A question very similar to the one I'm asking about has been asked before on here, but I'm uncertain about the reasoning behind the answer, and really need some clarification, so if anyone could help me that'd be grand.

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This question is from a past exam paper.

What I understood from books that I have read, and what the answer to this (Quick question on sketching wavefunction in well) suggests, is that the wavelength and the amplitude of the wavefunction are smaller when the potential is lower.

I'm fine with the wavelength part, that makes sense to me due to the second derivative of the wavefunction being greater when E - V is greater, but it is the amplitude that I am uncertain about, and in fact that examiner's solution to this question says that the amplitude is greater when the potential is lower.

So the argument that I have seen numerous times is that if E - V is lower in a certain region then the kinetic energy in that region is smaller, so the particle spends more time in that region, leading to a higher probability density, and hence a greater amplitude of the wavefunction.

But is this not essentially a classical argument that breaks down when you have a eigenfunction which doesn't represent a physically realisable state? I mean, for the case of the infinite well there would also then be points where the particle cannot exist, and this clearly doesn't represent a physical situation?

If you attack this problem by first order perturbation theory then the wavefunction does seem to tend towards having a greater amplitude when the potential is lower, rather than the other way round.

So, what is the correct answer to this?


2 Answers 2


As long as the energy is positive, there will be a wavelength - and the wavelength will be shorter when the energy is greater.

As for amplitude - that reflects the probability that the particle will be found in a particular location. There are two ways to think about this, and they give different results.

Method 1: think in terms of statistical mechanics. The lowest energy state would be most likely, with a $e^{-\Delta E/kT}$ type term describing the probability of occupying different levels.

Method 2: think in terms of speed: If the particle is moving along the potential well (let's think of it as being smooth for a moment), then it moves faster in some places and slower in others. If you observe at a random moment in time, you are more likely to find it where it goes more slowly (it spends relatively more time there).

It turns out that, on the QM scale, method 2 is the right way to think about this. You can find some of the mathematical detail in this lecture


The WKB approximation also implies that the amplitude of the "waves" is larger in regions where the potential energy is greater (but still less than the particle's energy.) I'll give a brief overview of the highlights; more detail can be found in many sources, including the textbooks by Liboff and/or Shankar.

We basically start by postulating a wavefunction of the form $$ \psi(x) = A e^{i S(x)/\hbar} $$ for some complex function $S(x)$. If we insert this into the time-independent Schrödinger equation, we obtain $$ - i \hbar \frac{\partial^2 S}{\partial x^2} + \left( \frac{\partial S}{\partial x} \right)^2 = p^2(x), \qquad \qquad (*) $$ where $$ p(x) = \sqrt{ 2 m (E - V(x))}. $$ We now express $S(x)$ as a power series in $\hbar$ (Shankar gives a good explanation of why we might want to do this, but I won't get into it here: $$ S(x) = S_0(x) + \hbar S_1(x) + \mathcal{O}(\hbar^2) $$ To zero order in $\hbar$, the equation (*) becomes $$ \frac{\partial S_0}{\partial x} = \pm p(x) $$ and to first order it is $$ \frac{\partial S_1}{\partial x} \frac{\partial S_0}{\partial x} = \frac{i}{2} \frac{\partial^2 S_0}{\partial x^2}. $$ These two equations can be solved to obtain $$ S_0 = \pm \int^x p(x) \, dx $$ (where the plus-or-minus corresponds to left- or right-going waves), and $$ S_1 = \frac{i}{2} \ln p(x). $$ The solution for right-going waves is then $$ \psi(x) \approx A \exp \left[ i (S_0(x)/\hbar + S_1(x)) \right] = A \exp \left[ \frac{i}{\hbar} \int^x p(x) dx \right] \exp \left[ -\frac{1}{2} \ln p(x) \right] \\ = \frac{A}{\sqrt{p(x)}} \exp \left[ \frac{i}{\hbar} \int^x p(x) dx \right]. $$ Thus, the probability amplitude is larger where $p(x)$ is smaller, and $p(x)$ is in turn smaller where $V(x)$ is larger (but not greater than $E$.)

That said, the whole WKB approximation hinges on being able to neglect higher-order terms in $\hbar$. In particular, the "effective wavelength" of the wave is determined by (effectively) the de Broglie condition: $\lambda(x) = \hbar/p(x)$. It can be shown that the higher-order terms in $\hbar$ will be negligible if $$ \left| \frac{d \lambda}{dx} \right| \ll 1 $$ which is not the case for a potential well with two discrete "steps." I'm not entirely sure how one would salvage the relationship between amplitude and potential energy in this case, or if there might be cases where it doesn't hold.


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