View Full Version : are operators always commutative with respect to operation +?
kakarukeys
Jul5-04, 08:37 AM
In general AB =/= BA, for example,
orbital angular momentum operators, L_x, L_y.
but is A + B = B + A always true?
Yes.
(A+B)f=Af+Bf=Bf+Af=(B+A)f, so A+B = B +A
You are simply working with functions (Af and Bf), so the addition is commutative.
kakarukeys
Jul5-04, 09:28 AM
I think so, but this is puzzling to me:
put hbar = 1
L_x + L_y = L_y + L_x
so, exp(-i theta L_x) exp(-i theta L_y) = exp(-i theta L_y) exp(-i theta L_x)
that means rotation about x-axis and rotation about y-axis are commutative?
but rotate a book around x-axis 90 deg followed by y-axis 90 deg is not same as the other way round.
Warning: for (non-commuting) operators A and B,
you cannot write:
exp(A) exp(B) = exp(A+B)
!!
cheers,
patrick.
kakarukeys
Jul5-04, 09:41 AM
I see,
things become clear when you expand the exponentials
Thank you.
provided the domains of the operators allow for the sum to be defined, linear operators will respect the addition operation.
You have to be careful with your order of operations (associativity). Exponentiation first and then addition vs. addition first and then exponentiation. The exponentiation makes different operators, say:
eA = Ω and eB = Λ
Not only do you have to worry about A + B = B + A, which is almost trivial in a physics context, you must also worry about [Ω,Λ] = 0, which is often enough untrue.
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