# separation of variables

I'm a math student who's dabbled a little in physics, and one thing I'm a little confused by is separation of variables. Specifically, consider the following simple example: I have a Hamiltonian $H$ which can be written as $H_x + H_y + H_z$ depending only on $x,y$, and $z$ , respectively, and I want to find the eigenfunctions. Now, it is clear that the product of eigenfunctions of $H_x, H_y$, and $H_z$ will be eigenfunctions for $H$. But why can all eigenfunctions be expressed this way?

I suspect this is not even strictly true, but can somebody give a physics-flavored plausibility argument for why we only need to concern ourselves with separable eigenfunctions?

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Actually, it's not true that all eigenfunctions are separable. Consider the 3D isotropic harmonic oscillator, whose Hamiltonian is a sum of three 1D SHO Hamiltonians,

$$H = \frac{p^2}{2m} + \frac{m\omega^2 r^2}{2} = \sum_{i\in\{x,y,z\}}\hbar\omega\biggl(a_i^\dagger a_i + \frac{1}{2}\biggr) = H_x + H_y + H_z$$

You can create eigenfunctions of $H$ by multiplying the eigenfunctions of the individual 1D SHOs. But because there are three identical dimensions you get degeneracies, for example

$$E_{0,0,1} = E_{0,1,0} = E_{1,0,0}$$

Since these eigenvalues are equal, any linear combination of the corresponding eigenfunctions will still be an eigenfunction of $H$. But an arbitrary combination like, say, $\frac{1}{\sqrt{3}}(\psi_{0,0,1} + \psi_{0,1,0} - \psi_{1,0,0})$, can't be factored into the form $X(x)Y(y)Z(z)$.

What you can say is that all eigenfunctions of $H$ with a given eigenvalue $E_{\{n\}}$ constitute a subspace of the Hilbert space, and for each such subspace, it is possible to choose a complete basis of functions which can be factorized into $X(x)Y(y)Z(z)$. If you're looking for a proof of that fact, I couldn't give it to you offhand, since it's something physicists tend to take for granted (when necessary), but I wouldn't be surprised if you could find it discussed on the math site. It'd definitely be in books on linear algebra or probably PDE analysis.

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The statement follows from the completeness of the eigenvectors of the self-adjoint Schrodinger operators (this is the spectral theorem--- self adjoint Schrodinger operators with sufficiently smooth potentials bounded below have a complete basis). If you know that the 1d problems individually have a complete basis of eigenvectors, it is easy to prove that the collection of products of eigenvectors span the whole of the space.

The reason is that you can write a three dimensional delta function as a product of the expansion for 1d delta functions.

$$\delta(x-a) = \sum_n A_n(a) \psi^1_n(x)$$ $$\delta(y-b) = \sum_n B_n(b) \psi^2_n(y)$$ $$\delta(z-c) = \sum_n C_n(c) \psi^3_n(z)$$

where the $\psi$ are the solutions of the one-d problem. You now multiply these together to get an expansion for a delta function, and this gives an expansion for any function by superposition. In order to turn this into rigorous mathematics, you only need to give a closeness bound for how good a delta function you get, and check that the product is close to a 3d delta function, which follows from the Fubini theorem for multiple integration.

$$\delta^3(r-q) = \sum_{n,m,p} A_n(q_x) B_m(q_y) C_p(q_z) \psi^1_n(r_x)\psi^2_m(r_y)\psi^3_p(r_z)$$

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