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Unformatted text preview: Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 5, September 17, 2010 Solved Homework (Homework for grading is also due today) We are told that the probability of a random variable taking on a value between x and x is P x ( ) = Ne ax 2 Such a function is called a &quot;gaussian&quot; function, incidentally, and we will see this function many, many times during this course. Since we want P to be normalized, we must have 1 = Ne ax 2 dx = N e ax 2 dx = N a in which case we demonstrate that N must be a / to normalize the function (the solution to the integral comes from looking it up in a table). A sketch for a = 0.3, just as an example. P 0.05 0.1 0.15 0.2 0.25 0.3 0.35- 6- 4- 2 2 4 6 P The points of inflection occur where the second derivative is equal to zero. 5-2 2 P x ( ) x 2 = x Ne ax 2 x = x 2 axNe ax 2 = 4 a 2 x 2 Ne ax 2 2 aNe ax 2 setting the r.h.s. equal to zero requires = 2 ax 2 1 ( ) so that x = 1 2 a For my graph above, where a = 0.3, that is x = 1.29. (Note that these points define one standard deviation for a random variable having normal distribution.) Now, to compute the mean value of x , we must solve x = a xe ax 2 dx while one could go through the math to determine the value of this integral, it is simpler to employ a bit of common sense. From the shape of the probability function, the probability of getting any value x is exactly equal to the probability of getting x . In such a case, the average value of x will necessarily be zero. (Later in the course, we will see that parity arguments can also be used to come to this conclusion without explicitly evaluating the integral.) The mean value of x 2 , on the other hand, is computed from x 2 = a x 2 e ax 2 dx Integral tables provide x 2 n e ax 2 dx = 1 3 5 2 n 1 ( ) 2 n a n a 1/ 2 In our case, n = 1, the square roots cancel one another, and we are left with simply &lt; x 2 &gt; = (2 a ) 1 . In probability, &lt; x 2 &gt; is called 2 and we indeed see that the square root is one standard deviation, consistent with our work thus far. 5-2 2 P x ( ) x 2 = x Ne ax 2 x = x 2 axNe ax 2 = 4 a 2 x 2 Ne ax 2 2 aNe ax 2 setting the r.h.s. equal to zero requires = 2 ax 2 1 ( ) so that x = 1 2 a For my graph above, where a = 0.3, that is x = 1.29. (Note that these points define one standard deviation for a random variable having normal distribution.) Now, to compute the mean value of x , we must solve x = a...
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