Representation theory and quantum mechanics tutorial Conserved quantities and the moment map Justin Campbell July 13, 2017 1 The free particle and Hamilton's equation 1.1 Consider the classical mechanical problem of a particle of mass m travelling through space, which we 3 model by R with coordinates labelled q1; q2; q3. The state of such a particle is completely determined by its position and velocity. These quantities can be represented by two 3-vectors, i.e. the phase space of the particle is R6. Rather than the trajectory of the particle taking place in the configuration space R3 which records only position, we think of it as rather in the phase space R6 whose points are states of the particle. For various reasons we typically use momentum p than velocity v, although the two are of course related by the equation p = mv. Suppose that the particle experiences a force dependent on its position, expressible as a gradient vector field F = −∇V = − @V ; − @V ; − @V ; @q1 @q2 @q3 where V : R3 ! R is a smooth function called the potential. The total energy of the particle is then given by jpj2 H(q; p) = + V (q); 2m where p 2 R3 is the momentum of the particle. Here the first term represents kinetic energy and the second \potential energy." The function H : R6 ! R which attaches to a state its energy is called the Hamiltonian of the system. According to Hamiltonian mechanics, the time evolution of such a system is given by Hamilton's equations dq @H dp @H i = and i = − : (1.1.1) dt @pi dt @qi If the potential vanishes, so that H is just the kinetic energy, we say that the particle is free. In this case Hamilton's equations become dq @H p dp @H i = = i and i = − = 0: dt @pi m dt @qi The first equation is simply the aforementioned relation p = mv. The second says that momentum is conserved in this system. Exercise 1.1.1. Prove the law of conservation of energy (for an arbitrary Hamiltonian) using equations (1.1.1). Using these equations, it is easy to see that if r : R ! R6 is the trajectory of a free particle, then p(0) r(t) = (q(0) + t; p(0)): m This is the best possible outcome for describing the time evolution of a physical system: an explicit formula for the state at a given time in terms of the initial state. 1 1.2 Thus we see that analyzing conserved quantities yields a great deal of information about the dynamics of a given system, although these quantities do not always completely determine possible trajectories as in the case of the free particle. What does all of this have to do with symmetry? Given Hamilton's equations, the reason that the momentum of a free particle is conserved is because its Hamiltonian is invariant under translations in space, i.e. the kinetic energy does not depend on position. Indeed, the second of Hamilton's equations (1.1.1) says that a chosen momentum coordinate is conserved if and only if the Hamiltonian does not depend on the corresponding position coordinate. Noether's theorem, which is often formulated in Lagrangian (rather than Hamiltonian) mechanics, says roughly that any symmetry of a physical system gives rise to a conserved quanty. In particular, the Hamil- tonian must be invariant under the symmetry in question to obtain a conserved quantity. We will state and prove a precise version of Noether's theorem, but first we introduce a more general Hamiltonian formalism. An observable of a physical system is a (say, smooth) real-valued function on the state space. For example, the Hamiltonian of any system is an observable, corresponding physically to the energy. The position and momentum coordinates of a particle are also observables. For concreteness, we suppose that the phase space of the physical system whose dynamics we wish to des- 2n cribe has the form R for some n ≥ 1, with position and momentum coordinates labelled q1; ··· ; qn; p1; ··· ; pn. Fix an observable f : R2n ! R, which we do not allow to explicitly depend on time (although this possibility can be incorporated). The time evolution of f can be described by the equation n df X @f @H @f @H = ff; Hg := − : (1.2.1) dt @q @p @p @q i=1 i i i i Taking f to be either a position or momentum coordinate, we recover equations (1.1.1). Remark 1.2.1. We should mention a peculiarity of the notation which we have thus far ignored. Despite df the fact that we do not allow an observable f to depend explicitly on time, the derivative dt can be nonzero. Why? Because this derivative is not simply the partial derivative of the function f with respect to time, @f 2n which would be denoted by @t and indeed must vanish under our assumption. Rather, if r :(−, ) ! R is a physical trajectory (i.e. a smooth path satisfying the equations (1.1.1)) such that r(0) = (q; p), then df d := f(r(t)) : dt (q;p) dt t=0 This makes sense because there exists a unique trajectory through any point in R2n by the Picard-Lindel¨of theorem on ordinary differential equations. Exercise 1.2.2. Derive equation (1.2.1) from equations (1.1.1). df We call an observable f a conserved quantity if dt = 0. According to equation (1.2.1), this is equivalent to ff; Hg = 0, i.e. f Poisson-commutes with the Hamiltonian. For example, f = H itself is a conserved quantity because any function Poisson-commutes with itself. That is, Exercise 1.1.1 is trivial given equation (1.2.1). 2 The moment map and Noether's theorem 2.1 Fix a real algebraic group G ⊂ GLm(R), which we remind the reader means a subgroup which is the vanishing locus of some polynomial equations. Definition 2.1.1. By an action of G on Rn we mean a smooth map n n G × R −! R ; written (g; x) 7! g · x, such that (i) we have I · x = x for all x 2 Rn, and 2 (ii) we have (gh) · x = g · (h · x) for all g; h 2 G and x 2 Rn. m For example, G acts on R by restricting the action of GLm(R), i.e. we can multiply an element of G (which is a square matrix) by a vector to obtain another vector. Another example is G = Rn acting on Rn by translations, i.e. g ·x = g +x. In most physical examples group actions fall into these two classes, or some combination thereof. In particular, actions in practice tend to be given by a polynomial map G × Rn ! Rn, as opposed to a transcendental smooth map. Exercise 2.1.2. Show that Rm is a real algebraic group, meaning find an injective group homomorphism n R ! GLm(R) for some m ≥ 1 whose image is the vanishing locus of polynomials. Hint: this is possible with m = 2n. An action of G on configuration space Rn induces an action on phase space R2n, given by the formula −1 > g · (q; p) := (g · q; (D(g )jq) p): −1 −1 n n Here D(g )jq 2 Matn×n(R) denotes the Jacobian matrix of the smooth map g : R ! R , evaluated at q 2 Rn. Remark 2.1.3. The following point may seem subtle, but it is fundamental in coordinate-free treatments of this material. The transpose and inverse appear in the above formula because phase space R2n is the cotangent bundle to Rn, as opposed to the tangent bundle. That is, for a given position q 2 Rn the space of momenta n 2n f(q; p) j p 2 R g ⊂ R is to be thought of as the linear dual n ∗ n Tq(R ) := HomR(Tq(R ); R) n n n ∼ n of the tangent space Tq(R ) of R at q. There is a canonical isomorphism Tq(R ) = R , and hence n ∗ ∼ n Tq(R ) = R using the dual basis. Under these isomorphisms dualization of a linear map corresponds to matrix transpose. 2.2 Let G be as above with Lie algebra g. Given an action of G on the configuration space Rn, we will now construct a linear transformation 1 2n µ : g −! C (R ) called the moment map. It is a Lie algebra homomorphism: µ([x; y]) = fµ(x); µ(y)g for all x; y 2 g, i.e. µ sends the Lie bracket on g to the Poisson bracket on observables. The moment map is a composition ' 1 n n 1 2n g −! C (R ; R ) −! C (R ); i.e. it factorizes through the space of vector fields on Rn. The map ' is defined as follows. Given x 2 g, we can find a smooth path γ :(−, ) ! G such that γ(0) = I and γ0(0) = x. Then the the vector field '(x) is defined by d q 7! γ(t) · q : dt t=0 Exercise 2.2.1. Verify that '(x): Rn ! Rn so defined is smooth, and that ' is R-linear. Proposition 2.2.2. The map ' is a Lie algebra antihomomorphism. The map is defined by the formula (ξ)j(q;p) := ξjq · p; where ξ 2 C1(Rn; Rn) and the right side is just the usual dot product of n-vectors. Exercise 2.2.3. Show that is a Lie algebra antihomomorphism, i.e. ([ξ; ζ]) = −{ (ξ); (ζ)g for all ξ; ζ 2 C1(Rn; Rn), by a direct calculation.