Linear potential February 8, 2008
Linear Potential  Gravity
The Schrodinger equation for the wave function of a bouncing ball is h2 d2 + mgx = E  2m dx2 (1)
where we assume a perfectly elastic collision of the ball with the floor. So V (x) = f
Physics 443 February 2008
Kaons
1 Quarks
According to the Standard Model there are 3 generations of quarks, with two quarks in each generation. Quarks have electric charge, color charge and weak charge. The electric charge couples the quarks by ele
Physics 4443 Problem 1
Solutions #9
Spring 2011
The firstorder correction to the ground state energy is E = 1 0 0H 1 0 0 (1)
where 1 0 0 is the ground state ket. To find H , we need to first find the potential inside and outside a uniform spherical ch
Physics 4443 Problem 1
Solutions #6
Spring 2011
In classical mechanics, the motion of a particle with energy E in a potential V (r) is constrained by the equation E = T + V . The maximum distance attainable from the center of an attractive potential is a
Physics 4443 Problem 1
Solutions #5
Spring 2011
Since the Hamiltonian is a sum of two independent terms in x and y, H = Hx + Hy , Hx = p2 x 1 + 2 m 2 x2 , 2m Hy = p2 y + 1 m 2 y 2 2 2m
the solution can be separated as a product of wavefunctions: (x, y) =
Physics 4443 Problem 1
Solutions #4
Spring 2011
The initial state is (x, t = 0) = (x, t = 0 ), the ground state of the harmonic oscillator. After t = 0, the state is a free particle, with general solution a superposition of plane waves (eqn. 2.100 in Gri
Physics 4443 Problem 1
Solutions #3
Spring 2011
cfw_a From the solution to Problem 2.5(b) (second problem in HW 1): " " , x 1 2x (x, t) = eit sin + sin e3it , a a a and so as in the solution to part (c) of that problem: Z a x2 = x2 (x, t)2 dx 0 , , Z
Physics 4443 Problem 1.4
(a) 1= A2 a2
a
Solutions #1
Spring 2011
x2 dx +
0
(b  a)
A2
b 2 a
(b  x)2 dx = A2 3 . b
1 a2
x3 3
a
+
0
1 (b  a)2

(b  x)3 3
b a
= A2 (b)
b a ba = A2 A = + 3 3 3
A
a
b
x
(c) At x = a. (d)
a
P =
0
2 dx =
A2 a2
a
Physics 4443 Problem 1
a
Solutions #13
Spring 2011
p(x) dx = n ,
0
with n = 1, 2, 3, . . . and
p(x) =
2m[E  V (x)] (Eq. 8.16).
a
Here
0
p(x) dx =
2mE
a + 2
2m(E  V0 ) 4 2m n a
2
a a = 2m 2 2
E+
E  V0 = n
E + E  V0 + 2 E(E  V0 ) =
0 = 4En ;
0 2 E(E
P443 WKB I D.Rubin February 18, 2008
Connection Formulae
The WKB approximation falls apart near a turning point. Then E  V 0 1 so p . And because the momentum goes to zero the wavelength gets very long and the approximation is only valid if the w