Re: Basic Harmonic Oscillator question

From: Theodore George Erler IV (terler_at_physics.ucsb.edu)
Date: 08/30/04 

Date: 30 Aug 2004 04:57:46 -0400

To see that
|p> = Exp[ (-1/4)p^2 + p a^\dag - (1/2) a^\dag a^\dag ] |0>
is the right expression for the momentum eigenstate, you need to show two
things:

1) p_op|p>=p|p>
2) <p|p'> = delta(p-p')

The first one is easy to show: just write p_op in terms of a, a^\dag and
act it on the above expression. There is nothing tricky about this, in
this context "a" just works like a a derivative "d/da^dag ".

Number 2) is not quite as easy. To prove it you need the following formula
for contracting squeezed states:

<0|exp( ta + (u/2) a^2)exp(va^dag + (w/2) (a^\dag)^2)|0>
              = (1-uw)^(-1/2)exp( (2tv + wt^2 + uv^2)/2(1-uw) )

The easiest way to prove this formula is to insert a resolution of the
identity in terms of an overcomplete set of coherent states (see for
example Shankar's textbook). However, this formula doesn't give us the
answer yet, since for our case u=w=1, and the expression is singular. This
of course is not a surprise, since the answer needs to be a delta
function. To prove that it is actually a delta function, we need to
regulate our expression for |p> and show that in the limit w->1 the
contraction formula converges, in the sense of distributions, to the delta
function. This is how you determine the normalization, perticularly the
exp(-(1/4)p^2) factor.