Solutions Homework 6
Physics 486, Fall 2014
University of Illinois at Urbana-Champaign
September 26, 2014
1
Problem 1: Harmonic Oscillator in Real Space
(a)
i t (x,t) = h 1/2
h
ma2 2it max it
e
+
e
(x,t)
2
h
h
1
m2
1
h2
xx (x,t) + m2 x2 (x,t) = h/2
(x a
UIUC Physics 486: Solutions for for HW 10
December 2, 2014
Problem 1: Matrix Elements of Ylm
(a): Using the relations, Lx = 1 (L+ + L ), and Ly =
2
1
2i (L+
L ), we can see that
Ylm | Lx | Yl m =
1
2
l(l + 1) m (m 1)l,l m,m 1 +
1
2
l(l + 1) m (m + 1)l,l
Homework 12
November 14, 2014
1. Grovers Algorithm
In this problem you will go carefully through the steps of Grovers algorithm
that we saw in class generalized to M marked elements. Assume the elements
you are looking for are in the set = cfw_f1 , f2 .fM
Homework 11
November 7, 2014
1. Additional degeneracy in the hydrogen atom
(a)Explicitly verify that
A=
1
( L
p
2m
L p)
e2
r
r
commutes with the coulomb Hamiltonian.
(b) Write A+ =
on 300 .
p
1/ 2(Ax + iAy ). What happens when you operator A+
(c) What
Solution to HW11
December 4, 2014
Problem I: Additional degeneracy in the hydrogen
atom
(a)
Introduce two operators:
1
i
K = K = (L r r L) = L r ir = [L2 , r]
2
2
1
i
M = M = (L p p L) = L p ip = [L2 , p]
2
2
(1)
(2)
Jacobi identity:
[L2 , p], H] + [p, H]
Homework 13
November 20, 2014
1. Particle in a box
For a particle in a box that extends from a to a, try (within the box), =
(x a)(x + a) and calculate E . There is no paramter to vary, but you will still
get an upper bound. Compare it to the true energy
UIUC Physics 486: Solutions for for HW 4
October 12, 2014
Problem 1: Evaluating operators
(a):
X =
dxdydz (x) x | y y y | z (z)
= dxdydz (x)(y x)y(z y)(z)
= dxdy (x)(y x)y(y)
= dx x|(x)|2
2
+
= dx xe2kx
= 0. (Since the integrand is an odd function.
(c):
Homework 14
December 8, 2014
1
1. Time evolve the state |(0) = 3 i|Y00 + ei/4 |Y11 + |Y11 |r
where the radial function r|r = (r r0 ) for time t = under the action of the Hamiltonian H = 2 . What is |() ?
2. Consider a hydrogen atom in the state
1
| = [|30
Solution to HW14
December 8, 2014
Problem 1
2
The energy of state |Ylm (r r0 ) is El = l(l + 1)/r0 . So E0 = 0 and
2 . The state evolves by multiying a factor eiEl t , therefore,
E1 = 2/r0
1
|() = i|Y00 eiE0 t + ei/4 |Y11 + |Y11 eiE1 t |r
(1)
3
1
2
= i|Y0
Solutions Homework 12
Physics 486, Fall 2014
University of Illinois at Urbana-Champaign
November 28, 2014
1
Problem 1: Grovers Algorithm
a) We know that i| j = i j . Then:
| =
O|O =
1
1
1
i| j = N i j = N N = 1
N i, j
i, j
1
1
i| j = N M (N M) = 1
N M i
Solutions Homework 9
Physics 486, Fall 2014
University of Illinois at Urbana-Champaign
October 24, 2014
1
Problem I: Commutators
(a)
[y, px ] f (x, y) = y(ix ) f (x, y) + ix (y f (x, y) = iy fx (x, y) + iy fx (x, y) = 0
(b) Let us write down all the commu
Homework 9
October 23, 2014
1. Commutators
In class, we worked through a number of commutators. In these problems you
are going to explicitly compute the commutators. You may start by assuming
that [, px ] = i . Start by showing that
x
(a) [, px ] = 0 .
Homework 6
October 1, 2014
1. Harmonic Oscillator in Real Space
(a) Show that
(x, t) =
m
1/4
exp
m
2
x2 +
a2
i t
(1 + e2it ) +
2axeit
2
m
satises the time-dependent Schrodinger equation for the harmonic oscillator
potential (Griths Eq. 2.43). Here a is
Homework 4
September 17, 2014
1. Evaluating Operators
Dene the Hermitian operator X as
= x|xihx|dx
X
and the operator P as
P =
and the state | i =
e
kx2
p|pihp|dp
dx|xi
Dene hx|pi = eipx and remember that I =
and hp|pi = (p p0 )
|xihx|dx and that hx|x0 i
Solution to HW5
September 30, 2014
Problem I: Trying to Communicate Faster than the
Speed of Light
(a)
The two states are |0 and |1 , with probability 50% for each state.
(b)
The density matrix is
1
2
0
(1)
1
2
0
(c)
50% and 50%.
(d)
1
U=
2
1 1
1 1
(2)
1
Homework 5
September 25, 2014
Problem 1: Trying to Communicate Faster then the Speed
of Light
In this problem, we are going to look at an explicit example where we will try
to use EPR pairs to communicate faster then the speed of light.
Consider the situa
HW7
October 9, 2014
1. Normalizing states
(a) Assuming cfw_|0 , |1 , |2 are a basis, normalize the state | = 0.3|0 + 0.7|1
(b) Normalize the state
cos(x) if /2 x /2
(x) =
0
otherwise
2. Unitary and Hermitian Operators
0 i
i
(a) Is the operator O = i 1 3
Homework 8
October 17, 2014
1. A Pendulum
A pendulum consists of a massless rod of length l = 50 cm with one end attached to a rigid support so that it is free to pivot about this point and the
other end attached to a mass m = 20g. The pendulum is oscilla
UIUC Physics 486: Solutions for for HW 7
October 27, 2014
1. Normalizing States
(a): The normalized form is
| =
1
(0.3 |0 + 0.7 |1
0.32 + 0.72
(1)
(b):
/2
cos2 (x)dx = /2.
(2)
/2
Hence, the normalized wave function is
2
(x) =
if /2 < x < /2
cos(x)
0
(3)
Solution to HW8
November 3, 2014
Problem I: A Pendulum
(a)
=
g
= 4.43Hz
l
(1)
(b)
1
E = m 2 A2 = 4.91 104 J
2
(2)
(c)
n
E
= 1.06 1030
h
(3)
(d)
= 4.65 1034 J
h
(4)
(e)
x
2A
= 9.43 1032 m
n
1
(5)
(f )
1
1
= m 2 x2
h
c
2
2
h
xc =
m
m 1/4 m x2
h
0 (x) =
e
Homework 10
October 29, 2014
Problem 1: Matrix Elements of Ylm
(a) Compute the matrix elements of Lx and Ly in the basis given by the spherical
harmonics. That is, calculate
Ylm |Lx |Yl m
and
Ylm |Ly |Yl m
(b) Prove that L2 = L2 = 0 is possible only for a
Solutions Homework 3
Physics 486, Fall 2014
University of Illinois at Urbana-Champaign
September 18, 2014
1
Problem I: Semi-innite square well
(a) In the region 0 x L, the potential is V = 0. Thus, we have to solve the following differential equation
( =
Physics 486: Spring 2017
Discussion 3. Solution
Thursday February 9th.
Problem 1, particle in a box
This problem is just a more formal version of the 1-D box problem we solved last week! 1
One modification from the last
weeks discussion is that will simp
V = Vo
E=
V=0
-0.1 nm
0
Vo
2
x
b
1. An electron with mass mc2 = 0.511 106 eV has the illustrated wave
function in the bound state for the well illustrated above. The total
energy of the electron is E = Vo /2. As you can see, the wavelength of
the electron
g ( x)
E
t exp( i k 2 x )
exp(i k 1 x )
V=V o
+ r exp( - i k1 x )
V=0
x
x=0
An electron of mass m with a wave function i = exp (ik1 x it) strikes
the step and function barrier depicted above. The function lies at the
origin with strength g. In the region