Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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.

share|improve this answer

There is absolutely physical meaning in it -- the sum is what gives you the final probability of the particle being at a particular location. But, often knowing just the single number is not illustrative enough to get much information from, which is why we look at the eigenfunctions individually.

Quantum mechanics isn't my area, but I'll tie it back to classical mechanics and vibrations. We can find the displacement of a beam in bending as the sum of the modes. That's great if we just want to know the displacement. But often we want to know natural frequencies, modal contributions (how much each individual mode contributes to the energy or the displacement) and so on. We're also often only interested in the lowest mode as that typically drives the solution.

So while we could look at the total response, that doesn't help us any. We might not need to compute all the modes, we might only want the first. Or we might not be able to compute them all due to expense, so we only sum over the ones we want. Or we might want to see at which points there are common nodes across the multiple modes, or we might want to see relative energy contents or...

Lots of things can be learned from modal decomposition. The sum of the modes is the total response, a single function, which is useful in some instances but often doesn't illustrate much of the physics of what's going on in detail. Just the "summary. Sort of like knowing the mean of a function -- sometimes good, but knowing more information is better.

share|improve this answer
    
the separation of variables only works when there is superposition and that definitely has physical meaning, so i am confused as to why you would assert, "There is absolutely physical meaning in it" –  mcodesmart Dec 19 '13 at 17:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.