
Why the i?
Something that just occurred to me, that I probably should have figured out back when I was in college. Oh well, better late than never.
As I said in a previous thread, to go from a classical system to a quantum system, simply promote observables to selfadjoint operators. For f,g two observables, define
[f,g] = i{f,g}
where {f,g} is the classically define Poisson bracket.
So why the i? It's because we want an operator to correspond to the observable quantity {f,g}. A natural choice is [f,g], but the problem is that [f,g] is actually skewadjoint:
Meanwhile, if {f,g} is observable, it must be promoted to a selfadjoint operator. So if we let
[f,g] = i{f,g}, then both sides are skewadjoint, so the equation makes sense. We can easily pick the sign of i at this point by observing that p = (1/i) d/dx in position space and that {x,p} = 1, so that as operators [x,p] = i = i{x,p}.