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Just a quick check: If given a time-independent wavefunction of the form $$\psi(x) = e^{ikx}f(x)$$, where $f(x)$ any arbitrary function of $x$ but one can't factor out another $e^{i\alpha x}$, $\alpha\in\mathbb R$.

Could I then immediately conclude that $k^2= \frac{2mE}{\hbar^2}$? Is there a name for this quantity? It seems to creep up a lot.

Suppose the wavefunction is $$\psi(x) = e^{ikx}(\tanh{x}-ik)$$. Am I justified to conclude by inspection that there is no reflection (since there is no $Re^{-ikx}$ term)? What else can I conclude by inspection?


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Please say "energy eigenstate" instead of "time-independent wave function". I got confused for a while. – Ron Maimon Aug 31 '11 at 22:15
How exactly do you define "can't factor out another $e^{i\alpha x}$"? It's not clear because you can factor a plane wave out of any function and get another function. Also, where does reflection come into this? (Also, "What else can I conclude by inspection?" is way too open-ended) – David Z Aug 31 '11 at 22:16
I understand what he means--- asymptotically, there is no other exponential factor. Reflection is because the wavefuncton is asymptotically free-particle, and is clearly describing a scattering event. "What else can I conclude by inspection" has a complete answer, because you can recognise this scattering form as the Paschl-Teller potential scattering eigenstate, and this is an extremely common exercise in quantum mechanics books. – Ron Maimon Aug 31 '11 at 22:29
OK, if you assume the wavefunction is an energy eigenstate. I forgot about that. But "What else can I conclude by inspection" is still too broad a question. If it was intended to evoke an explanation of the Paschl-Teller potential, the question should have asked about that specifically. – David Z Sep 1 '11 at 0:00
...actually, I'm not really convinced that "can't factor out another $e^{i\alpha x}$" is clearly defined. You can restate it as there being no other exponential factor asymptotically (or even not asymptotically) but that's equally unclear. $f(x) = \sin(k'x)$, for example, meets the criterion but I don't think that's the kind of wavefunction that was intended... I think "asymptotic momentum eigenstates" ($\psi(x) \to e^{ik_\pm x}$, i.e. $f(x) \to \text{const}$, as $x\to\pm\infty$) might be a better description. bra-ket, is that what you meant? – David Z Sep 1 '11 at 0:12

2 Answers 2

up vote 1 down vote accepted

You need to know that the $f(x)$ is asymptotically constant in order to conclude this, and that the potential is asymptotically zero to the left, and that the wavefunction is an energy eigenstate. Then you can conclude that the energy of the eigenstate is the same as the energy of the eigenstate far away and the only energy is kinetic. $k^2\over 2m$ is called the kinetic energy.

For your wavefunction, since $\tanh$ is asymptotically constant, you can conclude that there is no reflected wave, that the transmitted wave is phase shifted by the angle in the complex plane of the ratio of the two asymptotic values, $(-1+ik)\over (1+ik)$, and you can conclude that the potential of scattering is that which is required to make the Schrodinger equation work:

$$ mV(x) = - {1\over \cosh^2(x)}$$

Where m is the mass of the particle. This is a well-known exactly solvable potential, called the Paschl-Teller potential.

General Reflectionless Potential at Wavenumber k

You are looking for the most general potential with no reflection at wavenumber k when you make this ansatz for the wavefunction. For a general energy eigenstate of this form,

$${- \psi''\over 2m} = {k^2\over 2m}\psi + \psi( - {ik\over m} {f'\over f} - {1\over 2m}{f''\over f})$$

So that the potential you get is

$$ V(x) = -({f''\over 2m f} + ik {f'\over f})$$

But this is not an answer, because this potential is in general complex, while a physical potential is real. To find the constraint on f that $V$ is real, write $f$ as $e^\alpha$ and define $\beta=\alpha'$. Then the potential in terms of $\beta$ is

$$ V(x) = \beta' + 2ik\beta + \beta^2 $$

and in order for this to be real, if $\beta=P+iQ$ with P,Q real functions of x, then

$$ P = -{Q'\over 2(k+Q)}$$

This, for any choice of Q, with the appropriate boundary condition, give the class of potentials reflectionless at wavenumber k. The Paschl-Teller potential is exceptional for being reflectionless at all k simultaneously.

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Thanks, Ron, this is exactly what I was looking for by my perhaps ambiguous question. :) – bra-ket Sep 1 '11 at 7:48

Recall the Schrödinger equation:

$$\left(-\frac{\hbar^2}{2m}\Delta+V(x)\right)\psi(x) = E\psi(x)$$

which can be written as


when you rename the second factor $k(x)=\sqrt{\frac{2m(E-V(x))}{\hbar^2}}$ you get something like a wave equation if k is constant:


This is solved by the the plane wave $\psi(x)=e^{ikx}$. For a free particle V=0 and $k^2=\frac{2mE}{\hbar^2}$ which is why this term appears so often. $k$ is called the circular wave number

Now for your question,

$$\begin{align*}\Delta\psi(x) &= \Delta e^{ikx}f(x)\\ &=\frac{\partial^2}{\partial x^2}e^{ikx}f(x)\\ &=\frac{\partial}{\partial x}(ike^{ikx}f(x)+e^{ikx}f'(x))\\ &=\bigl(-k^2f(x)+2ikf'(x)+f''(x)\bigr)e^{ikx}\\ &=-k^2\psi(x)+\cdots\\ &= -\frac{2m(E-V(x))}{\hbar^2}\psi(x)\end{align*}$$

as you can see, in general this is not possible. Your function $f(x)$ generally is non constant thus the Laplacian has other terms besides $-k^2\psi(x)$. Thus in general the correspondence $k^2=\frac{2m(E-V(x))}{\hbar^2}$ does not hold, much less $k^2=\frac{2mE}{\hbar^2}$

There is an interesting special case, that is very relevant in solid state physics, called Bloch waves. These arise, when the potential is periodic $V(x)=V(x+a)$. The solutions of the Schrödinger equation for such a potential look as follows:

$$\psi(x)=e^{ikx}u_k(x)$$ where the $u_k(x)$ have the same periodicity as the potential.

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Thanks, luksen, great explanation. :) – bra-ket Sep 1 '11 at 7:47

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