- #1

raz

- 3

- 1

- TL;DR Summary
- How would be the most correct way to obtain the Bloch momentum-space wave functions?

Hello, I wonder if it is possible to write Bloch wave functions in momentum space.

To be more specific, it would calculate something like (using Sakurai's notation):

$$ \phi(\vec k) = \langle \vec k | \alpha \rangle$$

Moving forward in a few steps:

Expanding:

$$ \phi(\vec k) = \int d^3\vec r \langle \vec k | \vec r \rangle \langle \vec r | \alpha \rangle$$

Replacing the element ##\langle \vec k | \vec r \rangle## and considering that ##\langle \vec r | \alpha \rangle## will be the Bloch wave function:

$$ \phi(\vec k) = \frac 1 {(2\pi)^{3/2}} \int d^3\vec r e^{-i\vec k \cdot \vec r} u_{k'}(\vec r)e^{i\vec k' \cdot \vec r}$$

or:

$$ \phi(\vec k) = \frac 1 {(2\pi)^{3/2}} \int d^3\vec r e^{i(\vec k' - \vec k) \cdot \vec r} u_{k'}(\vec r)$$

Remembering that ##u_{k'}(\vec r)## may be represented as:

$$u_{k'}(\vec r) = \sum_{\vec G} c_{\vec k' - \vec G} e^{-i\vec G \cdot \vec r}$$

Being ##\vec G## a reciprocal lattice vectors family and ##c_{\vec k' - \vec G}## a parameter defined by the central equation.

From this point some doubts arise: if the step by step is correct; if ##\vec k' - \vec k = 0## or if ##\vec k' - \vec k = \vec G## may be considered. Note that if this last statement is correct, replacing ##u_{k'}(\vec r)## in the integral will cause the exponential terms to vanish.

Solving these questions, how would be the most correct way to calculate the integral and get a final answer?

To be more specific, it would calculate something like (using Sakurai's notation):

$$ \phi(\vec k) = \langle \vec k | \alpha \rangle$$

Moving forward in a few steps:

Expanding:

$$ \phi(\vec k) = \int d^3\vec r \langle \vec k | \vec r \rangle \langle \vec r | \alpha \rangle$$

Replacing the element ##\langle \vec k | \vec r \rangle## and considering that ##\langle \vec r | \alpha \rangle## will be the Bloch wave function:

$$ \phi(\vec k) = \frac 1 {(2\pi)^{3/2}} \int d^3\vec r e^{-i\vec k \cdot \vec r} u_{k'}(\vec r)e^{i\vec k' \cdot \vec r}$$

or:

$$ \phi(\vec k) = \frac 1 {(2\pi)^{3/2}} \int d^3\vec r e^{i(\vec k' - \vec k) \cdot \vec r} u_{k'}(\vec r)$$

Remembering that ##u_{k'}(\vec r)## may be represented as:

$$u_{k'}(\vec r) = \sum_{\vec G} c_{\vec k' - \vec G} e^{-i\vec G \cdot \vec r}$$

Being ##\vec G## a reciprocal lattice vectors family and ##c_{\vec k' - \vec G}## a parameter defined by the central equation.

From this point some doubts arise: if the step by step is correct; if ##\vec k' - \vec k = 0## or if ##\vec k' - \vec k = \vec G## may be considered. Note that if this last statement is correct, replacing ##u_{k'}(\vec r)## in the integral will cause the exponential terms to vanish.

Solving these questions, how would be the most correct way to calculate the integral and get a final answer?