Conditions for translational invariance in space/time for a 2-point function?

[AJS2011]

Hi,

Assume the following action:

\int d^4 x L[\phi,A]+ \int d^4 x A_{\mu} (x) J^{\mu}(x)

What are the conditions on the form of action to have space/time translational invariance for a two point function:

\left\langle J_{\mu}(x) J_{\nu}(y) \right\rangle = G_{\mu \nu}(x-y)

or in general for a function of different fields

\left\langle \quad f_{\mu}[\phi(x),J(x),A(x)] f_{\nu}[\phi(y),J(y),A(y)] \quad \right\rangle = H_{\mu \nu}(x-y)

I appreciate any help in this regard.

Thanks a lot in advance!

 

[dextercioby]

From the top of my head, the lagrangian density must not depend explicitely of x^{\mu} [/tex]. That would spoil translation invariance. I think that this is enough to ensure the invariance of the 2-point function (connected Green functional) wrt space-time translations.

 

[AJS2011]

From the top of my head, the lagrangian density must not depend explicitely of x^{\mu} [/tex]. That would spoil translation invariance. I think that this is enough to ensure the invariance of the 2-point function (connected Green functional) wrt space-time translations.

<br /> <br /> Thanks, dextercioby! Based on what you say, the presence of of non-constant gauge fields will ruin the the translational invariance. However in calculations of the linear response theory to electromagnetic field I see that people have<br /> <br /> \left\langle J_{\mu}(x) J_{\nu}(y) \right\rangle where in momentum space they write it as<br /> <br /> \left\langle J_{\mu}(p) J_{\nu}(-p) \right\rangle<br /> <br /> which obviously is true when we have translational invariance. I am confused!

 

[dextercioby]

Thanks, dextercioby! Based on what you say, the presence of of non-constant gauge fields will ruin the the translational invariance.

I didn't say that. The Langrangian functional dependence must be of fields Q and their space-time deriviatives up to a finite order and only that, not directly of x, but rather indirectly, through the fields and their derivatives.

\mathcal{L} = \mathcal{L}\left(\left\{Q_{a}\right\}_{a=1}^{n}, \left\{\partial_{\mu_{1}}...\partial_{\mu_{k}}Q_a\right\}_{a=1}^{n}, \not{x}^{\not{\mu}}\right), \, k&lt;\infty

\left\langle J_{\mu}(x) J_{\nu}(y) \right\rangle where in momentum space they write it as

\left\langle J_{\mu}(p) J_{\nu}(-p) \right\rangle

which obviously is true when we have translational invariance. I am confused!

x-y is assiociated to the momentum p through the Fourier transformation which contains the exponential as [...]p(x-y)[...].

 

[AJS2011]

I didn't say that. The Langrangian functional dependence must be of fields Q and their space-time deriviatives up to a finite order and only that, not directly of x, but rather indirectly, through the fields and their derivatives.

Thanks for the explanation! You are right. Well, I meant an external electromagnetic field for example. For magnetic fied in z direction for example in symmetric gauge we can have
A_{\mu}= \frac{B_0}{2}(0,y,-x,0)) \rightarrow \vec{B}= B_0 \hat{a}_z.

Then external A can cause the break down of translational invariance.

x-y is assiociated to the momentum p through the Fourier transformation which contains the exponential as [...]p(x-y)[...].

Well this means that translational invariance was assumed otherwise it should have been
\left\langle J_{\mu}(p) J_{\nu}(q) \right\rangle

This current-current correlation appears in linear response of a system to external electromagnetic fields which technically as I said above because of being external breaks the invariance, which to me seems incompatible.

 

[RedX]

For linear differential equations with constant coefficients and homogeneous boundary conditions, then the Green's function are translationally invariant.

Then I think it's easy to see that quadratic terms without mixing, such as ADA or BDB for fields A and B and quadratic differential operator D, will produce free Green's functions that are translationally invariant, since the Green's function G is just the solution to such an equation: DG=1 (the homogeneous boundary conditions will be that the fields vanish on the hypersurface at the infinities of spacetime - this is true for vacuum although not for statistical backgrounds) just from the integration of Gaussians.

However, if you have quadratic terms that mix fields such as ADB or BDA, then I think you might have problems. Fortunately I've never seen such quadratic mixes before (is there a reason for that?).

Anyways, if you establish that the free-field Green's functions are translationally invariant, then I think you can establish that the interacting Green's functions are also translationally invariant, since the interacting ones are made up of products of the free-field ones. That should be the case intuitively, and I don't remember any interacting propagators that weren't translationally invariant either.

Also, what is linear response theory?

 

[AJS2011]

Thanks RedX!

However, if you have quadratic terms that mix fields such as ADB or BDA, then I think you might have problems. Fortunately I've never seen such quadratic mixes before (is there a reason for that?).

In superconductivity such mixing terms appear, and the way around it is through using Bogolubov-Valatin transformation.

Also, what is linear response theory?

Linear response, aka Kubo formalism, studies near equilibrium properties of a system when a small external field is perturbing the system. For example, to study electrical conductivity of a system they apply an electric field and see what the current is due to that field. The term
"linear" reflects the assumption that the response of the system is proportional to the driving force that was assumed to be weak.