Problems:
- SM&M 5-16
Psi = Sqrt[2/.3] Sin[n Pi x /.3]
Probability = Int[Psi* Psi, {x,0,.1}]
a) = 2/.3 Int[ Sin[Pi x/.3]^2,{x,0,.1}]
= .1955
b) = 2/.3 Int[ Sin[99 x/.3]^2,{x,0,.1}]
= .3333
c) At very high energies, you would expect the particle to be
in each third of the box 1/3 of the time on the average.
Yes, this answer is consistent with that (in fact, too
consistent... n=98 --> .3347 and n = 100 --> .3320)
- SM&M 5-25
Psi = C x Exp[-a x^2]
a) E Psi = -hbar^2/(2 m) d^2 Psi/dx^2 + m w^2 x^2/2 Psi
d^2 Psi/dx^2 = (4 a^2 x^2 -6 a) Psi
--> E = -hbar^2/(2 m) (4 a^2 x^2 -6 a) + m w^2 x^2/2
= 3 a hbar^2/m + x^2 ( -2 hbar^2 a^2/m + m w^2/2)
This can only be true if:
E = 6 a hbar^2/(2 m) and
0 = -2 hbar^2 a^2 /m + m w^2/2
--> a^2 = m^2 w^2 / (4 hbar^2)
a = m w/(2 hbar)
--> E = 6 m w/(2 hbar) hbar^2/(2 m) = 3 hbar w/2
b) 1 = Int[Psi* Psi,{x, -Infinity,Infinity}]
= C^2 Int[x^2 Exp[-2 a x^2],{x,-Infinity,Infinity}]
= 2 C^2 Int[x^2 Exp[-2 a x^2],{x,0,Infinity}]
= 2 C^2 1/(8 a) Sqrt[Pi/(2 a)]
--> C^2 = 4 a Sqrt[2 a/Pi]
--> C = 2 a^(3/4) (2/Pi)^(1/4) with a from above.
- SM&M 5-33
Psi = C Exp[- m w x^2 /(2 hbar)]
Normalization:
C^2 = 1/Int[Exp[- a x^2]^2] = 1/Sqrt[Pi/(2a)]
--> C = (2 a/Pi)^(1/4)
where a = m w/(2 hbar)
<x> = Int[C Exp[ - a x^2] x C Exp[-a x^2],{x,-Infinity,Infinity}]
= 0 because integrand is an odd function of x.
<x^2> = Int[C Exp[-a x^2] x^2 C Exp[-a x^2], {x,-Infinity,Infinity}]
= C^2 Int[x^2 Exp[-2 a x^2],{x,-Infinity,Infinity}]
= 2 C^2 Int[x^2 Exp[-2 a x^2},{x,0,Infinity}] because funciton is even
= 2 C^2 1/(8a) Sqrt[Pi/(2 a)] = 1/(4 a) = hbar/(2 m w)
delta x = Sqrt[<x^2> - <x>^2] = Sqrt[1/(4 a)]
- SM&M 5-34
a) <p_x> = 0 because in a bound symmetric system, the particle
should go right as often as it goes left. Thus the average velocity
and hence momentum, should be zero.
b) <E> = <p^2>/(2m) + m w^2/2 <x^2>
hbar w/2 = <p^2>/(2m) + m w^2 /2 1/(4 m w/(2 hbar))
hbar w/2 = <p^2>/(2m) + hbar w/4
--> <p^2> = hbar w m/2
c) delta p = Sqrt[hbar w m/2]
- SM&M 5-35
delta_x delta_p = Sqrt[hbar/(2 m w)] Sqrt[hbar m w/2]
= Sqrt[hbar^2/4]
= hbar/2