How many of you guys actually like mathematics?

[Klockan3]

I like math, but I've noticed that some mathematicians don't like the way physicists use math.

I don't like the way many math students use maths either. When there are math grad students who can't even understand the basic geometric properties of some sets in R^2, for example the unit circle in some non-euclidean norms, I want to slam my head into the wall. Basically they just learned a bunch of theorems and have no clue of what it means.

I think that it just hurts people when they see others abuse the subject they love. One time I held a mini lecture for some girls who talked about linear algebra and they had totally misunderstood just about everything about it, can't just walk by listening to that :(

 

[jostpuur]

In order to reduce the Bloch problem to the Floquet one, we first write the Schrödinger equation

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

in a form where only first order derivatives are present, as follows:

<br /> \left(\begin{array}{c}<br /> \partial_x\psi(x) \\ \partial_x\phi(x)<br /> \end{array}\right)<br /> = \left(\begin{array}{cc}<br /> 0 &amp; 1 \\<br /> \frac{2m}{\hbar^2}(V(x) - E) &amp; 0 \\<br /> \end{array}\right)<br /> \left(\begin{array}{c}<br /> \psi(x) \\ \phi(x) \\<br /> \end{array}\right)<br />

Now by Floquet theory there exists a periodic function u(x) and a matrix M such that

<br /> \left(\begin{array}{c}<br /> \psi(x) \\ \phi(x) \\<br /> \end{array}\right)<br /> = \exp\Big(\left(\begin{array}{cc}<br /> M_{11} &amp; M_{12} \\<br /> M_{21} &amp; M_{22} \\<br /> \end{array}\right)x\Big)<br /> \left(\begin{array}{c}<br /> u_1(x) \\ u_2(x) \\<br /> \end{array}\right)<br />

Now if M was diagonal, we could arrive at the desired result that the wave function has a form e^{M_{11}x}u_1(x). However, there is no reason to assume that M is diagonal, and also there is no obvious reason to assume that

<br /> \big(e^{Mx}\big)_{11} u_1(x) + \big(e^{Mx}\big)_{12}u_2(x)<br />

could be written in a desired form. Consequently, I am not yet convinced Bloch's theorem would follow from Floquet theory.

For example it is not possible to find a number A and a bounded function g such that

<br /> e^{x}f_1(x) + e^{-x}f_2(x) = e^{Ax}g(x)<br />

if f_1 and f_2 are bounded too. The Floquet theory leads to a similar situation, but more complicated.

This problem is probably related to a similar problem which I encountered earlier when trying to prove Bloch's theorem. I explained about it here:

Bloch's theorem and diagonalization of translation operator

Logically, there seems to be two alternatives now. It could be that Bloch's theorem follows from Floquet theory in some way which I do not yet understand. Or then it could be that Bloch's theorem does not follow from Floquet theory, but people are not aware of it because physicists have protected Bloch's theorem from rigor examination by always writing the result unclearly with intention.

 

[Jack21222]

Reminds me of something someone here said (anyone remember who? was it Chi?): People today can be divided into 10 categories: those who understand binary, and those who don't.

I think attributing it to anybody at this point would be silly, since I've known that joke for 12 years.

 

[jostpuur]

So...

If physicists had stated very clearly how they believe that Bloch's theorem would follow from Floquet theory, then it would have been more likely that mathematicians would have pointed out why the argument doesn't really work.

If physicists had clearly called their "theorem" a Bloch's conjecture, Bloch's hypothesis, or Bloch's pseudotheorem, then it would have been clearer to those who are really interested in theorems, that there still remains something to be discovered in this problem.

However, now when physicists have left their claims and terminology deliberately ambiguous, no mathematician has shown interest to checking the physicists' claims. Consequently, now large amount of people believe that a Bloch's theorem is a proven theorem, while nobody really knows what the theorem even is.

This is how the physicists' policy of deliberate unclarity really works.

 

[cronxeh]

So...

If physicists had stated very clearly how they believe that Bloch's theorem would follow from Floquet theory, then it would have been more likely that mathematicians would have pointed out why the argument doesn't really work.

If physicists had clearly called their "theorem" a Bloch's conjecture, Bloch's hypothesis, or Bloch's pseudotheorem, then it would have been clearer to those who are really interested in theorems, that there still remains something to be discovered in this problem.

However, now when physicists have left their claims and terminology deliberately ambiguous, no mathematician has shown interest to checking the physicists' claims. Consequently, now large amount of people believe that a Bloch's theorem is a proven theorem, while nobody really knows what the theorem even is.

This is how the physicists' policy of deliberate unclarity really works.

Nicely said. Can we expand this to cover the string theory and et al delusionaries

 

[Klockan3]

If physicists had stated very clearly how they believe that Bloch's theorem would follow from Floquet theory, then it would have been more likely that mathematicians would have pointed out why the argument doesn't really work.

You are extremely out of line saying that, there is a complete spectrum of people between maths and physics and if something wasn't properly done they would have noted it somewhere, especially considering that as I said before just about everything before quantum field theory is properly done mathematically already. Just because a lot of physicists do not know about the mathematical structure do not mean that it is already done. There are a lot of physicists who know a ton more about maths than you do. There are a few terminology things that differ such as the Minkowski metric tensor but things like theorems and axioms are not any different.

Anyhow try this version of Floquet by the way, it is easier to understand:
http://mathworld.wolfram.com/FloquetsTheorem.html

It is quite easy to prove without as well. For a potential with period T the Hamiltonian commutes with the operator T' translating x with T. Thus they have a complete set of common orthogonal eigenvectors. Denote these f.
Now since f is an eigenvector of the translation it means that T'f=af, or basically if you have g in [0,T] then f(x+bT)=g*a^b for some g. When you got this you can just bake in some of the x dependence into g and you get for a T periodic function u: f(x)=u(x)*exp(ln(a)*x). Here we note that |a|=1 since we don't want unbounded solutions, also only the smallest solution to ln(a) is necessary since we are always taking everything from u anyway.

 

[jostpuur]

It is quite easy to prove without as well. For a potential with period T the Hamiltonian commutes with the operator T' translating x with T. Thus they have a complete set of common orthogonal eigenvectors.

I know that if you have two diagonalizable matrices H,T\in\mathbb{C}^{n\times n} which commute, then they can be diagonalized simultaneously.

It is not quite obvious how this result can be generalized to operators H:\mathcal{D}(H)\to L_2(\mathbb{R}) and T:L_2(\mathbb{R})\to L_2(\mathbb{R}), which do not have eigenvectors inside L_2(\mathbb{R}), and of which other one is unbounded.

 

[Klockan3]

It is not quite obvious how this result can be generalized to operators H:\mathcal{D}(H)\to L_2(\mathbb{R}) and T:L_2(\mathbb{R})\to L_2(\mathbb{R}), which do not have eigenvectors inside L_2(\mathbb{R}), and of which other one is unbounded.

Please, if you want to know every detail read up more on the theory, I can't go through everything here. Read Ballentine, it answers most of your questions and he do prove Blochs theorem but since he proves a ton of other things along the way which is used it is a bit extensive to write it all up here.

 

[jostpuur]

It is quite easy to prove without as well. For a potential with period T the Hamiltonian commutes with the operator T' translating x with T. Thus they have a complete set of common orthogonal eigenvectors.

I know that if you have two diagonalizable matrices H,T\in\mathbb{C}^{n\times n} which commute, then they can be diagonalized simultaneously.

It is not quite obvious how this result can be generalized to operators H:\mathcal{D}(H)\to L_2(\mathbb{R}) and T:L_2(\mathbb{R})\to L_2(\mathbb{R}), which do not have eigenvectors inside L_2(\mathbb{R}), and of which other one is unbounded.

Please, if you want to know every detail read up more on the theory, I can't go through everything here.

It might seem, that only after seeing your post, I quickly decided that I could find something minor that is wrong with it. This impression is false, however, as could be carefully deduced from my earlier posts.

Suppose you want to use a result that two operators can be diagonalized simultaneously, but you only know the result for finite matrices. What's the most natural thing to do then? IMO you should recall how the result is proven for finite matrices, and then check out if the same idea could be used to prove the Bloch's theorem.

Sometimes it happens that some idea can be used to prove some weak result, but then precisely the same idea can be used to prove a stronger result too.

So if A,B\in\mathbb{C}^{n\times n} are diagonalizable and they commute, what's the idea behind the proof? The idea is that we first move onto a basis where A is diagonal, and then if some eigenspaces of A have dimension higher than 1, we carry out further transformations in these subspaces to diagonalize remaining blocks of B.

I attempted to use this original idea to prove the Bloch's theorem in my earlier thread:

Bloch's theorem and diagonalization of translation operator

In this thread I first chose two linearly independent eigenstates \psi_1,\psi_2 of H with the same eigenvalue E, and then tried to find out, that if they are not readily eigenstates of translation operator, perhaps I could write new linear combinations of these eigenstates, so that I would get new states \phi_1,\phi_2 which would still be eigenstates of H, but also be eigenstates of translation at the same time.

You see, that's precisely the same idea that you use when you prove that two commuting diagonalizable matrices can be diagonalized simultaneously?

I'm not a kind of guy who simply looks at a physicist's "proof" and then spends rest of his life merely laughing at it. I have actually tried to fill the gaps in these proofs.

I'm interested to know if you have succeeded to figure out what the Bloch's theorem is. (That means that is it really a rigor theorem, or is it only a conjecture which holds under some conditions which are not known?)

This question from me was not taken seriously.

It could be that the Bloch's theorem is true roughly like it is stated.

But it could also be that it is only true for almost all energies in the spectrum, when the spectrum is equipped with some natural measure. So it could be that there are some special energies for which the eigenstates are not Bloch waves, but these energies are somehow exceptional.

Or then it could be that the Bloch's theorem is true only for almost all potentials, when the potential is chosen from some random ensemble, generated by equipping the Fourier coefficient space with some probability measure.

There are all kinds of mathematical conjectures that one can come up without of this physical problem.

I know, that I do not know, which one of these conjectures are right or wrong. (Remember Socrates )

 

[Klockan3]

By basic theory of DEs, there exists two linearly independent solutions \psi_1,\psi_2 to the Schrödinger's equation

This is wrong, it depends on the potential. The Schrödinger equation is not a normal DE and there are for most values of E no solution at all to a specific potential which is the whole deal with quantum mechanics and often there is just one solution for a specific E. According to you all energies would be viable for every system...

 

[Klockan3]

Ok, I found the real flaw in your argument. The thing is that you first assume that you got a solution that is not an eigenvector to translations and then translate it to get the orthogonal solution space. All fine so far right? The problem now is that it is possible that the whole solution space for this specific energy have the same translation eigenvalue! What would that mean?! That you can't find the function you first assume that you can find. The prohibited value you are talking about is one of the possible values where the initial function is already an eigenvector to the translation.

 

[jostpuur]

If at least somebody found my opinion, about using physics as a tool to sabotage mathematics, entertaining, then my contribution to this thread was not completely wasted, IMO

 

[Klockan3]

I love maths when I got something I want to think about, I appreciate it. I have also thought that Bloch's theorem was strange but now I am quite solid :p