Physics Friday 51

Quantum Mechanics and Momentum

Part 8: Commutators

An important mathematical entity in quantum mechanics is the commutator of two operators. For two operators A and B, their commutator, written with the symbol [A,B], is given by [A,B]=AB-BA. Note that commutator is itself an operator. The name “commutator” arises because two operators commute if and only if their commutator is zero.
As demonstrated in this post, if two operators have a non-zero commutator, then they obey an uncertanty relation: . In fact, it can be shown that a wavefunction can simultaneously be an eigenstate for two distinct observables if and only if the two operators commute, and thus have a commutator of zero.

Consider the one-dimensional position and momentum operators in position space, as discussed in earlier posts. Then their commutator is


The commutator is useful in examining the quantum operators for angular momentum. There are some important properties of commutators which we will note here. First, we see [B,A]=-[A,B] by immediate inspection.
Next, for (linear) operators A and B and scalar c,
[A,cB]=A(cB)-cBA=c(AB-BA)=c[A,B] and [cA,B]=cAB-B(cA)=c(AB-BA)=c[A,B];
and for operators A, B, and C:
[A,(B+C)]=A(B+C)-(B+C)A=AB+AC-BA-CA=AB-BA+AC-CA=[A,B]+[A,C]
[(A+B),C]=(A+B)C-C(A+B)=AC-CA+BC-CB=[A,C]+[B,C];
so the commutator is bilinear (linear in both arguements).
We see also [A,A]=0, as any operator commutes with itself. In addition, [A,An]=0, and in fact, for an operator f(A) that is a function of the operator A only, then [A,f(A)]=0.
One more useful relation is
[A2,B]=A2B-BA2
       =A2B-ABA+ABA-BA2
       =A(AB-BA)+(AB-BA)A
[A2,B]=A[A,B]+[A,B]A

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One Response to “Physics Friday 51”

  1. Physics Friday 52 « Twisted One 151’s Weblog Says:

    […] relations hold for the cyclic permutation of the coordinates: . And thus we also have . From our previous discussion of commutators, we then see that two different components of the angular momentum cannot be simultaneously fixed […]

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