Green's functions for translationally invariant systems

In summary: If you don't like that, then you can look at the variation in ##G## under an infinitesimal transformation ##(x,y)\rightarrow (x+a,y+a)##, which is$$ \delta G = G(x+a,y+a) - G(x,y) = G(x,y) + \delta x \partial_x G(x,y) + \delta y \partial_y G(x,y) - G(x,y) = a \left( \partial_x G(x,y) + \partial_y G(x,y) \right).$$To
  • #1
"Don't panic!"
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As I understand it a Green's function ##G(x,y)## for a translationally invariant differential equation satisfies $$G(x+a,y+a)=G(x,y)\qquad\Rightarrow\qquad G(x,y)=G(x-y)$$ (where ##a## is an arbitrary constant shift.)

My question is, given such a translationally invariant system, how does one show that ##G(x,y)=G(x-y)##?
Is it as simple as setting ##y=-a## such that $$G(x+a,y+a)=G(x-y,0)\equiv G(x-y)$$ or is it more to it than that? (I'm slightly uncomfortable with setting ##y=-a##, it doesn't seem correct?!)
 
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  • #2
Yes, it is as easy as that. Why are you uncomfortable with setting ##y = -a##?
 
  • #3
Orodruin said:
Yes, it is as easy as that. Why are you uncomfortable with setting y=−a?

I think it's because we're originally treating it as a function of x and y and then we're simply setting one of the variables to a particular value.
 
  • #4
But it is still a function of x and y, it just depends on them in a very particular combination. It is also not that you are setting y to a particular value, you are setting the translation amount ##a## to a particular value.
 
  • #5
Orodruin said:
But it is still a function of x and y, it just depends on them in a very particular combination. It is also not that you are setting y to a particular value, you are setting the translation amount aa to a particular value.

Isn't y more of a parameter now though as a is chosen to be a constant, right? Is it that the translation invariance places a constraint on the functional dependence?
 
  • #6
"Don't panic!" said:
Isn't y more of a parameter now though as a is chosen to be a constant, right? Is it that the translation invariance places a constraint on the functional dependence?

I do not understand your objection. You can select whatever value of a you want. In particular you can let it be -y. Whatever value of a you have chosen, the translation symmetry will remain true. In particular, it will remain true for a = -y regardless of the value of y.
 
  • #7
Orodruin said:
I do not understand your objection. You can select whatever value of a you want. In particular you can let it be -y. Whatever value of a you have chosen, the translation symmetry will remain true. In particular, it will remain true for a = -y regardless of the value of y.

I guess I'm looking for issues that aren't there, sorry.

So is the point that the translation symmetry is valid for arbitrary ##a## and so we can simply choose it such that ##a=-y##?

Also, I've read notes in which, in order to show that ##G(x,y)=G(x-y)##, they take a derivative with respect to ##a##, i.e. $$\frac{d}{da}G(x+a,y+a)=\frac{\partial G}{\partial x}\frac{\partial (x+a)}{\partial a}+\frac{\partial G}{\partial y}\frac{\partial (y+a)}{\partial a}=\frac{\partial G}{\partial x}+\frac{\partial G}{\partial y}=0$$ Now, at least initially, this doesn't make any sense to me since ##a## is a constant and so taking a derivative with respect to it doesn't make sense. Is the point though that we treat ##a## as a parameter and then "ask" how ##G## changes as we vary it, i.e. we take the derivative of G with respect to ##a##, then since G is translationally invariant this derivative equals zero?!
 
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  • #8
"Don't panic!" said:
I guess I'm looking for issues that aren't there, sorry.

So is the point that the translation symmetry is valid for arbitrary ##a## and so we can simply choose it such that ##a=-y##?

Also, I've read notes in which, in order to show that ##G(x,y)=G(x-y)##, they take a derivative with respect to ##a##, i.e. $$\frac{d}{da}G(x+a,y+a)=\frac{\partial G}{\partial x}\frac{\partial (x+a)}{\partial a}+\frac{\partial G}{\partial y}\frac{\partial (y+a)}{\partial a}=\frac{\partial G}{\partial x}+\frac{\partial G}{\partial y}=0$$ Now, at least initially, this doesn't make any sense to me since ##a## is a constant and so taking a derivative with respect to it doesn't make sense. Is the point though that we treat ##a## as a parameter and then "ask" how ##G## changes as we vary it, i.e. we take the derivative of G with respect to ##a##, then since G is translationally invariant this derivative equals zero?!

If you don't like that, then you can look at the variation in ##G## under an infinitesimal transformation ##(x,y)\rightarrow (x+a,y+a)##, which is
$$ \delta G = G(x+a,y+a) - G(x,y) = G(x,y) + \delta x \partial_x G(x,y) + \delta y \partial_y G(x,y) - G(x,y) = a \left( \partial_x G(x,y) + \partial_y G(x,y) \right).$$
To complete the proof nicely, you can change variables to ##x_\pm = x \pm y## and see that this is ##\partial_{x_+} G(x_+,x_-)##, so that ##G = G(x_-)##.

So here we are setting ##\delta G=0##. while the approach you mention is equivalent to requiring that ##\delta G## is independent of ##a##. In this case, they lead to the same constraint.
 
  • #9
fzero said:
If you don't like that, then you can look at the variation in GG under an infinitesimal transformation (x,y)→(x+a,y+a)(x,y)\rightarrow (x+a,y+a), which is
δG=G(x+a,y+a)−G(x,y)=G(x,y)+δx∂xG(x,y)+δy∂yG(x,y)−G(x,y)=a(∂xG(x,y)+∂yG(x,y)). \delta G = G(x+a,y+a) - G(x,y) = G(x,y) + \delta x \partial_x G(x,y) + \delta y \partial_y G(x,y) - G(x,y) = a \left( \partial_x G(x,y) + \partial_y G(x,y) \right).
To complete the proof nicely, you can change variables to x±=x±yx_\pm = x \pm y and see that this is ∂x+G(x+,x−)\partial_{x_+} G(x_+,x_-), so that G=G(x−)G = G(x_-).

So here we are setting δG=0\delta G=0. while the approach you mention is equivalent to requiring that δG\delta G is independent of aa. In this case, they lead to the same constraint.

Thanks for your help.
Going back to an earlier point, is it simply that ##a## is an arbitrary parameter and so we can always choose it such that ##a=-y## whatever the values of ##x## and ##y## are, hence $$G(x,y)=G(x+a,y+a)=G(x+(-y), y+(-y))=G(x-y,0)\equiv G(x-y)$$
 
  • #10
"Don't panic!" said:
Thanks for your help.
Going back to an earlier point, is it simply that ##a## is an arbitrary parameter and so we can always choose it such that ##a=-y## whatever the values of ##x## and ##y## are, hence $$G(x,y)=G(x+a,y+a)=G(x+(-y), y+(-y))=G(x-y,0)\equiv G(x-y)$$

The only objection that I have to this argument is that we can do this at a single point by choosing ##a=-y_1##, so that ##G(x_1,y_1)=G(x_1-y_1,0)##, but it does not immediately follow that for some other point ##G(x_2,y_2)=G(x_2-y_2,0)##. Therefore I think it is better to use an argument that involves the derivatives of ##G(x,y)##.
 
  • #11
fzero said:
The only objection that I have to this argument is that we can do this at a single point by choosing ##a=-y_1##, so that ##G(x_1,y_1)=G(x_1-y_1,0)##, but it does not immediately follow that for some other point ##G(x_2,y_2)=G(x_2-y_2,0)##. Therefore I think it is better to use an argument that involves the derivatives of ##G(x,y)##.
But the entire point is that it holds regardless of whether you select ##a=-y_1## or ##a=-y_2##. You can do the translation for any ##a##.
 
  • #12
fzero said:
The only objection that I have to this argument is that we can do this at a single point by choosing ##a=-y_1##, so that ##G(x_1,y_1)=G(x_1-y_1,0)##, but it does not immediately follow that for some other point ##G(x_2,y_2)=G(x_2-y_2,0)##. Therefore I think it is better to use an argument that involves the derivatives of ##G(x,y)##.

This was my first thought, but Orodruin is actually right. Remember that y is not a function; it's just a value. And if G(x,y) is translationally-invariant, then you're free to translate it by y.

Orodruin said:
But the entire point is that it holds regardless of whether you select ##a=-y_1## or ##a=-y_2##. You can do the translation for any ##a##.
 
  • #13
Orodruin said:
But the entire point is that it holds regardless of whether you select ##a=-y_1## or ##a=-y_2##. You can do the translation for any ##a##.

Yes, I realize that we can do a second transformation to prove it at another point. I didn't mean to imply that it was wrong, just that I prefer the other argument as a matter of taste.
 
  • #14
Orodruin said:
But the entire point is that it holds regardless of whether you select a=−y1a=-y_1 or a=−y2a=-y_2. You can do the translation for any aa.

So am I correct in thinking that as ##G(x,y)## is translationally invariant for each set of values ##(x,y)##, then we can simply choose ##a## such that ##a=-y## at each set of values ##(x,y)##?
 

FAQ: Green's functions for translationally invariant systems

1. What are Green's functions for translationally invariant systems?

Green's functions for translationally invariant systems are mathematical tools used in quantum mechanics to study the behavior of a system that is invariant under translations. They provide a way to calculate the response of a system to an external perturbation.

2. How are Green's functions for translationally invariant systems different from other types of Green's functions?

Green's functions for translationally invariant systems are different from other types of Green's functions because they are specifically designed to study systems that exhibit translational symmetry. This means that the properties of the system do not change when it is translated in space.

3. What are the applications of Green's functions for translationally invariant systems?

Green's functions for translationally invariant systems have various applications in physics, particularly in condensed matter physics. They are used to study the behavior of electrons in crystalline materials, as well as in other systems with translational symmetry, such as photonic and phononic crystals.

4. How are Green's functions for translationally invariant systems calculated?

Green's functions for translationally invariant systems are calculated using mathematical techniques, such as Fourier transforms and contour integration. These techniques involve transforming the problem from real space to momentum space, where the translational symmetry of the system is easier to analyze.

5. Can Green's functions for translationally invariant systems be used to study systems without translational symmetry?

No, Green's functions for translationally invariant systems are specifically designed to study systems with translational symmetry. They cannot be used to study systems without this symmetry, as their mathematical properties are different and they will not provide accurate results.

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