I Inadequate proof of Bloch's theorem?

vanhees71

Gold Member
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

Staff Emeritus
Gold Member
• vanhees71

PeterDonis

Mentor
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.

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

Last edited:

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]?

Last edited:

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]?

Last edited:

vanhees71

Gold Member
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.

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

vanhees71

Gold Member
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:

vanhees71

Gold Member
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

A minimum of group-representation theory helps a lot in quantum theory!

PeterDonis

Mentor
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

Gold Member
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.

"Inadequate proof of Bloch's theorem?"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving