Periodic potential: Bloch's theorem

1. Jul 5, 2007

tim_lou

for the solution to the time-independent Schrödinger's equation with a period potential,
$$V(x)=V(x+a)$$

one has:
$$\psi(x)=Af(x)+Bg(x)$$

and
$$\psi(x+a)=A'f(x)+B'g(x)$$

the coefficients are related by a matrix equation,

let
$$v=[A, B]^T$$
$$v'=[A', B']^T$$

then
$$v=Kv'$$
where K is some non-singular matrix.

hence we can choose a solution where v is the eigenvector to the matrix K, more specifically,
$$v'=Kv=\lambda v$$

and thus one can find solutions to the Schrödinger's equation that satisfy:
$$f(x+a)=\lambda f(x)$$ (1)

from there, one can show that there are band structures for the allowed energies.

However, how does one prove that these are the ONLY physical solution to the Schroedinger's equation? I mean without the constrain of (1), we can possibly have other wave functions with different energies (in the forbidden zone)? am I missing something?

Last edited: Jul 5, 2007
2. Jul 9, 2007

tim_lou

a humble bump for my thread.
does nobody knows the solution to the question?

3. Jul 9, 2007

olgranpappy

If the potential in the Schrodinger equation is invarient under translations by a lattice vector 'a' which are implimented via the unitary operator 'T' then one can show that for any solution |f) of the schrodinger equation that T|f) is also a solution with the same energy. Thus [T,H]=0 and thus T and H can be simultaniously diagonalized. The simultanious eigenkets of T and H are the Bloch functions.

See Ashcroft and Mermin "Solid State Physics" chapter 8 for more on Bloch's theorem and Bloch functions,

4. Jul 10, 2007

tim_lou

the trouble of my understanding comes from the last statement:

why must there exist simultaneous eigenfunctions of T, H if they commute?

one can prove the existence of eigenfunctions of T and H separately, but how can one prove the existence of an eigenfunctions for both T and H if T and H commute?

5. Jul 10, 2007

olgranpappy

This is a basic result that you can probably find in a linear algebra textbook, or probably most textbooks on quantum mechanics. I'll briefly try to motivate why the result is true.

Suppose that T and H commute then consider an arbitrary matrix element of the commutator (which, of course, is zero) in a basis where T is diagonal.
$$0=<n|[T,H]|m>$$

Next, expand the commutator and insert a complete set of states in each term and then use the fact that T is diagonal in this basic to find.
$$0=\left({<n|T|n>-<m|T|m>}\right)<n|H|m>$$
And so we see that $$<n|H|m>=0$$ for subspaces with different eigenvalues of T. On the other hand within a subspace that has the same eigenvalue of T, H need not be diagonal--in general it will not be--thus we can say that H is now "block diagonal." But, the "blocks" (subspaces) where H is not diagonal correspond to blocks were T is proportional to the unit matrix, so I can choose linear combinations of my original choice of basis vectors within these "blocks" in order to further diagonalize H without changing T. Thus, if T and H commute I can find a basis that diagonalizes both T and H.

6. Jul 13, 2007

jostpuur

Not all matrices are diagonalizable. Could it happen, that a such sub matrix of H, that is not diagonal in the beginning, is not diagonalizable. Then the fact a corresponding sub matrix of T is proportional to the unity wouldn't help.

7. Jul 13, 2007

jostpuur

hmhm... uh.. we of course can assume that T and H are both diagonalizable alone. Is it so, that when a matrix is diagonalizable, then these blocks are diagonalizable also?

8. Jul 13, 2007

olgranpappy

Yes, I believe so.

Consider implementing one or another proceedure to first diagonalize T (which we "of course can assume" is possible).

Next, consider one at a time each remaining subspace for which H is not yet diagonal. In each remaining subspace T is just a number times the identity and for these purposes I can completely forget about it.

So, forget everything about T--there is no T now--we no longer know anything about T.

Okay, now consider just the restriction of H to one of the remaining subspaces where it is non-diagonal. In that subspace H still has all the properties which it originally had, e.g. Hermitian. So I can diagonalize it.

Cheers.

9. Jul 14, 2007

tim_lou

hmmm, although one CAN find a complete basis for solutions to the time-independent Schrödinger's equation, how can we guarantee that other solutions wouldn't be physically meaningful?

it is possible to find a eigenfunction of the H operator that isn't an eigenfunction of the V operator. That way, band structures would not exist at all. is something missing?

10. Jul 14, 2007

olgranpappy

1. There are "other solutions" that are physically meaningful. They can of course still be written as linear combinations of the energy eigenstates.

2. Yes it is possible to "find a eigenfunction of the H operator that isn't an eigenfunction of the V operator."

3. Point 2 (above) does not at all imply that "band structure would not exist."

4. Yes, I believe that you are missing something.

Cheers.

Last edited: Jul 14, 2007
11. Jul 14, 2007

tim_lou

well, then in Griffith's book on periodic potential, the "proof" for the existence of band structures for periodic delta potential is no longer valid. I wonder how one would prove that in general, band structures exist for ANY periodic potentials.

Last edited: Jul 14, 2007
12. Jul 14, 2007

olgranpappy

...to further drive home the point that you are missing something:

Consider any state that *is* an eigenfunction of both H *and* V... this means that it is also an eigenfunction of T. This is very boring.

Consider a Bloch state. This state is *not* an eigenfunction of V, but it *is* an eigenfunction of H.

13. Jul 14, 2007

olgranpappy

check out Ashcroft and Mermin's book on solid state. or Peierls.

14. Jul 14, 2007

olgranpappy

Also, I can say a few more words about bands following from periodicity. We have already seen that if the potential is periodic the solutions are Bloch waves. The Bloch wave itself is not periodic, but it can be thought of as a plane-wave times a periodic function. E.g.,
$$\psi_k(x)=e^{ikx}u_k(x)$$
where $$u$$ is periodic.

When we solve the Schrodinger equation for a given "k-value" we plug in the Bloch wave and take some derivatives and then we get to an intermediate stage where we are only concerned with the Schrodinger equation for the periodic part of the Bloch wave $$u_k$$. This equation is to be solved inside a single unit cell since it just repeats throughout the crystal (it's the periodic part). Thus we solve the schro equation in a finite volume and it should be easy to see that the solutions will be labeled by an integer (think about particle in a box). Those integers label the bands!

15. Jul 15, 2007

tim_lou

RESOLVED
opps, sorry I meant T, not V....(I didn't mean the potential operator).

I guess I really didn't specify my question clearly... and I still couldn't figure out the answer, so bare with me and here it goes again (the complete problem restated):
edit: conceptually, there is a big hole in my understanding of the quantum mechanics, the problem has been generalized.

the ultimate goal:
find a complex function $$\Psi (x,t): \bold R\times \bold R \rightarrow \bold C$$
which satisfies the given initial conditions,
$$\Psi (x,0)=g(x)$$
and obeys the Schrödinger's equation:
$$\hat{E} \Psi=\hat H_t \Psi$$

for a potential V that is independent of t, the equation separates, and we can have:
$$\Psi(x,t)=\sum_n c_n\psi_n(x) e^{-i\omega_n t}$$
and $$\omega_n=\frac{E_n}{\hbar}$$
for any complete basis that we can choose according to the time-independent Schrödinger's equation.

now, if $$V$$ is periodic, or
(let $$Tf(x)= f(x+a)$$ for a constant a),

then one can find a complete basis that are eigenfunctions of both the H and the T operators.

1. and from the statistical interpretations, when measuring the energy we could get any of the $$E_n$$ with probability $$|c_n|^2$$

however, we COULD choose ANOTHER bases that are just eigenfunctions of H, but not T, and we will have:
$$\Psi'(x,t)=\sum_n d_n \psi '_n(x) e^{-i\omega_n' t}$$

by uniqueness of the solution to the Schrödinger's equation (i'll assume the truth of the uniqueness theorem, correct my assumption if I am wrong), the two wave functions are equal, $$\Psi(x,t)=\Psi'(x,t)$$ for all real x and t.

2. but now, the "statistical interpretation" gives a whole new meaning. When we measure the energy, we would get $$E'_n$$ with probability $$|d_n|^2$$

these two statements can't both be correct at the same time, so the second statement is probably false, but at which point was the logic flawed?

to have a consistent "statistical interpretation", then there must only be one unique basis (and the only one) that is normalizable and physically meaningful. How, then can one prove the uniqueness of the choice of basis? (of course, if the basis is unique, then there won't be any choice anymore).

Last edited: Jul 15, 2007
16. Jul 15, 2007

tim_lou

After thinking about the problem for some more time, I finally came to a resolution.

so, when the eigenvalues are distinct, the eigenstates of H are already the eigenstates of T, so uniqueness is preserved. ($$|c_n|^2=|d_n|^2$$ and $$E_n=E_n'$$ for non-degenerate levels)

when some eigenvalues are not distinct, let's say the first three energies levels, 1,2,3 are degenerate, then we have a choice issue. we can choose linear combinations of these states and get a basis.

However, the statistical interpretation is unaffected, for a measurement that yields a energy of the degenerate state, one cannot distinguish from which state the measurement came. The only thing that gives the probability of getting $$E_d$$, the energy of the degeneracy is $$|c_1|^2+|c_2|^2+|c_3|^2$$.

so as long as
$$|c_1|^2+|c_2|^2+|c_3|^2=|d_1|^2+|d_2|^2+|d_3|^2$$ we are fine (uniqueness for degenerate levels are not necessary), and indeed, this is the case with degenerate levels.

Last edited: Jul 15, 2007
17. Jul 15, 2007

olgranpappy

now you've introduced some $$\omega$$... I'm not sure why? Do you mean $$\omega_n$$? do the c_n have time dependence? this post is a little unclear.

18. Jul 15, 2007

tim_lou

oops...yeah, I mean $$\omega_n$$
and $$\omega_n=\frac{E_n}{\hbar}$$

Thanks for the help.

19. Sep 17, 2007

jostpuur

And this means that there could be eigenstates of energy, that are not Bloch waves, in contradiction with the Bloch's theorem?

I can see, that all eigenstates of the translation operator have the form

$$e^{Ax}u(x)$$

where u(x) is some periodic function. If we wish the solution to be physical, A must be purely imaginary. However, it is incorrect to say that all eigenstates of H would be also eigenstates of the translation operator, so I don't think it is clear where the Bloch's theorem really comes from.

20. Sep 17, 2007