Derivation of free fields... coefficient question

[topsquark]

Editing... Please be patient!

I came across this difficulty comparing several versions of the Feynman rules for QED. I traced the problem back to a statement of Weinberg's in "The Quantum Theory of Fields (Vol 1)", Chapter 5, page 195. Weinberg is busy setting up for deriving free field expressions for different particle types. He has got to the point where he is deriving the field operators as an integral expression. Specifically, he derives two equations:
$\displaystyle \psi_l^+(x) = \sum_{\sigma , n} (2 \pi )^{3/2} \int d^3p u_l(\textbf{p}, \sigma, n) e^{i p \cdot x} a(\textbf{p}, \sigma, n)$

and
$\displaystyle \psi_l^-(x) = \sum_{\sigma , n} (2 \pi )^{3/2} \int d^3p v_l(\textbf{p}, \sigma, n) e^{-i p \cdot x} a^{\dagger} (\textbf{p}, \sigma, n)$

Now, I understand the overall features here. But Weinberg continues on to say

The factors $(2 \pi )^{3/2}$ could be absorbed into the definition of $u_l$ and $v_l$, but it is conventional to show them explicitly in these Fourier integrals.

Now, it's a Fourier integral, so I do see the value in showing the $2 \pi$s. But by the time the Feynman rules are derived, practically everyone has redefined the field operators so that the $(2 \pi )^{3/2}$ factor disappears.

And when we get to the amplitude, Weinberg's insistence on keeping the $2 \pi$s alters the amplitude compared to other sources. As an example, see here, equations 2.1 and 2.2. Weinberg's set of Feynman rules gives an overall factor of $(2 \pi)^{-6}$ to these amplitudes.

How do we get rid of those extra $2 \pi$s?

Thanks!

-Dan

Addendum: Problem solved! Weinberg's extra $(2 \pi)^{-3/2}$ factor in his version of the Feynman rules is there to get rid of that extra $(2 \pi)^{3/2}$ factor in the Fourier integral. I saw it just as I posted this.

 

[hutchphd]

Congratulations on a very compact post.

 

[vanhees71]

This is a mess in the literature, and it drives you always nuts, if you like to compare results from different sources. Weinberg uses the non-relativistic convention for the anti-commutators (see p. 325 in Weinberg's book),
$$[\hat{a}(\vec{p},\sigma),\hat{a}^{\dagger}(\vec{p}',\sigma')]_+=[\hat{b}(\vec{p},\sigma),\hat{b}^{\dagger}(\vec{p}',\sigma')]_+=\delta^{(3)}(\vec{p}'-\vec{p}) \delta_{\sigma \sigma'}$$
and all other combinations of anti-commutators among annihilation and creation operators vanishing. That as the advantage that the corresponding momentum eigen states,
$$|a,\vec{p},\sigma \rangle=\hat{a}^{\dagger}(\vec{p},\sigma) |\Omega \rangle$$
and analogously for anti-particles are normalized to a $\delta$-distribution.

What's never changing is the equal-time (anti-)commutators between the field operators and the conjugate field momenta, which are always (using the canonical formalism as a shortcut in comparison to the more tedious but much more insightful from a fundamental point of view, where everything is built up from Lorentz invariance and microcausality, as explained in the chapters before) read
$$[\hat{\phi}_j(t,\vec{x}),\hat{\Pi}_k(t,\vec{y})]_{\pm} = \mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}) \delta_{jk}$$
for some fields $\phi_j$ and their canonical conjugated field momenta $\hat{\Pi}_k$ (for fermions/half-integer spin the upper and bosons/integer spin the lower sign).

 

[topsquark]

This is a mess in the literature, and it drives you always nuts, if you like to compare results from different sources. Weinberg uses the non-relativistic convention for the anti-commutators (see p. 325 in Weinberg's book),
$$[\hat{a}(\vec{p},\sigma),\hat{a}^{\dagger}(\vec{p}',\sigma')]_+=[\hat{b}(\vec{p},\sigma),\hat{b}^{\dagger}(\vec{p}',\sigma')]_+=\delta^{(3)}(\vec{p}'-\vec{p}) \delta_{\sigma \sigma'}$$
and all other combinations of anti-commutators among annihilation and creation operators vanishing. That as the advantage that the corresponding momentum eigen states,
$$|a,\vec{p},\sigma \rangle=\hat{a}^{\dagger}(\vec{p},\sigma) |\Omega \rangle$$
and analogously for anti-particles are normalized to a $\delta$-distribution.

What's never changing is the equal-time (anti-)commutators between the field operators and the conjugate field momenta, which are always (using the canonical formalism as a shortcut in comparison to the more tedious but much more insightful from a fundamental point of view, where everything is built up from Lorentz invariance and microcausality, as explained in the chapters before) read
$$[\hat{\phi}_j(t,\vec{x}),\hat{\Pi}_k(t,\vec{y})]_{\pm} = \mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}) \delta_{jk}$$
for some fields $\phi_j$ and their canonical conjugated field momenta $\hat{\Pi}_k$ (for fermions/half-integer spin the upper and bosons/integer spin the lower sign).

I know. I like his style of presentation, but occasionally it's useful to reference other sources and his conventions can get in the way. I've been writing some stuff on the Feynman rules for QED and his electron propagator is:
$\dfrac{\left ( \dfrac{-i}{(2 \pi )^4 } \right ) \dfrac{ -i \not \! q + m}{q^2 + m^2 - i \epsilon}$

I haven't found that anywhere else. Most everyone else seems to use
$\dfrac{i (\not \! q + m)}{q^2 - m^2 - i\epslion}$

This part I already knew: His convention for the $\gamma$ matrices is i times the "typical" choice, and he's using different metric. (Frankly, I use his metric, but anyway...)

I presume that there's a reason why he would choose this, and I can certainly work with it now that I know where the factors come from, but it can be really annoying to check my calculations with something I find online!

-Dan

Addendum: What did I do wrong with the LaTeX? That should be right!

 

[vanhees71]

I know. I like his style of presentation, but occasionally it's useful to reference other sources and his conventions can get in the way. I've been writing some stuff on the Feynman rules for QED and his electron propagator is:
$\left ( \dfrac{-i}{(2 \pi )^4 } \right ) \dfrac{ -i \not \! q + m}{q^2 + m^2 - i \epsilon}$

I haven't found that anywhere else. Most everyone else seems to use
$\dfrac{i (\not \! q + m)}{q^2 - m^2 - i\epsilon}$

This part I already knew: His convention for the $\gamma$ matrices is i times the "typical" choice, and he's using different metric. (Frankly, I use his metric, but anyway...)

I presume that there's a reason why he would choose this, and I can certainly work with it now that I know where the factors come from, but it can be really annoying to check my calculations with something I find online!

-Dan

Addendum: What did I do wrong with the LaTeX? That should be right!

That's the LaTeX-corrected posting :-). What puzzles me most is the i in front of the q-slash.

 

[topsquark]

That's the LaTeX-corrected posting :-). What puzzles me most is the i in front of the q-slash.

Weinberg defines the $\gamma$s as
$\gamma^0 = -i \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix}$

$\gamma^i = -i \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix}$

which makes his
$-i \not \! q + m = -i \gamma^0 q_0 - i \gamma^i q_i + mI = \begin{pmatrix} 0 & -imI -i q_0I - i \sigma^i q_i \\-m - i Iq_0 + i \sigma^i q_i & 0\end{pmatrix}$

$\hphantom{XXXXX} = - \begin{pmatrix} 0 & imI + iq_0I + \sigma^i q_i \\ imI + i Iq_0 - i \sigma^i q_i & 0 \end{pmatrix}$
vs the "usual" standard
$i ( \not \! q + m ) = \begin{pmatrix} im & i Iq_0 + i \sigma^i q_i \\ i Iq_0 - i \sigma^i q_i & -im \end{pmatrix}$

I'm still looking into this to see how this makes the amplitude come out right, but I'm thinking we essentially get an overall factor of i out of it. As we are going to square the modulus in the end, this won't matter.

-Dan