Feynman Interpretation of Perturbation

We start with just the perturbation expansion of any operator in quantum mechanics (derived here) :


H |n> = En |n>

(H + W) |n> = (En + wn) |n>

here, W is the perturbation of H, W << H.  Our goal is to find w.

wn = 

Where P = ( 1 - |n> +  +  + ...

If this expression of perturbatin bothers you, can just use
matrix multiplication to show that:

= Sum[m not= n] { ()^2 / (Em - En) } Any hoo, let's proceed with the interpretation. W is an energy - a coupling - an excitation - a disturbance. The operator W changes your system, raises its energy. You apply W to a state |n> and create W|n> - the state with the excitation in it. P is 1/(H - Eground) : 1/(delta E) : the inverse of the energy of the excitation, this makes the significance of the excitation inverse to how much it raises the energy. i.e. the change of your system's energy due to spontaneous fusion or creation of 10,000 photons is quite low; perhaps more accurately : you're more likely to excite slow vibration than fast. P is the propagator, it moves the excitation W through the "space". Let's look at the first couple of terms in the expansion: is : take |n> , the base : W creates an excitation : it falls back to is : take |n> , W creates an excitation, P propagates this excitation through space, another event occurs and W is absorbed back to may take a state with a par of photons, and spontaneously annihilate them, and create a new pair. : W may kick up the momentum of n by q, then P moves it along with a (1/q), then W kick the momentum back down. Now let's make it a bit more quantitative. |n> is a background |vac> with some excitations. Excitations of the vacuum are created with a creation-of-excitation operator, (a+q) which creates an excitation at q (that + is not addition) (a-q destroys an excitation, (a+q)(a-q) counts them). Now we can write the general form of a perturbation W: W = Wo Sum[q]{ q^-2 Sum[p1,p2]{ a+(p1+q) a+(p2-q) a-(p1) a-(p2) } } Wo is the coupling constant - the strength of this interaction Sum[q] is just the Fourier transform of the operator W [ a+(p1+q) a+(p2-q) a-(p1) a-(p2) ] destroys excitations of p1 and p2, and creates new ones at p1+q and p2-q (conserving total momentum). This is the same as speeding up p1 and slowing p2. The sum on p1 and p2 does this for all the states in the function.


Charles Bloom / cb at my domain
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