Download Problem Set 1 for Quantum Mechanics I at Massachusetts Institute of Technology, Fall 2002 and more Exercises Quantum Mechanics in PDF only on Docsity!
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
5.73 Quantum Mechanics I
Fall, 2002
Professor Robert W. Field
Problem Set #
DUE : At the start of Lecture on Friday, September 13.
Reading : C-TDL, pages 9-39, 50-56, 60-
Problems :
- This question is based on Figure 1 of Complement JI CTDL p84. This is a problem that involves many repeated calculations; thus it is best done using a computer. Since computers like pure numbers, it is convenient to express the problem in dimensionless quantities and then, if necessary, express the final result in the appropriate units. One very effective and widely used way to do this is to use “atomic units”:
mass = me mass of electron charge = e charge of electron length(Boh= 1a 0 r radius) action = h /2π = h (energy times time or length times momentum) energy = E (^) h (1 Hartree)
Every dimensioned quantity, Q, in an equation is converted to a pure number by dividing it by the appropriate atomic unit quantity (expressed in the same units as the quantity of interest), denoted by:
Q^ √ =Q /Q (atomic unit) m = m /m^ √ e q = q /e^ √ x = x/a√ 0 √h =h h = 1 E√ =E /Eh.
The 1-dimensional Schrödinger equation becomes (^1) d^2 1
−^ ( )^ −^ E√ψ( )^ =^0
2m (^) dx√^2
Eh V x√^ x√ Let m = 1 (i.e., an electron) V√^ = 1 0. × 10 −^3 (about kT at 300K).
Chemistry 5.73 Page 2 Problem Set #
A. Solve the problem for a Gaussian wavepacket of the generic form
( ; , ∆x) = π)−1 2
x
e−(x^ x^0 )^ [^2 (∆x)^
2 ]
G x x 0 (2 /^ − incident from the left on a step potential of step-height V√ 0. Choose the average energy of the wavepacket to be E√^ = 1 05V. √ 0 and the variance of the Gaussian wavepacket to be
( ∆x√)^2 = 1
4 (∆k√)
2 where 1 2
∆ = [0 1V√ 0 ]
/ k√^.. Convert this simple Gaussian in x√^ to a Gaussian wavepacket in k√^ which is localized at x√ 0 = 10 ∆x√ 0 to the left of the barrier. To get a sense of how this Gaussian wavepacket
scatters off of the potential step you would ideally make a movie of Ψ( x t^ √, √)^
2 vs. x√ sampled at a series of values of √t.
Ψ(^ x t√, √)^ = ∫ dkg k e√^ (^ √)^ iα^ (^ k√)^ eik√x√e−iE k t√ √ √(^ )
( ) α(^ √)
where g k√^ ei^ k^ is the k√^ - dependent complex amplitude of your Gaussian wavepacket
localized at x√ 0 = x√(t√= 0 )^. To make your task easier, sample this complex amplitude at
the 5 values of k : k√^ √ 0 , k√ 0 ± ∆k√, and k√ 0 ± 2 ∆k√^ and approximate Ψ(x t^ √, √)^ by a sum over the
five wavefunctions, Ψk√(x t^ √, √), rather than the integral. To make it even easier, sample
at 5 values of t√. These 5 values of t√^ should be t√= 0 , the time when the center of the wavepacket is at the barrier t√barrier , 0 9t. √barrier , 1 1t.√barrier , and 2t√barrier.
Now use your “movie” results to answer the following questions about the wavepacket incident on a step.
B. What is happening in part C of Figure 1? What are the fringes at x < 0? Why are there no fringes at x > 0? What change in E and/or m would cause the fringe spacing to double? What controls the modulation depth (define this any way you think reasonable) of the fringes?
C. What is happening in part D of the figure? What controls the relative areas of the two peaks? Are there any resonances in the ratio of the areas of the transmitted and reflected peaks? If there are no resonances, why not? If there is a resonance, what feature of the potential is responsible for it?
D. The left and right peaks in part D have different widths and are centered at different distances from x = 0. Why?
