Counter example to the Bloch's theorem?

[jostpuur]

<br /> V(x) = \left\{\begin{array}{ll}<br /> 0, &amp; \exists n\in\mathbb{Z},\; x\in [2nL, (2n+1)L]\\<br /> \infty, &amp;\exists n\in\mathbb{Z},\; x\in\; ](2n-1)L, 2nL[\\<br /> \end{array}\right.<br />

This is a periodic potential. L is some constant. Is a solution

<br /> \psi(x) = \chi_{[0,L]}(x)\;\sin\big(\frac{\pi x}{L}\big)<br />

of the Schrödinger's equation

<br /> \Big(-\frac{\hbar^2}{2m}\partial_x^2 + V(x)\Big)\psi(x) = E\psi(x)<br />

a counter example to the Bloch's theorem?

\chi_{[0,L]} is a characteristic function, 1 when x\in [0,L] and 0 otherwise.

 

[marcusl]

A potential that is infinite over finite distance will not permit wave propagation.

 

[reilly]

To add to marcusl's comment, the thrm requires the DE coefficients, potential for example, to be finite and well behaved.

If you make your barriers very high but finite, then Bloch will apply.
Regards,
Reilly Atkinson

 

[jostpuur]

I have never seen a rigorous formulation of the Bloch's theorem, that would actually say what we need to assume. But the idea behind the Bloch's theorem is that when two operators (satisfying some conditions, of which I'm not fully sure) commute, then there is a common eigenbasis (and I'm not sure what the basis means either. It's not necessarily Hilbert space basis for sure...). The smoothness of some functions doesn't seem to be the most important part in this reasoning, but instead more abstract properties, like self-adjointness and commutativity.

In my example the translation operator commutes with the H.

If the periodic part satisfies

<br /> u(x + 2L) = u(x),<br />

nothing yet would prevent it from satisfying

<br /> u(x) = 0,\quad x\in [(2n-1)L, 2nL]<br />

too. So one might think that in my example we should still have a basis of Bloch waves

<br /> e^{ikx}u(x),<br />

with this kind of periodic parts?

 

[jostpuur]

My first post doesn't make very clear what precisely is the Hamiltonian operator, because the infinity business is a little bit vague. I can formulate the problem more rigorously too. Basically we define the domain of the operators by hand like this

<br /> D = \{\psi\in\mathbb{C}^{\mathbb{R}}\;|\;\psi(x)=0,\;x\in [(2n-1)L,2nL],\quad \exists\psi&#039;&#039;(x),\;x\in ]2nL,(2n+1)L[\}<br />

Then the Schrödinger's equation is

<br /> -\frac{\hbar^2}{2m}\partial_x^2 \psi = E\psi,\quad\quad \psi\in D.<br />

The translation operator

<br /> \psi\mapsto T_{2L}\psi,\quad (T_{2L}\psi)(x) = \psi(x + 2L)<br />

commutes with the Hamiltonian operator, and on the other hand the intersection of the eigenstates of the T_{2L} with the D is

<br /> \{e^{Ax}u(x)\;|\; u(x+2L)=u(x),\quad u(x)=0,\; x\in [(2n-1)L,2nL]\}.<br />

Naturally we demand A to be purely imaginary for physical solutions.

Shouldn't we have a common basis for the Hamiltonian and translation operators now?

 

[marcusl]

Sorry, you've examined the spatial behavior of propagating waves if they exist but you haven't looked at whether they can propagate in this case. To do it properly, examine the wave function solutions for the periodic square potential as you carefully take the limit V_0\rightarrow 0. You'll find that the wave amplitudes, which you don't address in your posts, go to zero as soon as you cross the boundaries at 0 and L. In the region 0<x<L between the boundaries where V=0, the solutions are sinusoidal--this is just the familiar infinite well.

The only "exception" to this behavior is a periodic delta function potential, where Bloch waves do propagate. It's not truly an exception, however, because the delta "function" is very special--it's not a function, and although it is infinitely tall it has no width, for instance. Because it's area is unity, waves can tunnel through. They cannot tunnel through your potential.

Bloch waves are derived in Walt Harrison's book "Solid State Theory." I can look for other references in my bookshelf tomorrow at work.

 

[jostpuur]

It should be part of the definition of D that its members are required to be continuous also at the points nL. I forgot that part.

 

[marcusl]

You keep taking snippets of math out of context and saying "see, shouldn't that work?" E.g.,

...
If the periodic part satisfies

<br /> u(x + 2L) = u(x),<br />

nothing yet would prevent it from satisfying

<br /> u(x) = 0,\quad x\in [(2n-1)L, 2nL]<br />

too. So one might think that in my example we should still have a basis of Bloch waves

<br /> e^{ikx}u(x),<br />

with this kind of periodic parts?

Work the problem! From start to finish! If you still get propagating Bloch wave solutions instead of infinite square potential solutions, come back and we'll help you find your error.

 

[jostpuur]

I don't believe there are Bloch waves for these infinitely high periodic barriers (except the trivial ones with k=0).

The problem is that I don't understand why there are Bloch waves with finite periodic barriers either. If I ask why the solutions are Bloch waves, the answer is "because the translation and Hamilton's operators commute". Well now I have this Hamilton's operator with the infinite barriers, which is commuting with the translation operator, and I'm not getting Bloch waves as solutions. The standard response is "of course not, you have infinite barriers there". Well, why are we supposed to have the Bloch waves with finite barriers then? The proof was not supposed to be based on the assumption that the barriers are finite, but on the assumption that Hamilton's operator and translation operator commute!

I'm trying to understand the Bloch's theorem, and I don't want to fool myself into believing that I would be understanding something that I really don't. This example with the infinite barriers is making this very tricky. Not everything seems to be making sense...

 

[marcusl]

Ok, I think you are looking for some insight to complement the mathematics, which says that periodic Bloch wave solutions must result from the translational symmetry properties of the potential.

If we imagine turning the barrier height V0 smoothly down towards zero, then the solutions must smoothly turn into free electron plane waves. For non-zero but small V0, then, it is reasonable to expect that the wavefunctions are mostly plane waves [not any k but specifically those which are commensurate with the lattice spacing], modified by some periodic perturbation due to the potential. This is the wave

\{e^{ik\dotr}u(x)

where u is the Bloch function whose form depends on the shape of the periodic potential. Harrison has a nice discussion and an illustration on p. 59 of "Solid State Theory."

So we see intuitively that both parts of the solution have the periodicity that is predicted/required by group theory and symmetry. It is maybe plausible to expect the same behavior holds for strong potentials, too. This is where the mathematics you described provides a definitive yes answer.

 

[reilly]

Bloch's Thrm is like a subset of Floquet's Thrm, first published in the in 1880's .It tells us that any second order DE equation with periodic coefficients, will have a periodic solution: cofficients must be measureable and Lebesgue integrable within the basic periodic interval.

The infinite barriers don't obey the Floquet conditions( finite barriers do), but still admit many periodic solutions. Suppose that the potential is periodic,V(x) = V(x+d). Suppose it is 0 from x=0 to x=d, 2d to 3d,... and infinite otherwise. Let W(x) be a solution in the 0-d interval, say
W0= sin( pi x/d). Then for any integer N, W0((2N pi)d+x), 0<x<d , 0 otherwise. Thus there are many periodic solutions with the infinite barriers.

Lower the barriers, and the solutions under the barriers will be given terms of the form exp(+/- ax), where a is a constant, a will be the same for all the barriers -- provided said barriers are identical. Try to work out the solutions of the periodic finite barrier problem. The key is to formulate the continuity/ boundary conditions. Much will become clear.

Think about a cork floating in the waves of a speed boat, or whatever. There's an understandable periodic potential at work.
Regards,
Reilly Atkinson

 

[marcusl]

oops, typo above

e^{ikx}u(x)