Derivation in Ashcroft and Mermin (The Pseudopotential)

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In summary, on page 208 of Ashcroft and Mermin, the pseudopotential method is described as a recasting of the OPW approach. The exact wave function for a valence level can be written as a linear combination of OPW's, and the plane wave part of this expansion is denoted as ##\phi_{\vec{k}}^v##. The expansions (11.27) and (11.24) can be rewritten as (11.30) using the expressions (11.24) and (11.26) and substituting ## \phi_{\vec{k}}^v ## with ## \phi_{\vec{k}+\vec{K}}^v ##. However, it is possible that
  • #1
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On page 208 of Ashcroft and Mermin they write:

We describe the pseudopotential method only in its earliest formulation, which is basically a recasting of the OPW approach.
Suppose that we write the exact wave function for a valence level as a linear combination of OPW's, as in (11.27).
Let ##\phi_{\vec{k}}^v## be the plane wave part of this expansion:
$$(11.29)\ \ \ \ \ \ \ \phi_{\vec{k}}^v(\vec{r})=\sum_{\vec{K}}c_{\vec{K}}e^{i(\vec{k}+\vec{K})\cdot\vec{r}}$$
Then we can write the expansions (11.27) and (11.24) as:
$$(11.30)\ \ \ \ \ \ \psi_{\vec{k}}^v(\vec{r})=\phi_{\vec{k}}^v(\vec{r})-\sum_{c}\bigg( \int d\vec{r}' \psi_{\vec{k}}^{c*}(\vec{r}')\phi_{\vec{k}}^v(\vec{r}') \bigg)\psi_{\vec{k}}^c(\vec{r})$$

I tried to derive equation (11.30) but got stuck on something, I hope someone can help me with it.

I'll write below my attempt at deriving equation (11.30).

First, notice the following few equations on pages 206-207:
$$(11.24) \ \ \ \ \ \phi_{\vec{k}}^v = e^{i\vec{k}\cdot \vec{r}}+\sum_c b_c \psi_{\vec{k}}^c(\vec{r})$$
$$(11.26)\ \ \ \ \ \ b_c = -\int d\vec{r}\psi_{\vec{k}}^{c*}(\vec{r})e^{i\vec{k}\cdot\vec{r}}$$
$$(11.27)\ \ \ \ \ \ \ \psi_{\vec{k}} = \sum_{\vec{K}}c_{\vec{K}}\phi_{\vec{k}+\vec{K}}$$

Now for my attempt at solution:
$$\psi_{\vec{k}}^v(\vec{r})=\sum_{\vec{K}} c_{\vec{K}}\phi_{\vec{k}+\vec{K}}^v = $$
$$=\sum_{\vec{K}} c_{\vec{K}}\bigg[ e^{i(\vec{k}+\vec{K})\cdot\vec{r}}+\sum_c b_c \psi^c_{\vec{k}+\vec{K}}(\vec{r})\bigg]=$$
$$=\phi_{\vec{k}}^v(\vec{r})-\int d\vec{r}'\psi_{\vec{k}}^c(\vec{r})e^{i\vec{k}\cdot \vec{r}'}\sum_{\vec{K}}\sum_cc_{\vec{K}}\psi_{\vec{k}+\vec{K}}^c(\vec{r})$$Where in the third equality above I used (11.26), in the second equality I used equation (11.24).

Now, I am stuck, how to derive from the above last three equations, equation (11.30)?
 
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  • #2
I think the ## b_c ## should also be ## b_c (k+K) ##, (## b_c ## as a function of ## k+K ##), ## \\ ## [Note: the ## \phi_k ## that you substitute in is actually ## \phi_{k+K } ##], ## \\ ## which will make the first ## \psi_{k+K }^* ##, (you also omitted the ## * ## on the first ## \psi ## in your last line), and also give an ## e^{i(k+K)r} ##. ## \\ ## I don't know how Ashcroft and Mermin eliminated the sum over ## K ##. They could do it if it only involved the ## \phi ##, and it didn't also involve the ## \psi 's ##, but it does. Perhaps they made an error.
 
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  • #3
Charles Link said:
I think the ## b_c ## should also be ## b_c (k+K) ##, (## b_c ## as a function of ## k+K ##), ## \\ ## [Note: the ## \phi_k ## that you substitute in is actually ## \phi_{k+K } ##], ## \\ ## which will make the first ## \psi_{k+K }^* ##, (you also omitted the ## * ## on the first ## \psi ## in your last line), and also give an ## e^{i(k+K)r} ##. ## \\ ## I don't know how Ashcroft and Mermin eliminated the sum over ## K ##. They could do it if it only involved the ## \phi ##, and it didn't also involve the ## \psi 's ##, but it does. Perhaps they made an error.
Yes, I forgot the complex conjugation in my last equality.
 
  • #4
MathematicalPhysicist said:
Yes, I forgot the complex conjugation in my last equality.
I'd be curious to know if you also agree with me about the ## k+K ##. I think my algebra is correct, and I don't see how the sum over ## K ## can be eliminated by making it simply a ## \phi_k ##, using equation 11.29. The ## \psi_{k+K}^* ## and ## \psi_{k+K } ## terms still contain ## k+K ##, and the ## K ## doesn't simply go away.
 
  • #5
If the summation is over ##K## then this sum is the same as a sum over ##k+K##, I think this is because we are in a periodic lattice.

But I am not sure.
I don't think ##b_c## depends on ##k+K##, though if we are in the habit of correcting myself I think that ##b_c## should depend on ##K## and not on ##k## as I wrote; It's a pity I cannot edit my post after I submitted it.
 
  • #6
MathematicalPhysicist said:
If the summation is over ##K## then this sum is the same as a sum over ##k+K##, I think this is because we are in a periodic lattice.

But I am not sure.
I don't think ##b_c## depends on ##k+K##, though if we are in the habit of correcting myself I think that ##b_c## should depend on ##K## and not on ##k## as I wrote; It's a pity I cannot edit my post after I submitted it.
See equation (11.26). ## b_c ## is thereby a function of ## k ##. The ## \phi ## that gets used (substituted in) is in a modified form of equation 11.24, expressed as ## \phi_{k+K} ##, and thereby ## b_c ## will be the ## b_c(k+K) ## version.
 
  • #7
MathematicalPhysicist said:
If the summation is over ##K## then this sum is the same as a sum over ##k+K##, I think this is because we are in a periodic lattice.

But I am not sure.
I don't think ##b_c## depends on ##k+K##, though if we are in the habit of correcting myself I think that ##b_c## should depend on ##K## and not on ##k## as I wrote; It's a pity I cannot edit my post after I submitted it.
See equation (11.26). ## b_c ## is thereby a function of ## k ##. The ## \phi ## that gets used (substituted into 11.27) is in a modified form of equation 11.24, expressed as ## \phi_{k+K} ##. Thereby ## b_c ## will be the ## b_c(k+K) ## version.
 
  • #8
@Charles Link I can see what you mean, and I agree with your observation; then how do you think they derived equation (11.30)?
 
  • #9
MathematicalPhysicist said:
@Charles Link I can see what you mean, and I agree with your observation; then how do you think they derived equation (11.30)?
As I mentioned in post 2, perhaps they made an error. It is possible that some algebra is able to simplify the sum over ## K ##, but it is not obvious to me. Perhaps someone else can see a simplifying step which then generates 11.30, but I don't see one.
 
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  • #10
Thanks for your feedback, nonetheless.
 
  • #11
MathematicalPhysicist said:
Thanks for your feedback, nonetheless.
Is it possible that ## \psi_{k+K}=\psi_{k} ##? If that is the case, the result of 11.30 follows immediately. I haven't read through the part in Ashcroft and Mermin preceding these equations for quite a number of years, but if ## \psi_{k+K}=\psi_k ##, the problem is solved. ## \\ ## Editing: And yes, see equation 8.50: ## \psi_{k+K}=\psi_k ##. The problem is solved and equation 11.30 of Ashcroft and Mermin is correct.
 
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  • #12
Jesus and moses christ, thanks Charles.
:->
 

1. What is the pseudopotential used for in Ashcroft and Mermin's derivation?

The pseudopotential is used to simplify the mathematical description of the electron-ion interaction in a solid. It replaces the complicated potential due to the Coulomb interaction between the electrons and ions with an effective potential that captures the essence of the interaction.

2. Why is the pseudopotential useful in electronic structure calculations?

The pseudopotential allows for a more efficient and accurate calculation of the electronic structure of a solid. Since it focuses on the valence electrons, which are the most important for determining a solid's properties, it reduces the computational burden compared to including all electrons in the calculation.

3. How is the pseudopotential derived in Ashcroft and Mermin's approach?

The pseudopotential is derived by expanding the electron-ion interaction potential in terms of the electron wavefunctions and solving for the coefficients that minimize the difference between the exact potential and the pseudopotential. This results in an effective potential that reproduces the same behavior as the exact potential for the valence electrons.

4. What assumptions are made in the pseudopotential derivation in Ashcroft and Mermin?

The main assumptions made in the derivation are that the ions are stationary and the electron wavefunctions are smooth and slowly varying compared to the lattice spacing. These assumptions allow for a simplified description of the electron-ion interaction potential and the use of the pseudopotential.

5. How does the pseudopotential approach improve upon the Hartree-Fock method?

The pseudopotential approach improves upon the Hartree-Fock method by taking into account the effects of electron screening and exchange-correlation. This leads to a more accurate description of the electronic structure and properties of a solid, especially for materials with strong electron-electron interactions.

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