11
$\begingroup$

Most things like to occupy regions of lower potential. So the probability amplitude should be higher in a region of lower potential. I denote the potential by V.

However, we also know that the kinetic energy of a particle is given by E-V - the bigger this E-V difference (the lower the V basically), the bigger the kinetic energy, thus the bigger the velocity of that particle. By having a bigger velocity though, this means that less time is spent by the particle in the region, thus a lower wavefunction amplitude.

How do both of these statements reconcile with one another? By looking at the diagram below, we see that we have a decreasing amplitude as we go across to the left. But surely the particle will want to be in a region of lower potential?

enter image description here

$\endgroup$
3
  • $\begingroup$ Are you sure that diagram is accurate $\endgroup$
    – AfterShave
    May 26 at 21:28
  • $\begingroup$ Positively positive my friend $\endgroup$
    – jambajuice
    May 26 at 21:29
  • 6
    $\begingroup$ "Most things like to occupy regions of lower potential." You need to be careful about statements like this, because they usually only apply to thermodynamic systems, or, at least, systems where there is some sort of dissipation. The only way a system can "settle" into the state of lowest energy is if it can lose energy. However, especially in a first QM course, we rarely consider dissipative systems and instead consider systems where the energy is conserved. Then, the system can't be at the bottom of the well, because that's not the energy the system has. $\endgroup$
    – march
    May 27 at 15:43

3 Answers 3

17
$\begingroup$

Even classically, particles with a fixed total energy spend more time near the turning points since this is where the motion is the slowest. The probably of finding the particle in a small region near the bottom of the well is NOT the largest: it is in fact the smallest.

Thus your statement that the particle wants occupy a region of minimum potential energy refers to another situation, where you want to minimize the total energy of the system. The size of the region, i.e. the distance between the turning points, gets smaller as you decrease the total energy and the turning points get closer to the minimum.

Once it does occupy this region (bounded by the two turning points, for a given total energy), it does not occupy every part of this region with equal probability.

$\endgroup$
4
  • $\begingroup$ So just to also get this clear, if I were to have a particle in an INFINITE potential well between $x=0$ and $x=a$, and slightly perturb the inside, so that for $x>a/2$, we have a potential of $V_{0}$ - I expect the wavefunction to have a higher amplitude on the right? $\endgroup$
    – jambajuice
    May 27 at 8:35
  • $\begingroup$ If you look at this link : theory.physics.manchester.ac.uk/~judith/AQMI/… , seems to be saying the opposite. $\endgroup$
    – jambajuice
    May 27 at 8:36
  • 1
    $\begingroup$ The infinite well is a strange animal because the solution $\psi$ must be $0$ at the ends, and I would not trust a qualitative graphical method in this case. In cases other than the infinite well, $\psi(x)$ can extend smoothly and lazily in the classically forbidden region rather than be forced to $0$. In fact, if you solve (numerically) the finite well with a bump, I remember the probability would be larger in the shallower region (I have not done this in a while but that’s what I remember). $\endgroup$ May 27 at 10:41
  • $\begingroup$ Some of the difficulties with the momentum and kinetic energy of the infinite well are discussed here physics.stackexchange.com/q/362305/36194 or in this well-known paper arxiv.org/abs/quant-ph/0103153 (arXiv version). $\endgroup$ May 27 at 10:44
12
$\begingroup$

Imagine a perfectly elastic ball dropped vertically onto a flat surface. The ball heads for the point of lowest potential, ie the ground, but because of conservation of energy it bounces back to its initial position and continues to bounce up and down. It spends most of its time near the top of its bounce, because it is ravelling more slowly there, so the probability of finding it at a given instant is highest at the top of the bounce and lowest at the bottom, even though the bottom is the most attractive place to be in terms of PE.

$\endgroup$
3
$\begingroup$

You are probably looking at an energy eigenfunction, a wavefunction which has a definite energy. The statement "$E-V$ is the kinetic energy" does not apply to a single position but to the state as a whole.

This might be a confusing statement so let's compare this to the classical case of a similar well given by $$U(x)=\cases{\infty&$x\leq0$\\\frac{20}{50}x&$0<x\leq 50$\\20& $x<50$}$$ See picture below. I used large values for the width/depth of this well because for the quantum version some effects are shown more clearly.

Imagine you release a particle in this well and you forget about it. What is the probability (density) to find the particle at a position $x$? Well, it is proportional to the time it has spent in the region $[x,x+dx)$ so $$p(x)\propto dt=\frac{dt}{dx}dx=\frac{dx}{v(x)}$$ For a better explanation check out the first link. Like you mentioned, you can easily calculate $v(x)$ using $$v(x)=\sqrt{2(E-U(x))/m}$$. This is shown here:

enter image description here

As you can see, the probability density has a sharp peak (vertical asymptote) near the turning point, so it should be of no surprise that the quantum slanted well shows higher probability near the turning point.

Let's now compare this to the quantum well.

enter image description here

As you can see the wave functions are much better behaved compare to the classical probability function, very nice! Notice that only the ground state "minimizes" the energy. What is shown here are some of the eigenfunction which are all the possible states with definite energy. Since this is just a list there is no reason one of these states should minimize the energy. The energy eigenfunctions are special because they are the only wavefunctions whose probability amplitude is stationary in time.

A state you would find in nature would typically be many of these eigenfunctions added on top of each other, a "superposition" of different eigenfunctions. The probability density of such a state is not stationary and can slosh around in the well.

$\endgroup$
3
  • $\begingroup$ "The ground state minimises" the energy is just flat out wrong. You only see a pronounced bump because the energy is so low, that for the most part (x>5), the wavefunction is decaying. While there is a pronounced bump, One can see that it's shifted slightly to the right, towards the region of higher potential. If you can offer any other explanation as to why this ground function is so special, I'd be grateful but I can't see a reason. $\endgroup$
    – jambajuice
    May 27 at 21:12
  • $\begingroup$ @jambajuice Well, the ground state has the minimal amount of energy by definition and for that reason it is often considered special. It is not possible to have less energy than the ground state. It sounds like we are talking about two different definitions of minimizing the energy, could you clarify what you mean exactly? $\endgroup$ May 27 at 22:14
  • 1
    $\begingroup$ As in yes, the ground state has the least energy, but your answer was sort of implying that the ground state had this special characteristic of "minimising" the energy as I described in my question, when in reality it doesn't. Maybe we're talking about the same thing, but my understanding is that the ground state follows the same rules (which is why we see the peak shifted slightly to the right). $\endgroup$
    – jambajuice
    May 28 at 10:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.