I Inadequate proof of Bloch's theorem?

Happiness

No. Equation [5.48] is the eigenvalue equation for the Hamiltonian. Only eigenstates of the Hamiltonian satisfy it.
But there may exist eigenstates of the Hamiltonian that satisfy it (5.48 modified with 5.47) that are not eigenstates of D and so would not satisfy [5.49].

PeterDonis

Mentor
there may exist eigenstates of the Hamiltonian that satisfy it (5.48 modified with 5.47) that are not eigenstates of D and so would not satisfy [5.49]
No, there can't. Since $D$ and $H$ commute, every state that is an eigenstate of one must be an eigenstate of the other. (More precisely, this is true if each operator has a complete set of non-degenerate eigenstates; degeneracy complicates things somewhat but is not an issue for this discussion.) Griffiths' proof makes explicit use of this theorem. It is a basic theorem of QM; are you not aware of it?

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Happiness

No, there can't. Since $D$ and $H$ commute, every state that is an eigenstate of one must be an eigenstate of the other. (More precisely, this is true if each operator has a complete set of non-degenerate eigenstates; degeneracy complicates things somewhat but is not an issue for this discussion.) Griffiths' proof makes explicit use of this theorem. It is a basic theorem of QM; are you not aware of it?
I'm aware of it. Isn't degeneracy an issue here since we are talking about eigenstates of H with the same E? $\sin(kx)$ and $\cos(kx)$ have the same eigenvalue E. Doesn't this mean a degeneracy of 2?

PeterDonis

Mentor
Isn't degeneracy an issue here since we are talking about eigenstates of H with the same E?
Degeneracy complicates the proof of the theorem somewhat, but it still applies.

In the specific case you give, $\sin (kx)$ and $\cos (kx)$ are both eigenstates of $D$ for appropriate values of $k$, and only those values of $k$ will give eigenstates of $H$.

Happiness

Degeneracy complicates the proof of the theorem somewhat, but it still applies.
In cases with degeneracy, there may exist eigenstates of H that are not eigenstates of D, so Griffith's proof does not prove Bloch's theorem for these cases.
In the specific case you give, $\sin (kx)$ and $\cos (kx)$ are both eigenstates of $D$ for appropriate values of $k$, and only those values of $k$ will give eigenstates of $H$.
k can only take on certain values. This is correct. These values corresponds to the allowed energy values. For a certain allowed energy value, $\sin (kx)$, $\cos (kx)$, $2\sin (kx)+\cos (kx)$ and in fact all other linear combinations are all eigenstates of H. But only 2 of them (at the most) are eigenstates of D. They are the eigenvectors of the matrix in (4.189). Reference: QM 2nd ed., B&J, p183

Since the other eigenstates of H are not eigenstates of D, the following sentence must be false.
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].

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PeterDonis

Mentor
For a certain allowed energy value, $\sin (kx)$, $\cos (kx)$, $2\sin (kx)+\cos (kx)$ and in fact all other linear combinations are all eigenstates of H. But only 2 of them (at the most) are eigenstates of D.
You are ignoring the fact that the value of $k$ cannot be chosen arbitrarily; it must satisfy $k = n \pi / a$, where $n$ is an integer and $a$ is the "spacing" in the potential $V(x)$. Otherwise we won't have an eigenstate of $H$ (or $D$ for that matter). And for any given energy eigenvalue, all eigenfunctions must have the same $k$.

For any such value of $k$, we have $\sin k (x + a) = \sin (kx + n \pi) = \sin (kx)$, and similarly for $\cos k(x + a)$. Then any function $\psi (x) = a \sin (kx) + b \cos (kx)$ will give

$$\psi(x + a) = a \sin k (x + a) + b \cos k(x + a) = a \sin(kx + n \pi) + b \cos(kx + n \pi) \\ = a \sin (kx) + b \cos (kx) = \psi(x)$$

I think the source you reference is considering the more general case where $\psi_1$ and $\psi_2$ do not have the same energy eigenvalue and hence do not have the same $k$.

In cases with degeneracy, there may exist eigenstates of H that are not eigenstates of D, so Griffith's proof does not prove Bloch's theorem for these cases.
See above.

Happiness

it must satisfy k = $\pi$/a, where a is the "spacing" in the potential V(x)
Why? I know $K=\frac{2\pi n}{Na}$ cannot be arbitrarily chosen. For k, could you explain further? Is this only for this specific example?
I think the source you reference is considering the more general case where $\psi_1$ and $\psi_2$ do not have the same energy eigenvalue and hence do not have the same $k$.
They have the same energy: "corresponding to the energy E" (line 4). Last edited:

PeterDonis

Mentor
Because it has to be an eigenstate of $D$: any eigenfunction $\psi(x)$ of $D$ must be periodic with period $a$. I misspoke in my earlier post; I should have said "it won't be an eigenstate of $D$" instead of "it won't be an eigenstate of $H$".

PeterDonis

Mentor
Is this only for this specific example?
The logic I gave should work for any functions that are periodic with the same periodicity (which in this case will need to be $a$ divided by an integer).

PeterDonis

Mentor
I misspoke in my earlier post; I should have said "it won't be an eigenstate of $D$" instead of "it won't be an eigenstate of $H$".
Perhaps it will help to rephrase the argument I was making as follows:

Given any value of $k$ (not assuming any restriction on $k$ at the start of the argument), the functions $\sin(kx)$ and $\cos(kx)$ form a basis of a space of functions that all have the same periodicity. Assume that these functions are all eigenstates of $H$ (if any one of them is, they all are because they all have the same $k$, and if none of them are, we don't care about this value of $k$ anyway). In order for the conditions needed for Bloch's Theorem to apply at all, at least one of the functions must also be an eigenfunction of $D$. And if any one of them is, they all are, since they all have the same periodicity, and periodicity is the only criterion that determines whether a function is an eigenfunction of $D$. And the condition that at least one of the functions must be an eigenfunction of $D$ is what places the restriction on $k$; only certain values of $k$ allow any of the functions to be an eigenfunction of $D$.

Cthugha

This wasn't the issue I had.
Well, your issue already changed several times. This is the difference in the treatments between Griffith and Bransden/Joachain. Griffith first looks for common eigenstates of the Hamiltonian and the displacement operator for displacement by a single period and then introduces periodic boundary conditions, which is equivalent to looking for all solutions for all displacement operators that correspond to displacement by an integer number of lattice periods. Bransden/Joachain immediately assume periodic boundary conditions at the beginning of their discussion. It is a very different question whether you look for superpositions of functions that are eigenfunctions of one of the displacement operators and the Hamiltonian or superpositions of eigenfunctions of different displacement operators and the Hamiltonian.

There are two values of K for each energy E.
These differ only by sign. Your concern was that these states might result in different $\lambda$. For D(a) as treated by Griffith, you end up with $K=\pm \frac{2\pi}{a}$, which obiously yield the same $\lambda$. This will change when considering a different displacement operator.

Not quite clear what you mean. States that are eigenfunctions of D? States of the same energy?
States that are eigenfunctions of D(a) and the Hamiltonian.

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.
The approach given in the book is not simpler because what I noted was exactly the approach used in the book.

vanhees71

Gold Member
I think indeed, it's again the enigmatic writing by Griffiths. I don't understand why this book is so successful as it seems to cause so much confusion.

Bloch's theorem is about finding a convenient complete energy eigenbasis. The Hamiltonian is invariant under the unitary translation operators $\hat{T}(a)=\exp(\mathrm{i} \hat{p} a)$, where $a$ is the periodicity of the (1D) crystal lattice (that generalizes to the 3D case, where instead of $a$ you have the vectors $\vec{R}$ of the Bravais lattice defining the discrete tranlational crystal symmetries of the different classes of possible lattice). You have
$$\hat{H}=\frac{\hat{p}^2}{2m} + V(\hat{x}),$$
and the Hamiltonian obviously has the symmetry iff $V$ is a periodic function of period $a$, and then
$$\hat{T}(a) \hat{H} \hat{T}^{\dagger}(a)=\frac{\hat{p}^2}{2m} + V(\hat{x}+a \hat{1})=\frac{\hat{p}^2}{2m} + V(\hat{x})=\hat{H} \Rightarrow [\hat{T}(a),\hat{H}]=0.$$
Since unitary operators are normal operators, i.e., $\hat{T}(a) \hat{T}^{\dagger}(a)=\hat{1}=\hat{T}^{\dagger}(a) \hat{T}(a)$ they define a complete set of eigenvectors, and since it commutes with $\hat{H}$ you can find a common eigenbasis of $\hat{H}$ and all $\hat{T}(a)$. The eigenvalues of $\hat{T}$ are phase factors $\exp(\mathrm{i} k a)$.

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 $a$, $L=N a$ 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
$$\exp(\mathrm{i} k N a)=1 \; \Rightarrow \; k N a = 2\pi n, \quad n \in \mathbb{Z}.$$
Thus the possible
$$k=k_n=\frac{2 \pi n}{N}, \quad n \in \mathbb{N}.$$
Of course you can form arbitrary linear combinations of these basis vectors, allow to represent all the allowed Hilbert-space vectors (i.e., the $\mathrm{L}^2$ functions, fulfilling the Born von Karmann boundary conditions), but of course, as you rightfully realized superpositions of basis vectors with different $k_n$'s are not eigenvectors of $\hat{T}(a)$.

For the analog 3D case that's known as Born-von Karman boundary conditions, where you have in general a Bravais lattice in position space, defined by a primitive cell (Wigner-Seitz cell) and then a reciprocal lattice in momentum space, which formes again a Bravais lettice with the primitive cell known as the 1st Brillouin zone. If $\vec{a}_i$ are the vectors spanning up the Wigner-Seitz cell the three vectors $\vec{b}_i$ spanning the 1st Brillouin zone are conveniently chosen such that $\vec{b}_j \cdot \vec{a}_i=2 \pi \delta_{ji}$.

Mentor
the condition that at least one of the functions must be an eigenfunction of $D$ is what places the restriction on $k# Actually, on thinking this over, the condition for eigenfunctions of$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)$that is known to be periodic. vanhees71 Science Advisor Gold Member Why should all energy eigensolutions necessarily be periodic? If for some reason an energy eigenvalue is degenerate, i.e., there are Bloch eigenstates with the same$E$but different$k_n$, you can form linear combinations being still an energy eigenvector but not periodic with period$a$. Not all energy eigensolutions need to obey the symmetry of the underlying dynamics. Even the (then necessarily degenerate) ground states need not obey the symmetry. That's then known as "spontaneous symmetry breaking". PeterDonis Mentor Why should all energy eigensolutions necessarily be periodic? I don't think they need to be in general. In the subthread in question @Happiness had given a specific example of sines and cosines, and I was considering that specific case. PeterDonis Mentor if there are Bloch eigenstates with the same$E$but different$k_n$Is there a specific example of such a set of Bloch eigenstates? vanhees71 Science Advisor Gold Member For the 1D case, it's hard to imagine... I don't know. Happiness it must satisfy k = n$\pi$/ a, where n is an integer and a is the "spacing" in the potential V(x) This only covers the cases where the eigenvalue of D is$\lambda=1$(when n is even) and$\lambda=-1$(when n is odd). For complex values of$\lambda=e^{iKa}$,$k\neq\frac{n\pi}{a}$. So there are other values of$\lambda$that you did not consider, explicitly those complex values, which produce other values of$k$. any eigenfunction$\psi(x)$of D must be periodic with period a This is only true when$\lambda=1$. In general, it is the square of the amplitude of any eigenfunction$\psi_{\lambda}(x)$of D that must be periodic with period a. But not the eigenfunctions themselves. In fact, the eigenfunctions$\psi_{\lambda}(x)$of D acquire a phase factor$\lambda=e^{iKa}$when acted upon by D. Given any value of k (not assuming any restriction on k at the start of the argument), the functions$\sin(kx)$and$\cos(kx)$form a basis of a space of functions that all have the same periodicity. Assume that these functions are all eigenstates of H (if any one of them is, they all are because they all have the same k, and if none of them are, we don't care about this value of k anyway). In order for the conditions needed for Bloch's Theorem to apply at all, at least one of the functions must also be an eigenfunction of D. And if any one of them is, they all are, since they all have the same periodicity, and periodicity is the only criterion that determines whether a function is an eigenfunction of D. k is not the only criterion that determines whether a function is an eigenfunction of D: A and B also determine whether a function is an eigenfunction of D, via its effect on the derivative of the function (as the derivative needs to be periodic too), where A and B are as defined in [5.59]. The cause of my confusion is that I believe there are eigenstates of H that are not eigenstates of D. This is confirmed by B&J's paragraph following (4.190). Consider the simplest case of$\psi=A\sin (kx)+B\cos (kx)$. The eigenfunctions of D are given by$A=\pm iB$. So$\psi=\sin (kx)$is an eigenfunction of H but not of D. The eigenfunctions of D are$\psi_{\lambda_1}=e^{ikx}$and$\psi_{\lambda_2}=e^{-ikx}$. And since they have different eigenvalues, a linear combination of these two eigenfunctions will not be an eigenfunction of D in general, and so does not satisfy Bloch's theorem$\psi(x+a)=e^{iKa}\psi(x)$even though it is always an eigenfunction of H. Last edited: PeterDonis Mentor The eigenfunctions of D are given by$A=\pm iB$. So$\psi=\sin (kx)$is an eigenfunction of$H$but not of$D$. If both the sine and the cosine have the same$k$, I don't see this. The definition of$D$is  D \psi (x) = \psi(x + a)  For the values of$k$that I gave, we have$\sin k (x + a) = \sin (kx)$. So$D \sin(kx) = \sin k (x + a) = \sin (kx)$, and$\sin (kx)$is an eigenfunction of$D$. So is$\cos (kx)$for the same$k$, and so is any linear combination of them, by the argument I gave earlier. For other values of$k$, not meeting the condition I gave earlier that$k = n \pi / a$, obviously$\sin(kx)$won't be an eigenfunction of$D$, but I never claimed it would be. I understand that if you combine sines and cosines with different values of$k$, even if each of the$k$values meets the condition I gave earlier, you won't, in general, get an eigenfunction of$D$. But you won't, in general, get an eigenfunction of$H$either. PeterDonis Mentor This only covers the cases where the eigenvalue of D is$\lambda=1$(when n is even) and$\lambda=-1$(when n is odd). Actually, it only covers$\lambda = 1$, since$\sin (x + n \pi) = \sin x$for any integer$n$, and similarly for cosines. If I had put$2 \pi$instead of$\pi$in the formula just now, that would be different; but I didn't. I agree, though, that I did not cover other possibilities for$\lambda$. For that more general case, I think it's easier to use exponentials instead of sines and cosines; I'll take a look at that. Happiness If both the sine and the cosine have the same$k$, I don't see this. The definition of$D$is  D \psi (x) = \psi(x + a)  For the values of$k$that I gave, we have$\sin k (x + a) = \sin (kx)$. So$D \sin(kx) = \sin k (x + a) = \sin (kx)$, and$\sin (kx)$is an eigenfunction of$D$. So is$\cos (kx)$for the same$k$, and so is any linear combination of them, by the argument I gave earlier. For other values of$k$, not meeting the condition I gave earlier that$k = n \pi / a$, obviously$\sin(kx)$won't be an eigenfunction of$D$, but I never claimed it would be. I understand that if you combine sines and cosines with different values of$k$, even if each of the$k$values meets the condition I gave earlier, you won't, in general, get an eigenfunction of$D$. But you won't, in general, get an eigenfunction of$H$either. \psi_{\lambda_1}=B\cos(kx)+iB\sin(kx)=e^{ikx} D\psi_{\lambda_1}=e^{ik(x+a)}=e^{ika}e^{ikx} So$e^{ikx}$is an eigenvector of D with eigenvalue$e^{ika}$. \psi_{\lambda_2}=B\cos(kx)-iB\sin(kx)=e^{-ikx} D\psi_{\lambda_2}=e^{-ik(x+a)}=e^{-ika}e^{-ikx} So$e^{-ikx}$is an eigenvector of D with eigenvalue$e^{-ika}$. The k are all the same, so$\psi_{\lambda_1}$and$\psi_{\lambda_2}$have the same energy. But$\frac{\psi_{\lambda_1}-\psi_{\lambda_2}}{2i}=\sin(kx)$is not an eigenfunction of D, even though it is an eigenfunction of H. (B=1 after normalisation.) Happiness Actually, it only covers$\lambda = 1$, since$\sin (x + n \pi) = \sin x$for any integer$n$, and similarly for cosines. If I had put$2 \pi$instead of$\pi$in the formula just now, that would be different; but I didn't. I agree, though, that I did not cover other possibilities for$\lambda$. For that more general case, I think it's easier to use exponentials instead of sines and cosines; I'll take a look at that.$\sin (x + \pi) = -\sin x\cos (x + \pi) = -\cos x$PeterDonis Mentor$\sin (x + \pi) = -\sin x\cos (x + \pi) = -\cos x$Ah, you're right. <facepalm> So I did need to put$2 \pi$as the period in my earlier formula (which limits consideration to the case$\lambda = 1$). However, with that revised condition on$k$, the argument I gave in post #44 for$\sin (kx)$and$\cos (kx)$being eigenfunctions of$D$is still valid. In fact the argument can be stated even more simply:$\sin (kx)$and$\cos (kx)$are periodic, so if we choose$k$to have the appropriate relationship to$a$, they are obviously eigenfunctions of$D$. Happiness Ah, you're right. <facepalm> So I did need to put$2 \pi$as the period in my earlier formula (which limits consideration to the case$\lambda = 1$). However, with that revised condition on$k$, the argument I gave in post #44 for$\sin (kx)$and$\cos (kx)$being eigenfunctions of$D$is still valid. In fact the argument can be stated even more simply:$\sin (kx)$and$\cos (kx)$are periodic, so if we choose$k$to have the appropriate relationship to$a$, they are obviously eigenfunctions of$D$. Suppose you found that value of$k$such that$\sin (kx)$and$\cos (kx)$are both eigenfunctions of$D$, then$\sin (kx)+\cos (kx)$or$\psi_{\lambda_1}+\psi_{\lambda_2}$will not be an eigenfunction of$D$in general, because$\psi_{\lambda_1}$and$\psi_{\lambda_2}$have different eigenvalues in general. ($\psi_{\lambda_1}+\psi_{\lambda_2}$is always an eigenfunction of$H$.) It is proven by B&J that$\lambda_1=\frac{1}{\lambda_2}$. 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$D$. Cthugha Science Advisor This only covers the cases where the eigenvalue of D is$\lambda=1$(when n is even) and$\lambda=-1$(when n is odd). For complex values of$\lambda=e^{iKa}$,$k\neq\frac{n\pi}{a}$. So there are other values of$\lambda$that you did not consider, explicitly those complex values, which produce other values of$k##.
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).

"Inadequate proof of Bloch's theorem?"

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