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### Prob2_slt

Course: CHEM 562, Fall 2008
School: Wisconsin
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562. Chem Problem Set 2 Fundamentals of Quantum Mechanics Sept. 14. 2007; Due Sept. 21. 2007 1 Problems 1. Probability density Pb. 8.5 in Atkins: The ground state wavefunction of a hydrogen 1 atom is: (r, , ) = a3 er/a0 where a0 = 53pm (the Bohr radius). (a) Calculate 0 the probability that the electron will be found somewhere within a small sphere of radius 1.0pm centered on the nucleus; (b) Do the same when...

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562. Chem Problem Set 2 Fundamentals of Quantum Mechanics Sept. 14. 2007; Due Sept. 21. 2007 1 Problems 1. Probability density Pb. 8.5 in Atkins: The ground state wavefunction of a hydrogen 1 atom is: (r, , ) = a3 er/a0 where a0 = 53pm (the Bohr radius). (a) Calculate 0 the probability that the electron will be found somewhere within a small sphere of radius 1.0pm centered on the nucleus; (b) Do the same when the radius is a0 . 1/2 Ans: For a normalized wavefunction, the probability of nding the system in a region in 3D of volume V is simply given as: |(r, , )|2 d V (1) So for this problem, we have: P = 1 a3 0 R 2 r2 dr 0 0 sind 0 de2r/a0 (2) where R = 1.0pm for (a) and R = a0 for (b). 2. Superposition and expectation value Pb.8.18-19 in Atkins: Evaluate the kinetic energy of the particle with wavefunction: (x) = (cos)eikx + (sin)eikx where is a parameter. What form would the wavefunction have (i.e., the value of ) if it were 90 percent certain that the particle had linear momentum +k ? Ans: Clearly (x) is an eigenfunction of the 1D free particle Hamiltonian, which includes only the kinetic energy operator. If one acts the kinetic energy operator, 2 d2 2 k2 2m dx2 , on (x), the result is simply the value of the kinetic energy ( 2m ) multiplied by (x), 1 2 2 d2 (x) k (x) = 2 2m dx 2m conrming that (x) is an eigenfunction of the kinetic energy operator. 2 (3) Although (x) is an eigenfunction of the Hamiltonian operator, it is a superposition state for the momentum operator (eikx is an eigenfunction for the momentum operator with an eigenvalue of k). For a superposition state, recall that the square of the linear combination coecient (when properly normalized) gives the probability of nding the corresponding eigenvalue in each measurement. Therefore, if it were 90 percent certain that the particle had linear momentum +k , we have (recall cos2 + sin2 = 1), cos2 = 0.90 (4) thus cos = 0.95. 3. Hermitian Operator, Commutator (a) Conrm that the angular momentum opd erator z = ( /i) d where is an angle, is Hermitian; (b) Evaluate the commutator l [, ], a where a = ( + i)/21/2 and a = ( i)/21/2 . a x p x p Ans: (a) For a Hermitian operator, the relation is satised: k l dx = { l k dx} (5) We can substitute in the expression for z on both sides and see if they are equal (recall l that 1/i = i). l.h.s. = i while r.h.s. = {i l = {i k dl d d (6) dk d d} = {i (k l | k l d)} d d d dl k l d} = i k d d d (7) (8) where for r.h.s. we rst did integration by part (also see Pg. 265 in the text book, Justication 8.2), and then invoked the cyclic boundary condition that k l | must be zero). Clearly, l.h.s. = r.h.s., so z is a Hermitian operator. l (b). To evaluate a commutator, you act the commutator on a function, i.e., [, a ]f (x) = aa f (x) a af (x) a For our particular case, we have ( + i)/21/2 ( i)/21/2 f (x) ( i)/21/2 ( + i)/21/2 f (x) x p x p x p x p 2 (10) (9) Substituting the fact that p = i d/dx, we have: 1 1 (x + d/dx)(x d/dx)f (x) (x d/dx)(x + d/dx)f (x) 2 2 1 1 = (x + d/dx)(xf (x) f (x)) (x d/dx)(xf (x) + f (x)) 2 2 1 = (x2 f (x) xf (x) + (f (x) + xf (x)) 2 f (x)) 2 1 2 (x f (x) + xf (x) (f (x) + xf (x)) 2 f (x)) 2 = f (x) Thus [, a ] = . a 4. Particle in a box Calculate the expectation value of p, x, p2 , x2 for the n=3 state for the particle ...

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