3502_06_sept21_LG

3502_06_sept21_LG - Chem 3502/4502 Physical Chemistry II...

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Chem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 6, September 21, 2006 Solved Homework We are given that A " = a " and A * " = a " where a is a real number. As both A " and A * " are equal to the same thing, they must be equal to one another, i.e., A " = A * " . If we multiply each on the left and integrate, we have " * A " ( ) dr #$ $ % = " * A * " ( ) dr #$ $ % which is the first equality that we are tasked to prove. If we take the complex conjugate of both sides we have " * A " ( ) dr #$ $ % [ ] * = " * A * " ( ) dr #$ $ % [ ] * which is the last equality that we are asked to prove. To prove the central equality, let us simply evaluate the relevant integrals, thus " * A * " ( ) dr #$ $ % = " * a " ( ) dr #$ $ % = a " * " dr #$ $ % = a " 2 and " * A " ( ) dr #$ $ % [ ] * = " * a " ( ) dr #$ $ % [ ] * = a " * " dr #$ $ % [ ] * = a " 2 [ ] * but a is a real number, and the square modulus of a wave function is also a real number, so it must be true that a " 2 [ ] * = a " 2 Q.E.D.
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6-2 Stationary States Recall that the general solution to the time-dependent Schrödinger equation is the superposition-of-states wave packet defined by ! x , y , z , t ( ) = c n " n n = 1 # $ x , y , z ( ) e % iE n t / h (6-1) Let us consider what is required for the probability density at any position in space not to be varying with time. That is, the system is “stationary” or in a “stationary state”. The probability density at a particular position is ! * x , y , z , t ( ) ! x , y , z , t ( ) = c m * " m * m = 1 # $ x , y , z ( ) e iE m t / h c n " n n = 1 # $ x , y , z ( ) e % iE n t / h = c m * c n " m * x , y , z ( ) " n x , y , z ( ) e iE m t / h e % iE n t / h m , n = 1 # $ = c i 2 " i * x , y , z ( ) " i x , y , z ( ) i = 1 # $ + c m * c n " m * x , y , z ( ) " n x , y , z ( ) e i E m % E n ( ) t / h m & n # $ (6-2) Notice that we are not integrating over all space here, we are only asking about a particular position, so the various terms in the second sum of the bottom equality are not necessarily zero. It is these terms that include a time dependence, so for a system to have a stationary probability density, we must find a way to make every term in the second sum equal to zero. There is only one way to do this (unless all states are degenerate): every value of { c } must be zero except for a single one. That is ! x , y , z , t ( ) = c j " j x , y , z ( ) e # iE j t / h (6-3) Since both the time-dependent and time-independent wave functions are normalized over all space, note that ! x , y , z , t ( ) ! x , y , z , t ( ) = j " j x , y , z ( ) e # iE j t / h c j " j x , y , z ( ) e # iE j t / h = c j * e iE j t / h c j e # iE j t / h " j x , y , z ( ) " j x , y , z ( ) = c j 2 (6-4) Thus, the square modulus of c j must be one (since Ψ is normalized).
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6-3 Now, for any operator that does not depend on time, notice that for a stationary state we have ! x , y , z , t ( ) A ! x , y , z , t ( ) = " j x , y , z ( ) e # iE j t / h A " j x , y , z ( ) e # iE j t / h = e iE j t / h e # iE j t / h j x , y , z ( ) A " j x , y , z ( ) = " j x , y , z ( ) A " j x , y , z ( ) (6-5) That is, for a stationary state we can work exclusively with the time- independent spatial wave functions. Operators In lecture 3 we discussed a few operators, but let us make a list here of many which will prove useful as we go on. The table on the next page provides names, symbols, and operations for the most common operators in both 1 and 3 dimensions. Note that momentum in a single dimension is still a vector quantity, but since there is only one dimension it is often denoted as the scalar “component” times the vector basis for the dimension (e.g., the i vector of Cartesian space). Thus the parenthetical notations for this
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3502_06_sept21_LG - Chem 3502/4502 Physical Chemistry II...

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