Hamiltonian Mechanics, Poisson Brackets

Recommended reading: Taylor, 521-533 (again, although not directly applicable for this lecture) Complementary reading: Landau, 135-137

2.1 Recap of last lecture • We’ve seen that we can recast equations of motion as Hamilton’s equations.

• The extended the one-coordinate case to many coordinates and looked at a step-to-step example.

• We have seen again how conservation laws arise.

In this lecture, we will look more closely at conservation laws. But let’s start again with an example.

1 2.2 Example: Block on a wedge in the Hamiltionian formalism

Let’s look at a block sliding down a moveable wedge. In the exercises, we have derived (assuming that x, y as deﬁned by the picture are sensible coordinates) M m T = x˙ 2 + y˙2 +x ˙ 2 − 2x ˙y˙ cos φ (2.1) 2 2 U = −mgy sin φ (2.2)

Thus, since there is no explicity time dependence, we can directly write down the Hamiltonian M m H = T + U = x˙ 2 + y˙2 +x ˙ 2 − 2x ˙y˙ cos φ − mgy sin φ (2.3) 2 2 To get the Hamiltonian in terms of coordinates and momenta, we need to determine the latter. Since the potential is conservative (i.e. only depends on coordinates), we ﬁnd

∂T p = = (M + m)x ˙ − my˙ cos φ (⊗) (2.4) x ∂x˙ ∂T p = = my˙ − mx˙ cos φ (⊗⊗) (2.5) y ∂y˙ Looking at H, we see that it is independent of x, so that on the other hand ∂H m cos φ p˙x = − = 0 → x˙ = y˙ (2.6) ∂x |{z} M + m (⊗)

2 (where we’ve assumed that the block-wedge systems starts at rest with px = 0). We can then insert this into (⊗⊗) to get m cos φ m(M + m) − m2 cos φ mM + m2 sin φ p = my˙ − m y˙ cos φ = y˙ = y˙ (2.7) y M + m M + m M + m

With this, we can express the velocitiesy ˙ andx ˙ in terms of py M + m y˙ = p (2.8) mM + m2 sin φ y m cos φ M + m m cos φ x˙ = p = p (2.9) M + m mM + m2 sin φ y mM + m2 sin φ y Thus, the Hamiltonian becomes M m H = T + U = x˙ 2 + y˙2 +x ˙ 2 − 2x ˙y˙ cos φ − mgy sin φ (2.10) 2 2 p2 M m (M + m)2 m M + m = y cos2 φ + + cos2 φ − m cos2 φ − mgy(2.11)sin φ (mM + m2 sin φ)2 2 2 m2 2 m p2 (M + m)2 M + m = y − cos2 φ − mgy sin φ (2.12) (mM + m2 sin φ)2 2m 2 This can be further simplified by using angular theorems (M + m)2 M + m M + m M − cos2 φ = + 1 − cos2 φ (2.13) 2m 2 2 m M + m = M + m sin2 φ (2.14) 2m so that we in the end get p2(M + m) H = y − mgy sin φ (2.15) (mM + m2 sin φ) Rewriting in terms of the generalized momenta seem a lot of work, but note that through this, we explicitly see that H = H(y, py), and that the dependence on x and px is eliminated. Hamilton’s equations are, then, ∂H p˙ = − = mg sin φ (2.16) y ∂y ∂H py(M + m) y˙ = = 2 (2.17) ∂py m(M + m sin φ) We could in principle stop here, but to get something that can be compared to EL, let’s differ- entiate the second equation and insert the first:

∂H p˙y(M + m) g sin φ(M + m) y¨ = = 2 = 2 (2.18) ∂py m(M + m sin φ) (M + m sin φ) and from the relation betweenx ˙ and py, we ﬁnd g sin φ cos φ x¨ = (2.19) (M + m sin2 φ) just like in the exercises.

3 2.3 Conserved quantities and the Poisson bracket We have seen that it is easy to read oﬀ conserved quantities from Hamilton’s equations. We have also found that if the Hamiltonian has a cyclic coordinate, then it also does not depend on the generalized momentum for this coordinate. One subtlety of the Hamiltonian formulation is the relative sign in ∂H p˙ = − (2.20) ∂q ∂H q˙ = (2.21) ∂p We will now introduce a powerful tool – the Poisson bracket – that recasts Hamilton’s equations into a more symmetric form. The Poisson bracket will also extend our understanding of constants of motion. To introduce the Poisson bracket, note that any function f can vary with time either by explicitly depending on time, or through the variation of its parameters (say, coordinates and conjugate momenta):

N df ∂f X ∂f ∂f = + q˙ + p˙ (2.22) dt ∂t ∂q i ∂p i i=1 i i N ∂f X ∂f ∂H ∂f ∂H = + − (2.23) ∂t ∂q ∂p ∂p ∂q i=1 i i i i ∂f ≡ + [H, f] (2.24) ∂t where we’ve introduced the Poisson bracket [H, f]. Thus, the Poisson bracket measures the implicit time dependence of a function. Applied to the coordinates, we ﬁnd    ∂q ∂H ∂q ∂H  [H, q] =  −  =q ˙ (2.25)  ∂q ∂p ∂p ∂q  |{z} |{z} =1 =0    ∂p ∂H ∂p ∂H  [H, p] =  −  =p ˙ (2.26)  ∂q ∂p ∂p ∂q  |{z} |{z} =0 =1 (2.27) This is an equivalent form of Hamilton’s equations: [H, q] =q ˙ (2.28) [H, p] =p ˙ (2.29) (2.30) which is nicely symmetric. Also, note that for the coordinates and momenta, this means that the [H, q] operation is identical to taking the time derivative, so that we have found a new rep- d resentation of the “time detivate operator dt ”. Because of this, the Poisson bracket is intimately related to conservation laws.

4 For this, remember that df ∂f = + [H, f] (2.31) dt ∂t so that if f does not have an explicit time dependence, we get df = [H, f] (2.32) dt and if f is a constant of motion (i.e. a conserved current, i.e. has no time dependence), then

[H, f] = 0 (2.33)

So we can ﬁnd constants of motion by simply evaluating the Poisson bracket with H! That is often an easy way to check complicated functions. The Poisson bracket can actually help us even more. For that, let us deﬁne

N X ∂f ∂g ∂f ∂g [f, g] = − (2.34) ∂p ∂q ∂q ∂p i=1 i i i i From that, we immediately see

[f, g] = − [g, f] (2.35) [f, const] = 0 (2.36)

[f1 + f2, g] = [f1, g] + [f2, g] (2.37)

[f1f2, g] = f1 [f2, g] + f2 [f1, g] (2.38)

We also directly ﬁnd interesting relations between coordinates and momenta:

[qi, qk] = 0 (2.39)

[pi, pk] = 0 (2.40)

[pi, qk] = δik (2.41) (2.42)

These relations can be seen as a motivation for how to organize quantum mechanics, since they bear striking similarity with the canonical commutation relations, and the last eqn. is similar to the statement of the uncertainty principle. Note that this gives a hint how to construct QM, but does not mean that classical mechanics contains (all) QM features. It can be rigorously disproven that classical mechanics with many hidden parameters (for which Poisson brackets can be set up) contains all features of QM. For us here, the most relevant relation that we can prove is the Jacobi identity

[f, [g, h]] + [g, [h, f]] + [h, [f, g]] = 0 (2.43)

You can e.g. prove this by “brute-force”, but we’ll assume it here for now. This identity has a very interesting consequence, Poisson’s theorem: If two functions f and g are constants of motion, then [f, g] is also a constant of motion.

5 To prove this, simply replace h with H in the Jacobi identity, which makes the ﬁrst two Poisson brackets disappear, proving the statement. This then means that it is (sometimes, not always) possible to ﬁnd new constants of motion by using Poisson brackets. Most often, these new constants of motion will be trivial functions of the old constants of motion, or numerical constants (=numbers). However, sometimes the Poisson brachet can give interesting new constants of motion. For that, let’s look at an admittedly silly example. Say, we know, by some kind of measurement, that the z-component of the angular momentum vector Lz = xpy − ypx is conserved, and that the x-component of the momentum px is conserved,

[H,Lz] = 0 (2.44)

[H, px] = 0 (2.45)

Poisson’s theorem tells us that [Lz, px] is also a constant of motion. What is this quantity?

∂Lz ∂px ∂Lz ∂px ∂Lz ∂px ∂Lz ∂px ∂Lz [Lz, px] = − + − = = −py (2.46) ∂px ∂x ∂x ∂px ∂py ∂y ∂y ∂py ∂x

So we’ve found that py is also a constant of motion!