UVA PHYS 751 - Angular Momentum Operator Algebra

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Angular Momentum Operator AlgebraPreliminaries: Translation and Rotation OperatorsQuantum Generalization of the Rotation OperatorConsequences of the Commutation RelationsLadder OperatorsSummaryNormalizing J+ and J(Angular Momentum Operator Algebra Michael Fowler 10/29/07 Preliminaries: Translation and Rotation Operators As a warm up to analyzing how a wave function transforms under rotation, we review the effect of linear translation on a single particle wave function ()xψ. We have already seen an example of this: the coherent states of a simple harmonic oscillator discussed earlier were (at t = 0) identical to the ground state except that they were centered at some point displaced from the origin. In fact, the operator creating such a state from the ground state is a translation operator. The translation operator T(a) is defined at that operator which when it acts on a wave function ket ()xψ gives the ket corresponding to that wave function moved over by a, that is, ()()(),Ta x x aψψ=− so, for example, if ()xψis a wave function centered at the origin, T(a) moves it to be centered at the point a. We have written the wave function as a ket here to emphasize the parallels between this operation and some later ones, but it is simpler at this point to just work with the wave function as a function, so we will drop the ket bracket for now. The form of T(a) as an operator on a function is made evident by rewriting the Taylor series in operator form: ()() () ()()() ()2222!.dadxdadxa x a x xdx dxexTa xψψ ψ ψψψ−−= − + −==… Now for the quantum connection: the differential operator appearing in the exponential is in quantum mechanics proportional to the momentum operator (ˆ/piddx=− ) so the translation operator ()ˆ/.iapTa e−= An important special case is that of an infinitesimal translation, ()ˆ/ˆ1/ipTe ipεεε−==−. The momentum operator is said to be the generator of the translation. ˆp2(A note on possibly confusing notation: Shankar writes (page 281) ().Txxεε=+ Here x denotes a delta-function type wave function centered at x. It might be better if he had written ()00Txxεε=+, then we would see right away that this translates into the wave function transformation () ( ) ( )00Txxxxεδδ−= −−ε, the sign of ε now obviously consistent with our usage above.) It is important to be clear about whether the system is being translated by a, as we have done above or whether, alternately, the coordinate axes are being translated by a, that latter would result in the opposite change in the wave function. Translating the coordinate axes, along with the apparatus and any external fields by −a relative to the wave function would of course give the same physics as translating the wave function by +a. In fact, these two equivalent operations are analogous to the time development of a wave function being described either by a Schrödinger picture, in which the bras and kets change in time, but not the operators, and the Heisenberg picture in which the operators develop but the bras and kets do not change. To pursue this analogy a little further, in the “Heisenberg” case () ()[]ˆˆ1//ˆˆ ˆˆˆˆ,/ip ipxT xT exe xipx xεεˆεεε−−→ = =+ =+ε and is unchanged since it commutes with the operator. So there are two possible ways to deal with translations: transform the bras and kets, or transform the operators. We shall almost always leave the operators alone, and transform the bras and kets. ˆp We have established that the momentum operator is the generator of spatial translations (the generalization to three dimensions is trivial). We know from earlier work that the Hamiltonian is the generator of time translations, by which we mean ()()/.iHata e tψψ−+= It is tempting to conclude that the angular momentum must be the operator generating rotations of the system, and, in fact, it is easy to check that this is correct. Let us consider an infinitesimal rotation δθabout some axis through the origin (the infinitesimal vector being in the direction of the axis). A wavefunction initially localized at ()rψ0rwill shift to be localized at 00rrδ+, where So, how does a wave function transform under this small rotation? Just as for the translation case, 0.rδδθ=×0rr() ( )rrψψδ→−. If you don’t understand the minus sign, reread the discussion on translations and the sign of ε. Thus () () ()ˆ.irrrpψψ δψ→− r to first order in the infinitesimal quantity, so the rotation operator3()() ()()()ˆ1.ˆ1.ˆ1. .iRrrpirp riLrδθ ψ δθ ψδθ ψδθ ψ⎛⎞=− ×⎜⎟⎝⎠⎛⎞=− ×⎜⎟⎝⎠⎛⎞=−⎜⎟⎝⎠r  If we write this as ()() ()ˆ.iLRre rδθδθ ψ ψ−= it is clear that a finite rotation is given by multiplying together a large number of these operators, which just amounts to replacing δθby θ in the exponential. Another way of going from the infinitesimal rotation to a full rotation is to use the identity lim 1NANAeNθθ→∞⎛⎞+=⎜⎟⎝⎠ which is clearly valid even if A is an operator. We have therefore established that the orbital angular momentum operator ˆL is the generator of spatial rotations, by which we mean that if we rotate our apparatus, and the wave function with it, the appropriately transformed wave function is generated by the action of ()Rθ on the original wave function. It is perhaps worth giving an explicit example: suppose we rotate the system, and therefore the wave function, through an infinitesimal angle zδθ about the z-axis. Denote the rotated wave function by (),rotxyψ. Then () () ()() ()()()() ()()ˆ,1 ,1,1,,.rot z zzzzzixy L xyiddix y xydy dxddxy xydy dxxyyxψδθ ψδθ ψδθ ψψδθ δθ⎛⎞=−⎜⎟⎝⎠⎛⎞⎛⎞⎛⎞=− − −⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠⎛⎞⎛⎞=− −⎜⎟⎜⎟⎝⎠⎝⎠=+ − That is to say, the value of the new wave function at (x,y) is the value of the old wave function at the point which was rotated into (x,y). Quantum Generalization of the Rotation Operator However, it has long been known that in quantum mechanics, orbital angular momentum is not4the whole story. Particles like the electron are found experimentally to have an internal angular momentum, called spin. In contrast to the spin of an ordinary


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UVA PHYS 751 - Angular Momentum Operator Algebra

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