# Bloch electrons, periodicity

1. Sep 24, 2014

### Waxterzz

Hi

http://uw.physics.wisc.edu/~himpsel/551/Lectures/E_versus_k.pdf

Look at first picture

If you see this picture of a bloch wavefunction, you see that it has two types of periodicities involved, one with the lattice constant(the bloch part), but what is the periodicity of the enveloppe.

Has it something to do with Born-Von Karman cyclic boundary conditions?

If you see this video

They speak also of 2 periodicities, but I dont get it

big K is supposed to be 2*pi / a where a is the lattice constant

and small k is 2*pi / L

This "L" is the L from the Born-Von Karman cyclic boundary conditions and is the length of the total crystal?

Because a bloch function is a convolution between a free electron that is confined only bye the size L (and it has this periodicity because it repeats itself in a loop over the crystal again? ) of the metal and a periodic bloch part, periodic with the lattice (ions)

Im not sure why the Born-Von Karman conditions are cyclic over a crystal, why does it form a loop?

Is my overall interpretation correct?

2. Sep 24, 2014

### Waxterzz

I don't get this.

a bloch function is like

On the picture, u see that psi(r) and the exponential term and the u(r) term all have different period.

But bloch theorem states that

psi(r+ lattice vector) = psi(r)

and

u(r + same lattice vector) = u(r)

How can this all be if they all have a different period????

IF the exponential term corresponds to a free electron plane wave AND the periodic u-term corresponds to the lattice (it repeats itself between the ion-sites), how can Psi (the bloch wavefunction) has the same period as the u-term, as the lattice since it is getting modulated by the exponential term.

Clearly the Bloch wave hasn't the same period as the lattice. :(:(:(

The wavelength of the bloch wave is at least 5*a as u can see above ........

I don't get it!

Last edited by a moderator: Apr 19, 2017
3. Sep 25, 2014

### DrDu

That's not correct. Only the expectation values have to be invariant under translations by a lattice vector. Hence the wavefunction can get multiplied by a phase factor, namely, $\psi(r+R)=\exp(ikR) \psi(r)$.

4. Jan 12, 2016

### Waxterzz

Wow, I still don't get Bloch's Theorem.

I thought I knew it, but the second form always confuses me.
$\psi(r+R)=\exp(ikR) \psi(r)$.

The probability density is the same, because the absolute factor of the phase turns 1.

I just don't see where the probability density's are the same. If you would square the blue part, the Bloch wave, then the probability density should be the same at distances separated by R right? And R has the same periodicity as the lattice potential, so the probability density function should have the same periodicity as the lattice?

I just don't get the second form.

|$\psi(r+R)| ² =$ \|psi(r)|² The phase factor in absolute value squared is 1, so why is the probability density function (if you would square the blue part) not periodic with the black part (the lattice) It's both the same with R!

Sorry for bump, I just never seem to grasp Bloch. And if I think I understand, a year later or more or less, I seem to have forgotten it completely.

What does exp(ikR) really do visually, which periodicity, what effect has it on psi(r)?

5. Jan 12, 2016

### Waxterzz

Is it because the blue part is only the real part?

Another picture

If you would square the total Bloch wave (both real and imaginary part, not individually but the total (Abs(Real + Im)))^2 in the bottom would it then have the same periodicity as the top picture, the lattice? Because I've read somewhere that the probabilty to find the electron is the same at each lattice point. So while a Bloch wave be it the real or imaginary part doesn't have the same periodicity as the lattice, the square of the total Bloch wave does?

What does the phase factor do? Shift the Bloch wave with a value to the left?

What does the second form really implies? A Bloch wave is clearly periodic but with what value?

Last edited: Jan 12, 2016
6. Jan 19, 2016

### Waxterzz

?

Real part of a bloch wave is not periodic with the lattice, but a bloch wave squared is.

Correct?

7. Jan 19, 2016

### DrDu

Correct.

8. Feb 17, 2016

I am equally and massively confused by this theorem. Especially the second expression:

$\psi (\vec{r}+\vec{R}) = exp(i.\vec{k}.\vec{R}) \psi (\vec{r})$

I was trying to plot these in matlab and when I tried to match the real parts with this equation and it did not match at all and i wasted a day on that. So here shouldn't $\psi$ at least periodic with the period defined by $\vec{k}$ in both the real and imaginary parts. How will it necessarily obey the born-van karman conditions then ?Looking at the pictures with real and imaginary parts above, the wavefucntion doesn't even seem to obey the born-von karman conditions. Coudl someone please xplain whats going on here ? Why is the bloch theorem even useful ?

On another note, why should the wavefunction not be lattice periodic itself ? from intuition, it should not need any phase factors or anything. In a periodic lattice with periodic potentials, should the wavefunction not be lattice periodic (instead of just being crystal periodic) ?

9. Feb 18, 2016

### DrDu

Born von Karman will be satisfied as long as $k_i=2\pi j/L_i$, where j is an arbitrary integer and L_i the size of the crystal in direction i, so that the exponential will become one if $\vec{R}=\{L_i\}^T$.

Why should the wavefunction not be lattice periodic itself? For the same reason that also wavefunctions in free space aren't only constant. A general time dependent wavefunction may be chosen completely at will, e.g. localized inside some elementary cell. This function must be expressible in terms of eigenfunctions of the hamiltonian, hence these can't be only lattice periodic.

10. Feb 18, 2016

Dear DrDu, thanks for the reply. Still a lot of confusion though.

Born von Karman conditions are practically applicable for any value of $k_{i}$, isn't this true because the choice of $L_{i}$ is completely arbitrary. The Bloch theorem stands for a crystal that is ideally and infinitely periodic and that is why if we say the lattice parameter is $a_{i}$ and the crystal is composed of $N_{i}$ repetitions of the lattice then

$L_{i}=N_{i}a_{i}$

This means that if I keep increasing $N_{i}$, I can get an almost continuous series of allowed $k_{i}$ values from $0$ to $2\pi /a_{i}$. You're obviously right that the function must be expressible in terms of the eigenfunctions of the Hamiltonian, but my question is actually that if the Hamiltonian itself is lattice periodic then shouldn't its eigenfunctions be lattice periodic itself? Of course they are allowed to have other kinds of periodicities, but shouldn't they be at least lattice periodic. By looking up at a solution of a double potential square well form here (http://www2.chem.umd.edu/groups/alexander/chem691/double_well.pdf), I do get an intuitive answer why the wavefunction need not be lattice periodic, although this is not the perfect example because the problem is not infinitely periodic.

Nevertheless, your comparison to time dependent wavefunction makes its a lot more clear. That is to say that for e.g. in simple cases, the time dependent wavefunction is factorized into a time-only dependent phase factor (like the exponential term) and a space-only dependent function. This is kind of what we're saying here - that if we are not being too restrictive, then the wavefunction under a lattice translation undergoes only a phase factor change. Eventually the electron probability density remains still the same.

It would be great of you could please spare some time in clarifying this issue a bit more rigorously for starters like me. The amount of notation out there and the confusion that its creates is humongous.

Last edited: Feb 18, 2016
11. Feb 18, 2016

### Waxterzz

I think a Bloch wave is not lattice periodic. If you plot real part or imaginary plot it's not lattice it's not periodic, but the complete Bloch squared, so the probability density of finding electron is lattice periodic.

12. Feb 18, 2016

$\psi_{k_{i}}(r+n_{i}a_{i}) = exp^{i.k_{i}n_{i}a_{i}} \psi_{k_{i}}(r)$
Where $k_{i}$ goes from $0$ to $2\pi (N_{i}-1)/N_{i}$ due to the Born von Karman conditions and you should be able to choose $N_{i}$ arbitrarily (in my humble opinion).
If I choose $N_{i}=1$ then the only allowed $k_{i}$ is 0, which then makes the wavefunction lattice periodic by default because the lattice is the crystal here if i set $n_{i}=1$. You can see
$\psi_{0}(r+n_{i}a_{i}) = exp^{0} \psi_{0}(r)$
Whats so beautiful is that if you increase $N_{i}$, it hardly has any physical meaning because it doesn't effect the real configuration of the system, however, this now allows for solutions which are not necessarily lattice periodic but are necessarily only crystal periodic. My basic question here was that because I have demonstrated this for $N_{i}=1$ that the wavefunction has to be lattice periodic, shouldn't therefore there be at least one solution to the wavefunction which is lattice periodic?