Inadequate proof of Bloch's theorem?

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  • #71
vanhees71
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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 [itex]\lambda[/itex] for D(a) will be a real [itex]\lambda[/itex] 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.
 
  • #72
Happiness
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[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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  • #73
Cthugha
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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.
 
  • #74
vanhees71
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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##.
 
  • #75
Happiness
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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:
Screenshot 2019-08-14 at 6.14.38 PM.png


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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  • #76
vanhees71
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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"?
 
  • #77
George Jones
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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.
 
  • #79
Cthugha
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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?
 
  • #80
Happiness
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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]?
 
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  • #81
Happiness
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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]?
 
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  • #82
Happiness
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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.)

Screenshot 2019-08-14 at 11.31.55 PM.png

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]?
 
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  • #83
vanhees71
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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
 
  • #84
vanhees71
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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
 
  • #85
vanhees71
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  • #86
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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.
 
  • #87
Happiness
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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?
 
  • #88
vanhees71
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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.
 

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