Lecture 10: Poisson Brackets, Angular Momentum, and Symmetries

Lenny asked, “Hey George, can we hang fish on a Poisson Bracket?”

George smiled. “Only if they’re theoretical.”

An Axiomatic Formulation of Mechanics

Let’s abstract a set of rules that enable one to manipulate Poisson Brackets (from now on I’ll use the abbreviation PB) without all the effort of explicitly calculating them. You can check (consider it homework) that the rules really do follow from the definition of PB’s. Let A, B, and C be functions of the p’s and q’s. In the last lecture, I defined the PB:

(1)

•The first property is antisymmetry: If you interchange the two functions in the PB it changes sign:

(2)

In particular, that means that the PB of a function with itself is zero:

(3)

•Next is linearity in either entry. Linearity entails two properties. First, if you multiply A (but not C) by a constant k, the PB gets multiplied by the same constant:

(4)

Second if you add A + B and take the PB with C, the result is additive:

(5)

Equations (4) and (5) define the linearity property of PB’s.

•Next we consider what happens when we multiply A and B and then take the PB with C. To figure it out, all you need to do is go back to the definition of the PB and apply the rule for differentiating a product. For example,

The same thing is true for derivatives with respect to p. Here is the rule:

(6)

•Finally, there are some specific PB’s that you need in order to get started. Begin by noting that any q or any p is a function of the p’s and q’s. Since every PB involves derivatives with respect to both p’s and q’s, the PB of any q with any other q is zero. The same is true for the PB of two p’s:

(7)

But a PB of a q with a p is not zero. The rule is that {q_i, p_j} is one if i = j and zero otherwise. Using the Kronecker symbol,

(8)

Now we have everything we need to calculate any PB. We can forget the definition and think of Eq.s (2, 3, 4, 5, 6, 7, and 8) as a set of axioms for a formal mathematical system.

Suppose we want to compute

(9)

where for simplicity I have assumed a system with just one q and one p. I will tell you the answer and then prove it. The answer is

(10)

The way to prove this kind of formula is to use mathematical induction. That takes two steps. The first step is to assume the answer for n (assume the induction hypothesis, Eq. (10)) and show that it follows for n + 1. The second step is to explicitly show that the induction hypothesis holds for n = 1.

Thus, replacing n with n + 1, we can write Eq. (9) using Eq. (6):

Next use Eq. (8), which in this case is just {q, p} = 1:

We now use the induction hypothesis—Eq. (10)—and get

(11)

Equation (11) is exactly the induction hypothesis for n + 1. Therefore, all we need to do is show that Eq. (10) holds for n = 1. But all it says is that {q, p} = 1, which is of course true. Thus Eq. (10) is true.

We can write this example in another way that has farreaching consequences. Notice that n q^(n-1) is nothing but the derivative of qⁿ. Thus, for this case,

(12)

Now take any polynomial (even an infinite power series) of q. By applying Eq. (12) to each term in the polynomial and using linearity to combine the results, we can prove

(13)

Since any smooth function can be arbitrarily well approximated by a polynomial, this enables us to prove Eq. (13) for any function of q. In fact, it even goes further. For any function of q and p, it is easy to prove that

(14)

Exercise 1: Prove Eq. (14).

Thus we have discovered a new fact about Poisson Brackets: Taking the PB of any function with p_i has the effect of differentiating the function with respect to q_i. We could have proved that directly from the definition of the PB, but I wanted to show you that it follows from the formal axioms.

What about taking the Poisson bracket of F (q, p) with q_i? You may be able to guess the answer from the symmetric way in which the p’s and q’s enter all the rules. By now you may even guess the sign of the answer:

(15)

Exercise 2: Hamilton’s equations can be written in the form and . Assume that the Hamiltonian has the form . Using only the PB axioms, prove Newton’s equations of motion.

Angular Momentum

In Lecture 7, I explained the relationship between rotation symmetry and the conservation of angular momentum. Just to remind you, I will briefly review it for the case of a single particle moving in the x, y plane. We wrote the formula for an infinitesimal rotation in the form

(16)

Then, assuming that the Lagrangian is invariant, we derived a conserved quantity

with a change of sign, we call it the angular momentum L,

(17)

Now I want to go to three-dimensional space, where angular momentum has the status of a vector. Equation (16) is still true, but it takes on a new meaning: It becomes the rule for rotating a system about the z axis. In fact, we can fill it out with a third equation that expresses the fact that z is unchanged by a rotation about the z axis:

(18)

Equation (17) is also unchanged, except that we interpret the left-hand side as the z component of the angular momentum. The other two components of angular momentum are also easily computed, or you can guess them just by cycling the equation x → y, y → z, z → x:

As you might expect, each component of the vector is conserved if the system is rotationally symmetric about every axis.

Now let’s consider some Poisson Brackets involving angular momentum. For example, consider the PB’s of x, y, and z with L_z:

(19)

You can work out these PB’s using the definition Eq. (1), or you can use the axioms.

Exercise 3: Using both the definition of PB’s and the axioms, work out the PB’s in Equations (19). Hint: In each expression, look for things in the parentheses that have nonzero Poisson Brackets with the coordinate x, y, or z. For example, in the first PB, x has a nonzero PB with p_x.

Here are the results:

If we compare this with Equations (18) we see a very interesting pattern. By taking the PB’s of the coordinates with L_z we reproduce (apart from the ∊) the expressions for the infinitesimal rotation about the z axis. In other words,

where ∼ means “apart from the factor ∊.”

The fact that taking a PB with a conserved quantity gives the transformation behavior of the coordinates under a symmetry—the symmetry related to the conservation law—is not an accident. It is very general and gives us another way to think about the relationship between symmetry and conservation. Before we pursue this relationship further, let’s explore other PB’s involving angular momentum. First of all, it is easy to generalize to other components of L. Again, you can do it by cycling x → y, y → z, z → x. You’ll get six more equations, and you might wonder whether there is a nice way to summarize them. In fact there is.

Mathematical Interlude—The Levi-Civita Symbol

A good notation can be worth a lot of symbols, especially if it appears over and over. An example is the Kronecker delta symbol δ_ij. In this section I will give you another one, the Levi-Civita symbol, which is also called the ∊ symbol ∊_ijk. As in the Kronecker case, the indices i, j, k represent the three directions of space, either x, y, z or 1, 2, 3. The Kronecker symbol takes on two values: either 1 or 0, depending on whether i = j or i ≠ j. The ∊ symbol takes on one of three values: 0, 1, or −1. The rules for ∊_ijk are a little more complicated than those for δ_ij.

First of all, ∊_ijk = 0 if any two indices are the same—for example, ∊₁₁₁ and ∊₂₂₃ are both zero. The only time ∊_ijk is not zero is when all three indices are different. There are six possibilities: ∊₁₁₁, ∊₂₃₁, ∊₃₁₂, ∊₂₁₃, ∊₁₃₂, ∊₃₂₁. The first three have value 1, and the second three have value -1.

What is the difference between the two cases? Here is one way to describe it: Arrange the three numbers 1, 2, 3 on a circle, like a clock with only three hours (see Figure 1).

Figure 1: A circular arrangement of the numbers 1, 2, and 3.

Start at any of the three numbers and go around clockwise. You get (123), (231), or (312), depending on where you start. If you do the same going counterclockwise, you get (132), (213), or (321). The rule for the Levi-Civita symbol is that ∊_ijk = 1, for the clockwise sequences, and ∊_ijk = −1 for the counterclockwise sequences.

Back to Angular Momentum

Now, with the aid of the ∊ symbol, we can write the PB’s for all the coordinates and all the components of :

(20)

For example, suppose that you want to know {y, L_x}. Identifying 1, 2, 3 with x, y, z and plugging these into Eq. (20) we get

Since 213 is a counterclockwise sequence, ∊₁₂₃ = −1, so

Let’s consider another set of PB’s—namely, the PB’s of P_i with the components of . They are easy to work out, and with the aid of the ∊ symbol, we get

For example,

The thing to notice is that the PB’s of the p’s and L’s have exactly the same form as those of the x’s and L’s. That is interesting because the p’s and x’s transform exactly the same way under a rotation of coordinates. Just as δ x~− y for a rotation about z, the variation of p_x is proportional to −p_y.

The meaning of this is quite deep. It says that to compute the change in any quantity when the coordinates are rotated, we compute the Poisson bracket of the quantity with the angular momentum. For a rotation about the ith axis,

(21)

The angular momentum is the generator of rotations.

We will come back to this theme, and to the intimate relationship connecting symmetry transformations, Poisson Brackets, and conserved quantities, but first I want to explain how PB’s can be useful in formulating and solving problems.

Rotors and Precession

One thing we haven’t done yet is to compute the PB’s between different components of the angular momentum. The PB of anything with itself is always zero, but the PB of one component of with another is not zero. Consider

Either by using the definition of PB’s or by using the axioms, we will get.

Try it.

The general relation can be read off by cycling through x, y, z. Here it is using the Levi-Civita symbol:

(22)

That’s very pretty, but what can we do with it? To illustrate the power of relations such as Eq. (22), let’s consider a small, rapidly spinning ball in outer space. Call it a rotor. At any instant there is an axis of rotation, and the angular momentum is along that axis. If the rotor is isolated from all influences, then its angular momentum will be conserved, and the axis of rotation will not change.

Now suppose the rotor has some electric charge. Because the rotor is rapidly spinning, it behaves like an electromagnet with its north and south poles along the rotation axis. The strength of the dipole is proportional to the rate of rotation—or, better yet—to the angular momentum. This won’t make any difference unless we put the whole thing in a magnetic field . In that case, there will be some energy associated with any misalignment between and (see Figure 2).

Figure 2: A rotor aligned at an angle to a magnetic field.

That energy is proportional to the cosine of the angle between the two vectors and to the product of their magnitudes. In other words, the alignment energy is proportional to the dot product

(23)

I’ve used the notation H for energy because later we will identify it with the Hamiltonian of the system.

Let’s take the magnetic field to be along the z axis so that H is proportional to the z component of . Lumping the magnetic field, the electric charge, the radius of the sphere, and all the other unspecified constants into a single constant ω, the energy of alignment takes the form

(24)

Let’s pause for some perspective on what we are doing and where we are going. It’s obvious that without the magnetic field, the system is rotationally symmetric in the sense that the energy does not change if you rotate the axis of the rotor. But with the magnetic field, there is something to rotate relative to. Therefore, the rotational symmetry is ruined. Eq.s (23) and (24) represent the rotational asymmetry. But what is the effect? The answer is obvious: The angular momentum is no longer conserved—no symmetry, no conservation. That means the direction of the spin will change with time, but exactly how?

One can try to guess the answer. The rotor is a magnet—like a compass needle—and intuition suggests that the angular momentum will swing toward the direction of , like a pendulum. That’s wrong if the spin is very rapid. What does happen is that the angular momentum precesses, exactly like a gyroscope, around the magnetic field. (A gyroscope would precess about the gravitational field.) To see that, let’s use the Poisson Bracket formulation of mechanics to work out the equations of motion for the vector .

First, recall that the time derivative of any quantity is the PB of that quantity with the Hamiltonian. Applying this rule to the components of gives

or, using Eq. (24)

Now we can see the point. Even if we know nothing about the material that the rotor is made of, where the charge resides, or how many particles are involved, we can solve the problem: We know the PB’s between all components of . First we take the equation for . Since it involves the PB of L_z with itself,

The z component of does not change. That immediately precludes the idea that swings like a pendulum about .

Next we use Eq. (22) to work out and :

This is exactly the equation of a vector in the x, y plane rotating uniformly about the origin with angular frequency ω. In other words, precesses about the magnetic field. The magic of Poisson Brackets allows us to solve the problem knowing very little other than that the Hamiltonian is proportional to .

Symmetry and Conservation

Let’s go back to Eq. (21), the meaning of which is that the variation of any quantity, under the action of a rotation, is proportional to the PB of that quantity with L_i. Moreover, L_i happens to be the quantity that is conserved by virtue of invariance with respect to rotation. That’s an interesting connection, and one wonders how general it is. Let me give a couple of other examples of the same thing. Consider a particle on a line. If there is translation invariance, then the momentum p is conserved. Now take the PB of any function of x with p:

What is the change in F (x) under an infinitesimal translation by distance ∊? The answer is

or

Here’s another example: If a system has time-translation invariance, then the Hamiltonian is conserved. What is the small change in a quantity under a time translation? You guessed it—the PB of the quantity with H.

Let’s see if we can generalize the connection. Let G(q, p) be any function of the coordinates and momenta of a system. I use the letter G because I am going to call it a generator. What it generates is small displacements of the phase space points. By definition, we will shift every point in phase space by the amount δ _qi, δ _pi, where

(25)

Equations (25) generate an infinitesimal transformation of phase space. The transformation generated by G may or may not be a symmetry of the system. What exactly does it mean to say that it is a symmetry? It means that no matter where you start, the transformation does not change the energy. In other words, if δ H = 0 under the transformation generated by G, then the transformation is a symmetry. We can therefore write that the condition for a symmetry is

(26)

But Eq. (26) can be read another way. Since interchanging the order of the two functions in a PB changes only the sign, Eq. (26) may be expressed as

(27)

which is exactly the condition that G is conserved. One can say it this way: The same Poisson Bracket that tells us how H changes under the transformation generated by G also tells us how G changes with time.