# Thread: Why the i?

1. 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 self-adjoint 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 skew-adjoint:

Meanwhile, if {f,g} is observable, it must be promoted to a self-adjoint operator. So if we let

[f,g] = i{f,g}, then both sides are skew-adjoint, 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}.

2. ### Related Discussions:

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement