3502_05_sept18_LG

# 3502_05_sept18_LG - Chem 3502/5502 Physical Chemistry...

This preview shows pages 1–4. Sign up to view the full content.

Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 5, September 18, 2009 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 "gaussian" 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 0.05 0.1 0.15 0.2 0.25 0.3 0.35 - 6 - 4 - 2 0 2 4 6 P The points of inflection occur where the second derivative is equal to zero.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 0 = 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 L 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 < x 2 > = (2 a ) 1 . In probability, < x 2 > is called σ 2 and we indeed see that the square root is one standard deviation, consistent with our work thus far.
5-3 Finally, with respect to the question why is < x 2 > not equal to < x > 2 , that is fairly obvious. The former expectation value measures distance from the mean without respect to direction (which is to say, deviation to either left or right is weighted positive by virtue of squaring), while the latter is simply the square of the mean, not the deviation from it. In statistics, the difference between the two is referred to as the “dispersion” of the data. Necessary Mathematical Tools Complex numbers. Wave functions and operators can take on complex values, even though expectation values , i.e., the results of physical measurements, are always real numbers. Every complex number c can be represented as c = a + bi (5-1) where a and b are real numbers and i is the square root of 1 (the “unit” for the so- called imaginary part of complex numbers). We refer to the “complex conjugate” of c , which is written c *, as c * = a ! bi (5-2) The “square modulus” of the complex number c is c * c = c 2 = a + bi ( ) a ! bi ( ) = a 2 + b 2 (5-3)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern