I Inadequate proof of Bloch's theorem?

Happiness

Suppose a wave function is a linear combination of 2 stationary states: $\psi(x)=c_1\psi_1(x)+c_2\psi_2(x)$.

By [5.52] and [5.53], we have $\psi(x+a)=e^{iK_1a}c_1\psi_1(x)+e^{iK_2a}c_2\psi_2(x)$. But to prove [5.49], we need $K_1=K_2$. That means all the eigenvalues of the "displacement" operator D have to be the same. But why is it so?  Reference: Intro to QM, David J Griffiths, p224

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Cthugha

But to prove [5.49], we need $K_1=K_2$.
That is obviously incorrect. If you do not immediately see why, just plot both parts of the complex exponential function for several K, e.g. K=k, 2k and so on.

• vanhees71

Happiness

That is obviously incorrect. If you do not immediately see why, just plot both parts of the complex exponential function for several K, e.g. K=k, 2k and so on.
I don't understand you. What's the meaning of your k?

We need $K_1=K_2$ (mod $\frac{2\pi}{a})$.

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• vanhees71

Happiness

$K_1=-K_2$ (mod $\frac{2\pi}{a}$) according to another book, QM 2nd ed., Bransden & Joachain, p183, (see (4.196) below).

That would mean that $\psi(x+a)=D\psi(x)=D[c_1\psi_1(x)+c_2\psi_2(x)]=e^{iK_1a}c_1\psi_1(x)+e^{-iK_1a}c_2\psi_2(x)$. But this contradicts (4.198) of the same book itself.

Note that here $\psi_1(x)$ and $\psi_2(x)$ have the same energy, but could have different $\lambda$, the eigenvalues for D.  Last edited:

Cthugha

I don't understand you. What's the meaning of your k?
Just any wavenumber. Feel free to plot it for some arbitrary number, twice the number and so on.

$K_1=-K_2$ according to another book, QM 2nd ed., Bransden & Joachain, p183, (see (4.196) below).
This is not even close to what the book says. Note the restricted range of values chosen in 4.197 and especially the comment between 4.196 and 4.197. Is it even possible that in 5.49 just a single K exists that solves the equation, if exp(2 pi i n)=1?

Happiness

Just any wavenumber. Feel free to plot it for some arbitrary number, twice the number and so on.

This is not even close to what the book says. Note the restricted range of values chosen in 4.197 and especially the comment between 4.196 and 4.197. Is it even possible that in 5.49 just a single K exists that solves the equation, if exp(2 pi i n)=1?
The point about writing explicitly (mod $\frac{2\pi}{a}$)? Is that all you wanted to say?

(mod $\frac{2\pi}{a}$) doesn’t resolve the issue.

Cthugha

The point about writing explicitly (mod $\frac{2\pi}{a}$)? Is that all you wanted to say?

(mod $\frac{2\pi}{a}$) doesn’t resolve the issue.
It gives you an infinite number of eigenvalues (or rather: equivalent formulations) instead of just a single one. If that does not solve your problem , you should state your problem clearly. The proof in Griffith's book is pretty clear and straightforward.

Happiness

It gives you an infinite number of eigenvalues (or rather: equivalent formulations) instead of just a single one. If that does not solve your problem , you should state your problem clearly. The proof in Griffith's book is pretty clear and straightforward.
It’s possible for two linearly independent common eigenfunctions of the Hamiltonian and D that have the same energy to have different eigenvalues for D, meaning two different λ. But Griffiths assumes they must be the same (modularly). In fact, Bransden and Joachain show that they cannot be the same (modularly), proving Griffiths wrong (unless K=0, the trivial case).

It may be important to understand the implication of the bolden two above. There are exactly two linearly independent eigenfunctions of the same energy, since the Schrondinger equation is second order. And this allows the use of the Wronskian determinant of the 2X2 matrix in B&J's (4.193).

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Cthugha

It’s possible for two linearly independent eigenfunctions of the Hamiltonian that have the same energy to have different eigenvalues for D, meaning two different $\lambda$. But Griffiths assumes they must be the same (modulo-ly). In fact, Bransden and Joachain show that they cannot be the same (modulo-ly), proving Griffiths wrong (unless K=0, the trivial case).
Sorry, but what you claim is far from correct. It starts with your assumption that for a wave function $\psi(x)=c_1\psi_1(x)+c_2\psi_2(x)$, you will get something like $\psi(x+a)=e^{iK_1a}c_1\psi_1(x)+e^{iK_2a}c_2\psi_2(x)$. Of course you would get something similar to $\psi(x+a)=e^{iKa}(c_1\psi_1(x)+c_2\psi_2(x))$. Of course you could create a displacement on two individual functions already displaced to get something like: $\psi(x+a)=e^{iKa}(e^{iK_1a}c_1\psi_1(x)+e^{iK_2a}c_2\psi_2(x))$.

Any K that fulfills the periodicity condition in the proof by Griffith is a solution and of course this includes both signs. He never claims that all eigenvalues have to be the same.

Happiness

Of course you would get something similar to $\psi(x+a)=e^{iKa}(c_1\psi_1(x)+c_2\psi_2(x))$.
Why? The $K$ for $\psi_1(x)$ and $\psi_2(x)$ are different (modularly).
He never claims that all eigenvalues have to be the same.
If the eigenvalues are not the same, you cannot factorise out $e^{iKa}$!

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Cthugha

Why? The $K$ for $\psi_1(x)$ and $\psi_2(x)$ are different (modulo-ly).

If the eigenvalues are not the same, you cannot factorise out $e^{iKa}$!
You seem to have very severe misunderstandings about what Bloch's theorem is. I do not factorize anything.

There are no certain K for certain functions. If you have two functions, say $\psi_1$ and $\psi_2$, you can now look for values of K that leave them unchanged under a displacement of a. If you now find two possible values, say K1 and K2 that do this, you will find that $e^{(i K_1a)}\psi_1$ and $e^{(i K_2a)}\psi_2$ will be solutions. And so will be: $e^{(i K_2a)}\psi_1$ and $e^{(i K_1a)}\psi_2$. And of course $e^{(i K_1a)}(c_1\psi_1+c_2 \psi_2)$ and $e^{(i K_2a)}(c_1\psi_1+c_2 \psi_2)$ and $e^{(i K_1a)}(c_1e^{(i K_1a)} \psi_1+c_2 e^{(i K_1a)}\psi_2)$ and $e^{(i K_2a)}(c_1e^{(i K_1a)} \psi_1+c_2 e^{(i K_1a)}\psi_2)$ and $e^{(i K_1a)}(c_1e^{(i K_2a)} \psi_1+c_2 e^{(i K_2a)}\psi_2)$...and so on and so forth.

Any of these exponentials describes a translation that leaves the modulus squared unchanged. Of course you can stack as many as you want to and of course you can also act on functions that have already been displaced in this way.

Happiness

You seem to have very severe misunderstandings about what Bloch's theorem is. I do not factorize anything.
If you don't have a common factor to factorise, how is Bloch's theorem true?

Suppose $\psi(x)=\psi_1(x)+\psi_2(x)$ (ignore the normalisation coefficients), $D\psi_1(x)=\lambda_1\psi_1(x)$ and $D\psi_2(x)=\lambda_2\psi_2(x)$.

Then $D\psi(x)=D\psi_1(x)+D\psi_2(x)=\lambda_1\psi_1(x)+\lambda_2\psi_2(x)$, which cannot be written as a multiple of $\psi(x)$. So Bloch's theorem is yet to be proven, or it's wrong.

If you now find two possible values, say K1 and K2 that do this, you will find that $e^{(i K_1a)}\psi_1$ and $e^{(i K_2a)}\psi_2$ will be solutions. And so will be: $e^{(i K_2a)}\psi_1$ and $e^{(i K_1a)}\psi_2$.
You have not proved $K_1$ works for $\psi_2(x)$ as well, and $K_2$ works for $\psi_1(x)$ as well! $e^{i K_1a}$ is an eigenvalue for $\psi_1(x)$ but may not be for $\psi_2(x)$.

Cthugha

You have not proved $K_1$ works for $\psi_2(x)$ as well, and $K_2$ works for $\psi_1(x)$ as well! $e^{i K_1a}$ is an eigenvalue for $\psi_1(x)$ but may not be for $\psi_2(x)$.
It is hard to tell whether you are really interested or just thick on purpose. This is trivial and usually discussed way ahead of Bloch's theorem

For a really quick discussion: Consider a function $\psi(x)=e^{ikx}u(x)$, which is defined such that u itself has the periodicity of the lattice and the lattice periodicity is given by a. Now apply the displacement operator to it:
$D\psi(x)=\psi(x+a)=e^{ik(x+a)}u(x+a)$.
As u has the periodicity of the lattice, it remains unchanged:
$\psi(x+a)=e^{ika}e^{ikx}u(x)=e^{ika}\psi(x)$.

Obviously this does not depend on the function we choose as long as u(x) has the periodicity we need.

• romsofia and dextercioby

Happiness

As u has the periodicity of the lattice, it remains unchanged:
This is $u(x+a)=u(x)$ (4.200), which depends on Bloch's theorem! Not ahead of Bloch's theorem. You can't use a corollary of Bloch's theorem to prove the theorem itself!

Obviously this does not depend on the function we choose as long as u(x) has the periodicity we need.
This is a result of Bloch's theorem. But the theorem (4.198) is either not proven correctly, or it's wrong. Right now, the theorem only works for $\psi_1$ alone or $\psi_2$ alone, but not a linear combination of them. ($\psi_1$ and $\psi_2$ are common eigenfunctions of H and D that have the same energy but generally different $\lambda$.)

It is hard to tell whether you are really interested or just thick on purpose. This is trivial and usually discussed way ahead of Bloch's theorem
I really don't understand the proof. I find that it's wrong. It's not trivial to me.

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Cthugha

This is $u(x+a)=u(x)$ (4.200), which depends on Bloch's theorem! Not ahead of Bloch's theorem. You can't use a corollary of Bloch's theorem to prove the theorem itself!
Sigh...no, I have just gone the other way round and showed that assuming functions of this form work. Why should I not be allowed to choose that form of the wave function?

This is the result of Bloch's theorem. But the theorem (4.198) is either not proven correctly, or it's wrong. Right now, the theorem only works for $\psi_1$ alone or $\psi_2$ alone, but not a linear combination of them. ($\psi_1$ and $\psi_2$ are common eigenfunctions of H and D that have the same energy but generally different $\lambda$.)
What do you mean by result? The surprising part of Bloch's theorem is not that you arrive at a function, which has a factor that has the lattice periodicity, but that the total wave function does not need to have this periodicity.

Bloch's theorem is an equivalence statement. If I have a potential with some certain periodicity, I will get wave functions that are the product of a plane wave and a function that has the lattice periodicity. It does not matter, whether I start from the potential and show the form of the wave function or whether I start from the wave function and show that I arrive at the periodicity. Of course the first version is more thorough and would be more pedagogical. For this you just write down the potential in Fourier form, insert it into the Schrödinger equation and assume a sum over plane waves for the wave function. Doing the math results in a term that needs to vanish, which couples different wave vectors that differ by a reciprocal lattice vector. Ashcroft/Mermin and pretty much every other solid-state book covers this in lots of detail. This is absolutely standard textbook material that is covered everywhere. There is no need to repeat the calculations here.

Happiness

Bloch's theorem is an equivalence statement. I have just gone the other way round and showed that assuming functions of this form work. Why should I not be allowed to choose that form of the wave function?

If I have a potential with some certain periodicity, I will get wave functions that are the product of a plane wave and a function that has the lattice periodicity. It does not matter, whether I start from the potential and show the form of the wave function or whether I start from the wave function and show that I arrive at the periodicity. Of course the first version is more thorough and would be more pedagogical. For this you just write down the potential in Fourier form, insert it into the Schrödinger equation and assume a sum over plane waves for the wave function. Doing the math results in a term that needs to vanish, which couples different wave vectors that differ by a reciprocal lattice vector. Ashcroft/Mermin and pretty much every other solid-state book covers this in lots of detail. This is absolutely standard textbook material that is covered everywhere. There is no need to repeat the calculations here.
Ok, there may be other ways to prove Bloch's theorem, and Bloch's theorem is most likely correct. But let's now focus on Griffiths' and B&J's proofs. They did not show how the theorem (5.49) or (4.198) could be applied to a linear combination of eigenfunctions. One eigenfunction one at a time, yes. But otherwise, no. I don't see how. And I made clear where my confusion is: $\lambda_1\psi_1+\lambda_2\psi_2\neq k(\psi_1+\psi_2)$, for any constant $k$.

the total wave function does not need to have this periodicity.
I don't understand this. If the wave function contains a periodic function (in the form of a product), isn't it periodic? B&J's book says it itself: the wave function is periodic, with the same period as that of the crystal lattice. (the paragraph after 4.200)

Cthugha

Ok, there may be other ways to prove Bloch's theorem, and Bloch's theorem is most likely correct. But let's now focus on Griffiths' and B&J's proofs. They did not show how the theorem (5.49) or (4.198) could be applied to a linear combination of eigenfunctions. One eigenfunction one at a time, yes. But otherwise, no. I don't see how. And I made clear where my confusion is: $\lambda_1\psi_1+\lambda_2\psi_2\neq k(\psi_1+\psi_2)$, for any constant $k$.
I do not get your problem. In a nutshell you request a proof that shows that if cos(2 pi) is a solution to a problem, cos(4 pi) will be as well. Consider (5.49.), where you have $e^{iKa}$ as the eigenvalue to the translation operator. In 3D, a would be a linear combination of the three Brevais lattice vectors, in 1D it is just some multiple of the lattice periodicity B, so a=nB and n is an integer. Looking at K, in 3D it would be a linear combination of the reciprocal lattice vectors. In 1D it is just the reciprocal wavenumber G in the dimension of interest, so that K=mG, where m is an integer. Now by the definition of the reciprocal lattice in 3D:$\vec{B}_i\vec{G}_j=2\pi \delta_{i,j}$ and in 1D simply $BG=2\pi$, so your exponential becomes: $e^{imn2\pi}$, which is just the cosine of integer multiples of 2 pi. Why do you think it makes a difference which integer multiple of 2 pi I pick?

I don't understand this. If the wave function contains a periodic function (in the form of a product), isn't it periodic? B&J's book says it itself: the wave function is periodic, with the same period as that of the crystal lattice. (the paragraph after 4.200)
No, they do not say that. They say that the Bloch wave function has an amplitude with the periodicity of the crystal lattice. This is your function u. The wave function is u times an exponential which represents a plane wave. So you get a plane wave (with a wavelength longer than the lattice periodicity) that is modulated by a function that shares the periodicity of the lattice. It will roughly look as follows: Last edited:

Happiness

I figured out the problem. I'm going to write the resolution for the benefits of many others who have the same confusion as me and for my own future reference too.
They did not show how the theorem (5.49) or (4.198) could be applied to a linear combination of eigenfunctions. One eigenfunction one at a time, yes. But otherwise, no. I don't see how. And I made clear where my confusion is: $\lambda_1\psi_1+\lambda_2\psi_2\neq k(\psi_1+\psi_2)$, for any constant $k$.
This is the reason for confusion. Suppose $\psi_1$ and $\psi_2$ are separately a solution to the Schrondinger equation [5.48]. Then any linear combination $\psi=c_1\psi_1+c_2\psi_2$ will also be a solution to the Schrondinger equation [5.48].

Then by Bloch's theorem, $\psi=c_1\psi_1+c_2\psi_2$ will also satisfy [5.49]: $$\psi(x+a)=e^{iKa}\psi(x)$$ But in general, $\psi_1$ and $\psi_2$ may have different $K$. In other words, $\psi_1(x+a)=e^{iK_1a}\psi_1$ and $\psi_2(x+a)=e^{iK_2a}\psi_2$, but $K_1\neq K_2$ (mod $\frac{2\pi}{a}$). Then, there won't be a common factor to factorise. Instead, we will end up with $$\psi(x+a)=e^{iK_1a}c_1\psi_1(x)+e^{iK_2a}c_2\psi_2(x)\neq e^{iKa}\psi(x)$$
As a result, [5.49] cannot be satisfied, contradicting Bloch's theorem.

So it seems that Bloch's theorem only works for $\psi_1$ alone and $\psi_2$ alone but not a linear combination of them. This has been the main message in all my replies, so far.

And it turns out I was right in this all along. Bloch's theorem indeed doesn't work for any arbitrary linear combination: it works for some linear combinations but not for all linear combinations.

You may revisit Griffiths' words again: You could blame it on his phrasing. What Bloch's theorem is saying is that there exist some solutions to the Schrondinger equation [5.48] that satisfy the condition [5.49]. It is not saying all solutions to the Schrondinger equation [5.48] will satisfy the condition [5.49].

I have spent a lot of time thinking about the resolution to this problem, only to realise that I had the wrong interpretation of Bloch's theorem. I wish Griffiths could have phrased this better, like how the mathematicians do it clearly, with key words like "for every", "there exists one", etc. This would have communicated the right idea across to the readers the first time, and avoid ambiguities and confusion and wasting time trying to figure out what went wrong.

It's also important to introduce the notations $\psi_{\lambda_1}$ and $\psi_{\lambda_2}$, and differentiate them with $\psi_1$ and $\psi_2$, because usually $\psi_1$ and $\psi_2$ are reserved to denote orthonormal eigenfunctions of H. One question you may be pondering is "why do B&J express $\psi$ only as a sum of 2 linearly independent solutions (4.186), and why not as a sum of 3 or more terms?" Certainly, H generally have more than 2 orthonormal eigenfunctions. So why stop at 2? It's because we are solving the Schrodinger equation at a certain value of E. And for a fixed value of E, the solution space is spanned by at most two linearly independent functions, since the Schrodinger equation is a second-order differential equation and would hence introduce two constants to the general solution. These two constants can be determined from $\psi(x, 0)$ and $\frac{\partial\psi}{\partial t}(x, 0)$, giving us the solution to a particular scenario.

But why must we fix the value of E? It's because the K in Bloch's condition [5.49] may depend on E. So to make K truly constant, we should fix the value of E. (Of course, one needs to spend some time thinking over it to understand it for himself.)

Going back to the notations $\psi_{\lambda_1}$ and $\psi_{\lambda_2}$. They are the eigenfunctions of D, with eigenvalues $\lambda_1$ and $\lambda_2$ respectively.

When you look at Griffiths' proof: You may think that since D and H commute, an eigenfunction of H, such as $\psi_1$, is also an eigenfunction of D. But this is not true. Commutativity only tells us that there exist functions that are eigenfunctions of both D and H. So if $\psi_1$ are $\psi_2$ linearly independent eigenfunctions of the same energy, then $\psi_{\lambda_1}=c_1\psi_1+c_2\psi_2$, for some $c_1$ and $c_2$. But in general, $\psi_{\lambda_1}\neq\psi_1$. Similarly, $\psi_{\lambda_2}=c_1'\psi_1+c_2'\psi_2$, for some $c_1'$ and $c_2'$.

Take a look at the example following Griffiths' proof: Do you see why in [5.60] there is a common factor $e^{-iKa}$ even though I just said Bloch's theorem only works for one eigenfunction one at a time? In other words, could you explain why $\sin k(x+a)$ and $\cos k(x+a)$ have the same factor $e^{-iKa}$? Why not $e^{-iK_1a}$ for $\sin k(x+a)$ and $e^{-iK_2a}$ for $\cos k(x+a)$? How do we tell in what situation we should use the same $K$ and in what situation we should use $K_1$ and $K_2$, like B&J's (4.196) below? The key is to understand that we apply Bloch's theorem to an eigenfunction of D, not of H. In other words, $\psi(x+a)=e^{iKa}\psi_{\lambda_1}$ but $\psi(x+a)\neq e^{iKa}\psi_1$. In the above example, $\sin(kx)$ and $\cos(kx)$ are eigenfunctions of H, ie., $\psi_1=\sin(kx)$ and $\psi_2=\cos(kx)$. But since $\sin(kx)$ and $\cos(kx)$ are eigenfunctions of the same energy, $A\sin(kx)+B\cos(kx)$ is also an eigenfunction of the same energy. Then, since D and H commute, out of all the eigenfunctions of H, one of them must also be an eigenfunction of D. In other words, there exist some values of $A$ and $B$ such that $A\sin(kx)+B\cos(kx)=\psi_{\lambda_1}$. And this is [5.59] (see above), $\psi(x)$ being an eigenfunction of D, and $A$ and $B$ being certain values, not any arbitrary values. This is quite subtle, and requires some careful thought. [5.59] is presented by the book as the general solution. But when it is used with Bloch's theorem to form [5.60], we are searching from the pool of the general solution, a particular solution that is an eigenfunction of D, with A and B taking on certain values. Hence, after displacing to the left by $a$, we have $\psi_{\lambda_1}(x-a)=e^{-iK_1a}\psi_{\lambda_1}(x)=e^{-iK_1a}[A\sin(kx)+B\cos(kx)]$, which is [5.60] after substituting $x$ by $(x+a)$.

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Cthugha

You are still missing the point. The point lies in introducing the periodic Born-von Karman boundary conditions (the steps [5.55] and [5.56] you left out). While one assumes a symmetry for a shift by one single lattice constant a only, the value of K is already defined perfectly by the reciprocal lattice vector G and there is just this one discrete value of K, which works for all states.

However as you apply periodic boundary conditions, you also enforce shifts by integer multiples N of a to be a symmetry operation. Here, you get more allowed values for K. Every fraction $\frac{nG}{N}$ will now also form an allowed K. Starting from this point, you need to worry about phase shifts of less than 2 pi per unit cell, which cause wave functions of different periodicity to appear. However, this starts only as you take displacements along several multiples of a into account.
You can then go on to find the energy of each of these states. For large N, this means that k becomes quasi-continuous as the states are too close to be able to resolve the discreteness in k. For systems consisting of few unit cells, you can see the discreteness in experiments.

Happiness

You are still missing the point. The point lies in introducing the periodic Born-von Karman boundary conditions
This wasn't the issue I had.

there is just this one discrete value of K
There are two values of K for each energy E.

which works for all states
Not quite clear what you mean. States that are eigenfunctions of D? States of the same energy?

However as you apply periodic boundary conditions, you also enforce shifts by integer multiples N of a to be a symmetry operation. Here, you get more allowed values for K. Every fraction $\frac{nG}{N}$ will now also form an allowed K. Starting from this point, you need to worry about phase shifts of less than 2 pi per unit cell, which cause wave functions of different periodicity to appear. However, this starts only as you take displacements along several multiples of a into account.
Ok, there may be other ways to get [5.56]. But the book's approach is much easier. It's just one line. And this wasn't my issue. The priority should be on the accurate identifying of the cause of the confusion and addressing it using the simplest tool, without too much unnecessary information. Nonetheless, thank you for your advice.

PeterDonis

Mentor
What Bloch's theorem is saying is that there exist some solutions to the Schrondinger equation [5.48] that satisfy the condition [5.49]. It is not saying all solutions to the Schrondinger equation [5.48] will satisfy the condition [5.49].
No, Bloch's theorem does indeed say that all solutions of [5.48] given that the potential satisfies [5.47] satisfy the condition [5.49].

The priority should be on the accurate identifying of the cause of the confusion
What confusion? As far as I can see, the only confusion in this thread is that you keep making incorrect claims.

• Happiness

No, Bloch's theorem does indeed say that all solutions of [5.48] given that the potential satisfies [5.47] satisfy the condition [5.49].
Ok could you explain how do you know that it's impossible for a solution to Schrodinger equation with such a potential to not satisfy the condition [5.49]?

Consider this counter example. Let $\psi_1$ and $\psi_2$ be the solution with energy $E_1$ and $E_2$ to the Schrodinger equation with the Dirac comb potential respectively. Then, $\psi=\psi_1+\psi_2$ must be a solution too (if with the appropriate normalisation). But $\psi$ does not satisfy the condition [5.49], because $\psi(x+a)=e^{iK_1a}\psi_1(x)+e^{iK_2a}\psi_2(x)$ and in general, $K_1\neq K_2$ (mod $\frac{2\pi}{a}$), since $K_1$ depends on $E_1$ and $K_2$ depends on $E_2$.

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PeterDonis

Mentor
could you explain how do you know that it's impossible for a solution to Schrodinger equation with such a potential to not satisfy the condition [5.49]?
Sure, because your claim here...

Let $\psi_1$ and $\psi_2$ be the solution with energy $E_1$ and $E_2$ to the Schrodinger equation with the Dirac comb potential respectively. Then,$\psi=\psi_1+\psi_2$ must be a solution too.
...is incorrect. A linear combination of eigenstates of the Hamiltonian is not, in general, also an eigenstate. And Bloch's Theorem only applies to eigenstates of the Hamiltonian--more precisely, to states that are common eigenstates of both the Hamiltonian $H$ and the displacement operator $D$.

Happiness

And Bloch's Theorem only applies to eigenstates of the Hamiltonian--more precisely, to states that are common eigenstates of both the Hamiltonian H and the displacement operator D.
No, Bloch's theorem does indeed say that all solutions of [5.48] given that the potential satisfies [5.47] satisfy the condition [5.49].

PeterDonis

Mentor
No. Equation [5.48] is the eigenvalue equation for the Hamiltonian. Only eigenstates of the Hamiltonian satisfy it.

"Inadequate proof of Bloch's theorem?"

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