Work through the following relations involving commutators. Show all your work.

Question

Work through the following relations involving commutators. Show all your work. All operators have a hat over them. 1)Prove that [Px, x]f = -i*h-bar*f where Px is the operator of the x component of the linear momentum of a particle and x is the operator of x component of the position vector of the same particle. 2)Prove that [Px, T]f = 0 where Px is the operator of the x component of the linear momentum of a particle and T is the operator of kinetic energy. 3)Evaluate [H,T]f for a free particle moving under the potential V=0. 4)Evaluate [Lx, Lz]f 5)What are the implications of your results in (a-d)? Among these pairs or operators (a, c, and d), which pairs of corresponding observables can be accurately measured simultaneously?

Explanation / Answer

Commuting operators First, note that there is no fundamental reason why several quantities cannot have a definite value at the same time. For example, if the electron of the hydrogen atom is in a eigenstate, its total energy, square angular momentum, and -component of angular momentum all have definite values, with zero uncertainty. More generally, two different quantities with operators and have definite values if the wave function is an eigen­function of both and . So, the question whether two quantities can be definite at the same time is really whether their operators and have common eigen­functions. And it turns out that the answer has to do with whether these operators “commute”, in other words, on whether their order can be reversed as in . In particular, {D.18}: Iff two Hermitian operators commute, there is a complete set of eigen­functions that is common to them both. (For more than two operators, each operator has to commute with all others.) For example, the operators and of the harmonic oscillator of chapter 4.1.2 commute: This is true since it makes no difference whether you differen­tiate first with respect to and then with respect to or vice versa, and since the can be pulled in front of the -differen­tiations and the can be pushed inside the -differen­tiations, and since multi­plications can always be done in any order. The same way, commutes with and , and that means that commutes with them all, since is just their sum. So, these four operators should have a common set of eigen­functions, and they do: it is the set of eigen­functions derived in chapter 4.1.2. Similarly, for the hydrogen atom, the total energy Hamiltonian , the square angular momentum operator and the -component of angular momentum all commute, and they have the common set of eigen­functions . Note that such eigen­functions are not necessarily the only game in town. As a counter-example, for the hydrogen atom , , and the -component of angular momentum also all commute, and they too have a common set of eigen­functions. But that will not be the , since and do not commute. (It will however be the after you rotate them all 90 degrees around the -axis.) It would certainly be simpler mathemati­cally if each operator had just one unique set of eigen­functions, but nature does not cooperate. Key Points Operators commute if you can change their order, as in . For commuting operators, a common set of eigen­functions exists. For those eigen­functions, the physical quantities corre­sponding to the commuting operators all have definite values at the same time.

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