Graduate Derivation in Ashcroft and Mermin (The Pseudopotential)

[MathematicalPhysicist]

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)?

 

[Charles Link]

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.

 

[MathematicalPhysicist]

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.

 

[Charles Link]

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.

 

[MathematicalPhysicist]

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.

 

[Charles Link]

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.

 

[Charles Link]

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.

 

[MathematicalPhysicist]

@Charles Link I can see what you mean, and I agree with your observation; then how do you think they derived equation (11.30)?

 

[Charles Link]

@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.

 

[MathematicalPhysicist]

Thanks for your feedback, nonetheless.

 

[Charles Link]

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.

 

[MathematicalPhysicist]

Jesus and moses christ, thanks Charles.
:->