Electromagnetic current-current correlators

Let the free electromagnetic current $J_\mu(x)$ be = $:\bar{\psi}(x)\gamma_\mu Q \psi(x):$ where $::$ is the normal ordering.

• In this expression why is $Q$ thought of as a "charge operator" instead of just a number?...its quite pesky to keep track of this operator while doing the current-current OPEs though I don't see anything changing conceptually if I just thought of it as a number...

After a lot of (I found it very subtle!) calculations one can show that in the light-cone limit $x^2 \rightarrow 0$ the commutator, $[J_\mu(x),J_\nu (0)]$ has one of its terms (say X),

$$X = \frac{iTr[Q^2]}{\pi^3}\{\frac{2}{3}g_{\mu \nu}\delta''(x^2)\epsilon(x_0) + \frac{1}{6}\partial_\mu \partial_\nu [\delta'(x^2)\epsilon(x_0)]\}$$

Now one wants to compare the contribution of this term to two different situations,

$\sigma(e^+e^- \rightarrow hadrons) = \frac{8\pi^2\alpha^2}{3(q^2)^2}\int d^4x e^{iq.x}<0|[J_\mu(x),J^\mu(0)]0>$

• The inclusive hadronic tensor in deep inelastic lepton-nucleon scattering,

$W_{\mu \nu}(p,q) = \frac{1}{M} \sum _{\sigma} \int \frac{d^4x}{2\pi} e^{iq.x}<p,\sigma|[J_\mu(x), J_\nu(0)]|p,\sigma>$

In the derivation/argument for the expression for $W_{\mu \nu}$ it is kind of clear that the initial and final states have to be $|p,\sigma>$ - the initial state of the proton.

• But I am unable to pin down as to exactly why the initial and final states in the first case had to be vacuum $|0>$. It would be great if someone can explain this conceptual point about the difference in the initial and final states.

• Hence if someone can explain as to why the term $X$ contributes (and is infact the leading contributor!) to $\sigma$ but does not contribute to $W_{\mu \nu}$!?

(..my vague understanding is that this difference stems from the difference in the initial and final states..but can't make this precise..)

• Though initially $W_{\mu \nu}$ is defined in terms of the correlator $[J_\mu(x), J_\nu(0)]$, often I see that during calculations one is in practice evaluating $[J_\mu(\frac{x}{2}), J_\nu(-\frac{x}{2})]$. Why this change?
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The operator Q is a number for the electron--- the spinor fields in the expression are the quarks. The electron contribution is factored out, since it just creates a virtual photon that interacts with the hadrons by the quark current operator. Q is a diagonal 3 by 3 matrix with the charges of the light quarks. The derivations in Feynman's "Photon Hadron Interactions" are clearest, as this is as close to the original source (Bjorken and Feynman's circulating notes from the era) you will find. – Ron Maimon May 1 '12 at 3:49

The current $J_μ(x)= :\bar ψ(x)γ_μQψ(x):$ is the conserved current of the charge $Q$, obtained by integrating the charge density $J_0(x)$ over space. Therefore, the $Q$ is called a charge operator. It is typically an operator and not a number as it takes different values on states of different charge.

For example, in QED, $Q$ has the value $\pm e$ on positron states and electron states, respectively, and has an indefinite value on their superpositions.

You can treat $Q$ as a number only if you know that $\psi$ always has a fixed charge.

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On the last question: The two expressions are the same due to translation invariance.

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This is not true--- for the electron, "Q" is just the electric charge. The confusion is the OP's however, the current he is using is the QCD electric current, and the field "psi" is really all the relevant quark fields (up,down,strange), so that the question become moot. – Ron Maimon May 1 '12 at 3:47
@RonMaimon: For electrons only you are right, but I wass considering QED, which has electrons and positrons combined in the same $\psi$, so $Q$ there has two eigenvalues. – Arnold Neumaier May 2 '12 at 11:42
That's weird--- the electron and positron are both contained in the same relativistic field, so your operator $\bar\psi\gamma^\mu\psi$ can't separate the two. The operator Q in front of this Dirac current is just the electric charge. – Ron Maimon May 2 '12 at 15:56
@RonMaimon: The $Q$ is not in front of this but inside the expression. Thus things are well-defined. If one writes things out in terms of creation and annihilation operators, the signs get placed correctly. – Arnold Neumaier May 2 '12 at 16:00
They only get placed correctly for electrons and positrons if Q is a number. – Ron Maimon May 2 '12 at 16:07