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Equation (3) of https://arxiv.org/abs/cond-mat/0606800 is the current operator

$$j_i(x)=e\tilde{\psi}^\dagger(x')v_{x'x}^i\tilde\psi(x)-\frac{e^2}{c}\tilde\psi^\dagger(x')(m^{-1})^{ij}_{x'x}\tilde\psi(x)A_j\tag{3}$$ where $x'\rightarrow x$.

What does the $x'\rightarrow x$ mean? Is it some sort of integral transform?

The current $j$ is a function of $x$ and not $x'$. I feel like this is a notation I am unfamiliar with. The same notation is also mentioned in reference 5.

Qmechanic
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Zeeshan Ahmad
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2 Answers2

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This is known as point-splitting regularization, see e.g. this related Phys.SE post. Concretely, one inserts a $\lim_{x^{\prime}\to x}$ operation on the right-hand side of eq. (3).

Qmechanic
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  • What is the definition of the velocity operator $v_{x'x}$ in terms of a usual velocity operator $v_x$? (same question for mass operator) – Zeeshan Ahmad May 03 '20 at 18:37
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It means that you are taking a product of different operators at coinciding points. Typically taking such a product creates a divergence and the the product needs to be regularised. i.e. you need a procedure to tell you how to subtract the divergence. This is covered in most introductory text books on QFT.

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