Part I · The Action Principle — Chapter 1.3

Hamilton and Phase Space

Trade $n$ second-order equations for $2n$ first-order ones, and the geometry starts doing the work.

Tools you'll need  — Chapter 0.4: the determinant as a volume factor, the trace, and $\det(I+\epsilon A)=1+\epsilon\,\mathrm{tr}A+O(\epsilon^{2})$. Section 5 is that identity. Chapter 0.5: quadratic forms and positive definiteness, which is exactly the condition §1 needs. Chapter 0.6: the Hessian, convexity, the multivariable chain rule, and Clairaut's theorem on mixed partials. Sections 5, 6 and 7 each turn on Clairaut and nothing else. Chapter 0.7: divergence as the trace of the Jacobian, hence as the fractional rate of change of volume carried by a flow. Section 5 is one substitution into that result. Chapter 0.8: Picard–Lindelöf and the uniqueness of solutions, the harmonic oscillator, the phase-space ellipse, and the pendulum's separatrix. Chapter 1.2: Euler–Lagrange, the canonical momentum $p_{i}=\partial L/\partial\dot q^{i}$, and Problem 2's Beltrami identity, which, you are about to discover, was the Hamiltonian all along.

1 · The Legendre transform, properly

Chapter 1.2's Problem 2 asked you to show that when $L$ has no explicit time dependence the quantity

is conserved. We are about to promote that combination from a conserved quantity to the central object of a whole formulation of mechanics. It will be a great deal less mysterious if you first see what kind of operation it is.

It is not an ad hoc grouping of terms. It is an instance of a standard construction with a clean geometric meaning, and the same construction is what turns the internal energy of a gas into its free energy. So here is the plan for this section: §1 is mathematics, and there is no mechanics in it at all.

1.1 · Two ways to describe a curve

Take a smooth, strictly convex function $f:\R\to\R$. For now just picture a parabola. There are two ways to say which function it is.

By points. List the pairs $\big(x,f(x)\big)$. This is the description you grew up with.

By tangent lines. For each slope $p$, there is exactly one line of that slope tangent to the graph. Give me the family of all those lines, meaning for each $p$ its intercept, and I can reconstruct the curve. The reason is that a convex curve is the upper envelope of its own tangent lines. Every tangent line lies below the graph and touches it once, and the graph is what you get by taking, at each $x$, the highest of them.

Both descriptions contain the same information. The Legendre transform is the dictionary between them. That is the whole idea, and the formula is going to be an immediate consequence of it rather than a definition to memorise.

Let's set it up. A line of slope $p$ has the form $y = px - c$, where $-c$ is its $y$-intercept. We carry the minus sign because it makes every formula below come out with a plus.

Now push this line down from above, increasing $c$, until it first touches the graph. Touching means the vertical gap between line and graph has shrunk to zero somewhere. Before that moment the line sits above the graph everywhere. So the critical value of $c$ is the largest gap there ever was, which we write as

Read the bracket as: how far the line $y=px$ rises above the graph at the point $x$. Take the largest such gap. That is the amount you must lower $y=px$ to make it tangent, and it is therefore precisely $-(\text{intercept of the tangent line of slope }p)$.

1.2 · Where the supremum sits

Now let's do the calculus. The bracket in (1.3.2) is, for fixed $p$, an ordinary function of the single variable $x$. So we want its stationary point, and Chapter 0.6's rule for finding one applies without change:

So the maximising $x$ is the point where the graph's own slope equals $p$, which is exactly what "tangent line of slope $p$" meant. Nothing has been assumed yet except differentiability. Two questions remain, and they are what convexity is for.

Is the stationary point a maximum? The second derivative of the bracket is $-f''(x)$, which is negative wherever $f''\gt0$. So yes, provided $f$ is strictly convex.

Is there exactly one? The equation $f'(x)=p$ must have a unique solution for each $p$ in range. Since $f''\gt0$ means $f'$ is strictly increasing, $f'$ is injective, and so it is invertible on its image. Again convexity, and again it is doing genuine work rather than decorating the theorem.

That unique maximiser is a function of $p$, so give it a name and write $x(p)$. Substituting it back into the definition turns the supremum into an ordinary formula:

Put (1.3.4) beside (1.3.1) and the shape is already the same: slope times variable, minus the function. We will make that identification precise in §2.

1.3 · The transform is its own inverse

Here is the property that makes the construction worth having, and getting it takes one differentiation. We want to know what $g$ does when its own argument $p$ changes, so differentiate (1.3.4) with respect to $p$, using the product rule and the chain rule:

The last two terms cancel exactly, because $f'(x(p))=p$. That cancellation is not a coincidence, and it will happen again in §3 for the same reason: we are differentiating at a stationary point, so the motion of the stationary point itself contributes nothing at first order.

Now set the result beside the equation that defined the maximiser in the first place. Reading (1.3.5) next to (1.3.3) gives

Perfectly symmetric. The transform swaps the roles of the variable and the slope. And since $g''(p) = \dd x/\dd p = 1/f''(x) \gt 0$, the transform of a strictly convex function is strictly convex, so we are entitled to transform a second time. Let's do exactly that and see where it lands:

Its supremum sits where $g'(p)=x$, which by (1.3.6) is the $p$ with $x(p)=x$, which is $p=f'(x)$. So evaluate the bracket there and see what number comes out:

The Legendre transform is an involution on strictly convex functions. Do it twice and you are back where you started. That is what "dictionary" meant. No information is lost going from the point description to the tangent-line description, because you can always go back.

⚑ One statement here is quoted rather than proved. For a function that is not convex, the transform still exists, since the supremum in (1.3.2) is well defined whenever it is finite. What fails is the involution. The double transform $g^{*}$ returns the convex hull of $f$, meaning the largest convex function lying below it, and any non-convex wiggles are erased permanently. This is the Fenchel–Moreau theorem, and it is the precise sense in which "the tangent lines determine the curve" needs convexity. It matters because it tells us which Lagrangians the Hamiltonian formulation applies to, and we will need it for exactly that in §2.3.

1.4 · A worked scalar example

Abstract dictionaries are easier to trust once you have run one. Take $f(x)=\tfrac14 x^{4}$ on $x\ge0$, which is strictly convex there ($f''=3x^{2}\gt0$ for $x\neq0$). Then $p = f'(x) = x^{3}$, so $x = p^{1/3}$, and putting that into (1.3.4) gives

Now check the involution on this example, without appealing to the theorem. First the inverse relation: $g'(p) = \tfrac34\cdot\tfrac43 p^{1/3} = p^{1/3} = x$ ✓, matching (1.3.6). Then transform a second time and see whether $f$ comes back:

It does. And the supremum is a real number you can measure rather than a formal device. Numerically, at $p=2$ the supremum of $2x-\tfrac14x^{4}$ over a fine grid is $1.8898816$, and $\tfrac34\cdot2^{4/3} = 1.8898816$. The formula is measuring something real.

Grind box — Young's inequality, and what the supremum is really enforcing

Definition (1.3.2) says $g(p)\ge px-f(x)$ for every $x$, with equality only at the maximiser. Rearranged, that is

$$p\,x \;\le\; f(x) + g(p) \qquad\text{for all } x,p,$$

with equality precisely when $p=f'(x)$. This is Young's inequality, and it is not a separate theorem. It is the definition of the transform, read sideways.

The classic case is $f(x)=x^{a}/a$ with $a\gt1$ on $x\ge0$. Then $p=x^{a-1}$, so $x=p^{1/(a-1)}$ and

$$g(p) = p^{a/(a-1)} - \frac{1}{a}p^{a/(a-1)} = \frac{a-1}{a}\,p^{a/(a-1)} = \frac{p^{b}}{b}, \qquad \frac1a+\frac1b=1,$$

since $b\equiv a/(a-1)$ satisfies $1/b = (a-1)/a = 1-1/a$. So Young's inequality reads $px\le x^{a}/a + p^{b}/b$, the inequality from which Hölder's and Minkowski's inequalities are built, and the reason conjugate exponents come in pairs summing to $1$ in their reciprocals. Our worked example was $a=4$, $b=4/3$. ✓

What breaks without convexity. Take $f(x)=\tfrac14x^{4}-\tfrac12x^{2}$, the double well, which has $f''=3x^{2}-1\lt0$ on $\abs{x}\lt1/\sqrt3$. For each slope $p$ there can now be three points with $f'(x)=p$, and (1.3.2) selects whichever gives the largest gap, the global one. The information about the other two is discarded, and the double transform $g^{*}$ comes back as the convex hull: the double well with its non-convex middle replaced by a straight segment. This is not a defect. It is exactly the Maxwell construction of a first-order phase transition: the flat segment is the coexistence region, the tie-line joining liquid and gas, and its slope is the common tangent, which is the shared value of the chemical potential. Legendre-transforming a non-convex free energy and getting a flat piece is the phase transition. Chapter 6.6 meets the same picture again with the Higgs potential.

The condition in $n$ dimensions. With several variables, $p_{i}=\partial f/\partial x^{i}$ is invertible for $x$ precisely when the Jacobian of that map, the Hessian $\partial^{2}f/\partial x^{i}\partial x^{j}$, is nonsingular, by the inverse function theorem of Chapter 0.6. Strict convexity in the multivariable sense means the Hessian is positive definite (Chapter 0.5), which is more than enough. Hold that thought. In §2 the Hessian in question is $\partial^{2}L/\partial\dot q^{i}\partial\dot q^{j}$, which Chapter 1.2's grind box already showed is the positive-definite mass matrix whenever the kinetic energy is an honest quadratic form.

2 · Momentum and the Hamiltonian

2.1 · The canonical momentum is not $m\vv v$

Before we transform anything, we need to be clear about the variable we are trading for. Chapter 1.2 defined, for any Lagrangian and any coordinates,

the canonical momentum conjugate to $q^{i}$, and noted that this makes the Euler–Lagrange equation read $\dot p_{i}=\partial L/\partial q^{i}$. In Cartesian coordinates with $L=\half m\dot{\vv x}^{2}-V(\vv x)$, (1.3.11) returns $m\dot x^{i}$, which is why the name is not absurd. But that agreement is a special case, and it fails in two different ways that both matter.

It need not have the dimensions of momentum. For the pendulum with coordinate $\theta$, $L=\half m\ell^{2}\dot\theta^{2}+mg\ell\cos\theta$ gives $p_{\theta}=m\ell^{2}\dot\theta$. That is an angular momentum, with units of $\mathrm{J\,s}$ rather than $\mathrm{kg\,m\,s^{-1}}$. The word "conjugate" is the point. Whatever $p_{i}$ turns out to be, it is the thing that pairs with $q^{i}$ so that $p_{i}\,\dd q^{i}$ has the dimensions of action. Chapter 4.9's uncertainty relation $\Delta q\,\Delta p\ge\hbar/2$ is dimensionally consistent for every conjugate pair for exactly this reason.

It need not be $m\vv v$ even in Cartesian coordinates. This is the one worth being careful about, and seeing it costs one example. Chapter 2.6 will derive, from relativity, that a particle of charge $e$ in an electromagnetic field described by the scalar potential $\phi$ and vector potential $\vv A$ has the Lagrangian

Worked example 1 checks that this reproduces the Lorentz force, so it is the right Lagrangian. Our question is what its canonical momentum is, so apply (1.3.11) to it, component by component, remembering that $\vv A$ depends on position but not on velocity:

The canonical momentum and the mechanical momentum $m\vv v$ differ by $e\vv A$. This is the first time the vector potential has entered mechanics in this book, and it will not leave. Chapter 2.6 shows that $\vv A$ is the spatial part of a four-vector $A^{\mu}$ and that Maxwell's equations are its equations of motion. Chapter 6.3 shows that the entire structure of the Standard Model follows from demanding that the substitution $\vv p\to\vv p - e\vv A$ be forced on you by a symmetry principle. The name for that substitution is minimal coupling, and you have just watched it appear, three parts early, as an unremarkable partial derivative.

2.2 · The Hamiltonian

Now we have both halves of the trade: a function $L$ that depends on velocity, and a slope $p=\partial L/\partial\dot q$ we would rather use instead. So perform §1's transform on $L$, taking $\dot q^{i}$ as the variable and holding $q$ and $t$ fixed. The conjugate slope is $p_{i}=\partial L/\partial\dot q^{i}$ by (1.3.11), and the transform (1.3.4) reads

Every $\dot q^{i}$ on the right is to be eliminated in favour of the $p_{i}$ by inverting (1.3.11), and that instruction is the entire content of the definition. The result is a function of $q$, $p$ and $t$ only. This object is the Hamiltonian. Compare it with (1.3.1) and you can see that Chapter 1.2's Problem 2 was computing it: the conserved quantity of a time-independent Lagrangian is the Hamiltonian.

The inversion step is not free, and §1 told us exactly what it needs. Solving $p_{i}=\partial L/\partial\dot q^{i}$ for $\dot q$ requires that the map from velocities to momenta be invertible, and by the inverse function theorem that means asking about the Hessian

which must be nonsingular. Chapter 1.2's grind box on the double pendulum established that when the coordinate change from Cartesians is time-independent, $T=\half\sum_{ij}M_{ij}(q)\dot q^{i}\dot q^{j}$ with $M$ symmetric positive definite, so $W=M$ and the condition holds automatically.

⚑ It does not always hold, and the failure has a name. When $L$ is linear in some velocity, the corresponding row of $W$ vanishes, the transform is not invertible, and the system is called constrained. Dirac built a whole formalism for that case. It is quoted here and not developed, but you should know where it lives, because it is not an exotic corner. Gauge theories are there (Chapter 6.3: the Lagrangian of electromagnetism has no $\dot A_{0}$ at all), and so is the reparametrisation-invariant string action (Chapter 7.2). Every fundamental theory in Parts VI and VII is a constrained system. So the honest statement is this. The clean Legendre transform of this section is the easy case, and it is the one that teaches you what the hard case is deforming.

2.3 · When $H=T+V$, and when it isn't

You have surely been told that the Hamiltonian is the total energy. It often is. But that is a theorem rather than a definition, and the theorem has two hypotheses that fail in ordinary situations. Let's state it, prove it, and then break it.

Proof. With $L=T-V$ and hypothesis (i), $p_{i}=\partial L/\partial\dot q^{i} = \partial T/\partial\dot q^{i}$. Hypothesis (ii) says $T$ is homogeneous of degree $2$: replacing every $\dot q$ by $\lambda\dot q$ multiplies $T$ by $\lambda^{2}$. What we want is the combination $\sum p_{i}\dot q^{i}$, so differentiate that homogeneity identity $T(q,\lambda\dot q) = \lambda^{2}T(q,\dot q)$ with respect to $\lambda$ and set $\lambda=1$. The chain rule gives

which is Euler's theorem on homogeneous functions, derived rather than quoted. (The grind box does the general degree.) That is the whole input. Feed it into the definition of $H$ and every term is now known:

$\blacksquare$

Now let's break the hypotheses, using a system Chapter 1.2 already built. Problem 3 there put a bead on a hoop of radius $R$ spun about its vertical diameter at a rate $\Omega$ fixed by a motor, and found

The middle term is part of the kinetic energy, since it is the speed the bead has purely because the hoop is carrying it around. But it contains no $\dot\theta$. So $T$ is not homogeneous of degree $2$ in the velocity, and hypothesis (ii) fails. The theorem is therefore unavailable, and we must compute honestly instead. Since $p_{\theta}=mR^{2}\dot\theta$, we have $\dot\theta = p_{\theta}/mR^{2}$, and putting that into the definition gives $H$ alongside the true energy for comparison:

They differ by $mR^{2}\Omega^{2}\sin^{2}\theta$, which changes as the bead slides. $H$ is conserved, since it has no explicit $t$. $T+V$ is not, because the motor is doing work to hold $\Omega$ fixed. Conserved and "the energy" are different predicates, and $H$ is the first one.

Notice also what $H$ is here. It is the kinetic energy of the $\theta$ motion plus the effective potential $V_{\text{eff}}$ that Chapter 1.2 found. The centrifugal term has flipped sign on its way through the transform and become part of the potential. That is not sleight of hand. It is (1.3.16) refusing to apply to a degree-zero term.

Grind box — Euler's theorem in general, the three-piece decomposition, and the relativistic particle

Euler's theorem. Let $F$ be homogeneous of degree $k$ in the variables $u^{i}$, meaning $F(\lambda u) = \lambda^{k}F(u)$ for all $\lambda \gt 0$. Differentiate both sides with respect to $\lambda$ using the chain rule:

$$\sum_{i}u^{i}\,\pdv{F}{u^{i}}\bigg|_{\lambda u} = k\,\lambda^{k-1}F(u),$$

and set $\lambda=1$ to get $\sum_{i}u^{i}\partial F/\partial u^{i} = kF$. That is the whole proof. Degree $2$ gives (1.3.16), degree $1$ gives $\sum u^{i}\partial_{i}F = F$, and degree $0$ gives $\sum u^{i}\partial_{i}F = 0$.

The general Lagrangian. Split $L = L_{2}+L_{1}+L_{0}$ by degree in $\dot q$. Then $p_{i}\dot q^{i} = 2L_{2}+L_{1}$ by the three cases above, so

$$H \;=\; \big(2L_{2}+L_{1}\big) - \big(L_{2}+L_{1}+L_{0}\big) \;=\; L_{2}-L_{0}.$$

Every degree-one piece drops out of $H$ entirely. Three consequences, all checkable against what we have already done:

• Ordinary mechanics: $L_{2}=T$, $L_{1}=0$, $L_{0}=-V$, giving $H=T+V$. ✓
• Rotating hoop: $L_{2}=\half mR^{2}\dot\theta^{2}$, $L_{1}=0$, $L_{0}=\half mR^{2}\Omega^{2}\sin^{2}\theta+mgR\cos\theta$, giving exactly (1.3.19). ✓ The centrifugal term's sign flip is the minus in $L_{2}-L_{0}$.
• Charged particle: $L_{2}=\half mv^{2}$, $L_{1}=e\,\vv v\cdot\vv A$, $L_{0}=-e\phi$, giving $H = \half mv^{2}+e\phi$. The magnetic term vanishes from $H$, which is the statement that magnetic forces do no work, derived rather than asserted. It reappears the moment you express $H$ in terms of $\vv p$ rather than $\vv v$, because $\half mv^{2}=(\vv p-e\vv A)^{2}/2m$. Worked example 1 does this properly.

The relativistic free particle. Chapter 1.2's table listed $S=-mc^{2}\!\int\dd\tau$, i.e. in one dimension $L = -mc^{2}\sqrt{1-\dot x^{2}/c^{2}}$, which is not of the form $T-V$ and is not homogeneous of any degree. Turn the crank anyway:

$$p = \pdv{L}{\dot x} = -mc^{2}\cdot\frac{-\dot x/c^{2}}{\sqrt{1-\dot x^{2}/c^{2}}} = \frac{m\dot x}{\sqrt{1-\dot x^{2}/c^{2}}} = \gamma m\dot x.$$

Then, using $1/\gamma = \sqrt{1-\dot x^{2}/c^{2}}$,

$$H = p\dot x - L = \gamma m\dot x^{2} + \frac{mc^{2}}{\gamma} = \frac{m\big(\gamma^{2}\dot x^{2}+c^{2}\big)}{\gamma} = \frac{mc^{2}\gamma^{2}}{\gamma} = \gamma mc^{2},$$

where the middle step used $\gamma^{2}\dot x^{2}+c^{2} = c^{2}\gamma^{2}\big(\dot x^{2}/c^{2}+1/\gamma^{2}\big) = c^{2}\gamma^{2}$. So $H=\gamma mc^{2}$, the relativistic energy including the rest energy, produced by a Legendre transform and nothing else. Expressed in the right variable, which is what $H$ demands, we use $p^{2}c^{2}=\gamma^{2}m^{2}\dot x^{2}c^{2}$ and the same identity to get

$$H(p) = \sqrt{p^{2}c^{2}+m^{2}c^{4}}.$$

That is the relation Chapter 2.5 derives from four-vectors and Chapter 5.5 quantises into the Dirac equation. It fell out of §1's transform applied to a square root. Note in passing that $\partial^{2}L/\partial\dot x^{2} = m\gamma^{3} \gt 0$, so the Hessian condition (1.3.15) holds and the transform is legitimate. It is the same positivity Chapter 1.2's second-variation grind box used to show that a relativistic worldline maximises proper time.

3 · Hamilton's equations, derived

We have a new function $H(q,p,t)$. We do not yet have equations of motion in terms of it. There are two ways to get them, and they are worth seeing both, because they emphasise different things.

3.1 · First route: take the differential

Chapter 0.6's total differential is the statement that for any smooth function of several variables, the first-order change is the sum of the partial derivatives times the changes in their arguments. Our aim is to find out which variables $H$ genuinely depends on, so apply that rule to the right-hand side of (1.3.14), treating $q$, $\dot q$ and $t$ as the independent variables there:

Now look at the two terms containing $\dd\dot q^{i}$. Their coefficients are $p_{i}$ and $-\partial L/\partial\dot q^{i}$, and by the definition (1.3.11) those are equal and opposite. They cancel identically.

That cancellation is the entire trick, and it deserves a sentence of its own: the Legendre transform is engineered so that the variable you are trying to get rid of drops out of the differential. It is the same cancellation as (1.3.5), and the same one as $\dd F = \dd U - T\dd S - S\dd T$ in the thermodynamics callout. Not an analogy. One theorem, appearing for the third time.

With those two terms gone, what survives is

which contains only $\dd q$, $\dd p$ and $\dd t$. That confirms $H$ really is a function of $(q,p,t)$ and not secretly of the velocities. And once we know that, we may write its differential a second way, straight from the definition of a partial derivative:

Two expressions for the same differential, in terms of the same independent increments. Their coefficients must therefore match one by one. Comparing $\dd p_{i}$, then $\dd q^{i}$, then $\dd t$ gives three identities:

Only the middle one still mentions $L$, and we want equations in $H$ alone, so kill it with the Euler–Lagrange equation, which says $\partial L/\partial q^{i} = \dv{}{t}\big(\partial L/\partial\dot q^{i}\big) = \dot p_{i}$. And there they are:

Hamilton's equations. Count what happened. Our $n$ second-order equations became $2n$ first-order ones. Nothing was gained or lost in information, because a second-order equation always splits into two first-order ones, and the standard way to do that is to call $\dot q$ a new variable. What is special here is which new variable was chosen. Choosing $p=\partial L/\partial\dot q$ rather than $\dot q$ itself is what produces the near-symmetry of (1.3.24), and that symmetry is worth everything that follows.

One immediate dividend, free of charge. Suppose $H$ has no explicit time dependence, and ask how $H$ changes along a trajectory. The chain rule plus (1.3.24) answers in one line:

$H$ is conserved, by cancellation between the two halves of (1.3.24). That is Chapter 1.2's Beltrami identity again, now in three lines instead of five. Notice which feature did the work. The minus sign that made the two equations not symmetric is precisely what made the cancellation happen. Remember that, because §5 and §6 are both that minus sign.

3.2 · Second route: vary the action in phase space

The first derivation used the Euler–Lagrange equation as an input. Here is a derivation that does not, and that treats $q$ and $p$ on an equal footing from the start.

Rearrange (1.3.14) as $L = \sum_i p_{i}\dot q^{i}-H$ and substitute it into the action. Then declare $q^{i}(t)$ and $p_{i}(t)$ to be $2n$ independent functions, with no relation between them assumed. The object to be made stationary is

This is a functional on a space of paths in the $2n$-dimensional space of $(q,p)$, and we ask for its stationary points exactly as Chapter 1.2 did. So vary both: $q\to q+\epsilon\eta$, $p\to p+\epsilon\xi$. Note the asymmetry that is about to matter. In (1.3.26) the velocity $\dot q$ appears and $\dot p$ does not. Collecting the terms linear in $\epsilon$,

Only one term carries a derivative of a variation, and Chapter 1.2's move three handles it. We integrate $\int p_{i}\dot\eta^{i}\,\dd t$ by parts to get $\big[p_{i}\eta^{i}\big]_{t_{1}}^{t_{2}} - \int\dot p_{i}\eta^{i}\,\dd t$, so that every remaining term carries $\eta$ or $\xi$ as a plain factor. Collecting those coefficients separately,

Kill the boundary term by requiring $\eta^{i}(t_{1})=\eta^{i}(t_{2})=0$, which fixes the positions at both ends, as before. Note carefully that no condition on $\xi$ is needed. Since $p$ never appears differentiated, it generates no boundary term and is free at the endpoints.

That is not a technicality, and it is worth saying what it buys. It says the natural boundary-value problem in phase space is "specify where you start and where you finish", not "specify the whole initial state". Chapter 5.6's path integral sums over paths with fixed endpoints, and it inherits exactly this asymmetry.

Now apply the fundamental lemma (Chapter 1.2 §3.4) twice, once with $\xi$ arbitrary and $\eta=0$, and once the other way round. Each bracket must vanish separately, which gives

and that is (1.3.24) again. The two derivations agree, as they must, and each tells you something the other does not. Route one tells you the equations are the Legendre transform of Euler–Lagrange. Route two tells you they are themselves a variational principle, one in which position and momentum are independent variables of equal status.

3.3 · The minus sign, and what it is the seed of

Stare at (1.3.24) until the near-symmetry bothers you. The equations for $q$ and $p$ are the same equation with $q\leftrightarrow p$, except for one minus sign. That asymmetry is irreducible. You cannot rescale it away, because rescaling $p\to-p$ just moves the sign to the other equation.

Since we cannot remove it, the next best thing is to give it a home of its own where we can see exactly what it does. Start by stacking the coordinates into one list of $2n$ numbers,

so that a single letter $z$ now names the whole state. The minus sign is a relation between the first half of that list and the second half, so it should live in a $2n\times2n$ matrix. Define

with the three properties on the right all read straight off the block form. With that one matrix in hand, both halves of (1.3.24) collapse into a single line:

Check it: the top block of $\Omega\nabla H$ is $+\partial H/\partial p$ and the bottom block is $-\partial H/\partial q$. ✓

Now read that line geometrically, because it explains a fact we have already proved algebraically. The gradient of $H$ points uphill, at right angles to the level sets. $\Omega$ rotates that direction by a quarter turn. So the system moves along the level sets, which is why $H$ is conserved. The flow is everywhere perpendicular to the gradient, by construction.

$\Omega$ is called the symplectic form, and the antisymmetry $\Omega^{\mathsf T}=-\Omega$ is the minus sign, isolated. Compare it with the object that runs Chapter 3.3, the metric $g_{\mu\nu}$, which is symmetric and measures lengths. A symplectic form measures oriented areas instead, and it is the reason phase space has a geometry at all.

Every structural fact in the rest of this chapter is a consequence of the antisymmetry of $\Omega$: Liouville's theorem, the Poisson bracket, and the $[\hat q,\hat p]=\ii\hbar$ of quantum mechanics. So is the fact that phase space is always even-dimensional. An antisymmetric matrix in odd dimensions has $\det\Omega = \det\Omega^{\mathsf T} = \det(-\Omega) = (-1)^{2n+1}\det\Omega$, forcing $\det\Omega=0$, so no invertible $\Omega$ exists there.

4 · Phase space

The space whose points are the $2n$ numbers $z=(q,p)$ is phase space. Chapter 0.8 introduced it for the oscillator and promised it would come back. This is where it becomes the setting rather than a picture.

4.1 · One point fixes everything, and trajectories cannot cross

Hamilton's equations (1.3.32) are first order. That is the structural difference from the Lagrangian formulation, and it changes what a "state" is. Chapter 0.8's Picard–Lindelöf theorem says that a first-order system $\dot z = F(z)$ with $F$ Lipschitz has exactly one solution through each initial point. Applied here, that gives us the following.

Configuration space does not have this property. A pendulum passing through $\theta=0$ might be moving left or right, fast or slow, so "$\theta=0$" is not a state. Phase space is precisely the space of things that are states, which is why Chapter 0.8 said the list the existence theorem asks for is the list phase space is made of.

There is an immediate geometric consequence, and it is what makes phase portraits legible.

Claim. If $H$ has no explicit time dependence, two distinct trajectories can never cross.

Proof. Suppose the trajectory $z_{1}$ passes through the point $z_{\ast}$ at time $t_{1}$ and the trajectory $z_{2}$ passes through the same $z_{\ast}$ at time $t_{2}$. Because $H$ does not depend on $t$, the right-hand side of (1.3.32) depends only on the point, so shifting a solution in time gives another solution. Define $\tilde z_{2}(t) = z_{2}(t+t_{2}-t_{1})$, which also solves the equations and satisfies $\tilde z_{2}(t_{1}) = z_{\ast}$. Now $z_{1}$ and $\tilde z_{2}$ are two solutions with the same value at $t_{1}$, so by uniqueness they are the same solution. Hence $z_{2}$ is $z_{1}$ retimed, and it is the same curve. $\blacksquare$

So through every point of phase space there passes exactly one curve, and phase space is filled by these curves like the streamlines of a steady fluid, with no branching, no merging and no intersections. The only points where curves appear to meet are fixed points, where $\nabla H=0$ and the "trajectory" is a single stationary point. We will meet one shortly.

4.2 · The harmonic oscillator, again

The quickest way to get a feel for a flow is to draw one we already know. Take $H = \dfrac{p^{2}}{2m} + \half m\omega^{2}q^{2}$, which is $T+V$, since both hypotheses of §2.3 hold. Differentiating it in $p$ and in $q$ gives Hamilton's equations for the oscillator:

The first says $p=m\dot q$, recovering the mechanical momentum. Substituting it into the second gives $m\ddot q = -m\omega^{2}q$, the oscillator equation. Nothing new so far. But now read the pair as a flow instead of as a route back to a second-order equation.

The level sets of $H$ are the ellipses of Chapter 0.8 §4.4, and by §3.3 the motion runs along them. To see what the motion actually is, we would like the ellipses to be circles, so rescale the two axes by equal and opposite factors:

which has the virtue of preserving areas, since $\dd Q\,\dd P = \dd q\,\dd p$. That is a property §8 will name. In these variables $H = \tfrac{\omega}{2}\big(P^{2}+Q^{2}\big)$, and rewriting (1.3.33) in them gives

That is a rigid rotation of the whole plane at angular velocity $\omega$, clockwise. Every point, whatever its energy, goes round at the same rate, which is the phase-space statement of the fact that a harmonic oscillator's period is independent of amplitude. Chapter 0.1 observed that $\ii$ is a rotation by 90°. Here the rotation is generated by $\Omega$, which squares to $-1$ for the same reason. The oscillator flow is multiplication by $\ee^{-\ii\omega t}$, drawn.

4.3 · The pendulum, and the separatrix

Now a system where the flow is not uniform. Take the pendulum of Chapter 1.2 §6.2, with $q=\theta$ and $p=m\ell^{2}\dot\theta$. Both hypotheses of §2.3 hold, so its Hamiltonian is the total energy:

where the constant has been chosen so that $H=0$ at the bottom. Hamilton's equations are $\dot q = p/m\ell^{2}$ and $\dot p = -mg\ell\sin q$, reproducing $\ddot q = -(g/\ell)\sin q$. Write $\omega_{0}^{2}=g/\ell$.

Since $H$ is conserved, trajectories are its level sets, and there are three kinds.

Libration. For $0 \lt H \lt 2mg\ell$ the bob cannot reach the top. Setting $p=0$ in (1.3.36) gives a turning point at $\cos q = 1 - H/mg\ell$, which has a solution only when $H\le2mg\ell$. The level set is a closed curve encircling the origin, and the motion is back-and-forth swinging. Near the origin, expanding $1-\cos q\approx q^{2}/2$ recovers the oscillator of §4.2 and the level curves are Chapter 0.8's ellipses.

Rotation. For $H \gt 2mg\ell$ there is no turning point. Here $p$ never vanishes, so the bob goes over the top and keeps going. The level set is an open curve running across the whole range of $q$, and since $q$ and $q+2\pi$ are the same physical configuration, phase space here is really a cylinder and the curve is a loop around it.

The separatrix. Between those two behaviours sits one energy at which the bob arrives at the inverted position $q=\pi$ with exactly zero momentum. Putting $q=\pi$, $p=0$ into (1.3.36) gives its value:

We would like the curve itself, not only its energy, so solve (1.3.36) for $p$ at this energy, using $1-\cos q = 2\sin^{2}(q/2)$ to make the square root come out cleanly:

These are the two arcs Chapter 0.8 §8.2 found. They meet at $q=\pm\pi$, $p=0$, and that meeting point is a fixed point, since $\partial H/\partial q = mg\ell\sin\pi = 0$ and $\partial H/\partial p=0$. So §4.1's no-crossing theorem is not violated. To see what the flow does near it, linearise by writing $q=\pi+u$, so that $\sin q = -\sin u \approx -u$:

whose solutions are $\ee^{\pm\omega_{0}t}$, one growing and one decaying. This is a hyperbolic fixed point, or saddle. The level curves near it are hyperbolas, and the separatrix is the pair of straight lines they asymptote to. The decaying solution is the trajectory that creeps toward the inverted position and takes infinite time to arrive. The growing one is the same trajectory run backwards.

Compare the origin, where the same calculation gives $\ddot u = -\omega_{0}^{2}u$ and the level curves are closed. That is an elliptic fixed point, or centre. Two fixed points, two signs, two completely different local pictures, and the entire qualitative structure of the pendulum is the statement that they are joined by the separatrix.

4.4 · What the enclosed area means

For a closed orbit, meaning a libration, the natural number to attach to it is the area it encloses:

where the line integral around the closed curve equals the enclosed area by Green's theorem (Chapter 0.7 §5.1). That is what a line integral of $p\,\dd q$ around a loop is. Two facts make this the right quantity to care about.

First: its derivative is the period. To see it, ask how the area responds when the energy changes. Solve $H(q,p)=E$ for $p$ on the upper branch and differentiate (1.3.40) under the integral sign. Since $\partial p/\partial E$ at fixed $q$ is the reciprocal of $\partial H/\partial p$ at fixed $q$, and $\partial H/\partial p = \dot q$, the integrand turns into $\dd t$:

The area is the antiderivative of the period. For the oscillator, Chapter 0.8 computed $\mathcal{A}=2\pi E/\omega$, so $\dd\mathcal{A}/\dd E = 2\pi/\omega = T$ ✓, and since $T$ does not depend on $E$ there, the area is exactly linear in the energy. For the pendulum it is not. Numerical integration of (1.3.40) at $\omega_{0}=1$, $m\ell^{2}=1$ gives $\dd\mathcal{A}/\dd E = 6.449765$ at $E=0.2$ and $11.633349$ at $E=1.9$, against periods $4K(E/2)/\omega_{0}$ of $6.449765$ and $11.633349$ from Chapter 0.8's elliptic-integral formula. Agreement to ten digits, which is the sort of thing worth checking once.

Second: it is what quantum mechanics rations. ⚑ Quoted forward to Chapters 4.8 and 4.10. The Bohr–Sommerfeld condition states that in the semiclassical regime the allowed orbits are those with

with $h=2\pi\hbar$ Planck's constant. For the oscillator this reads $2\pi E_{n}/\omega = (n+\tfrac12)h$, that is $E_{n}=(n+\tfrac12)\hbar\omega$, which Chapter 4.8 will derive exactly, with ladder operators and no semiclassical approximation, and get precisely this answer. That is the one case in which the condition is exact rather than semiclassical, and it is the only part of this flag 4.8 discharges. The general statement, and the measurement of how far it is from the truth when the potential is not a parabola, is Chapter 4.10 §6.

So the classical phase-space ellipse is not deleted by quantum mechanics. It is quantised. The continuum of orbits becomes a discrete stack of them, one per unit of $h$ of enclosed area, and the ground state is the one with half a unit. Phase-space area is measured in units of Planck's constant. That sentence is the reason $h$ has the units it does, namely $\mathrm{J\,s}$, the units of $p\,\dd q$, and it is the single most useful thing to remember about phase space.

Grind box — the area–period relation done carefully, and the adiabatic invariant

The derivation, without hand-waving. Take a one-dimensional system with $H(q,p)$ and a closed orbit at energy $E$. On the upper half of the orbit write $p=p_{+}(q,E)$. The enclosed area is $\mathcal{A}(E)=\oint p\,\dd q$, traversed so the integral is positive. Differentiating with respect to $E$ moves the curve. The endpoints, which are the turning points, move too, but $p=0$ there, so the boundary contribution vanishes and we may differentiate under the integral:

$$\dv{\mathcal{A}}{E} = \oint\left(\pdv{p}{E}\right)_{\!q}\dd q.$$

Now, differentiating the identity $H\big(q,p(q,E)\big) = E$ with respect to $E$ at fixed $q$ gives $\big(\partial H/\partial p\big)\big(\partial p/\partial E\big) = 1$, so $\partial p/\partial E = 1/\dot q$. Hence the integrand is $\dd q/\dot q = \dd t$, and integrating once round the orbit gives one period. $\blacksquare$

The action variable. Define $J = \mathcal{A}/2\pi$. Then $\dd E/\dd J = 2\pi/T = \omega(E)$, the angular frequency. Problem 3 constructs a change of variables in which $J$ is literally the new momentum and the new coordinate is an angle advancing uniformly at rate $\omega$. These are action–angle variables, and they turn any one-dimensional bound motion into the trivial system $\dot\vartheta=\omega$, $\dot J = 0$.

⚑ One quoted fact, because it explains why (1.3.42) was ever a sensible guess. If a parameter of the system (a pendulum's length, say) is changed slowly compared with the period, then $E$ and $T$ both change but $J$ does not, to all orders in the slowness. $J$ is an adiabatic invariant. Ehrenfest's argument was that only an adiabatic invariant can sensibly be quantised, since if you could change $J$ by gently stretching a wire you could move a system off its allowed levels without a transition, and that is how Bohr and Sommerfeld chose which quantity to set equal to $nh$. The modern statement is Chapter 4.17's adiabatic theorem: a quantum system stays in the $n$-th eigenstate under slow change, which is the same sentence with $J$ replaced by $n$.

Numbers. For the pendulum at $m\ell^{2}=1$, $\omega_{0}=1$, quadrature of (1.3.40) against the elliptic-integral period gives

$E$	$\dd\mathcal{A}/\dd E$ (numerical)	$T=4K(E/2)$	relative difference
$0.2$	$6.449765$	$6.449765$	$2\times10^{-11}$
$0.8$	$7.110077$	$7.110077$	$1\times10^{-12}$
$1.4$	$8.301453$	$8.301453$	$4\times10^{-11}$
$1.9$	$11.633349$	$11.633349$	$3\times10^{-10}$

The period diverges as $E\to2$, since the separatrix takes infinite time, and (1.3.41) says the area therefore has infinite slope there while remaining finite. Both statements are visible in the interactive below. Near the separatrix, neighbouring orbits have wildly different periods, so a blob of initial conditions is sheared apart. That shearing is the whole show.

5 · Liouville's theorem

This is the centrepiece, and it costs almost nothing, because Chapter 0.7 already did the work.

Forget for a moment that (1.3.24) describes mechanics and read it as what it literally is: a formula assigning a velocity vector to every point of a $2n$-dimensional space. That is a vector field in the sense of Chapter 0.7 §1, and writing out its components gives

Chapter 0.7 spent a section establishing what the divergence of such a field means, and the answer was not "something to do with flux" but something much more useful. The divergence is the trace of the Jacobian, hence the fractional rate at which a blob carried by the flow changes its volume:

So the question "does a Hamiltonian flow squeeze or spread the blobs it carries" is now a question about one derivative. Compute the divergence of (1.3.43), which is the sum of $\partial(\text{component})/\partial(\text{corresponding coordinate})$ over all $2n$ coordinates:

by the equality of mixed partial derivatives, which is Clairaut's theorem, Chapter 0.6 §6.1. Every term cancels against its partner, and the cancellation happens because of the minus sign of §3.3 and nothing else. Now substitute that zero into (1.3.44) and read off what it says about volume:

Liouville's theorem: Hamiltonian flow preserves phase-space volume exactly. Not approximately, not on average, and not for special systems. For every Hamiltonian, including time-dependent ones, since nothing in the calculation used $\partial H/\partial t=0$, including chaotic ones, and including the entire universe if you are willing to write down its Hamiltonian.

Three payoffs, in increasing order of importance.

Statistical mechanics has a measure. To do statistical mechanics you must say what "equally likely" means for a continuous system, and that requires a measure on phase space that the dynamics does not distort. Liouville supplies exactly one, namely $\dd^{n}q\,\dd^{n}p$. The microcanonical ensemble, uniform on a surface of constant energy, is well defined because the flow is incompressible. Had the flow contracted volumes, an initially uniform ensemble would pile up and "uniform" would not be a stable notion. Every ensemble average you have ever seen rests on (1.3.46).

There is no classical attractor. A damped oscillator spirals into the origin, so a blob of initial conditions shrinks to a point and volume is destroyed. Liouville therefore says damping is never fundamental. It is what a Hamiltonian system looks like when you have thrown away the degrees of freedom that the energy leaked into: the air, the wire, the thermal bath. The full system, including the bath, conserves volume. This is the reason a fundamental theory can never contain a friction term, and why every dissipative equation in physics is an effective description.

Information is not destroyed, only stirred. This is the one to keep. Suppose you know the state to within a small blob of volume $V_{0}$, which is your measurement precision. After evolution the blob has some horrible filamentary shape, but its volume is still $V_{0}$. In principle, running the equations backwards recovers the initial blob exactly, so Hamiltonian evolution is reversible and loses nothing. Two forward pointers, both of which are this sentence in different clothes.

• Chapter 4.6: quantum time evolution is generated by $\ee^{-\ii\hat Ht/\hbar}$, which is a unitary operator. It preserves inner products, hence probabilities, hence distinguishability, so two distinct initial states stay distinct forever. That is the quantum statement of exactly this theorem, with "volume in phase space" replaced by "overlap of state vectors" and the symplectic form $\Omega$ replaced by the factor of $\ii$.
• Chapters 3.8 and 7.9: the black-hole information problem is what happens when the two statements appear to disagree. Hawking's calculation says a black hole evaporates into radiation that depends only on its mass, charge and spin, so the volume of states that could have formed it collapses to a point and information is destroyed. Something in that argument must be wrong, because unitarity is not a convenience you can trade away, and the search for what is wrong produced holography. The problem is a problem precisely because (1.3.46) and its quantum successor are load-bearing.

play reset

blob start p₀

p₀ = 1.55 (H at blob top = 1.428)

t = 0.00

boundary vertices = 400

area now = 0.0615727 initial = 0.0615727

drift = +0.0 ppm — exactly zero is the theorem

6 · Poisson brackets

We now extract from Hamilton's equations the structure that will survive quantisation. Everything in this section is elementary. Its importance is entirely in what it becomes.

6.1 · The equation of motion for any observable whatsoever

An observable is any function $f(q,p,t)$ on phase space: the energy, a component of angular momentum, the distance between two particles, the electric dipole moment, anything you could in principle measure. Ask how it changes along a trajectory. By the multivariable chain rule (Chapter 0.6),

which is true of any function on any space and says nothing about mechanics yet. The mechanics enters when we say what $\dot q$ and $\dot p$ actually are, so substitute Hamilton's equations (1.3.24) for both of them:

The bracketed combination has appeared without being asked for, so let's give it a name. For any two functions $f,g$ on phase space define the Poisson bracket

where the second form uses the symplectic matrix of (1.3.31). Check the block structure and you will find it reproduces the sum exactly. With that abbreviation, (1.3.48) shortens to

This is not a formula about a particular system. It is the equation of motion for every observable of every Hamiltonian system, and it contains Hamilton's equations as the special cases $f=q^{i}$ and $f=p_{i}$ (verify: $\{q^{i},H\}=\partial H/\partial p_{i}$ and $\{p_{i},H\}=-\partial H/\partial q^{i}$ ✓).

It also contains the master conservation law, immediately. Set $\partial f/\partial t=0$ in (1.3.50) and the left-hand side vanishes exactly when the bracket does:

Conservation has become an algebraic condition. You no longer solve the equations of motion to find out whether something is conserved. You compute one bracket. Taking $f=H$ gives $\{H,H\}=0$ by antisymmetry alone, recovering (1.3.25).

6.2 · The fundamental brackets

The first brackets to compute are the ones between the coordinates themselves, since everything else is built from those. For $\{q^{i},p_{j}\}$, note that $\partial q^{i}/\partial q^{k}=\delta^{i}_{k}$, $\partial q^{i}/\partial p_{k}=0$, $\partial p_{j}/\partial p_{k}=\delta_{jk}$ and $\partial p_{j}/\partial q^{k}=0$, so only the first term of (1.3.49) survives:

The last two vanish because in each case one of the two factors in every term is a derivative of a $q$ with respect to a $p$, or the other way round. These three lines are the canonical commutation relations of classical mechanics. Every coordinate commutes with every other coordinate. Every momentum commutes with every other momentum. And a coordinate fails to commute with its own conjugate momentum, by exactly $1$.

6.3 · The algebra

The bracket has four properties, and they are the properties rather than incidental features.

Bilinearity. $\{af+bg,h\} = a\{f,h\}+b\{g,h\}$ for constants $a,b$, and likewise in the second slot. Immediate from (1.3.49), since differentiation is linear.

Antisymmetry. $\{f,g\} = -\{g,f\}$, and hence $\{f,f\}=0$. Immediate from $\Omega^{\mathsf T}=-\Omega$. Swapping $f$ and $g$ in $(\nabla f)^{\mathsf T}\Omega\,\nabla g$ transposes a number, which does nothing, and transposing the middle factor flips the sign.

Leibniz. $\{f,gh\} = \{f,g\}h + g\{f,h\}$. Immediate from the product rule, since every term of (1.3.49) differentiates $g$ exactly once. This says $\{f,\,\cdot\,\}$ acts as a derivation, meaning it behaves like a directional derivative, and §7 will explain that by exhibiting the direction.

The Jacobi identity. The fourth is the only one that is not a line of inspection, and it says that the three ways of nesting a bracket inside a bracket sum to nothing:

This one is not obvious and is proved in the grind box. It is the compatibility condition that makes everything hang together. Bilinear, antisymmetric, and Jacobi is the definition of a Lie algebra, the structure Chapter 6.1 builds a whole part of the book on.

One concrete consequence is available right now: if $f$ and $g$ are both conserved, so is $\{f,g\}$. To see it, put $h=H$ in (1.3.53) and use $\{f,H\}=\{g,H\}=0$. The first two terms die, leaving $\{H,\{f,g\}\}=0$. This is Poisson's theorem, and it is a machine for manufacturing conservation laws from ones you already have. It occasionally produces something new and usually produces something you knew, and either way it does not cost you an integration.

Grind box — proving the Jacobi identity, and why no second derivatives can survive

Write everything in the compact notation of (1.3.49). Let $z^{a}$, $a=1,\dots,2n$, be the phase-space coordinates, $\partial_{a}=\partial/\partial z^{a}$, and $\Omega^{ab}$ the constant antisymmetric matrix of (1.3.31), so that

$$\{f,g\} = \Omega^{ab}\,\partial_{a}f\,\partial_{b}g$$

with repeated indices summed. The two facts we will use are that $\Omega^{ab}$ is constant, so it passes through derivatives, and antisymmetric.

The counting argument first, because it tells you why the identity is true. Each term of (1.3.53) is a bracket of a bracket, so it contains exactly two derivatives distributed over three functions. Every term therefore contains a second derivative of exactly one of $f,g,h$ and first derivatives of the other two. The claim is that the second-derivative terms cancel among themselves, separately for $f$, for $g$ and for $h$, and that once they are gone nothing is left, since every term has a second derivative somewhere. So it suffices to check one of the three cancellations, the other two following by relabelling.

The cancellation. Expand the first and third terms by the product rule:

$$\begin{aligned} \big\{f,\{g,h\}\big\} &= \Omega^{ab}\,\partial_{a}f\,\partial_{b}\big(\Omega^{cd}\partial_{c}g\,\partial_{d}h\big)\\[2pt] &= \Omega^{ab}\Omega^{cd}\Big[\partial_{a}f\,\partial_{b}\partial_{c}g\,\partial_{d}h \;+\; \partial_{a}f\,\partial_{c}g\,\partial_{b}\partial_{d}h\Big],\\[6pt] \big\{h,\{f,g\}\big\} &= \Omega^{ab}\Omega^{cd}\Big[\partial_{a}h\,\partial_{b}\partial_{c}f\,\partial_{d}g \;+\; \partial_{a}h\,\partial_{c}f\,\partial_{b}\partial_{d}g\Big]. \end{aligned}$$

Collect the two terms carrying second derivatives of $g$: they are

$$T_{1} = \Omega^{ab}\Omega^{cd}\,\partial_{a}f\,\partial_{d}h\,\partial_{b}\partial_{c}g, \qquad T_{2} = \Omega^{ab}\Omega^{cd}\,\partial_{a}h\,\partial_{c}f\,\partial_{b}\partial_{d}g.$$

(The middle term of (1.3.53), $\{g,\{h,f\}\}$, contains second derivatives of $h$ and of $f$ only, so it contributes nothing here.) Now relabel the dummy indices in $T_{2}$ by the substitution $a\to d$, $b\to c$, $c\to a$, $d\to b$:

$$T_{2} = \Omega^{dc}\Omega^{ab}\,\partial_{d}h\,\partial_{a}f\,\partial_{c}\partial_{b}g = -\,\Omega^{cd}\Omega^{ab}\,\partial_{a}f\,\partial_{d}h\,\partial_{b}\partial_{c}g = -\,T_{1},$$

using $\Omega^{dc}=-\Omega^{cd}$ once and the symmetry $\partial_{c}\partial_{b}g=\partial_{b}\partial_{c}g$ (Clairaut) once. So $T_{1}+T_{2}=0$. By the cyclic symmetry of (1.3.53), the second-derivative-of-$f$ terms and the second-derivative-of-$h$ terms cancel by the identical relabelling. Nothing remains. $\blacksquare$

Note what the proof used: antisymmetry of $\Omega$, constancy of $\Omega$, and equality of mixed partials. Nothing else, and in particular nothing about mechanics. The identity is a fact about any constant antisymmetric bilinear pairing of gradients.

Verification, since a proof by index gymnastics deserves a check. Taking the deliberately unstructured functions $f = x^{2}p_{y} + z\,p_{x}p_{y} + p_{z}\sin y$, $g = y\,p_{x}^{2} + xz\,p_{z} + p_{y}\cos z$, $h = z^{3}p_{x} + y\,p_{y}p_{z} + xyz$ on a three-degree-of-freedom phase space, symbolic differentiation gives a Jacobi residual of exactly $0$, and the Leibniz residual $\{f,gh\}-\{f,g\}h-g\{f,h\}$ is exactly $0$ as well.

6.4 · The payoff, and the reason Part I exists

Here is what the last three pages were for.

Watch what that does to the results we have. The fundamental brackets (1.3.52) become

which is the foundational relation of quantum mechanics, from which the uncertainty principle follows in three lines (Chapter 4.9) and for which $\hat p = -\ii\hbar\,\partial/\partial q$ is the standard realisation (Chapter 4.6). Do the same to the equation of motion (1.3.50) and it becomes

the Heisenberg equation of motion, which is the whole of quantum dynamics in the Heisenberg picture. And the conservation criterion (1.3.51) becomes "$\hat f$ is conserved if and only if it commutes with the Hamiltonian", the statement that gets used, without comment, in every quantum mechanics course from the second week onward.

The bracket properties go across too, and they are exactly the properties a commutator has. Bilinearity: immediate. Antisymmetry: $[\hat f,\hat g]=-[\hat g,\hat f]$ by inspection. Leibniz: $[\hat f,\hat g\hat h] = [\hat f,\hat g]\hat h + \hat g[\hat f,\hat h]$, which you can verify in one line by inserting and cancelling $\hat g\hat f\hat h$. Jacobi: holds for commutators identically, by expanding all twelve products and watching them cancel in pairs. The classical Poisson algebra and the quantum commutator algebra are the same Lie algebra. That is the sense in which the substitution "changes nothing else".

So the answer to "why did we spend a chapter on a change of variables" is that the change of variables is the thing that quantises. You cannot quantise the Lagrangian formulation directly, because there is no bracket there, only a variational principle. That is why Feynman had to invent an entirely different route, the path integral of Chapter 5.6, to quantise from an action. Canonical quantisation, by contrast, is a one-line substitution into a structure you now have.

7 · Symmetries generate motion

This is the deepest section in the chapter, and it is short. It answers a question you may not have thought to ask: why should conserved quantities and symmetries have anything to do with one another?

Chapter 1.4 will prove Noether's theorem, which says that every continuous symmetry of the action implies a conserved quantity. That is a one-way street, and it leaves the conserved quantity looking like a by-product, something the symmetry emits. The Hamiltonian picture reveals that the relationship is far tighter than that.

7.1 · Every observable generates a flow

The idea is to notice that Hamilton's equations never used any special property of $H$. So take any function $G(q,p)$ on phase space and put it where $H$ used to be. That is, define an infinitesimal change of every observable $f$ by

Applied to the coordinates themselves this reads $\delta q^{i} = \epsilon\,\partial G/\partial p_{i}$ and $\delta p_{i} = -\epsilon\,\partial G/\partial q^{i}$, which is the same structure as (1.3.24) with $\epsilon$ playing the role of a small time and $G$ the role of the Hamiltonian. So $G$ defines a flow on phase space, and (1.3.56) is how any observable changes along it. We say $G$ generates the transformation.

Before doing examples, we should check that these flows respect the structure of §3.3, and they do. Compute the fundamental bracket of the shifted variables, to first order in $\epsilon$, using bilinearity and then Clairaut:

So the transformation is automatically canonical, meaning it preserves the fundamental brackets. Mixed partials again, for the third time in this chapter, after (1.3.45) and the Jacobi identity. Whatever $G$ you pick, the flow it generates is a symmetry of the symplectic structure. Now the three cases that matter.

7.2 · The three canonical examples

The Hamiltonian generates time translation. Take $G=H$ and $\epsilon=\dd t$. Then (1.3.56) is $\delta f = \dd t\,\{f,H\} = \dd t\,\dv{f}{t}$ for any $f$ without explicit time dependence, by (1.3.50). So the flow generated by $H$ is the dynamics itself. Evolving the system forward by $\dd t$ and applying the transformation generated by $H$ are the same operation. The Hamiltonian is not merely the energy. It is the generator of translations in time.

Momentum generates spatial translation. Take $G=p_{x}$, the momentum conjugate to $x$, and work out its bracket with an arbitrary observable. Every term of (1.3.49) except one vanishes, because $p_{x}$ depends on nothing else:

And $\epsilon\,\partial f/\partial x$ is precisely the first-order change in $f$ when you move the system a distance $\epsilon$ in the $x$-direction, which is Chapter 0.1's linear approximation. So $p_{x}$ generates translations along $x$. In particular $\delta x = \epsilon$ and $\delta p = 0$, so the momentum shifts nothing and the position shifts by $\epsilon$, which is exactly what a translation does.

Angular momentum generates rotation. Take $G = L_{z} = xp_{y}-yp_{x}$. This one has four relevant coordinates rather than one, so compute the brackets with all four directly from (1.3.49):

Feeding those into (1.3.56) gives $\delta x = -\epsilon y$, $\delta y = +\epsilon x$, and identically $\delta p_{x} = -\epsilon p_{y}$, $\delta p_{y} = +\epsilon p_{x}$. Written out as a matrix acting on the position, the first pair is

which is a rotation by the angle $\epsilon$ about the $z$-axis, to first order, and the momenta rotate the same way, as vectors must. So $L_{z}$ generates rotations about $z$. Nothing was assumed about the system. This is a statement about the function $xp_{y}-yp_{x}$ and the bracket, valid for any Hamiltonian whatsoever.

7.3 · The two-way street

Now put the pieces together, and note that everything needed is antisymmetry.

Read it left to right and it is the converse of Noether's theorem. Read it right to left and it is Noether's theorem. And read as a single line it says something stronger than either. The conserved quantity is not merely associated with the symmetry, and it is not a by-product of it. The conserved quantity is the generator of the symmetry. They are one object. Momentum is spatial translation. Angular momentum is rotation. Energy is time translation. The bracket $\{G,H\}$ is simultaneously "the rate of change of $G$" and "the change in $H$ under $G$", and there is nothing else to say.

This is the single most portable idea in the book, and here is where it goes.

• Chapter 1.4 proves Noether's theorem for a general Lagrangian, including the field-theory version where the conserved quantity becomes a conserved current. This section is the Hamiltonian shadow of it, and it is the direction Noether does not give you.
• Chapter 4.2 repeats every word with operators. There, an observable $\hat G$ generates the one-parameter family of unitary transformations $\ee^{-\ii\epsilon\hat G/\hbar}$, and differentiating that at $\epsilon=0$ gives $\delta\hat f = \tfrac{\ii\epsilon}{\hbar}[\hat G,\hat f]$, which is (1.3.56) under §6.4's substitution. "Observables generate unitary transformations" is this section, quantised. It is why $\hat p=-\ii\hbar\partial_{x}$, the generator of translations, and why $\hat H$ appears in the exponent of the time-evolution operator.
• Chapters 6.1 and 6.2 take the set of generators seriously as an object in its own right. The brackets of generators with each other close into an algebra. Problem 2 computes $\{L_{i},L_{j}\}=\epsilon_{ijk}L_{k}$ and finds $\mathfrak{su}(2)$ arriving classically, and a Lie algebra with its representations is what a "symmetry group" concretely is. Chapter 6.3 then makes the parameter $\epsilon$ depend on position, discovers that this forces a new field into existence, and that field is the photon. The entire gauge principle is this section with $\epsilon\to \epsilon(x)$.

8 · Canonical transformations and Hamilton–Jacobi

This section is a signpost rather than a development. Two ideas, stated precisely enough to be usable, with the machinery left for the chapters that need it.

8.1 · Which changes of variable are allowed

Chapter 1.2 §7 proved that the Euler–Lagrange equations keep their form under any change of coordinates. Hamilton's equations are more demanding, because they involve the momenta, and the momenta are not free to be relabelled independently. They were defined by (1.3.11).

So we need a test for which relabellings are legitimate, and §6.2 has already supplied the thing to test. Call a transformation $(q,p)\to(Q,P)$ canonical if it preserves the fundamental brackets,

with the brackets still computed with respect to the old variables. There is a second way to say the same thing, in the notation of §3.3. If $M^{a}{}_{b} = \partial z'^{a}/\partial z^{b}$ is the Jacobian matrix of the transformation, then (1.3.61) is precisely the statement

which is the definition of a symplectic matrix. Such transformations preserve Hamilton's equations, the new equations being $\dot Q = \partial K/\partial P$, $\dot P = -\partial K/\partial Q$ for a suitable $K$. They preserve every Poisson bracket whatsoever, since (1.3.49) is built from $\Omega$. And taking determinants of (1.3.62) gives $(\det M)^{2}=1$. ⚑ The sharper statement that $\det M = +1$ always, never $-1$, is quoted here. It follows from a Pfaffian argument.

So canonical transformations preserve phase-space volume, and Liouville's theorem is the special case in which the transformation is "evolve for a time $t$". The flow map is canonical, which Problem 3 asks you to check.

Generating functions, in one paragraph. Knowing which transformations are allowed is not the same as knowing how to build one, and there is a practical recipe. Notice that the phase-space action (1.3.26) may differ between old and new variables by a total time derivative without changing the equations of motion (Chapter 1.2, Problem 4). So demand

and let $F=F_{1}(q,Q,t)$. Expanding $\dd F_{1}/\dd t$ by the chain rule and matching the coefficients of the independent quantities $\dot q^{i}$, $\dot Q^{i}$ and $1$ gives

Any $F_{1}$ you write down generates a canonical transformation. Three cousins $F_{2}(q,P,t)$, $F_{3}(p,Q,t)$ and $F_{4}(p,P,t)$ follow by Legendre-transforming $F_{1}$ in one or both arguments, which is §1 again in yet another costume. The grind box works the most useful one.

8.2 · Hamilton–Jacobi

Push the idea to its limit. Suppose you could find a canonical transformation making the new Hamiltonian $K$ identically zero. Then $\dot Q = \partial K/\partial P = 0$ and $\dot P = -\partial K/\partial Q = 0$, so all $2n$ new variables are constants and the system is completely solved.

The price is one equation for the generating function that does it. From (1.3.64), with the generating function called $S$ instead of $F_{1}$, $K=0$ means $H + \partial S/\partial t = 0$ with $p_{i}=\partial S/\partial q^{i}$ substituted into $H$:

This is the Hamilton–Jacobi equation: one first-order partial differential equation for one function $S(q,t)$, entirely equivalent to the $2n$ ordinary differential equations we started with. It is rarely the easiest way to solve a mechanics problem. It is on this page for two reasons.

First, $S$ is the action. Not a function that resembles the action. The action. To see it, compute its total time derivative along a trajectory, using $\partial S/\partial q^{i}=p_{i}$ and (1.3.65):

so $S=\int L\,\dd t$ along the path. The generating function that trivialises the dynamics is Chapter 1.2's action regarded as a function of the endpoint. That is why the letter is the same, and it closes a loop opened two chapters ago.

Second, it is the classical limit of the Schrödinger equation. ⚑ Quoted, with the derivation deferred to Chapter 4.10. The way to see it is to feed the wave equation a wave whose phase is $S/\hbar$, so substitute the ansatz

into $\ii\hbar\,\partial_{t}\psi = -\tfrac{\hbar^{2}}{2m}\nabla^{2}\psi + V\psi$ and separate the real and imaginary parts. The real part, at leading order in $\hbar$, is

which is exactly (1.3.65) for $H=\vv p^{2}/2m+V$ with $\vv p=\nabla S$, and the terms that were dropped are down by one power of $\hbar$. (The imaginary part gives $\partial_{t}\abs{A}^{2} + \nabla\cdot\big(\abs{A}^{2}\nabla S/m\big)=0$, which is Chapter 0.7's continuity equation for probability, with velocity $\nabla S/m$.)

So the phase of a quantum wavefunction is the classical action divided by $\hbar$, which is also what Chapter 1.2 §5 quoted about the path integral weight $\ee^{\ii S/\hbar}$, and what Chapter 5.6 will prove. Classical mechanics is the geometry of surfaces of constant phase, in exactly the way that ray optics is the geometry of surfaces of constant phase of a light wave. Hamilton knew this in 1834, which is ninety-two years before anyone wrote down a wave equation for matter.

Grind box — the $F_{2}$ generating function, and the oscillator solved by Hamilton–Jacobi

The transform between generating functions. $F_{1}(q,Q)$ is awkward because the identity transformation is not expressible in it. Legendre-transform the $Q$ argument: set $F_{2}(q,P,t) = F_{1}(q,Q,t) + \sum_i Q^{i}P_{i}$, treating $P_{i}=-\partial F_{1}/\partial Q^{i}$ as the conjugate slope. Then $\dd F_{2} = \sum_i p_{i}\dd q^{i} + \sum_i Q^{i}\dd P_{i} + (\partial F_{1}/\partial t)\dd t$, giving

$$p_{i} = \pdv{F_{2}}{q^{i}}, \qquad Q^{i} = \pdv{F_{2}}{P_{i}}, \qquad K = H + \pdv{F_{2}}{t}.$$

Now $F_{2}=\sum_i q^{i}P_{i}$ gives $p_{i}=P_{i}$ and $Q^{i}=q^{i}$, the identity, as required. This is the version used in practice, and the one Hamilton–Jacobi uses.

The oscillator, solved from scratch. Take $H = \dfrac{p^{2}}{2m}+\half m\omega^{2}q^{2}$. Since $H$ has no explicit $t$, look for a separated solution $S(q,t) = W(q) - Et$, so that $\partial S/\partial t = -E$ and (1.3.65) becomes

$$\frac{1}{2m}\left(\dv{W}{q}\right)^{2} + \half m\omega^{2}q^{2} = E \;\Longrightarrow\; \dv{W}{q} = \sqrt{2mE - m^{2}\omega^{2}q^{2}}.$$

So $W(q) = \int\sqrt{2mE-m^{2}\omega^{2}q^{2}}\;\dd q$, and $S = W(q)-Et$. Take the constant $E$ to be the new momentum $P$. Then the new coordinate is

$$Q = \pdv{S}{E} = \int\frac{m\,\dd q}{\sqrt{2mE-m^{2}\omega^{2}q^{2}}} - t = \frac{1}{\omega}\arcsin\!\left(q\,\omega\sqrt{\frac{m}{2E}}\right) - t,$$

and $Q$ is constant because $K=H+\partial S/\partial t = E - E = 0$. Setting $Q=-t_{0}$ and solving for $q$:

$$q(t) = \sqrt{\frac{2E}{m\omega^{2}}}\;\sin\big(\omega(t-t_{0})\big).$$

The oscillator solution, with the amplitude expressed through the energy exactly as Chapter 0.8 had it. Note what the method did. It turned solving a differential equation into evaluating an integral, and the integral was the one whose derivative is the period, (1.3.41) again, since $\oint\dd W = \oint p\,\dd q = \mathcal{A}$.

And the action variable. Instead of $E$, take the new momentum to be $J = \mathcal{A}/2\pi = E/\omega$. Then $K = \omega J$ and Hamilton's equations give $\dot\vartheta = \partial K/\partial J = \omega$ and $\dot J = 0$: an angle advancing uniformly and a constant. Every one-dimensional bound system can be brought to this form, which is the content of "action–angle variables" and the starting point for perturbation theory in celestial mechanics. It is also, ⚑ quoted forward, the starting point for the KAM theorem, which says what survives when you switch a small perturbation on and is the reason the solar system's stability is a hard question rather than an obvious one.

9 · Worked examples

10 · Your turn