Lec6 - Chem 3502/5502 Physical Chemistry II (Quantum...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 6, 20 September 2010 Solved Homework We are given that A Ψ = a Ψ and A * Ψ = a Ψ where a is a real number. As both A Ψ and A * Ψ are equal to the same thing, they must be equal to one another, i.e., A Ψ = A * Ψ . If we multiply each on the left and integrate, we have Ψ * A Ψ ( ) dr −∞ = Ψ * A * Ψ ( ) dr −∞ which is the first equality that we are tasked to prove. If we take the complex conjugate of both sides we have Ψ * A Ψ ( ) dr −∞ [ ] * = Ψ * A * Ψ ( ) 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 Ψ * A * Ψ ( ) dr −∞ = Ψ * a Ψ ( ) dr −∞ = a Ψ * Ψ dr −∞ = a Ψ 2 and Ψ * A Ψ ( ) dr −∞ [ ] * = Ψ * a Ψ ( ) dr −∞ [ ] * = a Ψ * Ψ dr −∞ [ ] * = a Ψ 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 Ψ 2 [ ] * = a Ψ 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 Ψ x , y , z , t ( ) = c n ψ n n = 1 x , y , z ( ) e iE n t / (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 Ψ * x , y , z , t ( ) Ψ x , y , z , t ( ) = c m * ψ m * m = 1 x , y , z ( ) e iE m t / c n ψ n n = 1 x , y , z ( ) e iE n t / = c m * c n ψ m * x , y , z ( ) ψ n x , y , z ( ) e iE m t / e iE n t / m , n = 1 = c i 2 ψ i * x , y , z ( ) ψ i x , y , z ( ) i = 1 + c m * c n ψ m * x , y , z ( ) ψ n x , y , z ( ) e i E m E n ( ) t / 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 Ψ x , y , z , t ( ) = c j ψ j x , y , z ( ) e iE j t / (6-3) Since both the time-dependent and time-independent wave functions are normalized over all space, note that Ψ x , y , z , t ( ) Ψ x , y , z , t ( ) = j ψ j x , y , z ( ) e iE j t / c j ψ j x , y , z ( ) e iE j t / = c j * e iE j t / c j e iE j t / ψ j x , y , z ( ) ψ 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 Ψ x , y , z , t ( ) A Ψ x , y , z , t ( ) = ψ j x , y , z ( ) e iE j t / A ψ j x , y , z ( ) e iE j t / = e iE j t / e iE j t / ψ j x , y , z ( ) A ψ j x , y , z ( ) = ψ j x , y , z ( ) A ψ j x , y , z ( ) (6-5) That is, for a stationary state we can work exclusively with the time- independent spatial wave functions.
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Lec6 - Chem 3502/5502 Physical Chemistry II (Quantum...

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