ANGULAR 4.1. MOMENTUM
4.1.3.1
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Solution anguc-a
Question: The general wave function of a state with azimuthal quantum number l and
magnetic quantum number m is = R(r)Ylm (, ), where R(r) is some further arbitrary
function of r. Show that the condition for this wave function to be normalized, so that the
total probability of nding the particle integrated over all possible positions is one, is that
r =0
R(r) R(r)r2 dr = 1.
Answer: You need to have | = d3 r = 1 for the wave function to be normalized.
Now the volume element d3 r is in spherical coordinates given by r2 sin drdd, so you must
have
2
R(r) Ylm (, ) R(r)Ylm (, )r2 sin drdd = 1.
r =0
=0
=0
Taking this apart into two separate integrals:
r =0
R(r) R(r)r2 dr
2
=0
=0
Ylm (, ) Ylm (, ) sin dd = 1.
The second integral is one on account of the normalization of the spherical harmonics, so you
must have
R(r) R(r)r2 dr = 1.
r anguc-b
Question: =0
4.1.3.2
Solution Can you invert the statement about zero angular momentum and say: if a particle
can be found at all angular positions compared to the origin with equal probability, it will
have zero angular momentum?
Answer: No. To be at zero angular momentum, not just the probability ||2 , but itself
must be independent of the spherical coordinate angles and . As an arbitrary example,
= R(r)ei sin would have a probability of nding the particle independent of and , but
not zero angular momentum.
4.1.3.3
Solution anguc-c
Question: What is the minimum amount that the total square angular momentum is larger
than just the square angular momentum in the z -direction for a given value of l?
Answer: The total square angular momentum is l(l + 1) 2 and the square angular z h
22
momentum is m h . Since for a given value of l, the largest that |m| can be is l, the dierence