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If a particle in a delta potential well has negative energy, why the particle will be bound in the well? And if it has positive well, why it is free to move in either half-space: x < 0 or x > 0?

I just read these in a wikipedia page: http://en.wikipedia.org/wiki/Delta_potential

I don't know how the energy of the particle influence its motion.

And by the way, why $e^{ikx}$ represents a wave traveling to the right, and $e^{-ikx}$ one traveling to the left?

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up vote 6 down vote accepted

I'm not totally sure I know what the first questions mean, so let me start with the last one. The wavefunction of a particle is a function of time as well as position. A lot of quantum mechanics is concerned with wavefunctions corresponding to "stationary states." You might think that such wavefunctions would have no time-dependence, but that's not the case. A stationary state corresponds to a particle with a certain amount of energy $E$, and the wavefunction has time dependence of the form $e^{-i\omega t}$, where $\omega=E/\hbar$.

People often leave out that time dependence, just to confuse you.

So when people refer to a wavefunction of the form $e^{ikx}$, they really mean $e^{ikx}e^{-i\omega t}$, or $e^{i(kx-\omega t)}$. This corresponds to a wave moving to the right. In particular, a point of any given "phase" moves to the right at speed $\omega/k$. For instance, the point $kx-\omega t=0$ is a point where the wavefunction is purely real. This point moves along according to the rule $x=(\omega/k)t$.

Another way to think about it: a particle has definite momentum if it is an eigenfunction of the momentum operator: $$ \hat p\psi = p\psi. $$ Here $\hat p=-i\hbar(\partial/\partial x)$ is the momentum operator (specifically its $x$ component), and the number $p$ is the eigenvalue. For a particle with such a wavefunction, a measurement of momentum is guaranteed to give the value $p$. The wavefunction $\psi(x)=e^{ikx}$ obeys this rule: $$ \hat pe^{ikx}=-i\hbar{\partial(e^{ikx})\over\partial x}=\hbar ke^{ikx}=\hbar k \psi, $$ so $p=\hbar k$. So a particle with positive $k$ will be observed to have positive momentum.

Now back to the first couple of questions. When you solve for the wavefunction of a particle in a delta function potential with negative energy, you find that the solutions are of the form $e^{-k|x|}$ for some $k$. These wavefunctions approach zero for large $|x|$. So physically, the particle has essentially zero probability of being far away from the origin. That's what "bound state" means.

On the other hand, with positive energy, there are solutions that look like $e^{\pm ikx}$. These wavefunctions do not die away to zero at large distances: $|\psi|^2$ remains constant as $|x|\to\infty$. So these particles have significant probability to be anywhere.

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