Re: wick's theorem question

On Sat, 10 Nov 2007 09:06:16 -0800, gans1973@xxxxxxxxxxxxxx wrote:
Let me quote your original statement
"Physically, if you do a change of basis, you can express the spinor in
terms of u^1(k), u^2(k), v^1(k), and v^2(k), where k is a four vector at
rest in your frame. Then you have four components, only one of which is
altered by creating a spin-up particle at rest, one by creating a spin-down
particle at rest, etc. But of course, creating a particle that isn't at
rest would affect more than one of the components."

Are you saying that (say) u^1(k) has 4 components, only one of which
is non-zero (in some basis)
And u^1(p) has more than one component non-zero.?

I (mistakenly) thought you said u^1(p) = c1.u^1(k) + c2.u^2(k) +
c3.v^1(k) + c4.v^2(k)

That's exactly what I said. But I disagree with the physical
interpretation you gave it. I think this is because you're treating the
Dirac equation as a quantum-mechanical wave equation instead of as an
ordinary classical field equation.

As an explicit example, if we let

k = (m, 0, 0, 0)

p = (5m/3, 4m/3, 0, 0)

and take

\xi^1 = \eta^1 = [1 0]^T
\xi^2 = \eta^2 = [0 1]^T

then in the chiral basis

u^1(k) = sqrt(m) [1 0 1 0]^T
u^2(k) = sqrt(m) [0 1 0 1]^T
v^1(k) = sqrt(m) [1 0 -1 0]^T
v^2(k) = sqrt(m) [0 1 0 -1]^T

and

u^1(p) = sqrt(m/3) [2 -1 2 1]^T.

We have

u^1(p) = 2/sqrt(3) u^1(k) - 1/sqrt(3) v^2(k).

We could do a change of basis making

u^1(k) = sqrt(2m) [1 0 0 0]^T
u^2(k) = sqrt(2m) [0 1 0 0]^T
v^1(k) = sqrt(2m) [0 0 -1 0]^T
v^2(k) = sqrt(2m) [0 0 0 -1]^T

and then we would have

u^1(p) = sqrt(2m/3) [2 0 0 1]^T.

This is another popular basis, called the Dirac basis. Peskin and
Schroeder use the chiral basis in their book because the chiral basis makes
it easier to break up a Dirac spinor into two Weyl spinors. This is
important because only one of the Weyl spinors couples to the W bosons.

But that doesn't mean that the particle state

|p> = sqrt(2 E_p) a^1_p' |0>

(where ' = herm. conj.) overlaps the antiparticle state

|k> = sqrt(2 E_k) b^2_k' |0>.

In fact,

<k|p> = sqrt(2 E_k) sqrt(2 E_p) <0| b^2_k a^1_p' |0>

= - sqrt(2 E_k) sqrt(2 E_p) <0| a^1_p' b^2_k |0>

= 0.

On Nov 10, 2:20 pm, Jim Black <fmla...@xxxxxxxxxxxxxxxx> wrote:
I think we may be using "contracting" to mean something different. I've
been using it in the sense introduced in the section onWick'sTheorem.
From the above, it sounds like you may be using it to mean "multiply and
sum over repeated indices or basis states." Which is it that you mean?

[gans1973@xxxxxxxxxxxxxx]
Unfortunately yes. A mistake. This probably came about since
contracting \psi(x) with |p>
ends up in being equal to dotting \psi(x) with |p> computationally.

Is it? \psi(x) |p> is a combination of a zero particle state and a bunch
of one-fermion-one-antifermion states.

[gans1973@xxxxxxxxxxxxxx]
I'm taking the ket |p> = a_p'|0> to be a single particle ket.
The ket |p1p2> = a_p1'.a_p2'|0> to be a multiparticle ket.
Now, lets take the contraction of \psi(x) with the ket |p>
Contraction = commuting a_k past a_p' in \Sigma_s {integral d^3k 1/
(sqrt(2E_k)). a_k u^s(k) exp(-ikx) sqrt(2E_p)a_p'|0> }
= \Sigma_s integral d3k 1/
sqrt(2E_k) u^s(k)exp(-ikx) sqrt(2E_p) \delta(k-p)|0>
= \Sigma_s exp(-ipx)u^s(p)|0>
So, here in the final result I am dotting u^s(p) with the ket |0>

But that isn't the final result. You're evaluating an expression of the
form

<0| ... annihilation ops ... interactions ... creation ops ... |0>.

After doing all the contractions, you'll not only have a |0> left over, but
you'll also have a <0|. The |0> gets "dotted" with the <0|, not the
u^s(p).

--
Jim E. Black
.

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