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Quantum Mechanics 2 Tutorial 9: the Lorentz Group

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Quantum Mechanics 2 Tutorial 9: The Lorentz Group Yehonatan Viernik December 27, 2020 1 Remarks on Lie Theory and Representations 1.1 Casimir elements and the Cartan subalgebra Let us first give loose definitions, and then discuss their implications. A Cartan subalgebra for semi-simple1 Lie algebras is an abelian subalgebra with the nice property that op- erators in the adjoint representation of the Cartan can be diagonalized simultaneously. Often it is claimed to be the largest commuting subalgebra, but this is actually slightly wrong. A correct version of this statement is that the Cartan is the maximal subalgebra that is both commuting and consisting of simultaneously diagonalizable operators in the adjoint. Next, a Casimir element is something that is constructed from the algebra, but is not actually part of the algebra. Its main property is that it commutes with all the generators of the algebra. The most commonly used one is the quadratic Casimir, which is typically just the sum of squares of the generators of the algebra 2 2 2 2 (e.g. L = Lx + Ly + Lz is a quadratic Casimir of so(3)). We now need to say what we mean by \square" 2 and \commutes", because the Lie algebra itself doesn't have these notions. For example, Lx is meaningless in terms of elements of so(3). As a matrix, once a representation is specified, it of course gains meaning, because matrices are endowed with multiplication. But the Lie algebra only has the Lie bracket and linear combinations to work with. So instead, the Casimir is living inside a \larger" algebra called the universal enveloping algebra, where the Lie bracket is realized as an explicit commutator. This is just a bunch of technicalities, with the sole intention to slightly de-mystify these concepts, and you are not expected to go and study abstract algebra. You only need to know that it is well defined within something very similar to the Lie algebra, and is actually even slightly less abstract (Lie bracket is explicitly the commutator), and its definition can be naturally extended to any actual representation of the algebra, e.g. when taking L2 as a 3 × 3 matrix. Now the implications. There is a famous lemma called Schur's lemma, which says the following. Suppose you have an irrep R, where for every element g in your algebra, R(g) is some operator (matrix) acting in your vector space V . Then any other operator T : V ! V that commutes with all R(g) must be a scalar operator. That is, if 8g 2 g :[T;R(g)] = 0, then T = λ · In×n. The implication is that if we have a Casimir C, then since it commutes with everything in the algebra, for every irrep it will act as a scalar operator. Therefore, we can label each irrep by the value of the Casimir operator. We already know this for angular momentum: L2 jl; mi = l(l + 1) jl; mi. We now just gave this idea a formal basis. As for the Cartan, since operators in the Cartan can be diagonalized simultaneously, they give us more labels for states within an irrep. For angular momentum, this is the m quantum number. For particles, it will be the conserved Noether charges: every particle is an irrep of the symmetry of our theory. Conserved charges commute with the Hamiltonian and can be diagonalized, thus labeling the different states of our particles (irreps). This applies both for internal symmetries as well as for spacetime. In the PS, you will use these concepts to construct all irreps of su(2). 1.2 Relationship between a Lie group and a Lie algebra Consider the following puzzle. You have three objects: A; B; C which act as generators for a real Lie algebra, satisfying [A; B] = iC ; [C; A] = iB ; [B; C] = iA : (1.1) What is the name for this Lie algebra? so(3)? su(2)? Something else? 1For more general Lie algebras, the definition is slightly more broad: instead of commuting, i.e. Lie bracket is zero [a; b] = 0, the Cartan is only nilpotent, which basically means that any chain of brackets will terminate after n steps for some fixed n: [a; [b; [:::[c; [d; e]]]]] = 0. 1 Answer: There's no concrete answer, because it's a silly question to ask in the first place. All of them are isomorphic, and such labels are only there as mental \bookkeeping" devices to remind us from which group we started. But the Lie algebra is a well defined structure irrespective of any Lie group it may be related to. Claim 1: A Lie group may contain more information than is available by exponentiating its corresponding algebra. For example, O(N) and SO(N) have the same Lie algebra, but O(N) also has reflections, which are discrete transformations and cannot be obtained from repeated infinitesimal transformations. SO(N), on the other hand, is entirely characterized by the algebra. It is then tempting to claim that exponentiating so(3) uniquely con- structs SO(N). But then, what do we make of our puzzle? The truth is, specifying an algebra doesn't uniquely identify a specific Lie group, even when we restrict ourselves to groups that are fully connected to the iden- tity like SO(3). To get an identification of a Lie algebra with a specific Lie group, we need a stronger restriction. Claim 2: Every Lie algebra is identified with exactly one simply-connected Lie group. This is a very important statement, which we're not going to prove, but we will get back to its implications in a bit. 1.3 SO(3) vs SU (2) We would like to ask the following question: which group is more \fundamentally spherical"? Is there a sensible notion through which we might prefer one over the other? Claim 3: SU(2) is equivalent, as a geometric object2, to a 3-sphere S3. a −b Proof: SU(2) can be defined as the set of 2×2 complex matrices of the form with jaj2 +jbj2 = 1. b∗ a∗ 2 2 2 2 Denoting a = a1 + ia2 and b = b1 + ib2, the last condition is written as a1 + a2 + b1 + b2 = 1, which is just the unit 3-sphere. Claim 4: SO(3) ' SU(2)=Z2. In words: SU(2) is a double cover of SO(3). In more words: there is a 2-to-1 surjective homomorphism from SU(2) onto SO(3). In even more words: there is a mapping between the two groups, where every element in SO(3) is identified with two elements in SU(2), and this mapping preserves the group structure. The geometric implication of this statement, is that if SU(2) can be identified with the 3-sphere, then SO(3) is also identified with S3, but with every pair of antipodal points identified as the same point as well! In a sense, SO(3) is only \half a sphere". Together, Claims 2-4 are telling us that SU(2) is a more complete, and perhaps more fundamental, descrip- tion of spherical symmetries: it is geometrically a sphere, it is uniquely identified by the Lie algebra, and it covers the rotation group SO(3). This teaches us a very important lesson: We may think we have an understanding of the symmetry of a system, but when we attempt to systematically construct its building blocks (the irreps of the algebra), it may lead us to a more fundamental symmetry that can cover the original one, and thereby reveal important physical objects that we'd otherwise miss! Therefore, when looking at the Lie algebra, we are encouraged to consider what the covering group has to say, that is { the fundamental, simply-connected Lie group the algebra is \really" coming from. Let's put this in the context of a concrete physical problem you are familiar with from undergrad. 1.4 Spherical symmetry in QM In non-relativistic QM, we sometimes want to solve systems with spherical symmetries, such as the hydrogen atom. Studying symmetry groups proves to be a useful tool to systematically construct the eigenfunctions of such potentials. When solving H = E , if H has rotational symmetry, then the space of eigenfunctions VE will be invariant under the action of SO(3). For each eigenenergy En, the space Vn can form various representations of SO(3), i.e. be constructed by objects that transform under SO(3) in different, well defined ways. As always with irreps: these are building blocks, in this case the building blocks for all wavefunctions that are solutions to the original Schrodinger equation. If we ignore the radial modes, these are the spherical harmonics. For every value of the angular momentum eigenvalue ` = 0; 1; :::; n−1, we have an 2`+1-dimensional 2Recall Lie groups are smooth manifolds. Thus, by equivalent we mean diffeomorphic. 2 space spanned by polynomials of degree `. For example, for ` = 1, we have V1 = span(x; y; z), and for ` = 2 we 2 2 2 2 have V2 = span(xy; xz; yz; x − y ; x − z ) with dimension 2 × 2 + 1 = 5. However, we know this isn't the whole story, since we also have fermions! Half-integer spin objects do not transform \natrually" under the rotation group. In particular, they don't return to themselves under a 2π rotation. Indeed, formally fermions don't form \ordinary" representations of SO(3), but instead they form something called projective representations, which is a slightly more general concept, which we will not define explicitly or worry about too much. We will, instead, see how it comes about in practice.
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