## Saturday, July 10, 2010

### Space Quantization & Further Simple QM Problems

We now want to venture a bit deeper into the quantum mechanics of the hydrogen atom with a view to doing more elaborate problems, ultimately leading to tangling with what we call "L-S Coupling", or combining the orbital angular momentum, L, with spin angular momentum, S, to arrive at the J value.

For now we just want to focus on basic principles of angular momentum for classical physics. In terms of an orbiting planet, for example, we can specify the orbital angular momentum as:

L = m v r

That is, assuming a circular orbit of radius r, and a body of mass m, with velocity r, the orbital angular momentum will be L.

Now, what about at the atomic level? Bohr, in his Bohr model, visualized electrons behaving like tiny planets and orbiting a central nucleus. For the simplest atom, hydrogen, this meant one electron circling a single central proton. Bohr specified the rule for orbital angular momentum - using the principal quantum number, n, as:

mvr = n (h/2 pi)

where again, h is Planck's constant (6.6254 x 10^-34 J-s) and h/2 pi the reduced Planck constant.

However, this model failed to correctly predict the orbital angular momentum for the hydrogen electron, yielding a value of 1 unit, which is wrong. In addition, if L = 0 one would find the electrons oscillating in a straight line through the proton nucleus - which is impossible!
Thus, it had to be modified. The modifications were possible once one removed the "planetary model" (which is classical and deterministic) and turned to the wave model. In this case, the orbital angular momentum assumes certain specific values such that:

L = [l(l + 1)]^1/2 (h/2pi)

(l = 0, 1, 2, ......., n-1)

In the above case, when l = 0 we find L = h/2pi

The fact L can be zero and is acceptable discloses why classical mechanical models fail at the quantum level.

Now, what if instead we have the angular momentum quantum number l = 2?

Then: L = [l(l + 1)]^1/2 (h/2pi) =

[2(2 +1)]^1/2 (h/2pi) = [6]^1/2 (h/2pi)

and a set of allowed projections for L are obtained (see diagram (a) ) as well as orientations. Of particular interest are the allowed values for the vertical component of the orbital angular momentum vector, which we designate: L(z). (L(z) is the projection of L along the z-axis and has discrete values, e.g. 0, h/2pi, h/pi, - h/2pi, - h/pi for the case given).

In general L(z) is specified according to:

L(z) = m_l(h/2pi)

where m_l is the magnetic quantum number. The key point here is that the direction of the orbital angular momentum quantum number L is quantized with respect to an external magnetic field. We call this "space quantization".

Now, recall that m_l can range from -l to l, so for l = 2, we have:

m_l = +2, +1, 0, -1, -2

When we multiply each of these by (h/2pi) we obtain the quantizations of L(z) as depicted in diagram (a) of the figure.

It should also be obvious to anyone who's taken trigonometry that one can obtain the angle between the vertical projection L(z) of the vector L, and L. ( See diagram (b) of figure).

The angle (theta) can indeed be found using the cosine relation (adjacent over the hypoteneuse) which yields (using 't' for theta):

cos (t) = L(z)/ L = m_l/ [l (l+1)]^1/2

Of course, it ought to be self-evident that we are obtaining allowed values for the angle, since obviously m_l is going to range from -2 to +2)

Thus, for the problem we've considered (with l = 2) we have the allowed cosines and angles:

m_l = 0 so cos(t) = 0 so t = 90 deg

m_l = 1 so cos (t) = 1/[6]^1/2 so t = 65.9 deg

m_l = -1 so cos(t) = -1/[6]^1/2 so t (theta) = 114.1 deg

m_l = 2 so cos(t) = 2/[6]^1/2 so t = 35. 2 deg

m_1 = -2 so cos(t) = -2/[6]^1/2 = 144.7 deg

Some simple problems:

1) Consider an electron for which n = 4, l = 3, and m_l = 3. Calculate:

a) the numerical value of L, the orbital angular momentum

b) the z-component of the orbital angular momentum.

For (a) we require: L = [l (l+1)]^1/2 (h/2pi)

then: L = [3(3+ 1)]^1/2 (h/2pi) = [12]^1/2 (h/ 2pi)

In part (b): L(z) = m_l(h/2pi) = 3h/2pi

2) For a hydrogen atom in the l=3 state, calculate the magnitude of the orbital angular momentum, and the allowed values of L(z) and theta (t).

We have L = [l (l+1)]^1/2 (h/2pi) = [12]^1/2 (h/ 2pi)

or L = 2[3]^1/2 (h/2pi)

Given l = 3, we have:

m_l = -3, -2, -1, 0, +1, +2, +3 so that:

L(z) = m_l (h/2pi)

and L(z) = -3h/2pi, -h/pi, - h/2pi, 0, h/2pi, h/pi, 3h/2pi

Finally, cos(t) = m_l/ [l (l+1)]^1/2

so (substituting each allowed value for m_l) =

cos(t) = +/- 0.866, +/- 0.577 +/- 0.289 and 0

so theta has values:

30 deg, 54.8 deg, 73.2 deg, 107 deg, 125 deg and 150 deg

Some more practice problems for next time:

1) An electron is in the n= 2 state of hydrogen. Find the possible values of: L, L(z) and theta.

2) Find the possible values of L, L(z) and theta for an electron in the 3d state of hydrogen.

3) The orbital angular momentum of the Earth in its orbit around the Sun is:

L = 4.83 x 10^31 kg-m^2/s

Assume it is quantized according to: L = [l (l+1)]^1/2 (h/2pi)

and find:

a) the value of l corresponding to the angular momentum, and

b) the fractional change in L as l changes from l to (l +1).