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Unformatted text preview: CHM 370 Handout One The Mean value of a Random Value_
The deﬁnition of the average (mean) value may be given as ' I . N
v = lim -1— [value of v obtained in the ith measurement]
1-! The rationale behind this is that the variable v is a random variable. This may result from
variations in experimental procedure or. as will be discovered later, from the very nature of the
system. In either case. the number of measurements N is extended to a large number in order to
deﬁne best the average of the observable measured. with the number extended to inﬁnity
providing the true mean value of the system. (That is. the larger the statistical sample, the more
accurate the measure.) In quantum mechanics. one will find that because of the nature of the system, data points will
occur several times. We now deﬁne the function N(v,.) to be the number of times data point v; occurs in the series of measurements. Let M be the number of different data points in the
statistical sample. Using the function N(v,.). the summation becomes '
N v )
ﬁve] 1-! N N M
2 [value of v obtained in the ith measurement] = 2v, a: I'Il ii] I"! Since the summation in brackets merely counts the number of times v.- shows up, the overall
summation simply becomes 7 M Nan is] j-l i-l Upon returning to the deﬁnition of the average value. we note that the limit operation on the
summation is linear. that is, the sum of several limits equals the limit of the sum Along with the
multiplication of l/N , then (hang-ﬁfe Ntv.)= igniv.[§§d]=iv[ﬁm&;2] is! in] . The definition of a probability function P(v,) is This is the fraction of all measurements that result in v.1. where extending the number of measurements to inﬁnity yields the most accurate fraction. (Example: flipping a fair coin has a
one-half probability of landing on the obverse [heads] side. meaning that as the number of coin
ﬂips becomes infinite. heads will show up half of the time.) Substituting, then CHM 370 HandoutL Two Proof that E(T,V) z kT2 6T _ V From equations in the lecture notes, 1 E(T, V) 2 Z(T,V)ZEexp[~ But Em} g] = W [aexpE-TE/kTUV. Comparing equations (1) and (2) gives 5W”): Z(1E,V)ZkT2[a—afexp[%ﬂy allE z 2(r1, V) kT2[aiT§exp[_—gﬂ]y However, Thus equation (3) becOmes or equivaiently the desired result. (1) (2) (3) (4) ...
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- Fall '08