# I Bloch functions and momentum of electrons in a lattice

#### dRic2

Gold Member
Hi, I'm a bit confused about Bloch functions. This is what, I think, I understood: can someone please tell me what's wrong?

From Bloch's theorem we know that the wave-function of an electron inside a periodical lattice can be written as $ψ_k(r)=u_k(r)e^{ik⋅r}$. We hope that far from a lattice point, i.e. far from the nucleus of an atom the $u_k(r)$ term will be more or less constant, in order to get the plane wave $e^{ik⋅r}$.

I think this is reasonable also in view of the effective mass theorem. In fact far from a lattice point the physics should be similar to the free particle: the potential is more or less constant and the energy band takes the form of a paraboloid $E(k)≈E(k_0)−\frac{ℏ^2k^2}{2m^*}$ where $m^*$ is the effective mass $m^*=\frac 1 {\frac 1 {ℏ^2} \left( \frac{∂^2E}{∂k^2} \right)}$. Thus the problem can be shown to be equivalent to solve the Schrodinger equation for the Hamiltonian $\hat H=−\frac {ℏ^2} {2m^*}Δ$ and the solutions are plane waves.

Now, assuming what I wrote is correct (very big assumption ) here it comes the critical part. The Bloch functions $ψ_k(r)=u_k(r)e^{ik⋅r}$ are not eigenfunctions of the momentum operator $\hat p$. This is clear. But now my professor said that , to avoid this problem, we note that the average velocity of the electron is given by the relation:
$$<v>=\frac 1 ℏ \frac {∂E(k)}{∂k} =\frac {<p>}m$$​
1st question: where does this relation come from ? Is it from the semi-classical treatment, substituting $p=kℏ$?
2nd question: if I take into account the Born-von Karman boundary conditions is my reasoning still correct (assuming it was correct in the first place)?
3rd question: far from a lattice point, where Bloch functions take the form of plane waves, they are eigenfunction of the momentum operator, right ? So the momentum (and velocity) is well defined and the above equation reduces to $p=ℏk$, right?

Ric.

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#### fluidistic

Gold Member
I am also studying the same topic, so take my comments with a grain of salt.

First, it isn't true that the wavefunction of an electron must satisfy Bloch theorem. When you assume that the solution to Schrödinger's equation satisfies Bloch theorem, you are looking at a subset of solutions, however they do form a basis. This means that you can write the general solution to the equation as a sum (or integral) of the eigenstates that satisfy Bloch theorem. So you can form a wavepacket made of an infinity of Bloch eigenstates, and it will still satisfy Schrödinger's equation, but not Bloch theorem. It's the same reasoning than in electrostatics, where you find solutions to Laplace or Poisson equation using separation of variables. You first find the form of the separable solutions (eigenfunctions), but the general solution to the equation with its boundary conditions isn't separable.
Now, if you focus only on eigenstates like you do, I wouldn't say that the wavefunction behaves as a plane wave far from a lattice point. The general form is a plane wave modulated by a periodic function, with periodicity of that of the lattice. $u_k(r)$ could be similar to a sine if you want (this model has a name, I'm too lazy to Google but feel free to investigate). So it doesn't matter if you're far or close to a lattice point, psi is going to be periodic as long as you're considering a crystal. Even if you make the crystal lattice constant tending to infinity, there would be little difference between being close and being far from a nucleus. It's like taking to infinity the period of a sine function, it isn't going to vary much differently near or far a lattice point.

1) This relation comes from assuming de Broglie's picture is valid, i.e. that we can assign a speed of a particle by assuming it behaves as a wave. Then, we consider the group velocity of this wave to be the velocity of the particle.

3) If there's no crystal, yes. I.e. if there's no potential energy term in the Hamiltonian, you fall back to the free particle case where $\hat p \psi = \hbar k \psi$. But in a crystal, I do not see what you mean by being far from a lattice point. Do you mean in between 2 lattice points and taking the limit of the lattice constant to infinity?

#### dRic2

Gold Member
I'm reading Ziman's bok "Principles of theory of solids" and it seems to me that the author implies that in between two lattice point the wave function of n electron looks like a plane wave.

In pag 91
The N.F.E. (near free electron) method looks at the wave functions outside the atomic cores, where they look very like plane waves.
Again in pag 95, discussing the limitations of L.C.A.O. method
... it cannot pretend to represent a Bloch state in the interstitial region, where it must behave like a simple combination of plane waves
In pag 178, discussing the equivalent hamiltonian:
If we know $E(k)$ in a single band, we can discuss all the dynamical effects of a slow perturbing fields by ignoring the lattice potential as such; we treat the electron as if it were free, but with a modified kinetic energy operator of the form $E_m(-i ∇)$.
This is what I'm thinking about

Moreover I think the electron as a particle in a box, rather than a free particle because, in the end, they are constraint to be inside the crystal. This is way I think the wave function looks like a plane wave.

PS: in the image I don't know why I wrote "the solutions of plane waves should be plane waves" I meant "the solutions of the Sch. equ. should be plane waves"

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#### dRic2

Gold Member
One more question. In a single conducting band there are $N$ possible allowed value of the wave-vector $k$, each corresponding to a precise energy level. Imagine a full band where all the state are occupied. Does it mean that every electron is known to have a precise value of energy ? If so, then it should have a huge uncertainty on position, but that doesn't seem the case.

#### fluidistic

Gold Member
One more question. In a single conducting band there are $N$ possible allowed value of the wave-vector $k$, each corresponding to a precise energy level. Imagine a full band where all the state are occupied. Does it mean that every electron is known to have a precise value of energy ? If so, then it should have a huge uncertainty on position, but that doesn't seem the case.
I might get back to your previous post. Meanwhile I hope the Solid State gurus will jump in your thread.
Alright, let's assume that each electron has a well defined value of k (earlier you said that since the psi of the Bloch electrons is an eigenstate of H, thus having a well defined energy, but not an eigenstate of p, thus having no well defined momentum. Now it seems that you assume that they have a well defined momentum? :) ). As you say, it implies that there is a huge $\Delta x$. But since any Bloch electron's wavefunction extends in the entire crystal, no matter the value of k, I guess that this means that this is indeed the case. One way to visualize what's going on is to think of these Bloch electrons as waves. Modulated plane waves extending over the whole (infinite) solid. Even the 2 electrons with k = 0, which have 0 velocity, are completely spread over the entire infinite solid. We commonly say that the electrons are delocalized. So, in your words, these electrons have a huge uncertainty on position.

#### fluidistic

Gold Member
Regarding your previous post, I'm sure you already took a look at this picture: https://en.wikipedia.org/wiki/Bloch_wave#/media/File:Bloch_function.svg. The dotted line represents the plane-wave envelope, the solid line represents the real part of the wavefunction. (Unfortunately we cannot see the complex part). It doesn't look like psi matches the plane wave in-between atoms.

Or take a look at https://en.wikipedia.org/wiki/Bloch_wave#/media/File:BlochWaves1D.svg where you can see both the real and complex parts of the wavefunctions as well as the plane wave. Essentially you're saying that the 2nd and 3rd row almost match when you're in between 2 atomic sites. I don't really see it.

#### dRic2

Gold Member
Now it seems that you assume that they have a well defined momentum? :) ).
No. Why is that ? I know k, I don't know anything about p. As I said k is not an eigenvalue of p, so knowing it won't contradict my previous statement.

My problem now is that if we have a huge $∆x$ how can we use the semi-classical limit where we consider the electron as a particle (very localized)? I thought $∆x$ was "medium", but then I wound have $∆k >0$ and I don't see how that is possible.

#### dRic2

Gold Member
It doesn't look like psi matches the plane wave in-between atoms.
Yeah I saw that picture and it really bugs me. I'm still a bit confused about this.

PS: thanks for the help, I really appreciate :D

#### fluidistic

Gold Member
No. Why is that ? I know k, I don't know anything about p. As I said k is not an eigenvalue of p, so knowing it won't contradict my previous statement."
You're right, I slipped.

My problem now is that if we have a huge $∆x$ how can we use the semi-classical limit where we consider the electron as a particle (very localized)? I thought $∆x$ was "medium", but then I wound have $∆k >0$ and I don't see how that is possible.
When we consider a Bloch electron we do not consider it very localized at all. We consider it spread over the entire infinite crystal, just like a plane wave. The only difference with a plane wave is the periodic modulation.
But as I wrote above, we could in principle consider an electron as (initially) a wavepacket made of a sum of Bloch wavefunctions to make it more localized. If I remember well, this is what Ashcroft and Mermin did in chapter 11 or something like that, in the semi-classical model. Note that such electron is not a Bloch electron.

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#### dRic2

Gold Member
I think I solved my problems. If you don't mind I'll try to explain them so that you can tell me if there is something wrong.

1st problem: my obsession for the semiclassical model.
Answer: well, I'm a bit unsatisfied, but I read in Ashcroft's book that
(...) it is extremely difficult to justify.
It might be well above my current possibilities. It just works well in predicting the motion of electrons in between collisions so I must get over it and content myself with that.

2nd problem: free electron vs nearly free electron vs tight binding
Answer: They are different approximations and depends on the particular crystal I'm dealing with. If the attractive potential energy of the ions is very strong, tight binding method is the most appropriate, while, as $V(r)$ decreases, the other approximations could also do the job.

3rd problem: does tight binding wave function reduce to plane wave ?
(Partial) Answer: Here I don't know if the wave function should resemble the plane wave one in between two lattice point (ions) , or if the "global" shape (the envelope, as you said) should be a plane wave.

Let me first consider the energy spectrum as a function of $k$:

In black you can see the free electron parabola, while in red the "corrections" applied by the nearly free electron model (from perturbation theory), and the green line is the behavior you'll get with the tight binding method (a cosine).

Here you can see that far from bragg planes the behavior of the tight binding method follows pretty well the free electron parabola. Also, when the potential is almost constant plane waves are the solutions of the Sch. eq.

Regarding your previous post, I'm sure you already took a look at this picture: https://en.wikipedia.org/wiki/Bloch_wave#/media/File:Bloch_function.svg. The dotted line represents the plane-wave envelope, the solid line represents the real part of the wavefunction. (Unfortunately we cannot see the complex part). It doesn't look like psi matches the plane wave in-between atoms.
Well, I honestly think this might be a bit exaggerate. If you take a closer look

it could look like a plane waves.

#### fluidistic

Gold Member
I think I solved my problems. If you don't mind I'll try to explain them so that you can tell me if there is something wrong.

1st problem: my obsession for the semiclassical model.
Answer: well, I'm a bit unsatisfied, but I read in Ashcroft's book that

It might be well above my current possibilities. It just works well in predicting the motion of electrons in between collisions so I must get over it and content myself with that.
Yeah it works well under some approximations, some of them are mentioned in that textbook.

2nd problem: free electron vs nearly free electron vs tight binding
Answer: They are different approximations and depends on the particular crystal I'm dealing with. If the attractive potential energy of the ions is very strong, tight binding method is the most appropriate, while, as $V(r)$ decreases, the other approximations could also do the job.
The free electron model does not even consider a crystal structure. It's just a pack of fermions forming a fermionic gas, with absolutely no underlying crystal structure (unless you consider a null potential to be periodic, which is valid of course, but not really useful).

The nearly free electron model is able to predict bandgaps (as you depict in your picture below) and thus is able to explain why there are semiconductors and insulators, as well as conductors.

In real materials I think that tight binding models are more accurate (DFT) to describe the band structure. The nearly free electron model might work reasonably well for potassium, and possibly a few other metals (it kind of fails for lithium which is.... extremely simple). It's the concepts that matters and how one can retrieve known quantities from the models.

3rd problem: does tight binding wave function reduce to plane wave ?
(Partial) Answer: Here I don't know if the wave function should resemble the plane wave one in between two lattice point (ions) , or if the "global" shape (the envelope, as you said) should be a plane wave.

Let me first consider the energy spectrum as a function of $k$:
View attachment 248882
In black you can see the free electron parabola, while in red the "corrections" applied by the nearly free electron model (from perturbation theory), and the green line is the behavior you'll get with the tight binding method (a cosine).

Here you can see that far from bragg planes the behavior of the tight binding method follows pretty well the free electron parabola. Also, when the potential is almost constant plane waves are the solutions of the Sch. eq.

Well, I honestly think this might be a bit exaggerate. If you take a closer look
View attachment 248883
it could look like a plane waves.
I'm not quite sure what to think. If we believe that this shape looks like a plane wave, then the same shape is seen for the wavefunction right at the nucleus, so we must also accept that the wavefunction looks like a plane wave at the nuclei too. What do you think?
I also do not see how you find, based on the parabolic picture, that when the potential is constant, plane waves satisfy Schrodinger's equation.

#### dRic2

Gold Member
Sorry, I had a full day...

I'm not quite sure what to think. If we believe that this shape looks like a plane wave, then the same shape is seen for the wavefunction right at the nucleus, so we must also accept that the wavefunction looks like a plane wave at the nuclei too. What do you think?
I don't really trust that picture. Here's another one taken from Ziman's book

in between the atoms it does look like a plan wave.

To be completely honest here the author is talking about orthogonalized plane waves which are a sort of improvement of the tight binding methods. As I understood reading the book, these functions are introduced to solve the "fundamental" problem of tight binding, i.e the wave function should look like a plane wave in between two lattice points.

based on the parabolic picture, that when the potential is constant, plane waves satisfy Schrodinger's equation.
Two reasons:
1) The solutions to $\left( -\frac {\hbar^2}{2m} \Delta + V \right) \Psi = E \Psi$ subtected to periodic boundary conditions are plane waves. You can verify by differentiation
2) (this is just speculation, but it is not a real proof) The eigenvalue of the Hamiltonian for plane waves in a constant potential are $E = V + \frac {\hbar^2 k^2}{2m}$ and if the eigenvalues of my wave-function match this trend, then I'm lead to the conclusion that its shape might resemble the plane wave.

PS: Of course these some "general" arguments, in a ionic crystal with strong internal forces, for example, the electrons might be very localized and the trend typical of the plane waves should be impossible to detect

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#### dRic2

Gold Member
Ehrenfest theorem says that $\frac {<p>} m = \left< -\frac {\partial V(x)}{\partial x} \right>$ how do you go from $\left< -\frac {\partial V(x)}{\partial x} \right>$ to $\frac 1 {\hbar} \frac {\partial E(k)}{\partial k}$ ?

#### Demystifier

2018 Award
Use
$$E(k)=\frac{\hbar^2 k^2}{2m}$$
and note that $\langle \; \rangle$ is missing, i.e. the right equation is
$$\left\langle \frac{1}{\hbar}\frac{\partial E(k)}{\partial k} \right\rangle= \left\langle -\frac{\partial V(x)}{\partial x} \right\rangle$$
Do you see it now?

#### dRic2

Gold Member
Use
$$E(k)=\frac{\hbar^2 k^2}{2m}$$
and note that $\langle \; \rangle$ is missing, i.e. the right equation is
$$\left\langle \frac{1}{\hbar}\frac{\partial E(k)}{\partial k} \right\rangle= \left\langle -\frac{\partial V(x)}{\partial x} \right\rangle$$
Do you see it now?
So and so. I mean, 2 things I didn't get.

1) In all the books the relation is without <>. I think there is a difference.

2) why should I use the relation $E=\frac {\hbar^2k^2}{2m}$ ? I know I was talking about plane waves, but the above relation I understood to be general for any Bloch electrons

#### Demystifier

2018 Award
1) In all the books the relation is without <>.
Such as? (Give also the relevant page numbers in the books.)

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#### dRic2

Gold Member
Such as? (Give also the relevant page numbers in the books.)

Here it makes the *assumption* that a giveb frequency is linked to a particula velocity as in classical wave mechanics. But the task to justify it, I think, is far more complicated. This is from Ziman "Principles of theory of solids", but the same formula is presented also in Ashcroft's book

"Bloch functions and momentum of electrons in a lattice"

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