# Can a normalizable function *always* be decompose into the discrete Hydrogen spectrum?

This question has been bothering me for a while now: can one reconstruct an arbitrary (normalizable) function $\phi(\mathbf r)$ in $\mathbb R^3$, with only the (discrete) set of Hydrogen wavefunctions (or, e.g., eigenfunctions of the 3D Harmonic oscillator), or must one include the free states as well? There are many sources (Griffiths, others) that suggest these are a 'complete' basis, from which we can construct any non-pathological function.

Yet, if they are complete, how can they be orthogonal to free states? I'll note there are very similar questions asked elsewhere on Physics SE, here and here: one is asking if continuum states must be included in perturbative expansion (though it's almost always omitted!), another suggesting that free states are an 'orthogonal complement' (somewhere in the comments), but there does not appear to be consensus. A clarification would be appreciated.

• "Yet, if they are complete, how can they be orthogonal to free states?" because free states are pathological. – AccidentalFourierTransform Feb 4 '16 at 21:52
• Can you not construct free states that are normalizable wavepackets? – anon01 Feb 4 '16 at 21:53
• I would encourage you to link to the examples you cite (mostly because this question would appear as Linked there). Good question. – Emilio Pisanty Feb 4 '16 at 22:07
• Closely related: physics.stackexchange.com/q/44839/50583; Also of use because it describes how to get normalizable scattering states: physics.stackexchange.com/q/90101/50583 – ACuriousMind Feb 4 '16 at 22:07
• @EmilioPisanty yes, I'm trying to find them. ACuriousMind already found one, I'll add a link to the question. – anon01 Feb 4 '16 at 22:10

## 2 Answers

In answer to the title question: no, you can't always decompose an $L_2$ function in terms of only the bound spectrum of hydrogen.

This is because there are orthogonal functions to all bound states, which naturally represent the free states of the electron. The quickest examples are of course the Coulomb-wave eigenfunctions $|\chi_{E,l,m}⟩$ of the continuous spectrum of hydrogen, and these are not normalizable, but you can take combinations of the form $$|\psi⟩=\sum_{l,m}\int f_{lm}(E)|\chi_{E,l,m}⟩\mathrm dE$$ which do fall in $L_2$, and which are orthogonal to all the bound states since the Coulomb waves $|\chi_{E,l,m}⟩$ are orthogonal to the bound states (since they're eigenfunctions of the same opretaor with different eigenvalues.

This is further clarified in this answer by Arnold Neumaier and in this question; a previous answer to this question states that the continuous Coulomb waves are not formally part of the Hilbert space, but that does not mean that their informal "span" is not.

For more information about these things (including formal proofs of orthogonality and (non)-completeness and so on), I quite like L. D. Faddeev and O. A. Yakubovskii's Lectures on quantum mechanics for mathematics students and L. A. Takhtajan's Quantum mechanics for Mathematicians (with a hat tip to Anatoly Kochubei here for the references).

If you want a concrete example of something that will sit outside of the span of the bound spectrum, simply take a gaussian wavepacket going fast enough, $$\psi(\mathbf r)=\left(\frac{2\pi}{\sigma^2}\right)^{3/2}\exp\left(-\frac{(\mathbf r-\mathbf r_0)^2}{2\sigma^2}+ikz\right), \tag1$$ project out the bound-state components, and normalize - from physical considerations, you must be left with a nonzero remainder.

As pointed out in Ruslan's comment, the wavefunction in $(1)$ must include a nonzero support outside of the bound-state manifold, because any combination of the form $\sum_{n,l,m}a_{n,l,m}|n,l,m⟩$ will always have a negative expectation value of the energy, whereas the state in $(1)$ will have positive expected energy whenever $k$ and $1/\sigma$ are large enough. That boils down to a routine calculation so I won't perform it here but the core is obvious from physical considerations and if you really care about it then it is a routine procedure to build it into a rigorous argument.

In any case, I'm not sure how any of this answer was unclear, but to be completely explicit:

• The Coulomb-wave continuum eigenstates of the hydrogenic hamiltonian are orthogonal to the bound states. (The eigenstates themselves are not in $L^2$, through the usual non-normalizable-state rigamarole, but linear combinations of them can be, and they will remain orthogonal to the bound states.)

• The bound states of the hydrogenic hamiltonian are not a complete basis for $L^2(\Bbb R^3)$.

Your question doesn't include any specific examples of sources that claim otherwise, so it's impossible to comment further on why you got the incorrect impression that there is not a consensus about either of the two points above.

• thanks for the answer. Is this true for real wavefunctions as well? I thought this would be equivalent to a gaussian wavefunction through local gauge transformation exp(i f(r))... – anon01 Feb 5 '16 at 17:25
• Obviously so. If $\mathrm{Re}(\psi(\mathbf r))$ and $\mathrm{Im}(\psi(\mathbf r))$ were in the span of the bound states, then so would $\psi$. – Emilio Pisanty Feb 5 '16 at 17:32
• Can you show mathematically your example has a nonzero remainder? It is not obvious to me that you could not decompose e.g. exp(-r^2)*cos(kr) into Hydrogen states for arbitrary finite value of k. I'm trying to avoid relying on physical considerations - they put me in this conundrum to begin with. – anon01 Feb 12 '16 at 20:46
• @ConfusinglyCuriousTheThird for any state its mean energy can't be larger than the energy of the most energetic basis state used for expansion. Thus if you take $\psi$ to have $E>0$, then no set of hydrogen bound-state wave functions will be able to represent it — there must be at least one function with $E>0$ in the expansion of $\psi$, but hydrogen bound states always have $E<0$. – Ruslan Jul 15 '17 at 6:05
• @EmilioPisanty thank you for the explicit added material. Some confusion comes from this perspective: I can express the negative energy eigenstates via Fourier transform as $\phi_i(\mathbf p)$. But this is precisely the state of free states (momentum basis of plane waves, in a cartesian decomposition) that they should be orthogonal to. Does one then obtain the "true free states" by "subtracting off" the bound states from plane wave states when a binding potential is present? – anon01 Jul 22 '17 at 5:26

If you integrate the collection of continuum states of the hydrogen atom, weighted by a smooth function, over an open and bounded set in the space of parameters in terms of which this set is parameterized, you get a normalizable state orthogonal to all discrete energy states.