Boundary Conditions for Wave Functions in Infinite Potential Wells

[Spinny]

Hi, I have a question about the mathematical requirements of a wave function in a potential that is infinite at $$x \leq 0$$. (At the other side it goes towards infinity at $$x = \infty$$.) Now, given a wave function in this potential that is zero for $$x = 0$$ and $$x = \infty$$. Does it matter what that wavefunction is at $$x = -\infty$$? I mean, I just figured you would have a wave function there that's zero all the way. Why will a wave function that goes to $$-\infty$$ at $$x = -\infty$$ not fit in the (time independent) Schrödinger equation, whereas one that goes to zero at $$-\infty$$ does? After all when we're normalizing it, we're just integrating from 0 to $$\infty$$ and doesn't really need to bother with it at negative x values. Or is that just some mathematical requirement that is independent of the physical properties? Can someone enlighten me, please?

 

[dextercioby]

Hi, I have a question about the mathematical requirements of a wave function in a potential that is infinite at $$x \leq 0$$. (At the other side it goes towards infinity at $$x = \infty$$.) Now, given a wave function in this potential that is zero for $$x = 0$$ and $$x = \infty$$. Does it matter what that wavefunction is at $$x = -\infty$$? I mean, I just figured you would have a wave function there that's zero all the way. Why will a wave function that goes to $$-\infty$$ at $$x = -\infty$$ not fit in the (time independent) Schrödinger equation, whereas one that goes to zero at $$-\infty$$ does? After all when we're normalizing it, we're just integrating from 0 to $$\infty$$ and doesn't really need to bother with it at negative x values. Or is that just some mathematical requirement that is independent of the physical properties? Can someone enlighten me, please?

Is this your problem:
"Solve the unidimensional SE for one particle in the the potential field:

$$U(x)=\left\{\begin{array}{c}+\infty,\mbox{for} \ x\in(-\infty,0]\\0,\mbox{for} \ x\in (0,+\infty)\end{array}\right$$

,because you didn't say anything about the potential in the positive semiaxis...

Daniel.

 

[Spinny]

Is this your problem:
"Solve the unidimensional SE for one particle in the the potential field:

$$U(x)=\left\{\begin{array}{c}+\infty,\mbox{for} \ x\in(-\infty,0]\\0,\mbox{for} \ x\in (0,+\infty)\end{array}\right$$

,because you didn't say anything about the potential in the positive semiaxis...

Daniel.

The potential is the harmonic oscillator on the positive semiaxis. The problem is what are the mathematical requirements for the wave function. Let's say you have a function $$\psi(x)$$, then what are the mathematical requirements that function need to meet in order to be a wavefunction for that potential?

 

[dextercioby]

Physical states are described by normalizable wavefunctions...

In your case,on the negative semiaxis the wave function is zero and on the positive semiaxis is a Hermite polynomial.So i'd say this is normalizable.

Then comes the continuity of the wavefunction.Both 0 & Hermite Polynomials are continuous,however,at the point 0,the continuity must be enforced.

The first derivative issue is rather tricky.U may want to consult a book how to deal with infinite potentials & the conditions imposed on the wavefunction.

Daniel.

 

[reilly]

Think about the following:

1. The eigenfunctions for the linear oscillator are strictly even or or odd.

2. For this problem, why should there be any boundary condition on the momentum, the first spatial derivative, at x=0, if two boundary conditions have already been imposed? (Think about a particle wave packet, in the oscillator well, moving toward the x=0 wall. What's going to happen at the wall?)

Regards,
Reilly Atkinson