Working my way through Sakurai problem 3.18.

I spent quite a bit of time thinking about how exactly to interpret the math that underlies this problem.

Here’s the problem as written in my notebook:

“18.) Consider an orbital angular momentum eigenstate |l=2, m=0>. Suppose this state is rotated by an angle beta about the y-axis. Find the probability for the new state to be found in m=0, +/-1, +/-2. (The spherical harmonics for l=0,1,2 are given in Appendix A and may be useful.)”

Going through these problems this time, I’ve not actually been very keen on yielding to the temptation to look up what the problem suggests. It certainly saves time to go to an appendix, but it doesn’t really help with understanding.

Fact is that earlier this year I derived the angular momentum commutation relations by hand and in this very notebook had crunched through all the math to produce the spherically symmetric form of the angular momentum ladder operators L+ and L- all on my own. I just decided to go back to that work and derive the spherical harmonics manually for l=0,1 and 2. It isn’t hard. The raising and lowering operators both have differential equations at either end of a given ladder to produce the eigenstates |l,l> and |l,-l>, which can then be subjected to raising and lowering to produce the other harmonics in the series. I solved the lowered equation at L-|l,-l> = 0, normalized the state and then raised through the other members of the ladder. You actually only need to raise about half way since the positive m values are symmetrical with the negative m, meaning that you find the whole series to within a minus sign just by raising halfway through the ladder to the m=0 value.

Once I remembered my calculus and rooted out a couple negative sign errors made in the normalization, my spherical harmonics matched the list found at wikipedia: The Ylms. When going through busy work like this, I typically sit back and smile and think, “Take that Jaden Smith!” After all, I actually know something about what Quantum Mechanics means and exercising knowledge that I gained from school is oddly satisfying.

With the spherical harmonics in hand, I turned my attention to the grist of the problem. How do you use these functions to rotate a given state? More than that, the question this problem really asks is, “What are the probabilities of finding the state in each of the angular momentum m-value eigenstates?” That means specifically to find how the rotated state overlaps with a given set of eigenstates and then report the squares of the overlap coefficients in order to find probability. It’s just your typical quantum mechanics probability question: after you rotate the state by an angle what is the probability of finding it again?

I had to think about this for quite a while. I do know that the Ylms can be used as rotation operators, but I don’t really remember all the details (since I basically never use them this way.) My cogitation involved rooting around in the chapter and studying early examples of rotations. There’s a fairly clear example of the process about mid-chapter that I finally fixated on and worked the math myself to make certain I understood it. In reality, this math ended up being by assertion rather than formal proof and I ended up doing flybys of the Jackson Ylm treatment and of another book in my possession that just gave general orthogonal series solutions.

The technique in question illustrated the whole problem using the Sakurai’s typical ket approach:

|n> = R|z>

Which simply states that a ket pointing in any direction can be obtained by rotating a ket that represents pointing in the z-axis direction. Sakurai then expands this by transferring to a particular representation. This means expanding the ket in terms of the representation free Ylms.

And then looking for a particular overlap coefficient by coming in from the left with a particular Ylm ket. This changes the rotated ket into a function of the Ylm space.

This was pretty interesting to me because the thing on the left is exactly what I need for finding the probability of that particular state. On the right, all the l values that are not equal to l’ cancel out and the entire coefficient pivots around the particular l’ value on the left that you’re looking for.

And Sakurai rewrites the rotation into a rotation matrix that transforms between m and m’ values for a given value of l.

And if the |z> ket contains only one harmonic of m, as above in the question where the ket we’re rotating is |2,0>, the sum again collapses to contain only one term.

I thought for a long time about this result. I didn’t quite understand why the Ylms should actually fit in as the rotation matrix in spherical harmonic form. As written here, the equality between l and l’ may not be completely clear, but they are: <l,0|z> is literally zero for every value of l that doesn’t overlap with the |z>… which is only one particular value of l here. Further, in this particular case, <l,m|z> is directly a Ylm already since this is simply the z-axis vector represented as a Ylm. It becomes a little more clear when you realize that the spherical projection of the cartesian coordinates onto spherical polar coordinates is really just the l=1 Ylms times x-hat, y-hat and z-hat.

The little equation above turns out to be a proof of the equality between the rotation matrix and the Ylms when you realize that <l,0|z> is actually just a constant, which you just divide across, making the equation into ‘rotation matrix = inverse constant*Ylm.’ The particular constant in question is dependent on the l you’re working with, but with the appropriate form, the overlap coefficient becomes extremely easy to find. The constant in question is just:

Which is just the full solid angle divided by the number of Ylm functions that describe that particular full solid angle.

I solved problem 3.18 by basically multiplying each l=2 Ylm by the constant and then squaring the result. To be certain I had it right, I added all the probabilities together to find if they equal 1 for every possible rotation angle. Interestingly, the outcomes would be roughly the same for rotating about the x-axis as by the same angle rotating about the y-axis. The spherical harmonics simply don’t care when rotating the m=0 states since this basically just requires the Legendre polynomials without modification –and these functions have polar symmetry.

Below is the probability for finding the rotated result in each unique state in the l=2 ladder and the line across the top is all the coefficients added together, showing that the probability sums to one since the overlap is not outside of l=2 at all.

After rotating the state, if you’re looking back at the m=0 state where you started, the probability is greatest for a rotation angle either small or nearing 180 degrees. This makes sense because the |2,0> state is symmetrical along z and you can’t distinguish pointing down from pointing up. The chances of falling into m=1 or m=-1 are maximum at about 45 degrees and 135 degrees. The chances of falling into m=2 or -2 are greatest at 90 degrees, while you have about an equal chance of showing up in m=0 at that angle.

For a bit of Quantum perspective, this shows a ridiculously weird behavior. Consider that the original ket might be considered the l=2 (quadrupoler) representation of an arrow pointed along the z-axis. If this thing starts out pointed along the z-axis, you have 100% probability of finding it pointing along the z-axis later if you don’t rotate it. Obviously. Now, as you rotate away from the z-axis, you’re liable to find it pointing in all sorts of crazy directions: for instance, if you rotate by 90 degrees so that this arrow is supposed to lie down classically in the x-y plane somewhere, you have about a 30% chance of finding it pointed back along the z-axis. Huh? Again, this is keeping in mind that the thing we’re rotating is quantum mechanical.

Edit: 3-28-17 Quick review of the significance of the ‘l’ and ‘m’ quantum numbers

I decided that a supplement which would add clarity to this post is to overview briefly what the ‘l’ and ‘m’ quantum numbers mean. I may have written this elsewhere, but I think it adds value here.

It is important to remember that the ket objects that this technique is rotating are themselves represented by the set of Ylms. Ylms are incredibly useful in describing orientational properties and dispositions of motion about the center of mass… not translational motion, but rotation. For a very simple object, like an electron orbiting a nucleus in an atom, the rotational state of the electron essentially breaks the symmetry on a spherically distributed probability of finding an electron located in a particular spot around the atom: it defines both the shape distribution and the overall motion that the electron is undergoing in a single function.

These quantum numbers, ‘l’ and ‘m,’ tell everything you can know about that distribution.

The ‘l’ number, which can be generalized to ‘J’ in representing any kind of rotation, or specialized to ‘S’ while talking about spin, is closely associated with the total angular momentum present in the system. This means that ‘l’ tells you how much rotation is present, but not about the axis around which the system is rotating. From ‘l,’ you know simply that there is rotation.

The ‘m’ number, which is almost always called ‘m’ under every circumstance where you find it, is the angular momentum about one particular axis. This axis can be chosen arbitrarily to be any axis, but it is usually the z-axis by convention. This defines an axis of convenience in the system from which you can define everything knowable about the rotation occurring in the system.

So, ‘l’ is the entire reservoir of rotational momentum while ‘m’ is the portion of rotation that is assigned to a particular axis. ‘l’ acts to confine the total number of values that ‘m’ can explore: angular momentum can be distributed entirely around the z-axis that ‘m’ represents, or it can be distributed entirely along the two other axes, x- and y-, that are independent of that conventional axis, or it can be distributed about a combination of different axes, where a portion is around the z-axis while the rest is hidden somewhere in x- and y-. Because of commutation, quantum mechanics hides the specifics of momentum occurring off the z-axis, but you can know that it’s present based on the total angular momentum of the system, from ‘l.’

The ‘m’ is often stuck as ‘m’ because it is frequently called the magnetization number. Within atoms, electrons don’t merely have angular momentum, they also have charge. The combination of (orbital) angular momentum with electric charge can be directly interpreted as an electric current: it’s a charge moving at some velocity… hence current. If the angular momentum is entirely distributed to the ‘m’ quantum number, that is ‘l’ = +/-‘m,’ then the axis to which ‘m’ is associated has “probability current” running around it, which is effectively the same thing as having a classical current loop. The sign, +m or -m, defines the right handed direction of that current loop, as if a magnetic dipole points either along (+) or against (-) the axis described by the quantum number. This current loop is a magnetic dipole with an associated magnetic dipole field!

You can therefore think about it as a magnetic compass needle which points along the axis defined by the magnetic quantum number. As the value of ‘m’ varies, the strength of the dipole along the axis varies correspondingly. In the plot shown above, the l=2 system has five possible ‘m’ values it can explore: -2,-1,0,1,2. Values 2 and -2 are (essentially) all the angular momentum either with or against the axis, 1 and -1 are some distributed with or against the z-axis while the rest is missing in the x-y plane and 0 is all of the angular momentum somewhere in the x-y plane and none around the z-axis. For the -2,2 and -1,1 distributions, the probability operation wipes out whether the direction is known to be up or down, more or less relating the two possibilities by a phase argument; the m=-2 Ylm is exactly the complex conjugate of the m=2 Ylm, to within a phase of factor of Pi.

So, for this system, the compass needle does not smoothly explore all directions: it only is known to explore certain directions.

## Leave a comment