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Ph 12b
Homework Assignment No. 5 Due: 5pm, Thursday, 18 February 2010 1. Minimal uncertainty I: particle in one dimension (10 points). If we measure the Hermitian operator A in the state vector | , the variance of the measurement outcomes is (A)2 = | A
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Ph 12b
Homework Assignment No. 4 Due: 5pm, Thursday, 4 February 2010 1. Weaker decoherence. In class we discussed the phase damping of a qubit that results when the qubit scatters a photon with probability p. The scattered photon is knocked into one of
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Ph 12b
Homework Assignment No. 3 Due: 5pm, Thursday, 28 January 2010 1. A watched quantum state never moves. Consider a simple model of an atom with two energy levels the ground state |g has energy Eg and the excited state |e has energy Ee > Eg , where
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Ph 12b
Homework Assignment No. 2 Due: 5pm, Thursday, 21 January 2010 1. Quantized rotor. A wheel spinning in a plane can be described as a Hamiltonian dynamical system with one degree of freedom: the coordinate is the angular orientation taking values i
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Ph 12b
Homework Assignment No. 1 Due: 5pm, Thursday, 14 January 2010 1. Uncertainty principle and the quantum harmonic oscillator. For any quantum state of a single particle moving in one dimension, there is a corresponding probability density P (x) tha
Ph 12b Midterm Exam Solutions
J. Preskill 10 February 2010 1. Two-state quantum dynamics 35 total points
Let |e1 and |e2 Hilbert space H: has Hamiltonian denote two normalized and mutually orthogonal states in a e1 |e1 = e2 |e2 = 1, e1 |e2 = 0. A certain
Ph 12b Midterm Exam Due: Wednesday, 10 February 2010, 5pm
This exam is to be taken in one continuous time interval not to exceed 3 hours, beginning when you rst open the exam. You may consult the textbook Introductory Quantum Mechanics by Libo, the textb
Ph12a Solution Set 9
December 1, 2009
9.4 The geometry of this problem is illustrated in Fig. 1. The light entering the two slits will be reasonably coherent when the path dierence to the two slits from an edge of the source is small compared to the wavel
Ph12a Solution Set 8
November 24, 2009
6.14 The dispersion relation for a system of coupled pendula is given by eq. 90 of section 2.4: 2 (k ) = The lower cuto frequency is cuto frequency is given by p.86 of Crawford).
g l g l
K ka g + 4 sin2 l M 2 corresp
Ph12a Solution Set 6
November 10, 2009
4.7 Use Gausss law to get the electric eld for r1 < r < r2 . Make a Gaussian pillbox of cylindrical shape, parallel to the axis of the transmission line: E da = |E |2ra = Q r. 2o ra Q ln(r2 /r1 ). 2o a Q , o
where a
Ph12a Solution Set 4
October 28, 2009
3.19 The absorptive and elastic amplitudes are given by Aab = F0 2 M 2 2 + (0 2 )2 2 F0 0 2 Ael = 2 M 2 2 + (0 2 )2 (1) (2)
The weak damping approximation is /2 < 0 . This implies that 0 , 2 so we have (0 2 )2 = ( + 0
Ph12a Solution Set 3
October 21, 2009
2.30 (a) Choosing the origin of time at the center of t, the sine terms all vanish (see eqn (52) on p. 66, with F (t) an even function). (b)
T1 / 2 t f (t)dt = 1 T T1 /2 T1 / 2 f (t) cos 2nt T1 T1 /2 t/2 2 cos 2nt T1
Ph12a Solution Set 2
October 16, 2009
2.12 In a given mode with wavenumber k , the displacement is given as = cos(t + )(A sin kz + B cos kz ), where = cs k and cs = T0 /0 . Since the ends of the string are free, it must have horizontal tangents at both en
Ph12a Solution Set 1
October 4, 2009
1.15 The equations of motion for two coupled oscillators can be put in the form d2 a = a11 a a12 b (1) dt2 d2 b = a21 a a22 b (2) dt2 Suppose we have two solutions to these equations, one for initial positions xa1 and
Solutions for 2009 Midterm Exam Phys. 12a
H. J. Kimble
(Dated: 3 November 2009)
PROBLEM 1: EIGENFREQUENCIES
The general solution to the one dimensional wave equation is (Textbook Chapter 2 Eq.(23) (z, t) = cos(t + )[A sin(kz ) + B cos(kz )]. (1)
We need
Phys. 12a 2009 - Solutions for Final Exam
H. J. Kimble 1. (15 points ) Simple oscillators Write an equation of motion and give the frequency for free oscillation of each of the systems shown in Figure 1 below.
a. 5 points g
b. 5 points
g fluid of density
PH12b 2010 Solutions HW#1
1.
a) hi = 0 = hi imply that ()2 = 2 2 + h i = 2 and ()2 = 2 Using this we get () 1 1 2 2 = + 2 ()2 2 2 2 1 + 2 ()2 2 8 ()
2 2
Now, from the uncertainty relation we know that 2, therefore h i b) 2
solving for we get
2 2 h i = +
Ph002a, Fall 2009
Quiz 4 Due: 24 November 2009
Preliminary note: You should be able to do this problem by hand, but you should feel free to use a calculator, Mathematica, or any other symbolicmanipulation program if you nd it easier to evaluate integrals
Quiz 3: Solutions
by Jonathan Arnold November 17, 2009
Problem 1
A phased-array of radio antennas can be electronically pointed by introducing regular phase delays between successive antennas. Assume that the system operates at frequency and wavelength .
PROBLEM 1
A "phased-array" of radio antennae can be electronically "pointed" by introducing regular phase delays between successive antennae. Assume that the system operates at frequency and wavelength . The antenna are "line" shaped, much longer than a s
Quiz 2 Solutions
Kevin Setter November 6, 2009
Note: The correct solutions to parts 3 and 4 involve a subtlety that a Ph2a student is not expected to be able to get. Therefore, full credit was awarded to the naive solution to 3 and 4 described below. The
P h002a, F all 2 009
Quiz 2 Due: 2 7 O ctober 2 009
A traveling wave is sent from position r : -@ and time f, : -oo along a string, as shown:
Yt
d'+
c:
y=o
Before the wave hits the right end, the string displacement is given by
u l :x .tl -
, .\
( + *,
<
Quiz 1 Solutions
Nate Bode October 31, 2009
Note: The rst thing we should do is take a look at the Hookes Law equation that the torsional springs obey. The rst thing we should note is that [i ] = Force Length , so we know we are dealing with a situation u
Quiz Problem 5 Solution Problem 3 Solution
a) The condition that the wavefunction is normalized is:
1=
0
| (x)|2 dx
=
0
|A|2
x xo
2b
e2x/xo dx
=
|A|2 xo 2b
x2b e2x/xo dx
0
Using the hint (with a = 2/xo ), this is equal to: = |A|2 (2b)! 2b (2/x )2b+1 xo o