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I have this potential

$$V(x) = \left\{ \begin{array}{ll} \infty & \mbox{if } x < -a \\ \frac{V_o}{a}x & \mbox{if } -a \leq x \leq a \\ V_o & \mbox{if } x \geq a \ \end{array} \right.$$

And I want to know, qualitatively, how the wave function would look like.

So, the particle cannot live at the left of the "wall" at $x=-a$, so the wave fucntion there is $0$. To the left of the ramp (i.e., for $x>a$), the potential is constant, so the particle will behave like a free one. Namely, the wave function will be constant in that zone.

But what happens in the middle? I'm not interested in the mathematical approach for this, I've already looked it up and it seems to be related with Airy functions or something like that. However, I want to understand what would happen, not just do the math. I think that the wave function in this zone will depend on the value of $E$ the particle has.

This is what I thought: for low values of energy, the particle will have a small probability of getting trough the ramp (tunneling?); on the other hand, for high values of energy ($E>V_o$ I suppose), the probability of the particle living in the zone with the constant potential would be higher, as the "box" in the middle wouldn't be able to contain it.

My guess is that if $E<V_o$ the wave function would look like a sine wave atenuated along the $x$-axis until it reaches $x=a$, where it would become constant. If $E>V_o$, it would be the same but with the sine wave increasing its amplitude this time.

Is this reasoning correct? Or any other form of thinking about it?

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  • $\begingroup$ A point for pedants like me: You're not interested in how a wavefunction would look, you're interested in how an energy eigenstate would look. $\endgroup$
    – user12029
    Commented Jul 18, 2016 at 23:33
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    $\begingroup$ (Pedants)^2 would say that you are interested in the wavefunction of the energy eigenstates. :) $\endgroup$
    – garyp
    Commented Jul 18, 2016 at 23:34
  • $\begingroup$ You're a little backwards in your reasoning, but you're on the right track. For $x>a$ you say the wave function will increase in amplitude. Think about normalizing that function. Once you get on the right track, you will notice that your picture for $E<v_0$ is not correct. $\endgroup$
    – garyp
    Commented Jul 18, 2016 at 23:39
  • $\begingroup$ I'd say the obvious first step is to solve the equation using the Airy functions, plot it, and see whether your intuition was right. Did you try that? $\endgroup$
    – David Z
    Commented Jul 18, 2016 at 23:40
  • $\begingroup$ @garyp I meant that for $x>a$ the function will be constant; the increasing-amplitude thing would happen just between $-a$ and $a$. And yes, it would be non-normalizable there, but I don't see how that could help :/ $\endgroup$
    – Tendero
    Commented Jul 18, 2016 at 23:43

1 Answer 1

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Qualitatively, the wave functions of the bound states in a triangular potential well like the one you described, look like this:

Triangular well

For $x<-a$, $\psi=0$ because of the infinite potential in that region.

Where the wave function crosses the potential line, quantum tunnelling occurs and $\psi \to 0$.

For particle energies above $V_0$, no bound states can exist (these so called scattered states are not shown).

The triangular potential well can be seen as a crude approximation of the Morse potential.

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  • $\begingroup$ Please do not post complete solutions! $\endgroup$
    – garyp
    Commented Jul 18, 2016 at 23:55
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    $\begingroup$ @garyp: this is hardly a complete solution. That would take some seriously hard calculations. I have the solutions somewhere but won't post them. $\endgroup$
    – Gert
    Commented Jul 19, 2016 at 0:00
  • $\begingroup$ It's nearly a complete answer to his question. Remember he or she said " I'm not interested in the mathematical approach for this" I will grant that you didn't include the $E>V_0$ case, so that does leave something for the OP to think about. $\endgroup$
    – garyp
    Commented Jul 19, 2016 at 0:07
  • $\begingroup$ In that graph, $\psi$ would be the probability function, right? Not the eigenstate itself, but the product of it with its conjugate $\endgroup$
    – Tendero
    Commented Jul 19, 2016 at 0:07
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    $\begingroup$ I don't see why it is constant for $x>a$ then Nobody says it is. $\psi$ goes to zero for $x \to +\infty$. In that region, $\psi$ is a decaying exponential function, something like $\psi=ae^{-\lambda x}$. $\endgroup$
    – Gert
    Commented Jul 19, 2016 at 0:37

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