I have seen two forms of linear response:

  1. One is in the calculation of susceptibilities using Green functions.

  2. The other is in the evaluation of response currents, say, London current of a superconductor. The latter is very neat. Provided an effective action $S$, simply we have $$j^\mu(x)=\frac{\delta S}{\delta A_\mu(x)},$$ where $A$ is the vector potential.

My questions:

  1. Are the two formalisms equivalent?

  2. Can I calculate the response current to some fields other than $A$ via the second formalism?

  • $\begingroup$ In which sense should the second be a a response? of what to what? $\endgroup$ – Arnold Neumaier Nov 14 '12 at 12:18
  • $\begingroup$ I mean the response to the gauge field $A$, say, the London current $j_i=m_A A_i$. $\endgroup$ – Machine Nov 14 '12 at 13:29

It is not completely clear what you mean by the approach 2. What one can do is to calculate the current via $$j^{\mu}(\phi,A)=\frac{\delta S_{eff}[\phi,A]}{\delta A_{\mu}}.$$ Here the effective action $S_{eff}$ is a functional of both the source $A$ and the phase $\phi$ of the condensate in the superconductor. Imagine now that you solve the linearized equation of motion for for the phase and get the solution $\phi(A)$. If you substitute now this solution into the current you get $j^{\mu}(A)$, i.e. the current now is the function of the source only. If you now functionally differentiate $j^{\mu}(A)$ with respect to $A$, you will get the response. This is equivalent to the Green's function calculation. This method is general and can be used not only for the response to the electromagnetic source, but to other (for example gravitational) sources.


In linear response theory, responses are changes of observables to controllable input.

But in your formula 2., the current is the rate of change of the action when changing the gauge field. As the action is not an observable, it is not really approriate to call this a response.

  • $\begingroup$ But the current is the response to the gauge field. It is observable. $\endgroup$ – Machine Nov 15 '12 at 7:58
  • $\begingroup$ @ChenChao: The response of what? The analogy is too imperfect to be useful. $\endgroup$ – Arnold Neumaier Nov 15 '12 at 8:46
  • $\begingroup$ OK, anyway, the second formula is actually the equation of motion of the background gauge field. It's maybe more useful to ask whether this formula can be applied to arbitrary background fields. $\endgroup$ – Machine Nov 16 '12 at 11:59

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