quantum mechanics - Separation of variables in various PDEs, physical meaning

The method of separation of variables produces an undetermined separation constant and a family of solutions indexed by the values of this constant.

For instance, in the case of an infinitely long rod along the positive x axis (with suitable boundary conditions, like a fixed temperature at the beginning, and a given temp. distribution at initial time), the separation constant, $-k^2$, can vary continuously, and the solutions are of the form

$$T_k(x,t)=C(k)\cdot \sin{kx} \cdot e^{-k^2\alpha t}$$

One is then supposed to integrate over $k$ to obtain the answer. In the case of a finitely long rod, the values of the constant are quantized and can be summed, rather than integrated.

But when we're dealing with Schroinger's equation for a particle in a box, we are only interested in the eigenfunctions corresponding to the values of the separation constant, which we associate with energy. But the general solution would be a superposition of these wavefunctions. Suppose we're dealing with a 3D box of length $L$, then the eigenfunctions are:

$$\psi_{k,\ell,m}(x,y,z)=\sqrt{\frac{8}{L^3}} \sin{\frac{\pi}{L}kx}\cdot\sin{\frac{\pi}{L}\ell y}\cdot\sin{\frac{\pi}{L}mz}, \quad k,\ell,m\in\mathbb{N}$$

My question: is there any physical significance to the function: $$\psi=\sum_k\sum_\ell\sum_m c_{k,\ell,m}\psi_{k,\ell,m},$$ where $c_{k,\ell,m}$ are constant coefficients. Clearly this function is a solution to the Schrodinger equation, by the superposition principle. But does this function describe anything at all? Or is it that, unlike in the case of heat transfer, only the eigenfunctions matter?

Answer

You've forgotten one crucial thing when you've written your superposition: the separate $\psi_{k,\ell,m}(x,y,z)$ are eigenfunctions of the Hamiltonian with different eigenvalues. The superposition will no longer be an eigenstate because of this. In fact, by taking an appropriate superposition, you would be able to get any function you like (in your case, with the boundary conditions of vanishing at the edge of the box).

But this is exactly the same as your first example, once you've remembered where the eigenvalue (energy) came from: it is when you do separation of variables on the time-dependent Schr\:odinger equation $$H\psi=i \hbar \frac{\partial \psi}{\partial t}.$$ If you include the appropriate time-dependent exponential, your final equation will tell you how the wavefunction evolves, given the initial state: $$\psi(x,y,z,t)=\sum_k\sum_\ell\sum_m c_{k,\ell,m}\psi_{k,\ell,m}(x,y,z)e^{\frac {-i E t}{\hbar}}.$$ In particular, at $t=0$ you get back your original equation, which we now know gives a completely arbitrary function of $x,y,z$. But that is just as well: this arbitrary function now appears as the initial condition, exactly as in your heat equation example.