# A Wrong integration in Sakurai-Napolitano(and Fetter-Walecka)?

1. May 27, 2016

### ShayanJ

In chapter 7 of "Modern Quantum Mechanics" 2nd edition by Sakurai and Napolitano, a treatment of the degenerate electron gas is given as an example for 2nd quantization. This treatments is mostly taken from the same material in chapter 1 of "Quantum theory of many particle systems" by Fetter and Walecka.

At first the Hamiltonian of the system is written as $H=H_{el}+H_b+H_{el-b}$ where the first part is the Hamiltonian for electrons and their interactions among themselves, the second part is the Hamiltonian for the interaction among the fixed background ions (no kinetic part because they're assumed to be fixed) and the last part is the Hamiltonian for the interaction between electrons and the background ions.

My question is about the last two parts. In the books I mentioned, they're written as:
$H_b=\frac 1 2 e^2 \int d^3 x' \int d^3 x'' \rho(\mathbf x')\rho(\mathbf x'') \frac{e^{-\mu|\mathbf x'-\mathbf x''|}}{|\mathbf x'-\mathbf x''|}$
$H_{el-b}=-e^2 \sum_i \int d^3x \rho(\mathbf x) \frac{e^{-\mu|\mathbf x-\mathbf x_i|}}{|\mathbf x-\mathbf x_i|}$
Where $\rho(\mathbf x )$ is describes the distribution of the background ions which is assume to be uniform($\rho(\mathbf x )=\frac N V$). Until now everything is fine. But then there are these two lines of calculation which are obviously wrong!

$\left. \begin{array}{l} z \equiv |\mathbf x'-\mathbf x''| \\ \rho (\mathbf x)=\frac N V \end{array} \right\} \Rightarrow H_b=\frac 1 2 e^2 (\frac N V)^2 \int d^3 x \int d^3 z \frac{e^{-\mu z}}{z}= \frac 1 2 e^2 \frac{N^2}{V} \frac{4\pi}{\mu^2}$
This is wrong because it ignores that $|\mathbf x'-\mathbf x''|$ depends on the angular coordinates too. The most you can do is choosing z axis to be along one of those vectors and then $|\mathbf x'-\mathbf x''|$ will depend only on $\theta$ but you can't completely ignore the dependence on angular coordinates. I get it, the term $H_b$ is a constant term in the Hamiltonian and can be ignored but it just annoys me when such well respected books simply don't care about such issues.

But then comes the part about $H_{el-b}$ which is more serious!
$H_{el-b}=-e^2 \sum_i \frac N V \int d^3 x \frac{e^{-\mu|\mathbf x-\mathbf x_i|}}{|\mathbf x-\mathbf x_i|} =-e^2 \sum_i \frac N V \int d^3 z \frac{e^{-\mu z}}{z}=-e^2 \frac{N^2}{V} \frac{4\pi}{\mu^2}$
Aside from the fact that this integration has the same problem as the above one, they simply draw the conclusion that this term is a c-number! But...come on! You can easily see that the position operators for the electrons appear in $H_{el-b}$. Whatever they do to that integral, they should end up with $H_{el-b}=\sum_i f(\mathbf x_i)$. If they want to simply conclude that this operator is a c-number, they should give really good reasons, not just...I don't even know what to call it! Actually this can't be true because an operator is an operator and a c-number is a c-number. I can only accept it as an approximation that they can actually ignore this operator but such approximations really need to be discussed!
Any objections or comments on this?

2. May 27, 2016

### ShayanJ

Sakurai-Napolitano actually writes that. But Fetter-Walecka seem to assume $z=|\mathbf x'-\mathbf x''|$. But the point is both of them take out a $4\pi$ and don't do any integration over angular coordinates which means both of them are actually assuming $z=|\mathbf x'-\mathbf x''|$ and simply ignoring the dependence on angular coordinates. Trust me, I did the calculation and the answer is not that simple if you assume $|\mathbf x'-\mathbf x''|=\sqrt{x'^2+x''^2-2x'x'' \cos\theta}$. Actually I couldn't do the integration!

3. May 27, 2016

### vanhees71

Let's see. It's all about the integral
$$I=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{\exp(-\mu|\vec{x}-\vec{x}'|)}{|\vec{x}-\vec{x}'|},$$
which is obviously divergent as it stands, because substituting $\vec{r}=\vec{x}-\vec{x}'$ in favor of $\vec{x}'$ in the inner integral yields
$$I=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{r} \frac{\exp(-\mu r)}{r}.$$
Now for the integration over $\vec{x}$ you can assume that in fact you have a finite volume $V$ as done in your posting above. Then you are left with
$$I=V \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{r} \frac{\exp(-\mu r)}{r}.$$
Introducing spherical coordinates in the remaining integral gives
$$I=4 \pi V \int_0^{\infty} \mathrm{d} r r \exp(-\mu r)=-4 \pi V \frac{\mathrm{d}}{\mathrm{d} \mu} \int_0^{\infty} \exp(-\mu r) =-4 \pi V \frac{\mathrm{d}}{\mathrm{d} \mu} \frac{1}{\mu} = +\frac{4 \pi V}{\mu^2}.$$
Multiplying with the remaining factors you precisely get $H_b$. So the (by the way both excellent) books you mention are right.

4. May 27, 2016

### ShayanJ

Maybe I'm just confused here, but it seems to me you're committing the same mistake @vanhees71 .
Because $r^2=\vec r \cdot \vec r=(\vec x-\vec x')\cdot(\vec x-\vec x')=x^2+x'^2-2\vec x \cdot \vec x'=x^2+x'^2-2xx'\cos\theta$(in coordinates where the z axis is along $\vec x'$).
So you can't simply write $\int_{\mathbb R^3} dr r d\Omega \exp(-\mu r) =4\pi \int_0^\infty dr r \exp(-\mu r)$ because r depends on $\theta$!

5. May 27, 2016

### vanhees71

I can, because I substituted $\vec{r}$ instead of $\vec{x}'$ first. Then the integrand of that integral only depends on $r$ not on the angles anymore. So the integration over the angles just gives a factor $4 \pi$.

6. May 27, 2016

### ShayanJ

Oh...I see it now...I was just confused!
Thanks guys.
And sorry for the blasphemy!

7. May 27, 2016

### blue_leaf77

An alternative way to comprehend this integral is probably the following. For the moment forget the left integral over $\mathbf x'$ so that you are left with
$$\int d^3 x'' \frac{e^{-\mu|\mathbf x'-\mathbf x''|}}{|\mathbf x'-\mathbf x''|}$$
At its state now, $\mathbf x'$ is just a constant vector, or a parameter if you like. Then make a change of variable $\mathbf{z} = \mathbf x'-\mathbf x''$. Computing this integral, do you think that the result will depend on $\mathbf x'$? Remember that you are integrating over all space, probably not really from the infinities but form significantly large volume such that the characteristic length $1/\mu$ is much smaller than the size of your container (I wonder if they mention anything about the condition for $\mu$).

8. May 27, 2016

### vanhees71

There is no blasphemy in science. If you have doubts about anything published in science you should carefully check it, and that's what you did, and that's the way science works! You must not and do not need to believe in authorities in science, but you should think yourself about it.

9. May 27, 2016

### ShayanJ

While we're at it...I have a question about the results.
The Yukawa potential has a singularity as $r\to 0$ which is basically the same singularity that the Coloumb potential has, i.e. the self energy of the particles. Now we're integrating the Yukawa potential over all of the solid and I see no procedure to avoid the self-energy of the ions. So why the result has only a singularity because of the infinite volume and not a singularity because of the self-energy?

10. May 28, 2016

### vanhees71

Where is a self-energy here? What we have calculated is the two-body interaction energy, but no self-energy. There's no IR divergence, because the Yukawa potential (btw. here I'd rather call it a Debye-screened Coulomb potential, which is the physics behind it) is of finite range of the order of the Debye radius $r_D=1/\mu$.

11. May 28, 2016

### ShayanJ

Both Coulomb and Debye-screened Coulomb potentials go to infinity as $\vec x' \to \vec x$. Maybe I shouldn't have used the term self-energy, but the during the integration nothing was done to prevent $\vec x'$ and $\vec x$ to become equal. So why the integral is not divergent?

12. May 28, 2016

### blue_leaf77

Because there is part of the volume element to multiply with, cancelling that term in the denominator. It's the same like calculating the expectation value of potential energy in a hydrogen atom. Even if the potential itself has a singularity the origin, you are saved by the multiplication with $d^3\vec r$.

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