I Inadequate proof of Bloch's theorem?

[Happiness]

You still emphasize the wrong point. D(a) is a pedagogical tool to get the intuition right. For a completely periodic potential, you will get a whole family of operators corresponding to D(na), where n is an integer. Any of them represents a symmetry of the system. Any complex \lambda for D(a) will be a real \lambda for D(na) for some n. The reason why you get wavefunctions that do not share the periodicity of the lattice is that all Bloch wave functions in a fully periodic potential need to be simultaneous eigenstates of the Hamiltonian and D(na) for one n, not necessarily of the Hamiltonian and D(a).

What you said is true. But what I'm trying to say is there exist eigenstates of H that are not eigenstates of D.

 

[Cthugha]

What you said is true. But what I'm trying to say is there exist eigenstates of H that are not eigenstates of D.

Which D? D(a)? Yes, indeed. I fully agree with that.

 

[Happiness]

Which D? D(a)? Yes, indeed. I fully agree with that.

Yes D(a).

This means [5.48] doesn't imply [5.49]. There exist solutions to [5.48] that are not solutions to [5.49]. This is what I have been saying all along:

You could blame it on his [Grifftihs's] 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].

in contrast to

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].

 

[Cthugha]

This means [5.48] doesn't imply [5.49]. There exist solutions to [5.48] that are not solutions to [5.49]. This is what I have been saying:

Huh? No, All solutions to [5.48] are still solutions to [5.49]. [5.49] does not require \psi to be periodic in a. e^{iKa} gives the phase shift between one unit cell and the next one.

 

[Happiness]

What you said is true. But what I'm trying to say is there exist eigenstates of H that are not eigenstates of D.

Which D? D(a)? Yes, indeed. I fully agree with that.

What you said here contradicts what you said below

All solutions to [5.48] are still solutions to [5.49].

because [5.48] is an eigenvalue equation for H and [5.49] is an eigenvalue equation for D.

 

[Cthugha]

Ok, sloppy language. My fault.
The eigenstates will not share the periodicity of a.
However, D is a unitary operator and not a Hermitian one, so the eigenvalues may be complex and as such do not need to share this periodicity. All solutions of the Hamiltonian are eigenstates of D(a), but they do not necessarily have real eigenvalues. However, all eigenstates will have real eigenvalues for some translation operator.

 

[PeterDonis]

all eigenstates will have real eigenvalues for some translation operator

Just to clarify, I assume you mean all eigenstates of the Hamiltonian will have real eigenvalues for some translation operator.

 

[Cthugha]

Just to clarify, I assume you mean all eigenstates of the Hamiltonian will have real eigenvalues for some translation operator.

Sorry, it is getting late. Of course you are right.

 

[PeterDonis]

But $\frac{\psi_{\lambda_1}-\psi_{\lambda_2}}{2i}=\sin(kx)$ is not an eigenfunction of $D$

It is if $k$ meets the condition I gave earlier, because for those values of $k$, $e^{ika} = e^{-ika}$, since $ka = n \pi$ and $e^{i n \pi} = e^{- i n \pi}$.

 

[Happiness]

All solutions of the Hamiltonian are eigenstates of D(a), but they do not necessarily have real eigenvalues.

This is false. Consider the simplest case of $\psi=\sin(kx)$. It is an eigenstate of H, but not of D. Even with complex eigenvalues, you cannot write $\sin(kx)$ as a complex multiple of $e^{ikx}$ alone or as a complex multiple of $e^{-ikx}$ alone.

 

[Happiness]

It is if $k$ meets the condition I gave earlier, because for those values of $k$, $e^{ika} = e^{-ika}$, since $ka = n \pi$ and $e^{i n \pi} = e^{- i n \pi}$.

This is because

If you only consider $\lambda$ to be real numbers, then you will mislead yourself into thinking that $\psi_{\lambda_1}+\psi_{\lambda_2}$ is always an eigenfunction of $D$.

 

[PeterDonis]

Consider the simplest case of $\psi=\sin(kx)$. It is an eigenstate of H, but not of D.

Why do you keep making this false statement even after I have shown that it is false?

At the very least, you should acknowledge that there are conditions on $k$ for which $\sin (kx)$ is an eigenstate of $D$, since I have explicitly shown what those conditions are. And you should also acknowledge that the $D$ you refer to here is what @Cthugha called $D(a)$, i.e., a specific translation operator out of multiple possible ones.

 

[Happiness]

@PeterDonis Do you agree that there exists an allowed value of k such that $\psi=\sin(kx)$ is an eigenstate of H, but not of D?

 

[PeterDonis]

Do you agree that there exists an allowed value of $k$ such that $\psi=\sin(kx)$ is an eigenstate of H, but not of D?

Have you shown one in this discussion?

 

[Happiness]

Have you shown one in this discussion?

Ok I didn't show an example explicitly as I thought B&J's paragraph following (4.190) is an adequate proof of what I have been saying.

Anyway, you can find one such example from Griffiths:

For each value of K in [5.56], we solve [5.64] graphically. For each K, we draw a horizontal line in Fig 5.6, and we may get several points of intersection. Each point corresponds to an allowed value of k. (K determines the value on the y axis; k is determined from the x coordinate of the point of intersection.) Each allowed k corresponds to two values of K, as you can see from [5.64] that if K is a solution for a particular k, then -K is also a solution (for the same k). Substitute the big K and small k that you get from a point of intersection into [5.63] to get A in terms of B. Substitute this A into [5.59], and after normalisation, we get a value of A and B that satisfy both [5.48] and [5.49] for an allowed value of k. (Actually, we get two values if you remember to include the case for -K.)

But for this allowed value of k, any A and B would have satisfied [5.48] (since k is the same means energy E is the same), but not [5.49].

Therefore, there exists an allowed value of k such that a solution to [5.48] is not a solution to [5.49].

 

[Cthugha]

Therefore, there exists an allowed value of k such that a solution to [5.48] is not a solution to [5.49].

That is not correct. In the band gap region, you arrive at a pair of values of \lambda that is real, but not 1, which corresponds to exponentially growing or decaying modes. Out of these, the decaying mode is usually the physically relevant one.

If you consider the optical equivalent of a periodic potential: a material with periodically structured refractive index, you will find optical band gaps equivalent to the electronic ones. If you shine a light beam in the optical band gap on a (obviously effectively finite size) material structured this way, you will find exactly this exponential decay.

Also, just as a general remark, one should consider that many of the possible superpositions of arbitrary Bloch states will not necessarily be stationary states.

 

[Happiness]

Therefore, there exists an allowed value of k such that a solution to [5.48] is not a solution to [5.49].

That is not correct. In the band gap region, you arrive at a pair of values of \lambda that is real, but not 1, which corresponds to exponentially growing or decaying modes. Out of these, the decaying mode is usually the physically relevant one.

What you said seems to contradict Griffiths's statement below:

Griffiths said states with energies in the gap region are physically impossible, but you said these states are physically relevant.

A definition of "growing mode" and "decaying mode" would be helpful, and also a concise explanation how they are "exponential".

But in any case, my following sentence

Therefore, there exists an allowed value of k such that a solution to [5.48] is not a solution to [5.49].

already set the discussion: we only consider an allowed value of k. But your reply is about disallowed values of k being physically relevant. But I wasn't making any claim about these disallowed values of k!

You seem to be very proselytising in sharing and showing your knowledge in numerous things, like decaying modes, optical materials, modular arithmetic, etc., which may be good in a way, but it may sometimes distract the attention away from the main issue in the discussion.

I am sure you didn't mean to contradict Griffiths. But to new learners unfamiliar with "decaying modes", it certainly seems so. Therefore, more mindfulness would be beneficial, more mindfulness about how much your readers could understand your statements, and how effective your statements are in communicating the ideas across to your readers. Also, would they cause more confusion to your readers? Are they really helpful and relevant to the core of the issue? Are they the simplest way to resolve the issue?

Simplicity and conciseness are very expensive gifts. (in tribute to Warren Buffett's quote on honesty)

 

[Cthugha]

What you said seems to contradict Griffiths's statement below:
View attachment 248117

Griffiths said states with energies in the gap region are physically impossible, but you said these states are physically relevant.

Then you should read up on the definition of forbidden modes. These refer to propagating modes. The "forbidden" modes are a synonym for being non-propagating or evanescent.

A definition of "growing mode" and "decaying mode" would be helpful, and also a concise explanation how they are "exponential".

Well, you already noted that \lambda_1=\lambda_2^{-1} and we had the definition of \lambda_1=e^{iKl} several times already, so it should be trivial to see what happens if the magnitude of \lambda_1 and \lambda_2 differs from unity.

But in any case, my following sentence

already set the discussion: we only consider an allowed value of k. But your reply is about disallowed values of k being physically relevant. But I wasn't making any claim about these disallowed values of k!

Well, yes, but @PeterDonis already responded to the whole issue beforehand. You then repeated the discussion in post #65 and made it more complicated. This conclusion:

But for this allowed value of k, any A and B would have satisfied [5.48] (since k is the same means energy E is the same), but not [5.49].
Therefore, there exists an allowed value of k such that a solution to [5.48] is not a solution to [5.49].

is not warranted as not any A and B would satisfy e.g. [5.62].
Anyhow, any periodic eigenfunction of the Hamiltonian will at some point match some periodicity of the periodic potential for an infinitely extended system, so it will be an eigenstate of D(na) with one of the eigenvalues equal to 1 for some n. This also determines the eigenvalues for D(a).

I am sure you didn't mean to contradict Griffiths. But to new learners unfamiliar with "decaying modes", it certainly seems so. Therefore, more mindfulness about how much your readers could understand your statements, how effective your statements are in communicating the ideas across to your readers, would be beneficial. Also, would they cause more confusion to your readers? Are they really helpful and relevant to the core of the issue? Are they the simplest way to resolve the issue?

Simplicity and conciseness are very expensive gifts. (in tribute to Warren Buffett's quote on honesty)

Over several years of teaching physics I learned that giving answers that match the simplicity and conciseness of the question is the best approach.

 

[Happiness]

This conclusion:

is not warranted as not any A and B would satisfy e.g. [5.62].

[5.62] is used to search for common eigenstates of H and D, since it is obtained from the eigenvalue equations of H [5.48] and of D [5.49]. Therefore, not any A and B would satisfy e.g. [5.62]. Isn't it true that any A and B would satisfy [5.48] for a particular E since [5.59] is its general solution?

You then repeated the discussion in post #65 and made it more complicated.

The post is a response to PeterDonis's request for an example. It is not complicated. You may need some time to read it, yes, but it's not complicated.

 

[Cthugha]

[5.62] is used to search for common eigenstates of H and D, since it is obtained from the eigenvalue equations of H [5.48] and of D [5.49]. Therefore, not any A and B would satisfy e.g. [5.62]. Isn't it true that any A and B would satisfy [5.48] for a particular E since [5.59] is its general solution?

The purpose of [5.62] is not to find common eigenstates, but to find the solutions to the Hamiltonian. The idea is to solve it in a single unit cell 0\leq x<a. [5.58] gives the free particle Hamiltonian in the absence of a potential for 0< x<a. [5.59] is its general solution. However you still need the value for 0. [5.59] does not tell you anything about this, so [5.59] is not the general solution to [5.48]. Usually you would have to solve the full Schrödinger equation including the potential at every point. As in this case the potential is 0 everywhere, except for 0 and a, you get away with the easier procedure of taking the general potential-free case for 0< x<a and adding the influence of the potential as a boundary value problem for 0 and a.

You also see from [5.57] that \alpha corresponds to the magnitude of the potential, so the solution to the full problem at hand must depend on it, which [5.59] does not. Accordingly [5.59] cannot the general solution to [5.48].

 

[vanhees71]

You still emphasize the wrong point. D(a) is a pedagogical tool to get the intuition right. For a completely periodic potential, you will get a whole family of operators corresponding to D(na), where n is an integer. Any of them represents a symmetry of the system. Any complex \lambda for D(a) will be a real \lambda for D(na) for some n. The reason why you get wavefunctions that do not share the periodicity of the lattice is that all Bloch wave functions in a fully periodic potential need to be simultaneous eigenstates of the Hamiltonian and D(na) for some n, not necessarily of the Hamiltonian and D(a).

No! $\hat{D}(a)$ is the crucial point and not a mere pedagogical tool. It selects the convenient Bloch energy eigenstates. As I showed in my previous posting the important point is the discrete translation symmetry of the (here simplified case of 1D) crystal lattice. Due to this symmetry there's a common eigenbasis of $\hat{H}$ and $\hat{D}(a)$.

The reason, why here common eigenstates of $\hat{H}$ with a unitary operator, representing a symmetry transformation, are considered and not common eigenvectors of $\hat{H}$ and some self-adjoint operator(s) representing observable(s) is that here we deal with a discrete symmetry group and not a continuous Lie group. In the latter case the generators of the corresponding Lie algebra define conserved observables.

For our example the continuous case is that all $\hat{D}(a)$ (i.e., for all $a \in \mathbb{R}$) are symmetry operators. After some (not too simple) analysis for a single particle in a Galilei-invariant theory it turns out that this is only fulfilled for a free particle, with the Hamiltonian fixed to $\hat{H}=\hat{p}^2/(2m)$.

Of course to have only a discrete translation symmetry a periodic potential for the particle (electron) is also allowed, i.e., there's a much larger class of Hamiltonian fulfilling the symmetry. This is almost trivial: The less operations are symmetry operations the less restricted is the Hamiltonian by these symmetries.

 

[Happiness]

[5.62] is used to search for common eigenstates of H and D, since it is obtained from the eigenvalue equations of H [5.48] and of D [5.49].

The purpose of [5.62] is not to find common eigenstates, but to find the solutions to the Hamiltonian.

Ok, let's hear the opinion of more people first.

[5.48] is Schrodinger equation. [5.49] is Bloch's theorem or Bloch's condition.
[5.62] is in post #65:

View attachment 248110
View attachment 248108
View attachment 248109

[5.58] gives the free particle Hamiltonian in the absence of a potential for $0<x<a$. [5.59] is its general solution. However you still need the value for 0. [5.59] does not tell you anything about this, so [5.59] is not the general solution to [5.48].

You are right that we need to take into account the Dirac-delta potential at $x=0$ and $x=a$. But it turns out that [5.59] is still the general solution to [5.48].

As shown in my post #65, for every allowed value of E, there are two solutions for A and B, representing the 2 common eigenfunctions to both H and D, ie., eigensolutions to both [5.48] and [5.49]. Let's denote these 2 common eigenfunctions as $\psi_{\lambda_1}$ and $\psi_{\lambda_2}$. Since H is a linear operator, any linear combinations of its particular solutions is also a solution (for a particular E). So $c_1\psi_{\lambda_1}+c_2\psi_{\lambda_2}$ is also a solution to [5.48] for any complex numbers $c_1$ and $c_2$. It can be shown that $\psi_{\lambda_1}$ and $\psi_{\lambda_2}$ are linearly independent, implying that the solution space of [5.48] is the full $\mathbb{C}^2$, where $\mathbb{C}$ is the field of complex numbers. As $\sin(kx)$ and $\cos(kx)$ in [5.59] are also linearly independent, a $\mathbb{C}^2$ solution space implies A and B can take on the value of any complex number.

So [5.59] is still the general solution to [5.48].

 

[Cthugha]

No! $\hat{D}(a)$ is the crucial point and not a mere pedagogical tool. It selects the convenient Bloch energy eigenstates. As I showed in my previous posting the important point is the discrete translation symmetry of the (here simplified case of 1D) crystal lattice. Due to this symmetry there's a common eigenbasis of $\hat{H}$ and $\hat{D}(a)$.

I think we do not disagree here. Indeed the full discrete translational symmetry is the key point here. Full discrete translational symmetry implies that D(na) is a symmetry operation for every integer n. This obviously includes also D(a). However, you can create systems with short range order that may become arbitrarily close to having perfect symmetry with respect to D(a), but not with respect to D(na). Consider amorphous materials or something like these famous quasicrystals with 5-fold rotational symmetry.

I just emphasized that the important point for the theorem is that you have FULL discrete translational symmetry.

 

[vanhees71]

Ok, let's hear the opinion of more people first.

[5.48] is Schrodinger equation. [5.49] is Bloch's theorem or Bloch's condition.
[5.62] is in post #65:
You are right that we need to take into account the Dirac-delta potential at $x=0$ and $x=a$. But it turns out that [5.59] is still the general solution to [5.48].

As shown in my post #65, for every allowed value of E, there are two solutions for A and B, representing the 2 common eigenfunctions to both H and D, ie., eigensolutions to both [5.48] and [5.49]. Let's denote these 2 common eigenfunctions as $\psi_{\lambda_1}$ and $\psi_{\lambda_2}$. Since H is a linear operator, any linear combinations of its particular solutions is also a solution. So $c_1\psi_{\lambda_1}+c_2\psi_{\lambda_2}$ is also a solution to [5.48] for any complex numbers $c_1$ and $c_2$. It can be shown that $\psi_{\lambda_1}$ and $\psi_{\lambda_2}$ are linearly independent, implying that the solution space of [5.48] is the full $\mathbb{C}^2$, where $\mathbb{C}$ is the field of complex number. As $\sin(kx)$ and $\cos(kx)$ in [5.59] are also linearly independent, a $\mathbb{C}^2$ solution space implies A and B can take on the value of any complex number.

So [5.59] is still the general solution to [5.48].

No! That's the very point of the misunderstanding. (5.48) is the solution of the time-independent Schrödinger equation, i.e., the eigenvalue problem for $\hat{H}$. Superpositions of eigenvectors with different eigenvalues don't solve the time-independent Schrödinger equation. Indeed, such a superposition is not reprensenting a stationary but a time-dependent state.

The same holds true for $\hat{D}(a)$: the superposition of two solutions with different eigenvalues is not a solution. If $\exp(\pm \mathrm{i}k a)$ are eigenvalues of $\hat{D}(a)$, the corresponding eigenvectors are $\exp(\pm \mathrm{i} k x)$, but obviously
$$\sin(k x) \propto \exp(\mathrm{i} k x)-\exp(-\mathrm{i} k x)$$
is not an eigensolution of $\hat{D}(a)$. Indeed
$$\hat{D}(a) \sin(k x) \propto \exp(\mathrm{i} k(x+a))-\exp(-\mathrm{i}k(x+a)),$$
and this is not $\lambda \sin(k x)$ with any number $\lambda$.

 

[Happiness]

No! That's the very point of the misunderstanding. (5.48) is the solution of the time-independent Schrödinger equation, i.e., the eigenvalue problem for $\hat{H}$. Superpositions of eigenvectors with different eigenvalues don't solve the time-independent Schrödinger equation. Indeed, such a superposition is not reprensenting a stationary but a time-dependent state.

$\psi_{\lambda_1}$ and $\psi_{\lambda_2}$ have the same E but different $\lambda$, same eigenvalue for H but different ones for D! (Different ones in general: $\lambda=\pm1$ are the exceptions.) This is proven by B&J:

The same holds true for $\hat{D}(a)$: the superposition of two solutions with different eigenvalues is not a solution. If $\exp(\pm \mathrm{i}k a)$ are eigenvalues of $\hat{D}(a)$, the corresponding eigenvectors are $\exp(\pm \mathrm{i} k x)$, but obviously
$$\sin(k x) \propto \exp(\mathrm{i} k x)-\exp(-\mathrm{i} k x)$$
is not an eigensolution of $\hat{D}(a)$. Indeed
$$\hat{D}(a) \sin(k x) \propto \exp(\mathrm{i} k(x+a))-\exp(-\mathrm{i}k(x+a)),$$
and this is not $\lambda \sin(k x)$ with any number $\lambda$.

YES! YES! YES! This is what I have been saying several times in this discussion, to @Cthugha and @PeterDonis in particular!

Consider the simplest case of $\psi=\sin(kx)$. It is an eigenstate of H, but not of D. Even with complex eigenvalues, you cannot write $\sin(kx)$ as a complex multiple of $e^{ikx}$ alone or as a complex multiple of $e^{-ikx}$ alone.

Why do you keep making this false statement even after I have shown that it is false?

 

[vanhees71]

I don't understand the confusion of this simple subject. It's not different from any behavior of eigenvectors and eigenvalues. Maybe it's imprecise notation? Well, we have these issues repeatedly with Griffths's QM textbook. As good as his E&M textbook is, as bad seems his QM textbook to be!

For a very nice intro to solid-state physics, I think a look in the classic by Ashcroft and Mermin is still very helpful. There everything with Block, Born, and Karman is presented in utmost clarity (even for the general 3D case).

The argument with the Wronskian is also clear, but what's "B&J"?

 

[George Jones]

what's "B&J"?

"Physics of Atoms and Molecules"

according to another book, QM 2nd ed., Bransden & Joachain, p183

 

[PeterDonis]

obviously
$$
\sin(k x) \propto \exp(\mathrm{i} k x)-\exp(-\mathrm{i} k x)
$$
is not an eigensolution of $\hat{D}(a)$.

It is for some particular values of $k$. That's what I was saying before. For certain values of $k$, $\exp (i k a) = \exp(- i k a)$, and therefore $\sin (kx)$ is an eigenstate of $\hat{D}(a)$.

YES! YES! YES! This is what I have been saying several times in this discussion, to @Cthugha and @PeterDonis in particular!

And you failed to recognize that what you said is not true for certain values of $k$, as above.

 

[Cthugha]

YES! YES! YES! This is what I have been saying several times in this discussion, to @Cthugha and @PeterDonis in particular!

Nobody doubted that. The issue was about these superposition states being a solution of the stationary Schrödinger equation.

Consider this simple scenario. A superposition of two plane waves of the same energy and modulus of the wavevector. Something like exp(ikx) and exp(-ikx) and their time dependence. Is the resulting standing wave a stationary state?

 

[Happiness]

I don't understand the confusion of this simple subject.

There exists an allowed value of k such that a solution to the time-independent Schrodinger equation [5.48] is not a solution to the Bloch's condition [5.49], but Griffiths said (according to @PeterDonis) this is impossible: all solutions to the Schrodinger equation is a solution to the Bloch's condition (for all allowed values of k).

This is the issue.

If anyone reading this thread understands what the issue is, please help to explain!

Could anyone explain how the periodicity of the potential energy function imply the Bloch's condition [5.49]?

 

[Happiness]

Consider this simple scenario. A superposition of two plane waves of the same energy and modulus of the wavevector. Something like exp(ikx) and exp(-ikx) and their time dependence. Is the resulting standing wave a stationary state?

It is not a stationary state. But this is not a counter example because [5.48] is the time-independent Schrodinger solution. There exists an allowed value of k such that a solution to the time-independent Schrodinger equation is not a solution to the Bloch's condition.

Could you explain how the periodicity of the potential energy function imply the Bloch's condition [5.49]?

 

[Happiness]

The argument with the Wronskian is also clear, but what's "B&J"?

Since you understood the argument with the Wronskian, could you explain to PeterDonis that it doesn't impose the following condition on $K$:

$e^{i K a} = e^{- i K a}$?

(Note that for the case of the free particle, K=k. K is defined by the Bloch's condition [5.49], while k is defined by the energy E.)

It is for some particular values of k. That's what I was saying before. For certain values of k, $\exp (i k a) = \exp(- i k a)$, and therefore $\sin (kx)$ is an eigenstate of $\hat{D}(a)$.

Could you explain how the periodicity of the potential energy function imply the Bloch's condition [5.49]?

 

[vanhees71]

There exists an allowed value of k such that a solution to the time-independent Schrodinger equation [5.48] is not a solution to the Bloch's condition [5.49], but Griffiths said (according to @PeterDonis) this is impossible: all solutions to the Schrodinger equation is a solution to the Bloch's condition (for all allowed values of k).

This is the issue.

If anyone reading this thread understands what the issue is, please help to explain!

Could anyone explain how the periodicity of the potential energy function imply the Bloch's condition [5.49]?

I have done that already in this thread. See #37

https://www.physicsforums.com/threads/inadequate-proof-of-blochs-theorem.975916/page-2#post-6219592

 

[vanhees71]

Actually, on thinking this over, the condition for eigenfunctions of $H$ do also restrict $k$ if we are talking about sines and cosines. For sines and cosines, the eigenvalue equation for $H$ can be written:

$$
\left( \frac{\hbar^2 k^2}{2m} + V(x) - E \right) \psi(x) = 0
$$

This can only be satisfied if $V(x)$ has the same constant value at every $x$ for which $\psi(x) \neq 0$. But that means that either $V(x)$ has the same constant value everywhere (which is just the trivial free particle case, not what we're discussing here), or we must have $\psi(x) = 0$ at some values of $x$, so that $V(x)$ can have a different value at those values of $x$. But if that happens for any value of $x$, it must also happen for all other values that differ from that one by an integer multiple of $a$, because $V(x)$ is periodic with period $a$. In other words, $\psi(x)$ must have zeros spaced by $a$ (or some integer fraction of $a$). And for sines and cosines, that is equivalent to the restriction on $k$ that I gave.

I think a similar argument will work for any function $\psi(x)$ that is known to be periodic.

Why do you want sines and cosines? To have definite parity? The usual choice of boundary conditions is those leading to the Born von Karman solutions:

https://www.physicsforums.com/threads/inadequate-proof-of-blochs-theorem.975916/page-2#post-6219592

 

[vanhees71]

Could you explain how the periodicity of the potential energy function imply the Bloch's condition [5.49]?

Concerning the Wronskian, just do the calculation yourself. It's not difficult.

Concerning the idea with the Bloch ansatz, see

https://www.physicsforums.com/threads/inadequate-proof-of-blochs-theorem.975916/page-2#post-6219592
A minimum of group-representation theory helps a lot in quantum theory!

 

[PeterDonis]

Why do you want sines and cosines?

Because I was responding to posts by @Happiness where he considers the specific case of sines and cosines. I understand that that's not the most general case.

 

[Happiness]

Concerning the idea with the Bloch ansatz, see

I guess you meant this part:

To get a concrete set of k-values you need to impose some boundary condition for the 1D crystal as a whole. Since for bulk properties the exact boundary conditions are of not too much importance, usually one chooses the length of the crystal to be an integer multiple of the primitive period a, L=Na and imposes periodic boundary conditions,
$$\psi(\vec{x}+N a)=\psi(\vec{x}).$$
For the Bloch energy-eigenstates you get for each k being an eigenvector of this kind

exp(ikNa)=1⇒kNa=2πn,n∈Z.​

All these I understand, but it does not mention how the condition $\psi(x+a)=e^{iKa}\psi(x)$ is motivated. Be aware that it is only by first assuming this condition to be true that we could get $\psi(\vec{x}+N a)=e^{iKNa}\psi(\vec{x})$.

If we start by not assuming Bloch's condition, we would consider $\psi(x+a)=\hat{L}\psi(a)$, where $\hat{L}$ is some linear operator. We want to know what are the possible $\hat{L}$ that could satisy the boundary condtion $\psi(\vec{x}+N a)=\psi(\vec{x})$. That is, we want to know what are all the possible $\hat{L}$ that satisfies $(\hat{L})^N=I$. Could you show that $e^{iKa}$ is the only $\hat{L}$ that has this property?

 

[vanhees71]

I explained how the Bloch ansatz is motivated. It comes from group theory, which tells that for a discrete translational symmetry the corresponding unitary symmetry operators are commuting with $\hat{H}$ (that's what defines a symmetry) and that's why there are common eigenstates of these operators and $\hat{H}$. It is convenient to choose that basis. That's all.

As I said, it's well worth to study group-representation theory in QT. Analysing the space-time symmetries it explains for both relativistic and non-relativistic QT, why the Hamiltonians look as they look, how typical states, including "elementary particles" (defined by irreducible representations of the quantum Galilei or Poincare group), are characterized etc.