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Home , Circle group

and quantum mechanics tutorial A little Lie theory

Justin Campbell July 6, 2017

1 Groups

1.1 The colloquial usage of the words “” and “symmetrical” is imprecise: we say, for example, that a regular pentagon

is symmetrical, but what does that mean? In the mathematical parlance, a symmetry (or more technically automorphism) of the pentagon is a (distance- and angle-preserving) transformation which leaves it unchanged. It has ten such : 2π 4π 6π 8π rotations through 0, 5 , 5 , 5 , and 5 radians, as well as reflection over any of the five lines passing through a vertex and the center of the pentagon. These ten symmetries form a set D5, called the dihedral of 10. Any two elements of D5 can be composed to obtain a third. This operation of composition has some important formal properties, summarized as follows. Definition 1.1.1. A group is a set G together with an operation

G × G → G, denoted by (x, y) 7→ xy, satisfying: (i) there exists e ∈ G, called the identity, such that ex = xe = g for all x ∈ G, (ii) any x ∈ G has an inverse x−1 ∈ G satisfying xx−1 = x−1x = e, (iii) for any x, y, z ∈ G the associativity law (xy)z = x(yz) holds. To any mathematical (geometric, algebraic, etc.) object one can attach its automorphism group consisting of structure-preserving invertible maps from the object to itself. For example, the automorphism group of a regular pentagon is the D5. For another example, let n ≥ 0 and consider Rn as a over R. Linear maps Rn → Rn can be thought of as n × n matrices with coefficients in R, a space we denote by Matn×n(R). Composition of linear maps goes over to multiplication. The linear automorphism group of Rn is

GLn(R) := {A ∈ Matn×n(R) | A is invertible},

the nth over R. Recall a square matrix A is invertible if and only if det A 6= 0. We also make frequent use of the related nth over R

SLn(R) := {A ∈ Matn×n(R) | det A = 1}.

1 1.2 Next we consider a geometrical figure more symmetrical than a regular pentagon, namely a

1 2 2 2 S := {(x, y) ∈ R | x + y = 1}. We call it more symmetrical because it has infinitely many symmetries, given by the second

> > O2(R) := {A ∈ GL2(R) | AA = A A = I}. 2 Equivalently, one can define O2(R) to consist of those A ∈ O2(R) which preserve the dot product on R , meaning 2 O2(R) = {A ∈ GL2(R) | (Av) · (Aw) = v · w for all v, w ∈ R }. This explains the term “orthogonal,” because the length of and angle between vectors can be expressed in terms of the inner product using the formulas √ v · w |v| = v · v and cos θ = . |v||w|

This means that O2(R) is precisely the group of linear transformations of the plane which preserve distance and angle, the basic quantities in Euclidean geometry. Any symmetry of a regular pentagon centered at (0, 0) is also a symmetry of the circle, so we have an inclusion D5 ⊂ O2(R) which realizes the former group as a of the latter. A physicist might say that the pentagon breaks the symmetry of the circle, reducing it from an infinite group to a group with ten elements. ∼ 4 Since the group O2(R) is infinite and it comes as a subset of Mat2×2(R) = R , we ought to think of it as a geometric figure in its own right, rather than an unstructured jumble of points. In fact, it consists of a disjoint union of two , which can be parameterized as cos θ − sin θ cos θ sin θ  θ 7→ and θ 7→ . sin θ cos θ sin θ − cos θ The first circle is the subgroup SO2(R) := SL2(R) ∩ O2(R) consisting of those A ∈ O2(R) satisfying det A = 1, and the other component (which is not a subgroup) consists of A ∈ O2(R) such that det A = −1. More geometrically, SO2(R) consists of all rotations about (0, 0), while a matrix in the other component of O2(R) is a reflection over a line through (0, 0). 1 1 In particular, for any v ∈ S , the map SO2(R) → S given by A 7→ Av is bijective. We also remark that our parameterization cos θ − sin θ θ 7→ sin θ cos θ

of SO2(R) sends addition of angles to matrix multiplication, i.e. cos(θ + ϕ) − sin(θ + ϕ) cos θ − sin θ cos ϕ − sin ϕ = . sin(θ + ϕ) cos(θ + ϕ) sin θ cos θ sin ϕ cos ϕ

1.3 Although the group GL2(R) is nonabelian, its subgroup SO2(R) consisting of rotations in the plane is abelian. On the other hand, rotations in three-dimensional space form a nonabelian group. First we introduce the third orthogonal group

> > O3(R) := {A ∈ GL3(R) | AA = A A = I}.

As in the case of O2(R), a matrix A ∈ O3(R) satisfies det A = ±1, and we put

SO3(R) := SL3(R) ∩ O3(R). Note that reflection through a plane containing (0, 0, 0) has −1, while a about a line containing (0, 0, 0) has determinant 1. If L is an oriented line (meaning it has a distinguished positive L direction) then we write Rθ ∈ SO3(R) for the rotation about L through θ radians, counterclockwise with respect to the given orientation of L according to the right-hand rule.

2 L Proposition 1.3.1 (Euler). Any element of SO3(R) has the form Rθ for some oriented line L through (0, 0, 0) and some angle 0 ≤ θ < 2π.

Proof. Fix A ∈ SO3(R), and let χA(λ) = det(λI − A) be its characteristic polynomial. Being a polynomial in three variables with real coefficients, χA necessarily has a real zero (reason: its three complex roots are permuted by complex conjugation, so there can be at most one pair of conjugate non-real roots). We claim that A fixes a nonzero vector, i.e. it has 1 as an eigenvalue. If v is an eigenvector of A with real eigenvalue λ, then |v| = |Av| = |λv| = |λ||v| implies that λ = ±1. Now if A has three real eigenvalues λ, µ, ν, then since det A = 1, either λ = µ = ν = 1, or −1 occurs with multiplicity two and 1 appears once. In either case one of the eigenvalues is 1. In the remaining case, A has one real eigenvalue λ = ±1 and two eigenvalues µ, µ. Then

λ|µ| = λµµ = det A = 1

implies that λ = 1. Let L be the line spanned by an eigenvector of A with eigenvalue 1, i.e. A acts as the identity on L. Since A is orthogonal, it preserves the plane H ⊂ R3 through (0, 0, 0) perpendicular to L. Moreover, since ∼ 2 det A = 1 and det A|L = 1, we must have det L|H = 1. Thus L|H is a linear transformation of H = R which preserves the dot product and has determinant one, i.e. a rotation.

In words, Euler’s proposition says that SO3(R) is precisely the subgroup of O3(R) consisting of rotations. It follows that any element of SO3(R) can be written as a composition Rx Ry Rz , θ1 θ2 θ3 where we use Rx to denote rotation counterclockwise around the x-axis, etc. Thus, in a sense we have not rigorously defined, the group SO3(R) is three-dimensional. We can explicitly write 1 0 0   cos θ 0 sin θ cos θ − sin θ 0 x y z Rθ = 0 cos θ − sin θ ,Rθ =  0 1 0  ,Rθ := sin θ cos θ 0 . 0 sin θ cos θ − sin θ 0 cos θ 0 0 1

2 Lie algebras

2.1 The group SO3(R) is geometrically somewhat complicated. It is useful to have a “linear approximation” to such a group, and the appropriate object is defined as follows.

Definition 2.1.1. A over R is a vector space g over R equipped with a bilinear operation [ , ]: g × g → g

called the Lie bracket, which is required to satisfy (i) skew-symmetry, meaning for any x, y ∈ g we have [x, y] = −[x, y],

(ii) the Jacobi identity, which says that for any x, y, z ∈ g we have

[x, [y, z]] + [z, [x, y]] + [y, [z, x]] = 0.

The most trivial example of a Lie algebra is a vector space g with the bracket which is identically zero. Such a Lie algebra is called abelian. For any n ≥ 1, one can equip Matn×n(R) with the commutator bracket [A, B] := AB − BA.

3 th The Lie algebra gln(R) so defined is called the n general linear Lie algebra; it is nonabelian for n > 1. We now explain the sense in which gln(R) is a “linear approximation” of GLn(R). First, since GLn(R) is an open subset of the vector space Matn×n(R) = gln(R), it is natural to view gln(R) as the tangent space of GLn(R) at any point, and we may as well take that point to be the identity matrix I ∈ GLn(R). To reiterate, gln(R) = TI (GLn(R)) where the notation on the right side indicates the tangent space of GLn(R) at I. Of course gln(R) contains more information than its underlying vector space, namely the commutator bracket. It is evident from the formula that this bracket measures the extent to which two matrices do not commute, and one can interpret this in terms of the group law on GLn(R) as follows. We can deform I in the direction of A ∈ gln(R) by considering the matrix I + tA for t ∈ R. If |t| is small enough then I + tA ∈ GLn(R), and in particular we have (I + tA)−1 = I − tA + t2A − · · · .

For |t| small the quantity t2 is negligible: after all, we are interested in a linear, i.e. first-order, approximation of GLn(R). Thus, first order we have (I + tA)−1 = I − tA.

To quantify the extent to which I + sA and I + tB do not commute, we consider their commutator

(I + sA)(I + tB)((I + sB)(I + tA))−1 = (I + sA)(I + tB)(I + sA)−1(I + tB)−1, which equals I if and only if I + sA and I + tB commute.

Exercise 2.1.2. Show that the coefficient of st in

(I + sA)(I + tB)(I + sA)−1(I + tB)−1 is [A, B]. Equivalently, we have   d d −1 −1 [A, B] = (I + sA)(I + tB)(I + sA) (I + tB) . (2.1.1) ds dt s,t=0

2.2 Now we explain a procedure by which we can attach a Lie algebra to a group other than GLn(R). First, we must understand tangent spaces more systematically. n Let f1(x1, ··· , xn), ··· , fr(x1, ··· , xn) be a collection of (infinitely) differentiable functions R → R, which for us will usually be polynomials. Put

n X := {(x1, ··· , xn) ∈ R | f1(x1, ··· , xn) = ··· = fr(x1, ··· , xn) = 0.}

n and suppose that x = (x1, ··· , xn) ∈ X. We define the tangent space Tx(X) ⊂ R to X at x to be the vector space of solutions to the linear equations

∂f1 ∂f1 x1 + ··· + xn = 0 ∂x1 x ∂xn x . .

∂fr ∂fr x1 + ··· + xn = 0. ∂x1 x ∂xn x

Note that x is not necessarily in Tx(x) but 0 is, so that x ∈ x+Tx(X). This “tangent affine space” x+Tx(X) is the one we learn to draw on curves and surfaces in a calculus class.

4 We will only consider those X ⊂ Rn which are smooth, i.e. their tangent space has the same dimension at every x ∈ X, which we may then meaningfully call the dimension of X. Suppose we are given a subgroup G ⊂ GLn(R) such that G = GLn(R) ∩ X for some

n2 X ⊂ Matn×n(R) = R defined by equations as above. Then G is automatically smooth, because for any x, y ∈ G, translation by −1 yx defines an isomorphism Tx(G)→ ˜ Ty(G). The Lie algebra g attached to the group G has underlying vector space TI (G) ⊂ gln(R), with the Lie bracket inherited from gln(R). To make sense of this, we must verify that g is closed under the commutator bracket.

Proposition 2.2.1. For A, B ∈ g where g ⊂ gln(R) is as above, we have [A, B] ∈ g. Proof. First we claim that for any C ∈ G, we have CBC−1 ∈ g. One can show (using the implicit function theorem, for instance) that for some  > 0, there exists a path β :(−, ) → G such that β0(0) = B. Since t 7→ Cβ(t)C−1 is a path in G sending 0 7→ G, its derivative at 0, which is CBC−1, lies in g. Now there exists δ > 0 and a path α :(−δ, δ) → G such that α(0) = A. A similar calculation to the one in Exercise 2.1.2 shows that   d −1 α(t)Bα(t) − B = [A, B]. dt t=0 Since t 7→ α(t)Bα(t)−1 − B is a path in g, its derivative at 0 lies in g as well.

× 2.3 The construction above is simple in the case G = SLn(R). The derivative of det : GLn(R) −→ R at I is tr : gln(R) → R, so sln(R) := {A ∈ gln(R) | tr A = 0} is the Lie algebra of SLn(R). In this case Proposition 2.2.1 is obvious, since tr(AB) = tr(BA) and hence tr([A, B]) = 0 for any A, B ∈ gln(R). For G = On(R), observe that

> on(R) = {A ∈ gln(R) | A = −A} is the Lie subalgebra of skew-symmetric matrices. Indeed, differentiating the relation A>A = I yields > A + A = 0. Since SOn(R) is the connected component of Øn(R) containing the identity matrix, we have

son(R) = sln(R) ∩ on(R) = on(R).

Let G = SO2(R) be the circle group. According to the previous paragraph so2(R) consists of those 2 × 2 matrices A such that A> = −A. Such matrices have the form

 0 a −a 0

∼ 4 for a ∈ R, and in particular they form a line in gl2(R) = R . Since SO2(R) is abelian, so is its Lie algebra, meaning the Lie bracket on so2(R) vanishes identically. Exercise 2.3.1. Show that a one-dimensional Lie algebra is abelian.

We now consider the rotation group G = SO3(R) in three dimensions and explain the structure of its Lie x y z algebra so3(R) = o3(R) in more familiar terms. As the angle θ varies, the matrices Rθ ,Rθ ,Rθ defined above make up three isomorphic to the circle group SO2(R). In particular, they form smooth paths through I, which can be differentiated to obtain tangent vectors

0 0 0   0 0 1 0 −1 0 x y z r := 0 0 −1 , r :=  0 0 0 , r := 1 0 0 ∈ so3(R). 0 1 0 −1 0 0 0 0 0

5 These matrices form a basis of the three-dimensional Lie algebra so3(R), i.e. the linear map

3 R −→ so3(R) (2.3.1) which sends (a, b, c) 7→ arx + bry + crz is an isomorphism. Since the commutator bracket is bilinear and skew-symmetric, it is determined by the three equations

[rx, ry] = rz, [ry, rz] = rx, [rz, rx] = ry.

Exercise 2.3.2. Verify these relations among rx, ry, rz.

Recall that R3 has a skew-symmetric bilinear product (v, w) 7→ v × w, called the cross product, uniquely characterized by e1 × e2 = e3, e2 × e3 = e1, e3 × e1 = e2. 3 Here e1, e2, e3 is the standard basis of R , sometimes denoted by i, j, k. Geometrically v × w is a vector of length |v||w| perpendicular to both v and w, with its orientation determined by the “right-hand rule”: as the fingers on your right hand sweep out the arc from v to w, your thumb points in the direction of v × w. x y z The isomorphism (2.3.1) sends e1 7→ r , e2 7→ r , and e3 7→ r , so the calculations above show that the 3 cross product in R corresponds to the commutator bracket in so3(R).

2.4 The Heisenberg Lie algebra is significant in both classical and quantum mechanics. As a vector space 2n+1 g = R , with the coordinates labeled as q1, ··· , qn, p1, ··· , pn, z. The Lie bracket is determined by the canonical commutation relations

[qi, qj] = 0, [, pj] = 0, [qi, pj] = 0, [qi, pi] = z, [z, qi] = 0, [z, pi] = 0, (2.4.1) where i 6= j. We remark that g can also be realized as the subalgebra of gln+2(R) consisting of matrices whose only nonzero entries lie in the first row and last column, with the (1, 1) and (n + 2, n + 2) entries equal to zero. For example, the 5-dimensional Heisenberg algebra can be viewed as the space of matrices of the form   0 q1 q2 z 0 0 0 p1   . 0 0 0 p2 0 0 0 0 Exercise 2.4.1. Show that the commutator bracket on matrices of this form satisfies the relations (2.4.1), whence this matrix Lie algebra is isomorphic to the Heisenberg Lie algebra. The Heisenberg algebra is contained within the following infinite-dimensional Lie algebra, which will appear later as the Poisson bracket of observables in . Let C∞(R2n) be the space of 2n smooth real-valued functions on R . We label the coordinates q1, ··· , qn, p1, ··· , pn. The Poisson bracket of f, g ∈ C∞(R2n) is defined by the formula

n X  ∂f ∂g ∂f ∂g  {f, g} := − . ∂q ∂p ∂p ∂q i=1 i i i i

Exercise 2.4.2. Check that the Poisson bracket makes C∞(R2n) into a Lie algebra, and that the Leibniz identity {f, gh} = {f, g}h + g{f, h} holds for f, g, h ∈ C∞(R2n). ∞ 2n We remark that the Lie subalgebra of C (R ) spanned by the coordinates q1, ··· , qn, p1, ··· pn and the constant function 1 is the (2n + 1)-dimensional Heisenberg algebra, with 1 corresponding to z. Indeed, it is easy to check that the relations (2.4.1) are satisfied. Here is another example of an infinite-dimensional Lie algebra occurring in geometry. The vector space C∞(Rn, Rn) of smooth functions Rn → Rn can be interpreted as the space of smooth vector fields on Rn.

6 ∂ n n For 1 ≤ i ≤ n, we denote by the constant vector field with value ei ∈ . A general vector field ξ on ∂xi R R can be written ∂ ∂ ξ = f1 + ··· + fn ∂x1 ∂xn n for some smooth functions f1, ··· , fn : R → R. Vector fields act on functions by the obvious formula, suggested by the notation: ∂f ∂f ξ · f := f1 + ··· + fn . ∂x1 ∂xn P ∂ P ∂ n The Lie bracket of vector fields ξ = fi and ζ = gi on is defined by the formula i ∂xi i ∂xi R

n X ∂ [ξ, ζ] := (ξ · g − ζ · f ) i i ∂x i=1 i n n   X X ∂gi ∂fi ∂ = f − g . j ∂x j ∂x ∂x i=1 j=1 j j i

Exercise 2.4.3. Show that C∞(Rn, Rn) is a Lie algebra under the Lie bracket of vector fields, and that the Leibniz identity [ξ, fζ] = (ξ · f)ζ + f[ξ, ζ] holds for all ξ, ζ ∈ C∞(Rn, Rn) and all f ∈ C∞(Rn).

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