# Wave function and spatial phase factor

1. Jan 8, 2009

### jostpuur

Suppose $\psi(x)$ is an eigenfunction of some Hamilton's operator $H$

$$(H\psi)(x)=-\frac{\hbar^2}{2m}\partial_x^2\psi(x) + V(x)\psi(x).$$

I've noticed that it seems to be true, that if the eigenvalue corresponding to this eigenfunction is isolated in the spectrum (I merely mean the set of eigenvalues, and not in Hilbert space sense, but non-normalizable functions are allowed too), then $\psi(x)$ does not have a spatial phase factor at all. In other words, the equation

$$\frac{\partial_x\psi(x)}{\psi(x)} \in \mathbb{R}$$

is true. This condition is not satisfied for the de'Broglie waves $e^{ipx/\hbar}$ for example, when $V=0$, but now the eigenvalue is not isolated, but is part of a continuous spectrum.

Is this hypothesis true in general? Anyone knowing how to prove it?

2. Jan 8, 2009

What potentials do you allow for V?

3. Jan 8, 2009

### jostpuur

I don't know.

4. Jan 8, 2009

The thing is. If you do particle in a box with periodic boundary conditions, you get the same de Broglie waves but only a discrete set, so your conjecture fails right away.

I have the feeling whenever you move things in circles, you should get some kind of rotating phase in the circumpherence direction. So if your potential has loops you will probably get a spacial phase factor. I am too lazy to check the hydrogen solutions, but it might already fail for m=1, since the electron is "spinning".

5. Jan 8, 2009

### jostpuur

I see.

I was so far thinking about one dimensional case only.

If we assume that $\psi$ is defined on the $\mathbb{R}$, another conjecture is that if $\psi$ satisfies condition $$\lim_{|x|\to\infty} \psi(x)=0$$, does it imply that $\psi$ has no spatial phase factor?

6. Jan 9, 2009

It doesn't seem to work. If you have a potential of the form:
$$V(x)= \left{ E\quad x\in [-L,L], V(x) = 0 \quad x \notin [-L,L]$$
Now in the free region in the middle you should get your old plane waves disobaying your conjecture, and on the borders you should get exponential decays. The eigenvalues of the potential well are discrete for bound states as well.

Just because we can make the eigenfunctions real doesn't mean we have to. You should check this though, because, I am just assuming that the continuity conditions on the potential boundaries can be met as easily in the complex case as in the real case.

7. Jan 9, 2009

### tim_lou

this hypothesis is wrong, since
$$\sin(kx)$$
is an equally valid solution.

in fact, you can choose all solutions to be real (or imaginary) by just doing
$$\psi \rightarrow \frac{1}{\sqrt{2}}(\psi+\psi^*)$$
since $\psi^*[/tex] is another solution with the same eigenvalue 8. Jan 10, 2009 ### jostpuur You did not understand the original hypothesis. I have made a remark, that if an eigenfunction has an isolated eigenvalue, or if the eigenfunction approaches zero at infinities (which seem to be equivalent conditions), then the eigenfunction has no spatial dependence in the phase factor. Or alternatively, if an eigenfunction has spatial dependence on the phase factor, then the eigenvalue is part of a continuous part of the spectrum and the eigenfunction does not approach zero at infinities. Now I'm asking, that are these properties true in general (or how general they are). The remark that there exist real eigenfunctions with eigenvalues in a continuous part of the spectrum, does not prove the hypothesis wrong. In order to prove the hypothesis wrong, you should succeed in the following: Find an eigenfunction that has an isolated eigenvalue, or that approaches zero at infinities, but which still has spatial dependence in the phase factor. Last edited: Jan 10, 2009 9. Jan 10, 2009 ### tim_lou I apologize for my misunderstanding. However, I still do not understand what your hypothesis is. What do you mean by phase factors? do you mean the phase factors δ when you write a function (or number) in the form re^{iδ}? if so, then there is no phase factor (δ can be chosen to be identically zero) since the eigenfunctions can be chosen to be real. Perhaps I am mistaken again, but I cannot think of any other well defined phase factors. Edit: perhaps you've defined the phase factor as ∂ψ/ψ. However, same as before, this can be chosen to be always real. Perhaps you want to show that ∂ψ/ψ is always real for bound states? I believe this is only true for nondegenerate states though. For that case, ψ and ψ* are both solutions of a differential equation. Since the state is nondegenerate, ψ*=aψ (a some complex constants). Hence we easily see that (∂ψ/ψ)*=∂ψ/ψ. Last edited: Jan 10, 2009 10. Jan 10, 2009 ### jostpuur Yes I mean the phase factor of a complex number. For example $$e^{-A x^2}$$ is the ground state of a harmonic oscillator. If I define a wave function [itex]\psi$ with a formula $$\psi(x) = e^{i\theta} e^{-A x^2}$$, so that $\theta\neq 0$, now $\psi$ is not real. In this case, the phase factor of $\psi$ does not depend on $x$, so it has no spatial dependence. Also, I can make this wave function real by multiplying with the phase factor $e^{-i\theta}$, because $e^{-i\theta}\psi(x)\in\mathbb{R}$.

The consider a wave function $\psi(x)=e^{iKx}$. This one is not real either, but can we now find a suitable phase factor $e^{-i\theta}$ so that $e^{-i\theta}\psi(x)\in\mathbb{R}$? The answer is no, because the phase factor of $\psi$ depends on $x$, so now it has spatial dependence.

This property divides all eigenfunctions into two classes. Some eigenfunctions have no spatial dependence in the phase factors, while some eigenfunctions do have it. My hypothesis deals with these classes.

11. Jan 10, 2009

### jostpuur

Hello. I wrote one answer to this, but I soon found some flawed mathematics in it, so I destroyed it. You probably got it into mail though... But I'll try to prove the same claim now again: I don't believe that this potential $V$ provides a counter example.

Suppose we have some eigenfunction

$$\psi(x) = Ae^{\alpha x} \chi_{]-\infty, -L]}(x) \;+\; \Big(Be^{i\beta x} + Ce^{-i\beta x}\Big)\chi_{]-L,L]}(x) \;+\; D e^{-\alpha x}\chi_{]L,\infty[}(x)$$

Here $A,B,C,D\in\mathbb{C}$, $\alpha,\beta\in\mathbb{R}$, and $\alpha > 0$. Without loss of generality, we can assume that $D\in\mathbb{R}$. The phase factor is constant on the interval $[L,\infty[$, so we can rotate the $\psi$ so that it becomes real at least on this interval. In order to have $\psi$ continuous and differentiable at $x=L$, we must have the following conditions met.

$$\lim_{\epsilon\to 0^+}\psi(L-\epsilon) = Be^{i\beta L} \;+\; Ce^{-i\beta L}$$
$$= (\textrm{Re}(B) \;+\; \textrm{Re}(C))\cos(\beta L) \;+\; (\textrm{Im}(C) \;-\; \textrm{Im}(B))\sin(\beta L)$$
$$+\; i\big((\textrm{Im}(B) \;+\; \textrm{Im}(C))\cos(\beta L) \;+\; (\textrm{Re}(B) \;-\; \textrm{Re}(C))\sin(\beta L)\big) \in \mathbb{R}$$

$$\lim_{\epsilon\to 0^+}\psi'(L-\epsilon) = i\beta B e^{i\beta L} \;-\; i\beta Ce^{-i\beta L}$$
$$= \beta\big((\textrm{Im}(C) \;-\; \textrm{Im}(B))\cos(\beta L) \;-\; (\textrm{Re}(B) \;+ \textrm{Re}(C))\sin(\beta L)\big)$$
$$+\; i\beta\big((\textrm{Re}(B) \;-\; \textrm{Re}(C))\cos(\beta L) \;-\; (\textrm{Im}(B) \;+\; \textrm{Im}(C))\sin(\beta L) \big) \in \mathbb{R}$$

This can be written in the following matrix form.

$$\left(\begin{array}{cc} \textrm{Im}(B) + \textrm{Im}(C) & \textrm{Re}(B) - \textrm{Re}(C) \\ \textrm{Re}(B) - \textrm{Re}(C) & -\textrm{Im}(B) - \textrm{Im}(C) \\ \end{array}\right) \left(\begin{array}{c} \cos(\beta L) \\ \sin(\beta L) \\ \end{array}\right) = \left(\begin{array}{c} 0 \\ 0 \\ \end{array}\right)$$

$\cos(\beta L)$ and $\sin(\beta L)$ cannot be both zero simultaneously, so the determinant of the 2x2-matrix must be zero. The determinant is

$$-(\textrm{Im}(B) + \textrm{Im}(C))^2 - (\textrm{Re}(B) - \textrm{Re}(C))^2 = 0$$

From this we get can solve $C=B^*$. So the $\psi$ is

$$Be^{i\beta x} + B^* e^{-i\beta x} = 2\textrm{Re}\Big( Be^{i\beta x}\Big) \in \mathbb{R}$$

on the interval $[-L,L]$. From demanding the boundary conditions at $x=-L$ it follows that also $A$ must be real. We can conclude, that the original non-rotated solution with $D\in\mathbb{C}$ did not have spatial dependence in the phase factor.

12. Jan 10, 2009

### jostpuur

Am I correct to guess that you thought that $V$ could provide a counter example, because on the limit $L\to\infty$ we get wave solutions $e^{iKx}$? It would seem natural to assume that already with very large $L$ we could have eigenfunctions which are approximately $e^{iKx}$? It is difficult to make sense out of this. I'm not sure what is happening.

One thing that might turn out to be relevant is that the eigenfunctions are not translation invariant with finite $L$, but they become translation invariant with $L=\infty$. If we have an eigenfunction

$$\cos(Kx),$$

with $L=\infty$, then we also get an eigenfunction

$$\cos(Kx) + i\cos\big(Kx - \frac{\pi}{2}\big) = e^{iKx}.$$

But if $L$ is finite, then no matter how large it is, this will not be possible.

It also could be, that with very large $L$ there exists functions which are approximately $e^{iKx}$, and approximately eigenfunctions, but still it would not be possible to approximate $e^{iKx}$ with precise eigenfunctions.

13. Jan 10, 2009

### jostpuur

comment on terminology

btw, in my few first posts I used a term "spatial phase factor", but then I changed this to "phase factor with spatial dependence". It could be that the original term wasn't a clearest possible, and caused some confusion.

14. Jan 10, 2009

### tim_lou

So, you wish to prove that for bound states, the phase factor is constant?
If the levels are non-degenerate, this is a direct consequence of the fact that
1. we can choose a real eigenfunction
2. any other solution is a multiple of this eigenfunction, hence its complex phase is constant

If the level is degenerate then, this cannot possibly be true as we can easily construct a
wave function
ψ+iφ
ψ≠φ so that the phase must depend on x.

However, I vaguely remember reading something that says one dimensional QM has no degeneracy. I can't think of a proof immediately, however, the statement does seem plausible.

15. Jan 10, 2009

### jostpuur

I see. This is right. If $\psi$ is an eigenfunction corresponding to a non-degenerate eigenvalue, it follows that $\psi$ and $\psi^*$ must be linearly dependent. So there exists a constant $\lambda\in\mathbb{C}$ so that $\psi^*(x) = \lambda \psi(x)$. Substituting a form $\psi(x) = r(x) e^{i\theta(x)}$, where $r(x)\geq 0$, we get a condition

$$e^{-2i\theta(x)} = \lambda\quad\quad\implies\quad\quad \theta(x) = \textrm{constant}\;(\textrm{mod}\;\pi).$$

No degeneracy with bound states, I suppose. The waves $e^{iKx}$ are degenerate at least.

Well, I'm interested of the proof of this degeneracy claim next

16. Jan 10, 2009

### tim_lou

Hmmm... actually, I think I might take back that statement.

Imagine a symmetric double well potential that is well separated. It could have two states (related by a reflection) that are concentrated in different wells. If this happens, a degeneracy will result.

And if such examples exist, it would not be easy to come up with explicit potentials and eigenstates though.

Last edited: Jan 10, 2009
17. Jan 10, 2009

### tim_lou

Anyway.. to prove that no degeneracy exists, we need to show that for some a, ψ'(a)=λ φ'(a) and ψ(a)=λ φ(a). Then by uniqueness of ODE, ψ(x)=λ φ(x).

If the potential tends to some finite limit at (either) infinity, then one could consider f(x)=ψ(tan^-1{x}) and f(x)=φ(tan^-1{x}) and see what kind of second order differential equations they satisfy. Then one using f'(a)=g'(a) and f(a)=g(a) at a=π/2, so that f(x)=g(x) by uniqueness of ODE.

This does not work for a potential like the harmonic oscillator though.

Last edited: Jan 10, 2009
18. Jan 11, 2009

Not quite. Sorry I don't have time right now to check your solution. I don't know if your definition of a phase factor is good. I just understand, that it doesn't seem to work for complex plain waves. So if I can make a plain wave with exponential decays on the borders it would be an isolated eigenfunction. (Only certain waves fit into the potential well) which goes to 0 at the borders and has a part that looks like a (complex) plain wave, thus colliding with your idea.

19. Jan 11, 2009

### jostpuur

I don't believe that this kind of a counter example exists. Of course if you put infinitely deep potential wells separated, then we get a counter example, but I'll exclude infinite potentials from the "domain of interest". One might think that if we have a counter example with infinitely deep potential wells, then we still get approximately the same eigenstates with finite, but extremely deep, potential wells. However, that would be a flawed conclusion. The mistake is similar as the one I described in the post #12:

Also a related remark is the counter example (with infinite potentials) to the Bloch's theorem (not a counter example to a real mathematical theorem, though, of course...) counter example to the Bloch's theorem? If the eigenfunctions with infinite potentials were approximately the same as the eigenfunctions with finite but large potentials, we would obtain a contradiction with the Bloch's theorem.

It is easy to draw picture of a seemingly reasonable wave function, which in the end cannot be made into an actual eigenfunction with any small corrections.

That is a big "if". I believe I already proved, in my post #11, that such plane waves with exponential decays don't exist. This is technical claim of course, you cannot avoid getting into the details. If you believe I'm wrong with this, you should try to go through the details of my proof.

20. Jan 12, 2009

### jostpuur

There is something tricky with the scaling.... it could be it works, but I'm not sure yet.

I thought about another way to think about this too. If $V$ becomes approximately constant at some domain $[L,\infty[$, it follows that there exists two linearly independent eigenfunctions, which are approximately $$e^{\pm Ax}$$, with some $A$, at this domain. If the eigenvalue (energy) is below the $V$, then $A$ is real, and the other one of the eigenfunctions is unacceptable. If there existed two linearly independent physical solutions $\psi_1,\psi_2$ with this eigenvalue, then they would span the same space as $$e^{\pm A x}$$ in the domain $[L,\infty[$, and therefore there cannot have been the linearly independent physical solutions $\psi_1, \psi_2$.

21. Jan 12, 2009

### jostpuur

0xDEADBEEF, actually it is not necessary for you to go through the all details in my post #11, because an alternative, less detail oriented, proof became available too It suffices to put some pieces together. First recall this:

In other words, provided that there are no infinities in the $V$, then for each fixed $E$ there are at most only two linearly independent solutions. The number of linearly independent physical solutions is 0, 1 or 2. (If $E$ is an eigenvalue, then 1 or 2.)

If the number is 1, then:

Then consider the wide potential well that you mentioned in the post #6. This holds then:

$$\implies$$ So the plane waves with exponential decays don't exist.

22. Jan 12, 2009

You are right that the counterexample fails, because the imaginary part would have to be a valid solution too, but that is not the case. If you would find one degenerate state though, than your idea doesn't work. Because you could always do $$\Psi = \Psi_1 + i Psi_2$$ if they are both real. Because the two functions are not the same the phase must vary. Second order equations normally allow twofold degeneracy. But I am coming to the believe that you are right. There is a conserved probability current
$$\mathbf{s} = \frac{\hbar}{2mi}(\Psi^*\mathrm{grad}\Psi - \Psi \mathrm{grad}\Psi^*)$$
$$\mathrm{div} \mathbf{s} + \frac{\partial \rho}{\partial t} = 0$$