Electromagnetic current-current correlators

Question

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,

• The total hadronic cross-section,

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

• 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?

Answer 1

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.