Bloch momentum-space wave functions

  • #1
raz
3
1
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?
 

Answers and Replies

  • #2
dRic2
Gold Member
862
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I think there is a misconception here. If you work with electrons in a periodic potential your Hamiltonian is invariant only with respect to a set of discrete translations so the momentum is not a good quantum number to label your states.

If you label your Bloch state as ##\ket{\mathbf \alpha}## then ##\braket{\mathbf r | \alpha}=e^{-i \alpha \cdot \mathbf r} u_{\alpha}(\mathbf r) = \psi_{\alpha}(\mathbf r)##. Here ##\alpha## is what you would usually call ##\mathbf k## and it is called crystal momentum (but it is not a momentum! it is just a quantum number that you use to label your state!).

In some context, it behaves as it were a momentum, but it is not (see Ashcroft, Mermin). You can clearly see it because ##-i \hbar \nabla \psi_{\alpha}(\mathbf r) \neq \hbar \alpha##.
 
  • #3
raz
3
1
Your explanation makes sense. In fact, the "momentum" depends on a discrete variable. Even if we approximate this discretization to a continuum (huge amount of electrons and possible states) still wouldn't it make sense to use a description via momentum?
 
  • #4
dRic2
Gold Member
862
219
To add new states you need to add another unit cell to your array of cells. But if you add another unit cell the symmetry of the hamiltonian doesn't change at all (it's still made of a set of cells, no continuous translational symmetry), hence momentum is still not a good variable to work with. You can add how many cells you want (meaning that you are working with a finer and finer k-mesh) but nothing can change.

If you really want a connection with the momentum you can show (check for yourself) that the velocity operator ## \mathbf{ \hat v} ## (defined as ## \mathbf {\hat v} = \frac {-i} {\hbar} [\mathbf{\hat r}, \hat H]##) is given by (in reciprocal space):
$$\mathbf{ \hat v_{\mathbf k}} = \frac 1 {\hbar } \nabla_{\mathbf k} \hat H_{\mathbf k}$$

That's as far as you can go, I guess.
 
Last edited:
  • #5
raz
3
1
Thank you dRic2, your answers were enlightening!
 

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