact that having too few chairs might create a very difficult eating atmosphere.) We might ask how many different ways, TÐR ± 8Ñ, there are to choose the nametags and place them by these 8 plates? As before, we saw there were R for the first plate, R 1 for the second, etc. until we reach the 8th plate. For this plate there are R ( 8 1) choices left. Thus, the number of possible permutations are given by TÐR ± 8Ñ R( R 1)( R 2) â( R [ 8 1]) (7.25) or TÐR ± 8Ñ R( R 1)( R 2) â( R 8 1) (7.26) which can be written TÐR ± 8Ñ R( R 1)( R 2) â( R 8 1) ( R 8)! R! ( R 8)! ( R 8)! (7.27) or TÐR ± 8Ñ TÐR ± RÑ ( R 8)! (7.28)

7 Chapter Seven: Probability for Small Systems 7 This is the number of different possible arrangements of the name tags on the tables. Now suppose we are not really interested in the number of different ways the name tags can be placed on the tables, but are really just interested in the number of different ways we can select (at random) 8 different people who would be allowed to eat at the state dinner from R people who had asked to be allowed to eat at the dinner. Would we care which persons name was placed by which plate? No! Therefore, after choosing the 8 names from the box who will be allowed to sit at the table, we could rearrange the nametags 8! different ways. When we determine the number of permutations, TÐR ± 8Ñ, we are counting the number of unique ways in which the people can be arranged. If we are only interested in the number of ways to choose 8 people, we have overcounted by the number of permutations, TÐ8 ± 8Ñ, of the 8 people which are chosen. We call the number of different ways to choose 8 different people from a pool of R people, without counting the permutations of the 8 people a Combination, GÐR ± 8Ñ, and it is given by TÐR ± 8Ñ R! GÐR ± 8Ñ TÐ8 ± 8Ñ ( R 8)! 8! Example1: Suppose you just received a beautiful ivory box for your birthday. The box has three cubical sections which are just the right size to hold three of your prize marbles. But you have 5 prize marbles. You place the three best marbles in the box and then gaze delightedly at your treasure. However, as you look at the ivory box with your three marbles you realize that one of the other marbles would actually look better, because of its unique coloring. As you change out the marbles, it becomes more and more difficult to decide just which marbles you should place in the box. How many different possibilities are there for you to consider as you try to make up your mind? Since the marbles are distinguishable (i.e., they are not all identical) you are asking how many unique ways you can place 5 marbles into 3 spaces. This is the same problem as the R people &x &x being seated at 8 different plates. There are TÐ&l$Ñ! unique permutations of the Ð& $Ñx x marbles among the three available spaces. However, of these 60 permutations, some have the same three marbles in the available spaces, but simply arranged in different orders within the ivory box. If you are not concerned with just how the three marbles in the ivory box are oriented, then the actual number of different ways of putting 5 marbles in 3 spaces is 60/3!! combinations. [Note: You also get the same answer if you consider the two marbles which are left outside the &x &x &x ivory box, since GÐ&l$Ñ GÐ&lÑÞÓ Ð& $Ñx$x x$x Ð& Ñxx Example 2: Suppose you have a club with 50 members and you are about to elect four (4) officers, a president, vice-president, secretary, and tresurer. How many different possible outcomes are there to this selection process? In selecting the president, you have 50 choices. But once he is selected there are now only 49 choices left for vice-president. After choosing him, there are only 48 ways to choose the secretary, and then only 47 ways to choose the treasurer. Thus, there are &!x T &!l% &ß&(ß!! &! % x different possible outcomes if the selection process were purely random. Example 3: Now suppose that this same club of 50 members wishes to choose a four member committee to consider the initiation rites of the club. How many different combinations are there for selecting this committee? The difference between this situation and the one in the last example is that all the members of the committee are assumed to be equal (which would be true provided a chairman were not selected in advance). This means that the ordering in which the four were selected would not be important. (7.29)

8 Chapter Seven: Probability for Small Systems 8 This is an example of a combination &!x &ß&(ß!! G &!l% $!ß$!! &! % x%x % The Binomial Expansion and Probability In cases where the outcome of some process can be categorized as either successful or unsuccessful, we can make use of the binomial expansion to determine the probability of success of failure of that process. To illustrate this, we will first consider tossing two coins. We know that the probability of getting heads for a single toss is and the probability of getting heads is both %. But what about the other possible outcomes? The possibility of getting both tails is likewise, so that the possibility of getting one head and one tail is when % tossing two coins. A table of all the possible outcomes is shown in table 7.1. TABLE 7.1 Tabulation of the outcomes of tossing two coins simultaneously. System State Coin 1 Coin 2 Heads 1 H H 2 2 H T 1 3 T H 1 4 T T 0 Now for each coin the probability of getting either heads or tails is unity. Mathematically, we write this as ( p i ; i ), where : 3 is the probability of obtaining heads, ; 3 is the probability of obtaining tails, and where the subscript 3 denotes which coin. If we toss two coins, then, the probabilty that the experiment has an outcome of any sort must be given by which gives, upon expansion, (: 1 ; 1) (: 2 ; 2) 1 (7.30) : : : ; ; : ; ; (7.31) Notice that we obtain four different terms, one for each of the possible outcomes: one with two heads, one with two tails, and two with one head and one tail! Therefore, it appears that this binomial expansion can be used to discribe our coin tossing experiment. Notice that this expansion designates which coin is heads and which is tails! This is a result of the fact that we numbered each : and ;, which implies that the different coins are distinguishable (i.e., one may be older and therefore less shiny). If we are not interested, however, in which coin is heads and which is tails, we might simply designate this problem by the equation 2 2 (: ;)(: ;) (1) : (2) :; (1) ; (7.32) where the number appearing in parentheses on the right hand side of the equation indicates the number of ways of obtaining the desired result. Thus, there are two distinct ways of obtaining one head and one tail, but only one way to obtain two heads or two tails. This same process can be extended to the case of tossing three coins. If we want to distinguish each coin we could write : ; : ; : ; $ $

9 Chapter Seven: Probability for Small Systems 9 or ::: $ : :; $ : ;: $ ; :: $ : ;; $ ; : ; $ ; ;: $ ; ; ; $ This way of expressing the probabilities specifically indicates which coin is a head and which coin is a tail. ÒNote: This listing, however, does not include all possible permutationsß as we defined them earlier. The permutations would be equivalent to specifying the ordering of the coins. For example, ::; $ would be a different arrangement from :; $ : Þ] If we are not interested in distinguishing the three coins (i.e., if we dont care which coin is heads and which tails), then we have (: ;)(: ;)(: ;) (: ;) : 3: ; 3 :; ; (7.33) Here we see that there are three ways to obtain two heads and one tail, three ways to obtain one head and two tails, and only one way to obtain all heads or all tails! The binomial expansion, then, is a useful method of determining the probability of multiple events when each individual event is independent and can be expressed in terms of only one of two possibilities with constant probability, and when we are considering indistinguishable particles. The general expression for the binomial expansion is RÐR Ñ RÐR ÑÐR ) Ðp ; Ñ p R p ; p ; p ; â ; x $x R R R R R $ $ R (7.34) Notice that the coefficients of the binomial expansion have the form of a combination GÐR ± 8Ñ which we introduced earlier, since RÐR ÑÐR ) RÐR ÑÐR ) ÐR $Ñx Rx $x ÐR $Ñx$x ÐR $Ñx$x Using this last equation, we can write the binomial expansion as Each term of the form R (7.35) R Rx Ð: ;Ñ 8 R 8 : ; 8 R 8 GÐR l8ñ: ; (7.36) ÐR 8Ñx8x 8! 8! R c Rl8 GÐRl8Ñ: ; 8 R 8 represents the probability of obtaining 8 desired results in R attempts. The number G Rl8 is the number of 8 R 8 different ways of obtaining the same result, : is the probability of 8 successes, and ; is the probability of R 8 failures. Example 1. Consider tossing 4 die. What is the probability of obtaining 3 aces (ones)? The probability of success (an ace) is 1/6, while the probability of failure is 5/6. The $ probability of obtaining 3 aces is then Î, but there are G %l$ % different ways of obtaining 3 aces if four throws ( EEER EERE EREE REEEÑ, so we have $ % $ &! & c %l$ GÐ%l$Ñ: ; % Œ * $% Example 2. What is the probability of getting 3 heads in four tosses of a balanced coin? This is expressed as c %l$ G %l$ Œ Œ G %l$ Œ % $ $ %

10 Chapter Seven: Probability for Small Systems 10 You might think that the binomial distribution is a very special case which would be of little real interest in physics. However, in the case of objects having spin, these particles can exhibit either spin up or spin down and this type of reasoning is just what we need! Similarly, we can often express a desired outcome as one of two choices. Using the binomial expansion for determining the probability of 8 successes in R attempts is useful when dealing with a small number of events, but when R becomes large this process becomes very cumbersome. In fact, when R becomes large we may not really be so interested in the probabilty of getting a particular outcome, but rather in the most probable outcome and the size of the fluctuations about that most probable outcome. This is what we will begin to examine in the next chapter.