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In considering a delta potential barrier in an infinite well, I can just enforce continuity at the potential barrier-it doesn't have to go to zero. Why then does it need to go to zero at the walls of the infinite well? These two cases seem to be very similar to be, I even feel like the well wall is equivalent to a summation of delta functions... Where is my logic faulty?

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  • $\begingroup$ Indeed any wall admits a generic boundary condition $\Psi(x)= A \Psi'(x)$: the current $\Psi^* \Psi' - \Psi \Psi'^*$ is proportional to $A-A^*$ and so it is null for any real $A$ $\endgroup$ – arivero Aug 23 '15 at 20:07
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One may view both (i) the infinite wall

$$\tag{1} V(x)~=~\left\{\begin{array}{ccc}\infty & \text{for} & x>0, \\ 0 & \text{for} & x\leq 0, \end{array} \right. $$

and (ii) the delta function potential

$$\tag{2} V(x)~=~A\delta(x),$$

as an appropriate limit of a finite barrier wall

$\tag{3} V(x) ~=~ V_0 1_{[0,a]}(x)=\left\{\begin{array}{cc}V_0 & \text{for } 0<x<a \\ 0 & \text{otherwise} & \end{array} \right. $

by letting (i) $a=\infty$ and (ii) $a=A/V_0$, respectively, and taking the limit $V_0\to \infty$.

For energies $E<V_0$ less than the barrier height (3), the second-order ODE TISE yields exponentially decaying and exponentially growing solutions inside the barrier (3).

(i) On one hand, for an infinitely thick wall $a=\infty$, the exponentially growing solution is physically unacceptable, and hence discarded, such that $\psi(\infty)=0$. Thus we learn that the wave function is exponentially decaying when entering the potential wall at $x=0$. The inverse characteristic penetration depth is proportional to $\sqrt{V_0-E}$. Thus in the limit $V_0\to \infty$, the characteristic penetration depth is zero, i.e. we get the sought-for boundary condition $\psi(x=0)=0$, if we assume that the wave function is continuous$^1$.

(ii) On the other hand, for a wall of finite thickness $a<\infty$, the exponentially growing solution may also be relevant, and $\psi(x=0)=0$ does not have to be satisfied.

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$^1$ Continuity of the wave function $\psi$ can be justified for a wide class of potentials $V$ via a bootstrap argument on the TISE in integral form. See also e.g. my Phys.SE answer here.

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  • $\begingroup$ Certainly it does not need to be zero inside one of them. But taking infinitely many of them, wouldn't this be an infinite well? I guess that doesn't matter. The width becomes near trivial for one, or two... So then what DOES happen with infinitely many? $\endgroup$ – user24082 Oct 22 '13 at 2:54
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The infinite potential well is an idealization in which we imagine that, as Kyle Kanos has already said in his answer, we are completely confining the particle to an interval $(-a,a)$ on the real line. Since the squared modulus $|\psi(x)|^2$ of the wavefunction at a point $x$ on the real axis represents the probability (density) for finding the particle at a given position $x$, a particle that is confined to the region $(-a,a)$ must have zero probability of being found in that region, so we impose the boundary condition $|\psi(x)|^2 = 0$ for all $x\geq a$ and for all $x\leq-a$ which implies that $\psi(x) = 0$ for all these values of $x$.

So far, this is essentially just a restatement of what Kyle has already said, but let's dig a bit further to understand what's going on physically.

You might think to yourself "ok, given the idealization described above of confinement of the particle, the boundary conditions make intuitive sense, but in the real world, there is no such thing as an infinite potential. We can, however, produce very large potentials in the real world. I would be more convinced of the infinite square well boundary conditions if we could show that given a finite potential well of strength $V_0$, the behavior of the wavefunction in the regions of high potential get's closer and closer to that of its behavior for the infinite potential well as $V_0$ is taken larger and larger.

In fact, we can show that this is the case for the energy eigenvectors of the finite potential well.

Consider a bound state of a particle in a finite potential well of strength $V_0$, namely \begin{align} V(x) = \left\{\begin{array}{cl} 0 & ,|x|<a \\ V_0 & ,|x|\geq a \end{array}\right. \end{align} A bound state occurs when the energy of the particle is less than the maximum potential energy, namely when $E<V_0$, then for the regions $|x|\geq a$ the energy eigenvalue equation (aka time-independent Schrodinger equation) gives \begin{align} \psi''(x) = \kappa^2\psi(x), \qquad \kappa^2 = -\frac{2m(E-V_0)}{\hbar^2} \end{align} The general solution is either a growing or decaying exponential $\psi(x) = A e^{\pm \kappa x}$. For the wavefunction to be normalizable, this tells us that the general solution outside of the well is \begin{align} \psi(x) = \left\{\begin{array}{cl} A_-e^{\kappa x} & ,x\leq -a \\ A_+ e^{-\kappa x} & ,x\geq a \end{array}\right. \end{align} Now here's the cool thing. For a fixed $E$, if we take $V_0\to\infty$, then $\kappa\to\infty$, and this means that the wavefunction decays more and more quickly outside of the well as we increase the potential. In fact, for a given $E$ and $x$, we have $\psi(x)\to 0$ as $V_0\to \infty$ for all $|x|\geq a$.

From this perspective, we can view the infinite square well boundary conditions as being obtained from a limit of the finite square well in which the well is "infinitely deep."

Notice, also, that from this point of view, a pair of delta functions is not equivalent to the infinite square well because that would correspond to an infinite limit of a pair of finite potential spikes at the positions $\pm a$, and that is simply not the physical situation we are attempting to model when we discuss the potential well.

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  • $\begingroup$ What about infinitely many deltas? $\endgroup$ – user24082 Oct 21 '13 at 2:57
  • $\begingroup$ @Anonymous I'm not sure how that would help. The important point is that the infinite square well is a well-motivated idealization. What matters is that there are physically sensible reasons for considering it. In particular, there is no need to consider delta functions, especially an infinite number of them. $\endgroup$ – joshphysics Oct 21 '13 at 6:31
  • $\begingroup$ I still fail to understand why a dirac delta doesn't pull the wave to zero but the wall of the well does. If it's because it's infinitely thin, wouldn't a sum of deltas have a nontrivial width? $\endgroup$ – user24082 Oct 21 '13 at 16:04
  • $\begingroup$ @Anonymous Each of your deltas in your sum will just introduce a discontinuity in the first derivative of the wavefunction at the location of the delta function. I don't think there is any way (that isn't completely pathological) to imitate the exponential decay of a wavefunction using any number of them. $\endgroup$ – joshphysics Oct 21 '13 at 18:03
  • $\begingroup$ Wouldn't the sum of delta functions be, effectively, the Heaviside step function? $\endgroup$ – Kyle Kanos Oct 21 '13 at 19:17
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When you consider a potential barrier, you will see that the wave function of a particle with definite energy will decay exponentially at a rate that depends on the difference of the potential and the energy. Precisely, the exponent is proportional to $\sqrt{V - E}$.

In the limit, for $V\to\infty$, it will decay to 0 over a width 0.

A delta function doesn't exist as a function, but it is a limit of functions (in a suitable topology). For this problem it is easiest to consider block functions: functions of width w and height 1/w. The delta distribution is the limit as $w\to 0$. You can not first take the limit of the height going to infinity, followed by the width going to 0, something that you implicitly are doing.

Now for a fixed $w$ you get a value for the wavefunction on the other side of the barrier depending on the value on this side. When the height goes up, the decay will be faster, but it will also fall over a shorter interval, and in fact the decay will be less because the decay exponent contains a square root. In the limit there is no decay in height at all, in particular you do have continuity, but no reason to end at 0.

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You are confining the particle into a region of $|x|< a$ with an infinite potential that extends infinitely far for $|x|\geq a$: enter image description here Image source

Since we are confining the particle to a particular region (by applying a potential outside this region), you will never find the particle outside this region, so the wave function, $\psi$, must be zero starting at $|x|=a$.

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Why then does it need to go to zero at the walls of the infinite well?

Because the proper way to find $\psi$ is to solve Schr. equation for finite potential well first and find how $\psi$ depends on the parameters of the potential. Then try to make the limit to the infinite potential well and look what happens to the $\psi$ function.

One cannot solve Schr. equation for something like "infinite potential" directly, because "infinite potential" is not a valid function.

Due to the requirement of normalizability of $\psi$, in the case of finite well, the $\psi$ function decays to zero for large $x$, and the limiting procedure leads to continuous $\psi$ even in the limit and to boundary condition where $\psi$ is zero outside the well

I even feel like the well wall is equivalent to a summation of delta functions...

True, the constant potential $V_0$ outside the finite well can be written as

$$ V(x) = V_0 \int_{(-\infty,-a)+(a,\infty)} \delta(x-x_0)dx_0. $$

However, in general the solution of the Schr. equation, $\psi$, is not given by linear operator acting on the potential $V(x)$ figuring in the Schr. equation. There is no reason to expect that the $\psi$ function for $V(x)$ will be sum of functions $\psi_{x_0}$ that are solutions to Schr. equation with delta potential $V_0 \delta(x-x_0)$.

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