A.20 Angular momentum of vector particles

This addendum is concerned with vector particles, particles whose wave
functions are vectors. To be sure, the wave function of an electron
can also be written as a vector, chapters 3.1 and
5.5.1:

But that is not a normal vector. It is a two-dimensional vector in three-dimensional space, and is known as a spinor. This addendum is concerned with wave functions that are normal three-dimensional vectors. That is of importance for understanding, for example, the spin angular momentum of photons. A photon is a vector particle, though a special one. It will be shown in this addendum that the spin of a vector particle is 1. The parity of such a particle will also be discussed.

To really appreciate this addendum, you may want to read the previous addendum {A.19} first. In any case, according to that addendum angular momentum is related to what happens to the wave function under rotation of the coordinate system. In particular, the angular momentum in the -direction is related to what happens if the coordinate system is rotated around the -axis.

Consider first the simplest possible vector wave function:

Here is a constant vector. Also is the distance from the origin around which the angular momentum is defined. Finally, is any arbitrary function of . The big question is now what happens to this wave function if the coordinate system is rotated over some angle around the -axis.

The factor does not change under rotations because the distance
from the origin does not. But the vector does
change. Figure A.8 shows what happens. The vector in the
rotated coordinate system has components

Now consider three very special vectors:

The vector has only a -component, so it does not change under the rotation. Phrased differently, it changes by a unit factor . That makes its magnetic quantum number zero, as the superscript in says. Then the angular momentum in the -direction of this state is zero too. Finally, the vector changes by a factor under rotation, so it has 1, as its superscript says, and the angular momentum in the -direction is .

To get at square angular momentum, first the operator of spin
angular momentum around the -axis is needed. The relation between
angular momentum and rotations shows that this operator takes the
general form, {A.19} (A.76),

Here is the operator that describes the effect of the rotation of the coordinate system on the wave function. Applied on the vector in the unrotated coordinate system, that means

Plugging in the components of as given earlier, (A.82), and taking the limits using l’Hôpital, that produces

So the operator drops the -component and swaps the other two components, changing the sign of the first, and then adds a factor . If the same operations are performed another time, the net result is:

So the square operator just drops the -component and adds a factor .

Of course, the operators and are defined
similarly. There is nothing special about the -axis. The operator
of square spin angular momentum is defined as

Since each of the operators in the right hand side drops a different component and adds a factor to the other two, the total for any vector is,

So the square spin operator always produces a simple multiple of the original vector. That makes any vector an eigenvector of square spin angular momentum. Also, the azimuthal quantum number , the spin, can by definition be found from equating the coefficient of in the right hand side above to . The only nonnegative value that can satisfy this condition is 1.

That then means that the spin of vector particles is equal to 1. So a vector particle is a boson of spin 1. The subscript on the special vectors indicates their spin 1.

You can write the most general vector wave function in the form

Then you can put the coefficients in a vector much like the wave function of the electron, but now three-dimensional:

Like the electron, the vector particle can of course also have orbital
angular momentum. That is due to the coefficients in the
wave function above. So far it has been assumed that these
coefficients only depended on the distance from the origin.
However, consider the following more general component of a vector
wave function:

(To check this, take a dot, or rather inner, product of the wave function with itself. Then integrate over all space using spherical coordinates and the orthonormality of the spherical harmonics.)

The wave function (A.84) above has orbital angular momentum in the -direction equal to in addition to the spin angular momentum . So the total angular momentum in the -direction is . To check that, note that under rotations of the coordinate system, the vector changes by a factor while the spherical harmonic changes by an additional factor . That makes the total change a factor .

In general, then, the magnetic quantum number of the net
angular momentum is simply the sum of the spin and orbital ones,

However, the situation for the azimuthal quantum number of the net angular momentum is not so simple. In general the wave function (A.84) above will have uncertainty in the value of . Combinations of wave functions of the form (A.84) are usually needed to get states of definite .

That is a complicated issue best left to chapter 12. But a couple of special cases are worth mentioning already. First, if 1 and , or alternatively, if 1 and , then is simply the sum of the spin and orbital azimuthal quantum numbers .

The other special case is that there is zero net angular momentum.
Zero net angular momentum means that the wave function is exactly the
same regardless how the coordinate system is rotated. And that only
happens for a vector wave function if it is purely radial:

Here is the unit vector sticking radially outward away from the origin. The final constant is the spherical harmonic . It is needed the satisfy the normalization requirement unless you change the one on .

The above state has zero net angular momentum. The question of
interest is what can be said about its spin and orbital angular
momentum. To answer that, it must be rewritten in terms of Cartesian
components. Now the unit vector has Cartesian components

The spatial factors in this expression can be written in terms of the spherical harmonics , chapter 4.2.3. That gives the state of zero net angular momentum as

To check this, just plug in the expressions for the of (A.83), and for the of table 4.3.

The bottom line is that by combining states of unit spin 1, and unit orbital angular momentum 1, you can create a state of zero net angular momentum, 0. Note also that in each of the three terms in the right hand side above, and add up to zero. A state of zero angular momentum 0 must have 0 without uncertainty. Further note that the values of both the spin and orbital angular momentum in the -direction are uncertain. Each of the two has measurable values , 0, or with equal probability .

The above relation may be written more neatly in terms of “ket
notation.” In ket notation, an angular momentum state with
azimuthal quantum number and magnetic quantum number is
indicated as . Using this notation, and dropping
the common factor , the above relation can be written as

Here the subscripts , , and indicate net, spin, and orbital angular momentum, respectively.

There is a quicker way to get this result than going through the above algebraic mess. You can simply read off the coefficients in the appropriate column of the bottom-right tabulation in figure 12.6. (In this figure take to stand for spin, for orbital, and for net angular momentum.) Figure 12.6 also has the coefficients for many other net spin states that you might need. A derivation of the figure must wait until chapter 12.

The parity of vector wave functions is also important. Parity is what happens to a wave function if you invert the positive direction of all three Cartesian axes. What happens to a vector wave function under such an inversion can vary. A normal, or “polar,” vector changes sign when you invert the axes. For example, a position vector in classical physics is a polar vector. Each position coordinate , , and changes sign, and therefore so does the entire vector. Similarly, a velocity vector is a polar vector; it is just the time derivative of position. A particle with a vector wave function that behaves like a normal vector has negative intrinsic parity. The sign of the wave function flips over under axes inversion. Particles of this type turn out to include the photon.

But now consider an example like a classical angular momentum vector, . Since both the position and the velocity change sign under spatial inversion, a classical angular momentum vector stays the same. A vector that does not change under axes inversion is called a “pseudovector” or “axial” vector. A particle whose wave function behaves like a pseudovector has positive intrinsic parity.

Note however that the orbital angular momentum of the particle also has an effect on the net parity. In particular, if the quantum number of orbital angular momentum is odd, then the net parity is the opposite of the intrinsic one. If the quantum number is even, then the net parity is the intrinsic one. The reason is that spherical harmonics change sign under spatial inversion if is odd, but not when is even, {D.14}.

Particles of all types often have definite parity. Such a particle may still have uncertainty in . But if parity is definite, the measurable values of will need to be all even or all odd.