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Suppose we have a typical finite square well where $\lim_{x \to \pm\infty} V(x)=0$ and $V(x)=-V_0$ for all $x\in[-a,a]$ where $V_0>0$. The finite square well admits both bounds state solutions (with $E<0$) and scattering state solutions ($E>0$). For a wide and deep well the bound state solutions have energies $$E_n \approx \frac{n^2\pi^2\hbar^2}{2m(2a)^2}-V_0 \,\,\,\,\,\,\,\,\,n=1,2,3,... \tag{1}$$ Now for bound states, we require that the energies ($E_n$) be negative. This means that for valid bound states, $n$ must be below some maximal integer $N$ and hence we will only have a finite amount of bound state solutions. Moving over to the scattering state solutions (we now take $E>0$), we are able to determine the transmission coefficient for the well which takes the form $$T=\frac{1}{1+\frac{V_0^2}{4E(E+V_0)}\sin^2(\frac{2a}{\hbar}\sqrt{2m(E+V_0)})} \tag{2}$$ This equation clearly implies that the well becomes perfectly transparent ($T=1$) when $$\frac{2a}{\hbar}\sqrt{2m(E+V_0)})=n\pi \,\,\,\Rightarrow E_n=\frac{n^2\pi^2\hbar^2}{2m(2a)^2}-V_0 \tag{3}$$ Now my first problem is that for a general $n\in Z$, the above expression will not result in positive $E_n$ values. That is, for all $n\in Z$ below the previously mentioned maximal integer $N$, equation (3) will yield negative $E_n$ values. These are not scattering states. So am I correct in thinking that the above formula for $E_n$ such that $T=1$ only applies for those integers ($n>N$) whereby the resultant energies are positive?

My second question is regarding why this perfect transmission actually occurs at these energies (which are exactly the same form as the energies for the bound state energies). My initial thinking is that it must be some type of resonance phenomenon however I can't seem to reconcile this with the fact that photons will be much more likely to be absorbed by atoms when their energies match the bound state energies of the atomic electrons and hence the atoms become opaque in this case. Not transparent. But in the finite square well we have the opposite behaviour. The well becomes transparent when the incoming particles energy matches those of the bound states. Why is this the case? Why are photons absorbed when their energy equals the difference between two bound state energies of an atomic electron while a particle moves straight through a finite potential well if the if its energy equals a bound state energy plus some constant? Should maximal reflection occur here? or maximal absorption?

Any help on this would be most appreciated!

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  1. OP is observing that the infinite tower of reflectionless scattering energies $$\begin{align}E_n~=~& \frac{1}{2m}\left(\frac{n\pi\hbar}{2a}\right)^2-V_0~>~0, \cr n~\in~&\{N\!+\!1, N\!+\!2, \ldots \},\end{align}\tag{2.153}$$ of the finite potential well is the natural continuation of the wide & deep well approximation $$\begin{align} E_n~\approx~& \frac{1}{2m}\left(\frac{n\pi\hbar}{2a}\right)^2-V_0~<~0, \cr n~\in~&\{1,2,\ldots, N\},\end{align}\tag{2.139}$$ to the $$ N ~:=~\lceil \sqrt{2mV_0}\frac{2a}{\pi\hbar}\rceil~\in~\mathbb{N}. \tag{wiki} $$ bound state energies. OP is asking: Why?

  2. More generally, away from the wide & deep well approximation (2.139) the bound states are given by 2 transcendental equations$^1$ $$\kappa~=~\ell \tan(\ell a) \quad\vee\quad \kappa~=~-\ell \cot(\ell a), \tag{2.136}$$ which can be written as a single equation $$\begin{align} 0~=~&\left(\ell \tan(\ell a)-\kappa\right)\left(\ell \cot(\ell a)+\kappa\right)\cr ~=~&\ell^2-\kappa^2 - 2\kappa\ell \cot (2\ell a). \end{align}\tag{2.136'} $$ So OP's observation is not an exact statement for the general finite potential well.

  3. In fact OP's observation can be violated more dramatically: There exist so-called reflectionless potentials, which are transparent $T\equiv 1$ at all scattering energies $E>0$, but have only a finite number of bound states, cf. e.g. this Phys.SE post. So OP's observation does not hold for a general potential.

  4. Remarkably, there is another intimate relation between bound states & scattering states that does hold in general: If we analytically continue the real $k$ into the complex plane $\mathbb{C}$, then the scattering reflection & transmission coefficients will have poles at positions $k=i\kappa$ along the imaginary axis in the complex $k$-plane whenever $\kappa>0$ corresponds to one of the discrete bound states, cf. e.g. Ref. 2.

    Let us for fun check this relationship for the finite potential well. There the transmission coefficient is related to $$ \frac{F}{A}~=~\frac{e^{-2ika}}{\cos (2\ell a)- i\frac{2k\ell}{\ell^2+k^2} \sin (2\ell a)},\tag{2.150}$$ which precisely has poles along the upper imaginary $k$-axis at positions given by the bound state solutions (2.136').

References:

  1. D. Griffiths, Intro to QM, 1995; section 2.6.

  2. P.G. Drazin & R.S. Johnson, Solitons: An Introduction, 2nd edition, 1989; section 3.3.

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$^1$ See Ref. 1 for notation and details: $\ell>0$ is the angular wavenumber inside the well. $k>0$ (resp. $\kappa>0$) is the oscillatory (resp. exponentially decaying) wavenumber outside the well in the case $E>0$ (resp. $E<0$). See also e.g. this related Phys.SE post.

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The OP makes multiple correct observations of properties of the finite square well. I will try to provide some intuition and give statements on which of these properties are generic.

So am I correct in thinking that the above formula for 𝐸𝑛 such that 𝑇=1 only applies for those integers (𝑛>𝑁) whereby the resultant energies are positive?

First some clarification on this question. What are scattering and bound states in the context of the 1D-Schrödinger equation? Roughly speaking, scattering states are solutions which are wave-like outside the potential region. Bound states are localized, such that they correspond to solutions which decay to zero far from the potential region.

The transmission coefficient is then a relation between the asymtotic waves on the left/right of the potential. I.e. if we have something like $e^{ikx} + re^{-ikx}$ on the left, we have $te^{ikx}$ on the right. $T=|t|^2$ is then the transmission coefficient. Note that $k$ is a function of energy, indicating the wave dispersion. For the Schrödinger equation, we have $E \propto k^2$.

We see that scattering states can then only have positive $E$, since imaginary $k$ does not correspond to a wave-like solution, as the OP correctly observed. This means that the transmission coefficient is, strictly speaking, only defined for positive $E$, which resolves the first question.

Why is this the case? Why are photons absorbed when their energy equals the difference between two bound state energies of an atomic electron while a particle moves straight through a finite potential well if the if its energy equals a bound state energy plus some constant? Should maximal reflection occur here? or maximal absorption?

Note that nothing is absorbed here. The finite square well is a unitary problem that does not feature absorption at all. The trade-off is between transmission and reflection. The OP's concern that when light is resonant with an atomic transition, we would usually expect absorption, is therefore largely a different problem.

That said, I would like to give some intuition on what is happening here.

Why do the transmission peaks roughly coincide with the continuation of the bound states above the energy threshold?

Consider the following scenario: We start with an infinite square well. There, we have the bound state formula given by the OP, with all $n$ corresponding to bound states. Due to the infinite potential, there is no energy cutoff.

Now consider what happens if we lower the potential barrier to slowly transition to the finite square well. First, the states will obtain a small probability to be inside the energetically forbidden region. When that is the case, they are still bound states. However, eventually they will lie above the energy threshold and oscillate outside. When this happens, they also obtain an energetic width, since a continuum of oscillating solutions is allowed around them. This is what is called a resonance, a scattering state that is still tightly confined.

On the level of the transmission coefficient, this corresponds to a pole appearing at complex energies, where the imaginary part corresponds to the width of the resonance peak. From this background, we can see that the peak at real energies will roughly correspond to the original location of the bound state in the infinite well, where the pole lies on the real axis. Note that this correspondence is, generally speaking, not necessarily exact, since the pole can be shifted from the peak location.

Why maximum transmission at resonance?

The finite potential well features a potential jump on its left. Generally, this jump will cause a strong reflection. However, at the energy of a resonance state, the external wave can efficiently transfer into one of the strongly confined states and subsequently has a high probability of going through to the other side. Note that if the peak goes all the way up to one, we have a sitution that is referred to as critical coupling in resonance theory. This is also not a generic feature, but happens rather frequently in systems without absorption and narrow resonances.

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    $\begingroup$ Make sure to check out the answer by @Qmechanic, which gives a rigorous mathematical description for some of the points that I only explained intuitively. $\endgroup$ May 26, 2021 at 10:16

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