Here is the paper I found it on. Page 5,6.

So, the arguments given are:

"In order to keep track of such dipole motion, one’s first instinct might be to define a (non-symmetric) current tensor Jij representing the current of the i directed dipoles in the j direction. However, there is a fundamental ambiguity in this definition. For example, consider the close-packed charge configuration in Figure 1. There is not a unique way of defining either the dipole density or Jij in such a case. Microscopically, an operator hopping an i dipole in the j direction is the same operator hopping a j dipole in the i direction, so the true microscopic current operator is actually a symmetric tensor Jij."

This is the other explanation they gave.

I don't see how the ij and ji hopping operators must be the same and I still feel like the tensor should have 9 independent components. If someone could explain the argument to me, it would really be helpful.

Sorry for the bad formatting.


1 Answer 1


Look at the first two dipoles in the top row. If they were far apart, it would be obvious that the dipoles are oriented in the $y$ direction. However, when those dipoles get that close together, the charge arrangement is exactly the same as what you'd have if you thought of them as dipoles oriented in the $x$ direction, as you can see by the first two dipoles in the bottom row.

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Consider the operator which swaps the first two dipoles in the top row. What it really does is flip the sign of each one of the charges in the square. This is essentially $J_{yx}$ - the operator which swaps $y$-oriented dipoles in the $x$-direction.

Now consider the operator which swaps the two horizontal dipoles in the square on the bottom row, $J_{xy}$. It does precisely the same thing - meaning that it must be the same operator.

  • $\begingroup$ Okay, I see how it works in the case where net dipole moment is zero, but how do you show this is true in general? If we just had dipoles along x, pushing them along y would be completely different from pushing y dipoles along x, right? $\endgroup$ May 31, 2019 at 7:22

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