Derivation of the velocity of bloch electrons

[patric44]

Homework Statement

Derivation of the velocity of bloch electrons

Relevant Equations

In the picture below

Hi guys
I saw that equation of the velocity of electrons In a periodic potential $$ v = (1/h) grad E(k) $$ in my textbook we use in our solid state physics course without any proof or any thing and when I searched for it I found that its derived in Ashcroft book appendix E :

Is there is any other relatively simpler proof, I mean how did he even expand the first term it's doesn't look like the Taylor series that I know!, and it uses perturbation theory, can I drive it using less heavier quantum mechanics?

 

[Dr Transport]

Nope, that is the proof.

 

[patric44]

Nope, that is the proof.

ok. perhaps someone could help me with this points in the derivation :

• (1) - how did he expand the $$ ε(k+q) $$ term i suppose its just Taylor series , but why the summation tho !?
• (2) - how the effect of the perturbed Hamiltonian led me to the eigenvalues on E.4 , its not very clear to me ?
• (3) - is there a derivation that involves Dirac notation ?

 

[dRic2]

1)

suppose its just Taylor series , but why the summation tho !?

It's just a Taylor series, but $\epsilon$ is a function of the vector $\mathbf k$ so the expansions involves all the coefficients $k_i$. If you like a compact form try $f( \mathbf x) = f(\mathbf 0) + grad( f) \cdot \mathbf x + ...$ (The second term is not very intuitive to write with the grad operator as the first one is, that is why most of the time the index notation is adopted). You can see here http://www.math.ucdenver.edu/~esulliva/Calculus3/Taylor.pdf

2) That's just perturbation theory. https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics). Otherwise try to find a copy of Griffith's book "Introduction to QM". There is a very nice and easy chapter on perturbation theory.

3) If you like dirac's notation, remember this:
Suppose you have the expectation value of an operator $A$, taken on the states $|n_1>$ and $|n_2>$, that is $<n_1|A|n_2>$. Now insert the "indentity" $\int dx |x><x| = 1$. You end up with $$<n_1|A|n_2> = \int dx <n_1|x> A <x|n_2>$$. Now remember that $<x|n_2>$ is by definition the wave function $\psi_2(x)$, so at the end you have $\int dx \psi_1^*(x) A \psi_2(x)$. If you revert these steps you can switch back from that notation to dirac's notation.

 

[patric44]

1)

It's just a Taylor series, but $\epsilon$ is a function of the vector $\mathbf k$ so the expansions involves all the coefficients $k_i$. If you like a compact form try $f( \mathbf x) = f(\mathbf 0) + grad( f) \cdot \mathbf x + ...$ (The second term is not very intuitive to write with the grad operator as the first one is, that is why most of the time the index notation is adopted). You can see here http://www.math.ucdenver.edu/~esulliva/Calculus3/Taylor.pdf

2) That's just perturbation theory. https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics). Otherwise try to find a copy of Griffith's book "Introduction to QM". There is a very nice and easy chapter on perturbation theory.

3) If you like dirac's notation, remember this:
Suppose you have the expectation value of an operator $A$, taken on the states $|n_1>$ and $|n_2>$, that is $<n_1|A|n_2>$. Now insert the "indentity" $\int dx |x><x| = 1$. You end up with $$<n_1|A|n_2> = \int dx <n_1|x> A <x|n_2>$$. Now remember that $<x|n_2>$ is by definition the wave function $\psi_2(x)$, so at the end you have $\int dx \psi_1^*(x) A \psi_2(x)$. If you revert these steps you can switch back from that notation to dirac's notation.

thanks for helping , i guess i am trying to skip a head a whole course in QM which we didn't take yet
to understand this derivation .
i think the best way now to stick to this simple crude derivation that i came up with to convince my self that the equation holds :
$$E = \frac{p^2}{2m} $$
$$ ⇒ E = \frac{ħ^2κ^2}{2m} $$
$$ ⇒ \frac{dE(k)}{dk} = \frac{ħ^2κ}{m} $$
$$ ⇒ \frac{dE(k)}{dk} \frac{1}{ħ}= \frac{ħκ}{m} = \frac{p}{m} $$
$$ ⇒ \frac{dE(k)}{dk} \frac{1}{ħ}= \frac{mv}{m} $$
$$ ⇒ v= \frac{dE(k)}{dk} \frac{1}{ħ} ∴ v =\nabla E(k) \frac{1}{ħ} $$
what you think of that derivation :)

 

[dRic2]

In my personal, but probably wrong, opinion I don't think you need a full course in QM to tackle Ashcroft's book. But (I think) it's supposed to be a grad-level textbook for a reason! You have to know at least something about QM. Perturbation theory as treated in Griffith's book doesn't require a profound knowledge of QM so if you can't understand it you are not probably enough prepared to read Ashcroft's book as well.

That's just my 2 cents of advise from a guy who's used to "skip" the prerequisites (and regret the decision every time!).

Regarding you attempt. What you did it's just a hint that should convince you that it might work, but why do you say that $E = \frac{\hbar^2 k^2} {2m}$? That is the dispersion relation for a free electron (that is, a plane wave), but in a crystal the electron is subject to a period potential and thus $E$ is different from $\frac{\hbar^2 k^2} {2m}$.