Representation theory and quantum mechanics tutorial Spin and the hydrogen atom

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1 Representation theory and quantum mechanics tutorial Spin and the hydrogen atom Justin Campbell August 3, Representations of SU 2 and SO 3 (R) 1.1 The following observation is long overdue. Proposition The groups SU n and SO n (R) are compact. Proof. Recall that the Heine-Borel theorem says that a subset of Euclidean space is compact if and only if it is closed and bounded. First, notice that SU n and SO n (R) are closed in Mat n n (R) = R n2, being defined by the equations AA = I = A A and AA = I = A A respectively. They are bounded because all columns of a unitary or orthogonal matrix have unit length (for the Hermitian and ordinary dot products respectively). In particular, all representations of SU n and SO n (R) are completely decomposable. Thus the problem of classifying representations of these groups up to isomorphism reduces to the problem of classifying their irreducible representations up to isomorphism. Consider the case G = SU 2. We will now construct a family of irreducible SU 2 -representations, and then prove that our list is complete. Namely, let V n := {degree n homogeneous polynomials in two variables with C-coefficients}. More concretely, as a C-vector space V n has a basis consisting of the monomials x i y j for i + j = n, with i, j 0. In particular V n is (n + 1)-dimensional over C. Thinking of f(x, y) V n as a polynomial function on C 2, the action of A SU 2 is defined by the formula A f(x, y) = f(a 1 (x, y)). It is not hard to see that this defines a smooth (even R-polynomial) homomorphism SU 2 GL(V n ). For example, the representation V 0 = C 1 is trivial, and V 1 is the standard representation of SU 2 on C 2. Theorem The SU 2 -representations V n for n 0 are all irreducible, and any (n + 1)-dimensional irreducible representation of SU 2 is isomorphic to V n. The following results demonstrate the utility of Lie algebras for studying group representations. Proposition Suppose that G is connected and that V is a G-representation. Then a subspace of V is G-stable if and only if it is g-stable. Corollary A representation V of a connected group G is irreducible as a G-representation if and only if it is irreducible as a g-representation. Moreover, it is not hard to show that we can replace g by g R C in these statements. 1

2 1.2 In this section we prove Theorem By Corollary , in order to show that V n is irreducible as an SU 2 -representation it suffices to prove that it is irreducible as an su 2 -representation, or equivalently as an sl 2 (C)-representation because su 2 R C = sl 2 (C). Recall that sl 2 (C) has the standard C-basis e, f, h. The basic relations are [h, e] = 2e, [h, f] = 2f, and [e, f] = h. One computes that, as operators on polynomials in two variables x, y, we have e = x y, f = y x, and h = x x y y. In particular, on a basis vector x i y j for i + j = n, we have e x i y j = jx i+1 y j 1, f x i y j = ix i 1 y j+1, and h x i y j = (i j)x i y j. Thus h is diagonalizable with eigenvalues n, n 2,, n + 2, n. For this reason we call (C-multiples of) x n and y n highest weight and lowest weight vectors respectively, and e and f are called raising and lowering operators respectively. Irreducibility of V n is immediate from this description. If g V n and d is the y-degree of g, then f d g is a heighest weight vector, i.e. is proportional to x n. Since e j x n is proportional to x i y j for any 0 j n, it follows that g generates V n as an sl 2 (C)-representation. Suppose w W is an eigenvector of h with eigenvalue λ C. Then we have h (e w) = [h, e] w + e (h w) = (λ + 2)e w, so that e w is either zero or an eigenvector of h with eigenvalue λ + 2. Similarly f w is either zero or an eigenvector of h with eigenvalue λ 2. Let λ be the eigenvalue of h acting on W with the greatest real part, and fix an eigenvector w λ W of h with eigenvalue λ. Define w λ 2j := 1 j! f j w λ for j 0. By the computation just performed w λ 2j is either zero or an eigenvector with eigenvalue λ 2j. Moreover, by induction we have e w λ 2j = (λ j + 1)w λ 2j+2. Let k 1 be the least positive integer such that w λ 2k = 0. Then 0 = e w λ 2k = (λ k + 1)w λ 2k+2, and since w λ 2k+2 0, it follows that k = λ + 1 (in particular λ is a nonnegative integer). Moreover, our calculations show that w λ, w λ 2,, w λ span an sl 2 (C)-invariant subspace of W, which by irreducibility must be W itself. Since the w λ, w λ 2,, w λ have distinct eigenvalues they are independent and hence form a basis. In particular λ = n. Now it is easy to check, using the calculations already performed, that the linear map V n W defined by x i y j w n 2j is an sl 2 (C)-equivariant isomorphism. 1.3 Next, we classify representations of G = SO 3 (R). The groups SO 3 (R) and SU 2 are related as follows. Consider the adjoint action of SU 2 on its Lie algebra su 2, given by conjugation: A B = ABA 1 for A SU 2 and B su 2. We take as a basis of su 2 the skew-hermitian Pauli matrices iσ 1, iσ 2, iσ 3, which yields an isomorphism R 3 = su 2. Thus we obtain a three-dimensional real matrix representation of SU 2, i.e. a smooth group homomorphism π : SU 2 GL 3 (R). Proposition The image of π : SU 2 GL 3 (R) is SO 3 (R), and its kernel is {±I}. 2

3 Proof. To see that π(su 2 ) O 3 (R), we first observe that the dot product in R 3 corresponds to the pairing, : su 2 su 2 R given by B, C = 1 2 tr(bc). But SU 2 preserves this inner product by conjugation-invariance of trace: AB 1 A 1, AB 2 A 1 = 1 2 tr(ab 1B 2 A 1 ) = 1 2 tr(b 1B 2 ) = B 1, B 2. Now we show that det π(a) = 1 for A SU 2. For this, recall that O 3 (R) is the disjoint union of two components, and SO 3 (R) is the component containing I. On the other hand SU 2 is connected and π is continuous, so π(i) = I implies π(su 2 ) SO 3 (R) as desired. For the surjectivity of π, recall that SO 3 (R) is generated by rotations Rθ x, Ry θ, Rz θ about the x-, y-, and z-axes respectively. We show that Rθ z is in the image of π, the other axes being similar. For this, put [ ] e iθ 0 A θ := 0 e iθ. We compute directly that A θ iσ 1 A 1 θ = cos(2θ)iσ 1 sin(2θ)iσ 2, A θ iσ 2 A 1 θ = sin(2θ)iσ 1 + cos(2θ)iσ 2, and A θ iσ 3 A 1 θ = iσ 3. This implies that π(a θ ) = R2θ z, whence the surjectivity. Finally, we show that ker π = {±I}. If π(a) = I, then Aσ 1 = σ 1 A, which immediately implies that A = A θ for some θ R. Now σ 1 = Aσ 1 A 1 = cos(2θ)σ 1 sin(2θ)σ 2 implies that sin(2θ) = 0, so A = ±I as desired. From now on we view π as a surjective homomorphism π : SU 2 SO 3 (R). In particular, any representation of SO 3 (R) gives rise to an SU 2 -representation by restriction along π. A representation of SU 2 comes from an SO 3 (R)-representation if and only if I SU 2 acts trivially on the representation. Proposition The Lie algebra homomorphism π : su 2 so 3 (R) is an isomorphism which sends iσ 1 2r x, iσ 2 2r y, and iσ 3 2r z. Proof. The calculation π (iσ 3 ) = 2r z was essentially done in the proof of Proposition 1.3.1, and the others are similar. In particular, it follows that so 3 (R) R C is canonically isomorphic to sl 2 (C). Proposition The SU 2 -representation V n comes from SO 3 (R) by restriction along π if and only if n is even, i.e. V n has odd dimension. Any irreducible representation of SO 3 (R) is isomorphic to one of this form, and in particular has odd dimension. Proof. The action of SU 2 on V n factors through π if and only if ker π = {±I} acts trivially on V n. Since I SU 2 acts by ( I f)(x, y) = f( x, y), this endomorphism of V n is trivial if and only if n is even (so that f(x, y) has even degree). Suppose W is an irreducible representation of SO 3 (R), hence of so 3 (R). Since so 3 (R) R C = sl 2 (C), we can view W as an irreducible sl 2 (C)-representation. But any such is isomorphic to V n for some n 0. 3

4 1.4 The irreducible representations V 0, V 2, V 4, of SO 3 (R) from the previous sections are simple to construct as vector spaces, but the action of SO 3 (R) is not manifest. Rather, the SU 2 -action on V n is obvious, and then one checks that it is induced by an SO 3 (R)-action for n even. We now construct the irreducible representations of SO 3 (R) in a way better-adapted to this group. Namely, we realize these representations using functions on a space with SO 3 (R)-action. To this end, consider the tautological action of SO 3 (R) on R 3 by rotations. This induces an SO 3 (R)-action on the space C (R 3, C) of smooth functions R 3 C via the formula We write (A f)(x, y, z) = f(a 1 (x, y, z)). ρ : SO 3 (R) GL(C (R 3, C)) for the action homomorphism. Differentiating the action of SO 3 (R) on R 3 gives rise to a Lie algebra homomorphism ϕ : so 3 (R) C (R 3, R 3 ) into smooth vector fields on R 3. Recall that we introduced basis elements r x, r y, r z so 3 (R), and one computes that ϕ(r x ) = z y y z, ϕ(ry ) = x z z x, and ϕ(rz ) = y x x y. Now we observe that so 3 (R)-representation obtained from ρ is the composition ρ : so 3 (R) ϕ C (R 3, R 3 ) gl(c (R 3, C)), where the second Lie algebra homomorphism is the action of vector fields on functions. That is, we can compute the action of so 3 (R) on a function R 3 C by applying the differential operators written above. In order to find an SO 3 (R)-subrepresentation of C (R 3, C) isomorphic to V 2l where l 0, it suffices to do the same for the action of so 3 (R), or equivalently so 3 (R) R C = sl 2 (C), on C (R 3, C). Based on our understanding of sl 2 (C)-representations, we should look for a highest weight vector, i.e. an eigenfunction Y l l : R3 C for h with eigenvalue 2l such that e Y l l = 0. It is convenient to work with the spherical coordinates r, θ, φ on R 3. In fact, the space of functions on R 3 is somewhat too large for our present purposes, because the radial coordinate r is SO 3 (R)-invariant. Therefore we consider only functions f(θ, φ) of the two angular coordinates, which can be viewed as smooth functions f : S 2 C on the 2-sphere The equations h Y l l = 2lY l l The first of these implies that S 2 := {(x, y, z) R 3 x 2 + y 2 + z 2 = 1}. and e Y l l = 0 become the differential equations i φ Y l l (θ, φ) = ly l l (θ, φ) and e iφ ( θ + i cot θ φ Y l l (θ, φ) = e ilφ F l (θ) for some smooth function F l (θ), and then the second gives The solutions of the latter equation have the form θ F l(θ) = l cot(θ)f l (θ). F l (θ) = C ll sin l θ for C ll C, which means our desired highest weight vector is Y l l (θ, φ) = C ll e ilφ sin l θ. ) Y l l (θ, φ) = 0. 4

5 We remark that the elements r ± := r x ± ir y of so 3 (R) are raising and lowering operators, i.e. up to scaling they correspond to e and f under the isomorphism so 3 (R) R C = sl 2 (C). Define W l C (S 2, C) to be the subspace spanned by the functions ( ( Yl m (θ, φ) := C lm (r ) l m Yl l (θ, φ) = C lm e iφ θ + i cot θ )) l m Yl l (θ, φ) φ for l m l. Note that r z Yl m = myl m. By our classification of representations it follows that W l is SO 3 (R)-stable, and that it is moreover isomorphic to V 2l. The functions Yl m are called spherical harmonics. One can show that the space of functions C (S 2, C) almost decomposes into the direct sum W l. Namely, an arbitrary smooth function f : S 2 C can be written uniquely as an infinite sum f(θ, φ) = C lm Yl m (θ, φ) l 0 l m l for some C ml C. Compare to the Fourier series decomposition of C (S 1, C). 2 Quantization and the hydrogen atom l Before analyzing the hydrogen atom, we need to discuss the process of passing from classical mechanical systems to quantum systems, called quantization. Consider the classical mechanical situation of a particle of mass m > 0 moving in space R 3, so that the phase space is R 6 with the usual coordinates q 1, q 2 q 3, p 1, p 2, p 3. The Hamiltonian of this system generally takes the form H(q, p) = p 2 2m + V (q), where V : R 3 R is the potential function. How do we produce a quantum mechanical system analogous to this one? The first question is how to construct the Hilbert space space of states. We propose H = L 2 (R 3 ), the space of measurable (on a first pass, ignore this condition) functions ψ : R 3 C such that R 3 ψ(q) 2 dq <. Moreover, if ψ : R 3 C is such that R 3 ψ(q) 2 dq = 0 (equivalently, ψ vanishes outside a set of measure zero) we identify ψ with the zero function. We call the state vectors ψ L 2 (R 3 ) wavefunctions of the particle. For any ψ 1, ψ 2 L 2 (R 3 ), we can define ψ 1, ψ 2 := ψ 1 (q)ψ 2 (q) dq, R 3 which converges by virtue of the L 2 condition. This Hermitian inner product makes L 2 (R 3 ) into a Hilbert space, which we view as the state space of the particle. 5

6 2.2 The next step is to quantize observables, meaning pass from functions R 6 R to self-adjoint operators on L 2 (R 3 ). The most basic observables for our classical system are the position and momentum coordinates, so we quantize these first. The position operators Q i, corresponding to measurement of the position coordinate q i, are defined by Q i (ψ)(q) := q i ψ(q). Unfortunately, the resulting function R 3 R need not belong to L 2 (R 3 ), but we simply ignore this issue. The operators Q i are evidently self-adjoint to the extent that it makes sense. A related problem is that the eigenstates of Q i do not belong to L 2 (R): they are the so-called delta distributions δ q for q R 3. These objects are characterized by the formula R 3 ψδ q = ψ(q), and clearly there no functions with this property. They form an eigenbasis of L 2 (R) in the sense that any function ψ : R 3 R can be written as an integral ψ = ψδ q, q R 3 so that the coefficient of δ q in this expansion of ψ is ψ(q). The upshot of this discussion is that the probability of observing a particle with wavefunction ψ at the position q is ψ(q) 2. The momentum operators are given by the formula P i := i q i. Like the position operators, they are not well-defined on a general wavefunction, but we will not comment on this. Another similarity with position is that the eigenstates of P i do not belong to L 2 (R): they have the form e ikq for k R. The eigenvalue of Q i acting on e ikq is k, the so-called de Broglie momentum. Exercise Use integration by parts to check the the momentum operators are self-adjoint, insofar as they are defined. The transition map between the position and momentum eigenbases is the Fourier transform. In particular, if ˆψ : R 3 C is the Fourier transform of ψ, suitably normalized, then the probability of measuring the momentum of a particle with wavefunction ψ to be p is ˆψ(p) 2. Another observable we must quantize is the Hamiltonian. Having done this for position and momentum coordinates, this is fairly straightforward: the Hamiltonian operator on L 2 (R 3 ) is H = 1 2m (P P P 2 3 ) + V (Q 1, Q 2, Q 3 ). Here V (Q 1, Q 2, Q 3 ) is the operator of multiplying by V (q), so there is not much harm in simply writing V (q) as the potential term in the Hamiltonian operator. It is convenient to use the notation := 2 q q q3 2 for the Laplace operator. The Hamiltonian can then be rewritten as H = 2 2m + V (q). 2.3 Now we introduce SO 3 (R) symmetry. The action of SO 3 (R) on R 3 defines a representation ρ : SO 3 (R) GL(L 2 (R 3 )) via the usual formula (A ψ)(q) = ψ(a 1 q). 6

7 This is a unitary representation because rotating a function in this manner evidently leaves its integral unchanged. One then computes that the induced Lie algebra representation ρ : so 3 (R) gl(l 2 (R 3 )) sends ( ρ (r x ) = i(q 2 P 3 Q 3 P 2 ) = ( ρ (r y ) = i(q 3 P 1 Q 1 P 3 ) = ( ρ (r z ) = i(q 1 P 2 Q 2 P 1 ) = q 2 q 3 q 1 ) q 3, q 3 q 2 ) q 1, q 1 q 3 ) q 2. q 2 q 1 It therefore seems reasonable to define the angular momentum operators by the formulas L 1 := Q 2 P 3 Q 3 P 2, L 2 := Q 3 P 1 Q 1 P 3, L 3 := Q 1 P 2 Q 2 P 1. These self-adjoint operators correspond to measurement of angular momentum with respect to the x-, y-, or z-axes respectively. Since ρ is a Lie algebra homomorphism, we automatically have the relations [ il 1, il 2 ] = il 3, [ il 2, il 3 ] = il 1, [ il 3, il 1 ] = il 2. (2.3.1) The Casimir operator for so 3 (R) acting on L 2 (R 3 ) is defined by L 2 := L L L 2 3. Exercise Use the relations (2.3.1) to show that L 2 commutes with ρ (x) for any x so 3 (R). Thus L 2 also commutes with the action of SO 3 (R). One can show that in spherical coordinates, ( 1 L 2 = sin θ θ ( sin θ ) + 1 θ sin 2 θ 2 ) φ 2. Then we rewrite the Laplacian as = 2 r r r 1 r 2 L2. Since the action of SO 3 (R) leaves the radial coordinate r = q invariant, Exercise implies that commutes with the SO 3 (R)-action. In particular, if the potential V : R 3 R only depends on r, then H commutes with the SO 3 (R)-action, and consequently angular momentum is conserved. 2.4 Finally, we come to the hydrogen atom. More precisely, we consider the case of an electron moving in a Coulomb potential produced by a proton fixed at the origin. In this case the Hamiltonian is H = 2 2m e2 r, where e is the electron s charge. Since the potential is rotation-invariant, we know by the previous section that the SO 3 (R)-action commutes with H, and therefore that L 1, L 2, and L 3 are conserved quantities. The upshot of this discussion is that any eigenspace of H, consisting of states with a definite energy, will be SO 3 (R)-stable. In particular, we may ask how they decompose into irreducible SO 3 (R)-representations, which we previously classified. When searching for copies of the irreducible representation V 2l = Wl, the following will be helpful. 7

8 Proposition If ϕ : so 3 (R) gl(w l ) is the action map, then the operator acts on W l by the scalar l(l + 1). ϕ(r x ) 2 + ϕ(r y ) 2 + ϕ(r z ) 2 Therefore wavefunctions satisfy the differential equation L 2 ψ = l(l + 1)ψ if and only if they generate a copy of W l in L 2 (R 3 ) as an SO 3 (R)-subrepresentation. In particular, to find wavefunctions with energy (i.e. H-eigenvalue) E which belong to W l, we can first look for their radial component by solving the equation Hg l,e (r) = ( ( 2 d 2 2m dr r Then, for any l m l, we will have and the wavefunctions d l(l + 1) dr r 2 Hg l,e (r)y l m(θ, ψ) = Eg l,e (r)y l m(θ, ψ), ψ(r, θ, ψ) = g l,e (r)y l m(θ, ψ) ) ) e2 g l,e (r) = Eg l,e (r). r will form a basis for a copy of W l in the E-eigenspace of H. We assume that E < 0, which only allows bound states corresponding to electron orbitals, as opposed the scattering states which have E > 0. It turns out that for fixed l 0, there is precisely one solution g l,en (r) for each n > l, where E n := me4 2 2 n 2. In particular, at a fixed energy E n there are n solutions g l,en (r) corresponding to 0 l < n. Thus the E n -eigenspace of H decomposes as W 0 W 1 W n 1. Using this decomposition, we can explain three of the so-called quantum numbers which classify states of an electron in a hydrogen atom. Namely, n is the so-called principal quantum number, which determines the energy. The orbital quantum number l is responsible for the letter in the spectroscopic notation for orbitals: s means l = 0, p means l = 1, etc. The magnetic quantum number is m, the eigenvalue of L 3. Finally, we have until now ignored the spin of the electron. This doubles the number of states and introduces a new spin quantum number s = ±