# Uniqueness Theorem and the 1D Infinite Square Well

Consider the 1D infinite square well problem: $$\frac{d^2\psi (x)}{dx^2} = -k^2\psi (x)\tag{1}$$ along with the boundary conditions $$\psi (0) = \psi (L) = 0$$. This seems to be a well posed problem with enough information to have a unique solution but the solution $$\psi (x) = A\sin(kx)$$ has an ambiguity in $$A$$. Only with the normalization condition is the problem uniquely solved. On top of all this, I could not have specified any value of $$k$$ at the beginning as that would be an over-posed problem with potentially no solution (Eg: in retrospect, if I had specified a value of $$k$$ corresponding to an energy that is not allowed, there would not exist any solution).

This is in sharp contrast to what happens in the case of a simple harmonic oscillator from classical mechanics: $$\frac{d^2\psi (t)}{dt^2} = -\omega^2\psi (t)\tag{2}$$ along with the initial conditions $$\psi (t = 0) = 0, \psi^{\prime} (t = 0) = 1$$. This has a unique solution $$\psi (t) = \frac{1}{\omega}\sin(\omega t)$$. Not only is some normalization condition not required to fix the value of the amplitude but this solution is valid for any value of $$\omega$$ that could have been specified at the start of the problem.

I can guess that this difference in behavior comes from the difference in boundary conditions but I want to know what exactly it is that makes the first one different? It seems like you need more information to pin down a unique solution for the first problem. Does the uniqueness theorem not hold due to some technicality?

• "On top of all this, I could not have specified any value of k at the beginning as that would be an over-posed problem with potentially no solution (Eg: in retrospect, if I had specified a value of k corresponding to an energy that is not allowed, there would not exist any solution)" But that's not how it works: $k^2=\frac{2mE}{\hbar^2}$ is not arbitrary. – Gert Apr 19 at 22:00
• @Gert I know that the quantization falls out of the boundary conditions but that's my point. To draw a contrast between the IVP and the BVP. The IVP is agnostic about the parameter whereas the BVP isn't. I could specify the value of the parameter for the IVP and it won't care and the solution still continues to exist whereas specifying the parameter for the BVP makes it over-posed. – PhCguy Apr 19 at 22:24
• Sorry but I don’t get your point. The boundary conditions on the infinite well are not in $t$ or its derivative but in $x$: $\psi(0)=\psi(L)=0$, which of course forces restrictions on $k=\sqrt{2m E/\hbar}$, as per the usual standing wave conditions. Because the boundary conditions force a function to be $0$ at specified points, of course multiples of such a function will also be $0$ at said points. Moreover, normalization does not resolve the ambiguity as any $e^{i\varphi}\psi(x)$ is also a valid solution. – ZeroTheHero Apr 19 at 22:41
• @ZeroTheHero Good point on the phase of the amplitude $A$. That makes the problem worse. – PhCguy Apr 19 at 22:54
• There is not "the uniqueness theorem," there are many even in differential calculus alone. Which exact theorem are you referring to? – kkm Apr 19 at 23:13

The 1D square well is quite different from the classic harmonic oscillator. You need to remember that the wave function in the square well is actually a function of $$x$$ and $$t$$. While it is normal to compute the amplitude from normalization, we could compute it from specifying initial conditions.
If you want to compare it to a classical system, the 1D heat conduction equation is closer. For instance, finding the temperature in a 1D bar as a function of $$x$$ and $$t$$ while specifying the temperature at the end points as well as some initial condition. Separating variables in the heat conduction equation will give you the same form in the $$x$$ equation as you get in the square well problem. You will similarly get a discreet set of eigenvalues.