# 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

Mentor
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

Mentor
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

Mentor
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

Mentor
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.

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

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#### 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).

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.

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#### 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

Gold Member
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:
[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].

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#### 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

Gold Member
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?

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