Phys 3810, Spring 2009
Problem Set #3, Hint-o-licious Hints
1. Griffiths,2.29 Actually, this one isn’t so bad. You just need to go through the finite–well
derivation in the book and make the appropriate changes for the asymmetric state.
The main results you should get to are the results of applying the boundary conditions:
−κ=`cot(`a)
and, using the same definitions of zand z0, the new condition for finding the energies
(graphically, perhaps)
−cot(z) = q(z0/z)2−1
which is not guaranteed to have a root since the right side is positive and the left side starts
off being negative for small z.
2. Griffiths,3.7
3. Griffiths,3.12 Some steps: Start with
hxi=Z∞
−∞
Ψ(x, t)∗xΨ(x, t)dx
and substitute (3.55),
Ψ(x, t) = 1
√2π¯hZ∞
−∞
eipx/¯hΦ(p, t)dp
Watch for complex conjugates and be sure that the dummy variable of pintegration is
different in the two integrals. When you do this you have a big integral over three variables:
x,pand p0.
Do what Griffiths says, use
xeipx/¯h=¯h
i
d
dpeipx/¯h
and then do an integration by parts. Be careful with with variable names to know what the
derivatives are acting on.
Use (and review) the result of problem 2.26,
Z∞
−∞
dx eiqx = 2πδ(q)
Use this for one of the integrations, but as usual be careful with the variables.
The next integration collapses the delta function and gives the desired result.
4. Griffiths,3.23 As discussed in class, from considering the action of this Hon the state
|1iand |2i, one can show that
H= E E
E−E!
With all of the √2’s flying around, the normalization is a little funky.
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