Thursday, August 28, 2014

An Introduction to Quantum Mechanics (2)

2.  Electron Spin and the Stern-Gerlach Experiment

So far, we've seen three different quantum numbers: n (principal), (angular momentum), and m (magnetic). We now examine how the fourth quantum number (ms) or electron spin came about.

By the early 1920s, quantum mechanics had developed to the point that theorists realized an electron magnetic moment ought to exist. In any external magnetic field B, the magnetic moment ought to experience a force similar to what a magnetic compass needle experiences in Earth's magnetic field. According to quantum mechanics, the assumed values should be quantized, so the magnetic moment can assume only certain values, given by:

+ 1

Thus, the orientation number will depend on the second or azimuthal quantum number,
. The total angular momentum (L) would therefore be:

L =
ħ  [ ( +1)]1/2

    In 1922, the German physicists Otto Stern and Walther Gerlach passed a beam of electrons through a non-uniform magnetic field B as shown in the top sketch of the diagram (C denotes collimator and D, detector plate). A non-uniform field meant the field was stronger on one side of the beam than the other. As predicted from theory, the force on the magnetic moment of the electrons is such that the field ought to deflect the beam according to the orientation of its moment.

   The field should therefore split the beam into 2
+ 1 parts (according to 2 +1 orientations). Stern and Gerlach found their beam of hot silver atoms split into two parts. At first this appeared surprising and at odds with theory but later work showed the conflict could be resolved if the electrons going into the opposing (up and down) beams, had their own spins, or intrinsic angular momenta.

No automatic alt text available.

Fig. 7. Illustrating the Stern-Gerlach Experiment.

In this case, the upper beam would feature electrons with "spin up" angular momentum of (+1/2) while the lower beam would feature electrons with "spin down" angular momentum of (-1/2) . These are shown at the lower section of the diagram. Note that in each case, spin up or down, the orientation accords with the direction of rotation: clockwise or anti-clockwise.

The refining experiment was actually done by Phipps and Taylor using a beam of heated H-atoms. They began by defining the magnetic dipole moment such that:

m z = - g() u B  m

where, as before, m
= -, - +1, 0, ....+

B is the Bohr Magneton: u B = eh/4p m

where m is the mass of the H-electron.

Meanwhile, g(
) is known as the "orbital g-factor".

From their low oven temperature, Phipps-Taylor recognized that
= 0 for the electrons coming off and entering the magnetic field. Therefore, m = 0. Since this was so, then:

m z  = 0

Phipps -Taylor therefore assumed the beam would be unaffected or not split at all. Yet, it split into two symmetric components, e.g. deflected symmetrically, as shown in the graphic below. Given the earlier Stern-Gerlach experiment, plus their own, they therefore had to expect the electron had its own magnetic dipole moment..

They assumed this to be a spin magnetic moment 
m s, due to the electron having an intrinsic angular momentum S  analogous to L (angular momentum) so that:

S = [s(s + 1)]1/2  
ħ  and S z = m s (ħ)

Where S z   is the z-component of spin angular momentum

Then m
s is the electron spin or spin quantum number, 1/2 or - 1/2,

Then the electron magnetic dipole moment would be:

m s = - g(s) u B  / ħ [S]

To nail down the basis quantitatively, Phipps-Taylor knew that the net force felt by the electrons traversing the field would be:

F z = - (dB z /dz) u B  g(s) m

where m
s is the putative spin quantum number.

Since they knew the Bohr magneton u B = 9.27 x 10-24 J/T

and dB/dz could be measured (e.g. the difference in the B-field over the collimation width dz)

then :

g(s) m
s =  F z / - (dB z /dz) u B = + 1

as measured.

Since g(s) = 2, then by deduction:

s = +  1/2

Thus, Phipps and Taylor discovered that the splitting effectively showed two possible values for the orientational potential energy, or:

 D(E) = - m s B = +g(s) u B  B/2.

As an example, let's compute S for the value m s = 1/2.

S = [s(s+ 1)]1/2
ħ  = [1/2(1/2 + 1) 1/2 ħ  =

[1/2 (3/2)] 1/2
ħ  = [3/4] 1/2 ħ  =  (Ö3/2) ħ  

This result can be shown in diagrammatic form on a spin diagram, e.g.

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Fig. 8. Illustrating spin quantization

Note in the diagram how the magnitudes for  S z   are first laid out on the vertical axis with lines drawn to a semi-circle, then the arrows are drawn to intersect with the  S z values  of + ½.

Example Problem: (See the Solution for Problem (2) from Previous set posted yesterday)

3. Space quantization and L-S Coupling:

 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 (

where again, h is Planck's constant (6.6254 x 10 -34 J-s) and
ħ 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 = [
( + 1)]1/2 (ħ)

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

In the above case, when
= 0 we find L = h/2p  = ħ  = 1. 054 x 10 -34 J-s

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
= 2?

Then: L = [
( + 1)] 1/2 (ħ) =  [2(2 +1)] 1/2 (ħ) =  Ö6 (ħ)

and a set of allowed projections for L are obtained as shown in the  diagram (a) as well as orientations.

Fig. 9. Space quantization and the quantum number, L

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/2p  , h/p  , - h/2p  , - h/p  for the case given).

In general L z is specified according to:

L(z) = m

where m
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".

Recall that m
 can range from - to , so for = 2, we have:

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

When we multiply each of these by (
ħ) we obtain the quantization of L z as depicted in diagram (a) of Fig. 9.

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 9).

The angle (
q) can indeed be found using the cosine relation (adjacent over the hypoteneuse) which yields:

cos (
q) = L z  / L = m  / [ ( +1)]1/2

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

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

= 0  so cos(q) = 0 so t = 90 deg

= 1  so cos (q) = 1/ Ö6 so q = 65.9 deg

= -1  so cos(q) = -1/ Ö6  so  q = 114.1 deg

= 2 so cos(q) = 2/Ö6  so q  = 35. 2 deg

= -2 so cos(q) = -2/Ö6 so  q= 144.7 deg

Question: Say an electron in an atom (e.g. hydrogen) has zero orbital angular momentum (
= 0) does that mean it has zero total angular momentum?

NO, because in quantum mechanics we find that for every electron in a given atom we have to process two kinds of angular momentum: orbital (L) and spin (S). Even if the electron experiences no precession or torque it must still exhibit a total angular momentum.

Earlier, we saw for the total orbital angular momentum:

L = [
( + 1)] 1/2  ħ  

We also saw that each electron has a spin angular momentum:

S = [s(s + 1)] 1/2  

for which s assumes one or other of the electron spin quantum numbers, m
s = +1/2 or -1/2.

It can be shown  that S is always:        [3/4] 1/2  

Now, we reckon in what we call the total angular momentum or J, such that:

J = [j(j + 1)] 1/2  

and j =
+ s (note the common letters apply to different quantities than the capital ones!)

Thus, for an electron with zero orbital angular momentum (l=0) we have:

j =
+ s = 0 + 1/2 or + s = 0 - 1/2

so: j = +1/2 or -1/2

Then we have:

J = [j(j + 1)] 1/2  
ħ  = [3/4] 1/2  ħ  

for either j (which readers can also verify)

We can also find (as we did with L for Lz, the projection of the total angular momentum quantum number on the z-axis (Jz  ) which will be:

J(z) = m
J ħ

where m
J = -j, -j+1, ......+j

Readers with an intuitive grasp of vectors, or if they've worked with vectors - say in high school or college physics, will quickly see that the name of the game is to obtain a vector sum such that:

V = V(1) + V(2)

In this case, L plays the role of V(1), and S plays the role of V(2)

Then we obtain for J:

J = L + S

As any physics student knows, the way to obtain the vector sum is via the law of cosines and this is demonstrated in the diagram below along with the computations. This is for the case:

L = 3

S = 1/2

J = 5/2

and the angle (
q) can also be obtained as shown in the diagram below:

Note that in a weak magnetic in which the atom is situated, the L-S coupled system depends on j, in other words this very angle between the vectors L and S. Meanwhile, the orientation of the atom on the whole, depends on m

With detailed descriptions using the quantum numbers and proper solutions of the Schrodinger equation, we can assemble a table of the respective numbers and the corresponding wave state solutions, e.g.


Another useful activity is to draw energy level diagrams using the given electron configurations for a particular atom. For example, consider the schematic energy diagram for the 2p 3s configuration for the carbon atom (12 C 6) with each level labeled  with spectroscopic notation:

No automatic alt text available.

In the 2p3s configuration, we know  =1 corresponds to p: so n=2, =1, m = +1, m s = + ½.   For the electrons in the 3s shell: n=3, =0, m = 0 and m s = ½ (and (- ½), for second).]

Each level of the diagram can then be sketched in, i.e. for the higher s-level s= s1 + s2 = 1/2 + 1/2 = while for the lower: s= s1 - s2 = 1/2 - 1/2 = 0. The j’s are found by taking the appropriate ℓ + s combinations, thus ℓ1 + s1 = 1 + 1 = 2, ℓ1 - s1 = 1 – 0 = 1 and so on.


1) State which values of
, s, and j would apply to the case in the diagram, such that the assorted angular momentum vectors have the values identified.

2) Enumerate the possible values of j and m J for the states in which = 1 and s = 1/2 and draw the associated vector diagrams.

3) Consider an electron for which n = 4, = 3, and m = 3. Calculate:

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

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

4) A two-electron atom for which the orbital angular momentum quantum numbers are ℓ1 =3 and ℓ2 = 2 can have what values for the total orbital angular momentum number L?  Determine the possible values of the total angular momentum quantum number of single f  electron.

5) For 4s 3d  configuration we have:

1 = 2, s1 = 1/2, 1 = 0 and s1 = 1/2.

Using the assorted combinations, for '= 0 and ' = 2, to get the respective j' values (in combination with s' = 0), and then further for s' = 1, sketch the energy configuration diagram

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