Mathematical Methods for Physics and Engineering - Matematica.NET

19.2 PHYSICAL EXAMPLES OF OPERATORSConsider firstNow consider(L x L y = − 2 y ∂ ∂z − z ∂ )(z ∂∂y ∂x − x ∂ )∂z(= − 2 y ∂ ∂2 ∂2+ yz − yx∂x ∂z∂x ∂z 2 − ∂ 2z2 ∂2+ zx∂y∂x ∂y∂z(L y L x = − 2 z ∂∂x − x ∂ )(y ∂ ∂z ∂z − z ∂ )∂y(= − 2 zy ∂2∂x∂z − ∂ 2z2 ∂2− xy∂x∂y ∂z 2 + x ∂ ∂2+ xz∂y ∂z∂yThese two expressions are not the same. The difference between them, i.e. thecommutator of L x and L y , is given by[ ]Lx ,L y = Lx L y − L y L x = (x 2 ∂∂y − y ∂ )= iL z . (19.26)∂xThis, and two similar results obtained by permutting x, y and z cyclically,summarise the commutation relationships between the quantum operators correspondingto the three Cartesian components of angular momentum:[ ]Lx ,L y = iLz ,[ ]Ly ,L z = iLx , (19.27)[ L z ,L x ] = iL y .As well as its separate components of angular momentum, the total angularmomentum associated with a particular state |ψ〉 is a physical quantity of interest.This is measured by the operator corresponding to the sum of squares of itscomponents,).).L 2 = L 2 x + L 2 y + L 2 z. (19.28)This is an Hermitian operator, as each term in it is the product of two Hermitianoperators that (trivially) commute. It might seem natural to want to ‘take thesquare root’ of this operator, but such a process is undefined and we will notpursue the matter.We next show that, although no two of its components commute, the totalangular momentum operator does commute with each of its components. In theproof we use some of the properties (19.17) to (19.20) and result (19.27). We begin659