Translation-invariant systems and binary functions of spatial variables

[hokhani]

TL;DR Summary

the correctness of the relation f(r,r')=f(r-r') in translation-invariant systems

My question is about a statement in the book, Many-Body Quantum Theory in Condensed Matter Physics: An Introduction, by Henrik Bruus, Karsten Flensberg in appendix A:
In a translation invariant system, any physical observable $f(\vec r,\vec r')$ of two spatial variables $\vec r$ and $\vec r'$ can only depend on the difference between the coordinates and not on the absolute position of any of them, $$f(\vec r,\vec r')=f(\vec r - \vec r').$$ However, I think in one unit cell if we shift identically both $\vec r$ and $\vec r'$ then this statement does not seem necessarily correct!

 

[BvU]

I think in one unit cell if we shift identically both $\vec r$ and $\vec r'$ then this statement does not seem necessarily correct!

What do you mean with that ? The statement doesn't mention unit cells.
And what is 'shift identically' ?

$\$

 

[anuttarasammyak]

TL;DR Summary: the correctness of the relation f(r,r')=f(r-r') in translation-invariant systems

coordinates and not on the absolute position of any of them, f(r→,r→′)=f(r→−r→′). However, I think in one unit cell if we shift identically both r→ and r→′ then this statement does not seem

For any position vector R
$$f(\mathbf{r}, \mathbf{r'})=f(\mathbf{r}+\mathbf{R}, \mathbf{r'}+\mathbf{R})$$
choosing R=-r or R=-r'
$$=f(\mathbf{r}-\mathbf{r'},0)=f(0, \mathbf{r'}-\mathbf{r}):=F(\mathbf{r}-\mathbf{r'})$$

 

[div_grad]

@hokhani This follows from considering infinitesimal symmetry transformations (in this case, spatial translations).

Take any function $f(\mathbf{r}, \mathbf{r}')$ and perform an infinitesimal translation by a vector $\mathbf{\varepsilon}$. Since $\mathbf{\varepsilon}$ is small, you can use Taylor expansion (up to first-order terms):
$$
f(\mathbf{r}+\mathbf{\varepsilon}, \mathbf{r}'+\mathbf{\varepsilon}) = f(\mathbf{r}, \mathbf{r}') + \mathbf{\varepsilon}\cdot\frac{\partial f}{\partial \mathbf{r}} + \mathbf{\varepsilon}\cdot\frac{\partial f}{\partial \mathbf{r}'} + \mathcal{O}(\varepsilon^2) \approx f(\mathbf{r}, \mathbf{r}') + \mathbf{\varepsilon}\cdot\left(\frac{\partial f}{\partial \mathbf{r}}+\frac{\partial f}{\partial \mathbf{r}'}\right) \rm{.}
$$
Now, if your physical system is translation-invariant then $f(\mathbf{r}+\mathbf{\varepsilon}, \mathbf{r}'+\mathbf{\varepsilon}) = f(\mathbf{r}, \mathbf{r}')$, by definition. Then the above equation tells you that $\mathbf{\varepsilon}\cdot\left(\frac{\partial f}{\partial \mathbf{r}}+\frac{\partial f}{\partial \mathbf{r}'}\right)=0$, and since this must hold for any $\mathbf{\varepsilon}$ you obtain that
$$
\frac{\partial f}{\partial \mathbf{r}} = - \frac{\partial f}{\partial \mathbf{r}'} \rm{,}
$$
which is true if $f(\mathbf{r}, \mathbf{r}') = f(\mathbf{r}-\mathbf{r}')$.

To quickly convince yourself that this is the case, consider a (simpler) function of just two variables $x$ and $y$ which depends on their difference, $f(x-y) \equiv f(t)$, and calculate partial derivatives. Then the result for
$$
\frac{\partial f}{\partial{x}} = \frac{\partial f}{\partial{t}} \frac{\partial t}{\partial{x}} = \frac{\partial f}{\partial{t}}
$$
can be used in obtaining the above-mentioned property
$$
\frac{\partial f}{\partial{y}} = \frac{\partial f}{\partial{t}} \frac{\partial t}{\partial{y}} = -\frac{\partial f}{\partial{t}} = -\frac{\partial f}{\partial{x}} \rm{.}
$$

 

[div_grad]

For any position vector R
$$f(\mathbf{r}, \mathbf{r'})=f(\mathbf{r}+\mathbf{R}, \mathbf{r'}+\mathbf{R})$$
choosing R=-r or R=-r'
$$=f(\mathbf{r}-\mathbf{r'},0)=f(0, \mathbf{r'}-\mathbf{r}):=F(\mathbf{r}-\mathbf{r'})$$

@anuttarasammyak This explanation is wrong, as it amounts only to renaming the variables $\mathbf{r}$ and $\mathbf{r}'$. It does not show that $f(\mathbf{r},\mathbf{r}')=f(\mathbf{r}-\mathbf{r}')$ in a homogenous system. Simply choosing the origin of an arbitrary coordinate system does not imply that all points in space are equivalent in this regard.

 

[anuttarasammyak]

Now, if your physical system is translation-invariant then f(r+ε,r′+ε)=f(r,r′), by definition.

I do not find a difference between my $\mathbf{R}$ and your $\epsilon$.

@anuttarasammyak This explanation is wrong, as it amounts only to renaming the variables $\mathbf{r}$ and $\mathbf{r}'$. It does not show that $f(\mathbf{r},\mathbf{r}')=f(\mathbf{r}-\mathbf{r}')$ in a homogenous system. Simply choosing the origin of an arbitrary coordinate system does not imply that all points in space are equivalent in this regard.

BTW Using same alphabet f for RHS and LHS seems confusing to me because LHS is a two variable function but RHS is a single variable function.

 

[div_grad]

@anuttarasammyak I admit that when I wrote "by definition" in the following

Now, if your physical system is translation-invariant then $f(\mathbf{r}+\mathbf{\varepsilon}, \mathbf{r}'+\mathbf{\varepsilon}) = f(\mathbf{r}, \mathbf{r}')$, by definition.

I might have introduced confusion by not spelling out one additional step in the reasoning explicitly. Namely, I consider a function $f(\mathbf{r}, \mathbf{r}')$ defined in a reference frame in which the space is homogeneous. I then translate the system by an infinitesimal amount in the direction $\mathbf{\varepsilon}$, which changes the function $f(\mathbf{r}, \mathbf{r}')$ accordingly:
$$
f(\mathbf{r}+\mathbf{\varepsilon}, \mathbf{r}'+\mathbf{\varepsilon}) = f(\mathbf{r}, \mathbf{r}') + \delta f \rm{,}
$$
where the small change $\delta f$ can be calculated using Taylor expansion, as I did in my original reply. Now, since I translated the system by an infinitesimal amount $\mathbf{\varepsilon}$ (and not by an arbitrary finite vector $\mathbf{R}$) from my original reference frame, I can still assume that the space is homogeneous and therefore require that the change $\delta f = 0$ (this point was not explicitly mentioned in my reply). In this case one obtains, as in my original reply, that the function $f$ must depend on its arguments only through their difference,
$$
f(\mathbf{r}, \mathbf{r}') \equiv g(\mathbf{r}-\mathbf{r}') \rm{,}
$$
and since in showing this I did not specify a particular reference frame, then this result holds true in any reference frame - provided, of course, that space is homogeneous in such frame. Thus, knowing that the function must depend on the difference of its arguments, I may now readily conclude that for any $\mathbf{R}$
$$
f(\mathbf{r}+\mathbf{R}, \mathbf{r}'+\mathbf{R}) = g(\mathbf{r}+\mathbf{R}-\mathbf{r}'-\mathbf{R}) = g(\mathbf{r}-\mathbf{r}') = f(\mathbf{r},\mathbf{r}') \rm{.}
$$
This result, however, was the starting point when you wrote

For any position vector R
$$f(\mathbf{r}, \mathbf{r'})=f(\mathbf{r}+\mathbf{R}, \mathbf{r'}+\mathbf{R})$$

so you already implicitly assumed that the function $f$ depends on its arguments $\mathbf{r}$ and $\mathbf{r}'$ only through their difference, while this was supposed to be demonstrated by assuming the homogeneity of space. In fact, the reasoning given in the second part of your original reply

choosing R=-r or R=-r'
$$=f(\mathbf{r}-\mathbf{r'},0)=f(0, \mathbf{r'}-\mathbf{r}):=F(\mathbf{r}-\mathbf{r'})$$

shows that somehow there are only two "preferred" reference systems - one having origin at $\mathbf{r}$, the other at $\mathbf{r}'$ - in which you can define a function $F$ such that it depends on the difference of the arguments of the original $f$. This then demonstrates that space is inhomogeneous, contradicting the premise of the question.

So in regard to your comment

I do not find a difference between my $\mathbf{R}$ and your $\epsilon$.

the difference is that I chose a reference frame and performed the infinitesimal symmetry transformation
$$
f(\mathbf{r},\mathbf{r}') \rightarrow f(\mathbf{r}+\mathbf{\varepsilon},\mathbf{r}'+\mathbf{\varepsilon})
$$
while you simply chose another reference frame (that is, you renamed the variables) but did not study the translational symmetry of $f$ in this frame which would be achieved, again, by considering infinitesimal transformation of the form
$$
f(\mathbf{r}+\mathbf{R},\mathbf{r}'+\mathbf{R}) \rightarrow f(\mathbf{r}+\mathbf{R}+\mathbf{\varepsilon},\mathbf{r}'+\mathbf{R}+\mathbf{\varepsilon}) \rm{.}
$$
Besides, studying the symmetries of the system by means of infinitesimal coordinate transformations is a standard and indispensable method, as all the important integrals of motion - momentum, energy, spin, etc. - are obtained as the generators of such transformations.

@hokhani As for your suspicion regarding translations in one "unit cell" (I assume you mean here the crystal lattice):

However, I think in one unit cell if we shift identically both $\vec r$ and $\vec r'$ then this statement does not seem necessarily correct!

If the system is homogeneous then it means that its physical properties remain the same if you parallel-translate all particles in the system by the same amount, that is, if you add a constant vector to all the particle's position vectors. Therefore even if you consider $f(\mathbf{r},\mathbf{r}')$ to be defined in a "unit cell" of a periodic crystal lattice, translational symmetry means that you add the same vector (and not two different ones) to both $\mathbf{r}$ and $\mathbf{r}'$. I showed above that for translation-invariant systems $f(\mathbf{r},\mathbf{r}') = g(\mathbf{r}-\mathbf{r}')$, so this operation does not affect your function $f$ which, in such cases, always depends only on the difference of its arguments. Observe, however, that in the case of a periodic crystal lattice whose unit cells are defined by fixed basis vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$, only translations by vectors of the form
$$
\mathbf{R} = \alpha\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c} \rm{,}
$$
where $\alpha$, $\beta$ and $\gamma$ are integers, leave the system invariant. In particular, this "discrete" nature of the translational symetry of a lattice leads to the quantization of the momentum/wave vector inside a crystal.

 

[anuttarasammyak]

@div_grad You pointed out that I set R to choose only two "preferred" reference systems. I am afraid your claim can be applied to your way also.

Now, if your physical system is translation-invariant then f(r+ε,r′+ε)=f(r,r′), by definition. Then the above equation tells you that ε⋅(∂f∂r+∂f∂r′)=0, and since this must hold for any ε you obtain that
∂f∂r=−∂f∂r′,
which is true if f(r,r′)=f(r−r′).

which is also true if $f(r,r')=g(r-r'+c)$ where c is constant. Do you have "preference" that c=0 ? As for me I prefer $f(r-r',0)$ to $f(r-r'+c,c)$ in order to define one variable function g, believing that it does not harm the generality.

 

[div_grad]

@div_grad You pointed out that I set R to choose only two "preferred" reference systems. I am afraid your claim can be applied to your way also.

@anuttarasammyak Then I am afraid that you are wrong. In posts #4 and #7 I have shown that if one requires that the function $f(\mathbf{r},\mathbf{r}')$ does not change under the infinitesimal translation of the vectors $\mathbf{r}$ and $\mathbf{r}'$ - this is the assumption of translation-ivariance - then this function must depend on its arguments through their difference, $f(\mathbf{r}, \mathbf{r}') = g(\mathbf{r}-\mathbf{r}')$, a fact which you have not proved. You are absolutely right that another function $h(\mathbf{r}-\mathbf{r}' + \mathbf{C})$, where $\mathbf{C}$ is a constant vector, also satisfies the condition on partial derivatives of $f$ given in my post #4, and so one could also write that $f(\mathbf{r}, \mathbf{r}') = h(\mathbf{r}-\mathbf{r}' + \mathbf{C})$. Observe, however, that this implies that $g(\mathbf{r}-\mathbf{r}') = h(\mathbf{r}-\mathbf{r}' + \mathbf{C})$, so in my approach there is no need to distinguish the rest frames of either of the two "particles" (i.e., points located in $\mathbf{r}$ or $\mathbf{r}'$) as the preferred frames, as in your approach - here, the function $f$ always depends on the difference of its arguments. In fact, one may very well use the function $h$ to demonstrate the same property that I showed in post #7:
$$
f(\mathbf{r}+\mathbf{R}, \mathbf{r}'+\mathbf{R}) = h(\mathbf{r}+\mathbf{R}-\mathbf{r}'-\mathbf{R}+\mathbf{C}) = h(\mathbf{r}-\mathbf{r}'+\mathbf{C}) = f(\mathbf{r}, \mathbf{r}') \rm{.}
$$
There. No preference is given to the constant vector $\mathbf{C}$. No matter the reference frame, the two-variable function $f(\mathbf{r}, \mathbf{r}')$ depends on its arguments through their difference. Because of this, the above property - which you assumed already true in post #3 - holds.

What you did now in your post #8 is you used the same name $g$ for two different functions (one of which was introduced in my post #7), despite that earlier in your post #6 you remarked that one should use different names for different functions (I agree, that's why I introduced different names - $g$ in post #7 and $h$ above):

BTW Using same alphabet f for RHS and LHS seems confusing to me because LHS is a two variable function but RHS is a single variable function.

You then equated these two fuctions like so
$$
g(\mathbf{r}-\mathbf{r}') = g(\mathbf{r}-\mathbf{r}'+\mathbf{C})
$$
and based on this you made a wrong conclusion that somehow this implies that $\mathbf{C}=0$. This is not true! If the above equality was valid, then it would imply only that $g$ is a periodic function of the difference $\mathbf{\varrho} \equiv \mathbf{r}-\mathbf{r}'$, so it would take the same values for all $\mathbf{\varrho}$ that are separated by a constant vector $\mathbf{C}$. Now, since $\mathbf{C}$ can be any arbitrary vector, this would mean that $g$ takes the same value for all $\mathbf{\varrho}$ in space. Therefore it would mean that $f(\mathbf{r},\mathbf{r}') = g(\mathbf{\varrho}) = \text{const.}$, and in this way one would "prove" that in a translation-invariant system all two-variable functions $f(\mathbf{r},\mathbf{r}')$ are just constant numbers, which is false. Your claim that in my method there is a preferrence for choosing $\mathbf{C}=0$ is therefore based on invalid reasoning and is simply not true, as I have demonstrated above by using the function $h(\mathbf{r}-\mathbf{r}'+\mathbf{C})$ with arbitrary $\mathbf{C}$ - which, by the way, is equivalent to using $g(\mathbf{r}-\mathbf{r}')$.

As for me I prefer $f(r-r',0)$ to $f(r-r'+c,c)$ in order to define one variable function g, believing that it does not harm the generality.

As far as I'm concerned, there is no generality in your proof, only circular reasoning - in order to show that a certain function depends on the difference of its arguments you begin by implicitly assuming that this is already the case.

 

[hokhani]

@hokhani
Now, if your physical system is translation-invariant then $f(\mathbf{r}+\mathbf{\varepsilon}, \mathbf{r}'+\mathbf{\varepsilon}) = f(\mathbf{r}, \mathbf{r}')$, by definition.

Thank you for your thorough answer and also anuttarasammyak for contribution. By the above definition of the translation invariant system, it seems that physically for such a system the single-particle potential must be constant throughout the system and the inter-particle interaction is something like Coulomb interaction which is a function of inter-particle distant! Isn't it?

P.S. Sorry if returning so late to the discussion. I surely follow your interesting posts.

 

[div_grad]

@hokhani Yes, basically something along these lines. There is a profound result in mechanics known as the Noether's theorem, which states that if a physical system exhibits some continuous symmetry, then to this symmetry corresponds some conservation law. What is important is that this theorem is not just a "theological" proof that certain conserved quantities exist, it actually gives you a concrete recipe for how to calculate these quantities. And so you can show that the most general symmetries of space and time lead to the following conservation laws:

1) (time-translation invariance) If there are no external fields acting on the system, or if the existing external field is independent of time, then the energy of the system is constant.

2) (homogenity of space) If there are no external fields acting on the system, then all three components $p_x$, $p_y$ and $p_z$ of the momentum $\mathbf{p}$ are constant. If there is an external field directed along some unit vector $\mathbf{n}$, then only those components of $\mathbf{p}$ that are perpendicular to $\mathbf{n}$ are constant.

3) (isotropy of space) If there are no external fields acting on the system, then all three components $M_x$, $M_y$ and $M_z$ of the angular momentum $\mathbf{M}$ are constant. If there is an external field directed along some unit vector $\mathbf{n}$, then only the component of $\mathbf{M}$ that is parallel to $\mathbf{n}$ is constant.

Observe that any boxes, constrains, etc. imposed on the system (such that it is not just particles moving in free space) are equivalent to some external fields, so for example closing a system in some rectangular box also breaks the translation-invariance of space.

it seems that physically for such a system the single-particle potential must be constant throughout the system

The single-particle potential is precisely the external field imposed on the system (particle). If it is constant throughout the system, then you have "full" translation-invariance and all three components of the particle's momentum are conserved. However it may equally well not be constant throughout, but define some direction $\mathbf{n}$ - then you will still have "restricted" translation-invariance, in that the system would exhibit symmetry with respect to translations in the plane perpendicular to $\mathbf{n}$ and the momentum components lying in this plane would be conserved.

and the inter-particle interaction is something like Coulomb interaction which is a function of inter-particle distant!

The inter-particle interaction, as the name implies, is not the external field imposed on the system (tough you may, however, formulate the problem of two interacting particles as the equivalent problem for one fictitious particle whose mass is given by the reduced mass of the two original particles, and which moves in a constant external field). In post #4 I have demonstrated that if a system is homogeneous, then the two-variable function, such as the inter-particle potential energy, depends only on the difference of its arguments - that is on the relative distance between the particles and their relative orientation. If the system was also isotropic, then this function would depend only on the distance between the particles and not on their relative orientation - which is the case for Coulomb interaction. But you could also consider cases such as diatomic molecule interacting with an atom - then their interaction is not isotropic (depends on the angles) but still depends on the relative distance between their centers of mass, that is it depends on the difference $\mathbf{r}-\mathbf{r}'$ of the center-of-mass position vectors.

As far as applications in quantum physics are concerned, the translation-invariance of a system allows you to neatly apply Fourier transforms to quantities such as $f(\mathbf{r},\mathbf{r}')$, since momentum is conserved for a homogeneous system and you can therefore expand in the basis of momentum eigenfunctions (plane waves),
$$
f(\mathbf{r},\mathbf{r}') = \int \frac{\text{d}^3p}{(2\pi\hbar)^3}\, \tilde{f}(\mathbf{p})\, e^{\frac{i}{\hbar}\mathbf{p}\cdot(\mathbf{r}-\mathbf{r}')} \rm{.}
$$
Note that in the exponential function there is a difference $\mathbf{r}-\mathbf{r}'$ which allows for using one conserved momentum vector $\mathbf{p}$ in the expansion. If the system was not translation-invariant and the function $f$ would not depend on the difference of its arguments, then the above expansion would produce nastier formulas, bit harder to deal with,
$$
f(\mathbf{r},\mathbf{r}') = \int \frac{\text{d}^3p}{(2\pi\hbar)^3}\, \int \frac{\text{d}^3p'}{(2\pi\hbar)^3}\, \tilde{f}(\mathbf{p},\mathbf{p}')\, e^{\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r}}\, e^{\frac{i}{\hbar}\mathbf{p}'\cdot\mathbf{r}'} \rm{.}
$$