# Are negative energy eigenstates orthogonal to plane wave states?

Orthogonality in discrete Hilbert spaces is straightforward - those encountered by typical examples of infinite wells of any type, spin systems etc. Continuous Hilbert spaces are fine too - we usually talk about free particles and such.

My confusion comes when I try to 'put the two together'. In systems that have both bound and free states (finite potentials), are bound and free states always orthogonal? Can this be shown easily?

My confusion stems from the fact that the continuous momentum eigenspace is the basis of plane waves, and yet bound wave functions exist in this space. How can this be reconciled? I feel as though there is something being swept under the rug.

• – user83548
Commented Jun 7, 2016 at 22:37
• If bound and free states are both eigenvectors of the Hamiltonian with different eigenvalues, they are orthogonal. But I do not see how the "momentum space" (what exactly do you mean?) comes into play in your question? In the end, the choice of the basis does not affect the Hilbert space per se ... Commented Jun 7, 2016 at 22:40
• @Sanya Yes, that all makes sense, except that bound states have momentum space representations, ie $\psi(k)$; but this is exactly the continuous space of (free) plane waves. This is the source of confusion. Commented Jun 7, 2016 at 22:44
• @anon0909 That they have a representation in that basis (which is not in $L_2$ to complicate the matter) does not mean that they have all its properties. Think of a harmonic oscillator and a superposition of different eigenvectors. Even though this is a wavefunction with a eigenfunction representation, it is not an eigenfunction of the oscillator. Commented Jun 7, 2016 at 22:49
• @Sanya I don't understand your point. The harmonic oscillator has a unique decomposition; my point is that \delta(k-k_0) corresponds to both a plane wave, and the Fourier coefficient of a bound state. I don't see how to reconcile the orthogonality here. Commented Jun 7, 2016 at 22:56

They are not orthogonal because planar waves do not belong to hilbert space. It is true that you can have hilbert spaces of any cardinality, but the one in quantum mechanics is $L_2$, the space of square integrable functions, which has a countable cardinality, $\aleph_0$. A consistent way to do the math would be to have atoms in a box of length $l$, where $l$ can be as large as you like, and in such a case the unbounded states will be countable, as close as you like to planar waves, and orthogonal to any bound state. Using the dirac delta (a generalized function, or distribution, that does not belong to $L_2$) is a slight of hand that works well in practice but is not mathematically rigorous.

• If they are not orthogonal, how is one supposed to determine probability amplitudes of these free/bound states? This answer is helpful, if you care to add more details I would be interested. Using your box example, I guess you have excluded nonperiodic functions that have a fourier representation. Commented Jun 7, 2016 at 23:44
• What I meant is that you cannot even compare them, it is not meaningful because it is like comparing apples and oranges, as they do not even belong to the same space (a delta dirac does not belong to $L_2$). The only way is to compare them is to make the bound states belong to $L_2$, which is the case if the atom is in a "box"
– user83548
Commented Jun 7, 2016 at 23:50
• Of course, they are orthogonal. If you don't like non-normalizable states then just construct a wave-pocket. Any wave-packet composed of only positive energy states will be orthogonal to any negative energy state. Commented Jun 8, 2016 at 1:03
• There is no Hilbert space that has countable cardinality, apart from the one consisting only of the vector zero. This is because, as for every real or complex vector space, there are at least $2^{\aleph_0}$ elements in the set ($\lambda x$ for $\lambda\in\mathbb{R}$, where $x$ is a non-zero vector). It is true, however, that in physics we usually restrict to separable Hilbert spaces, and the countability is that of the corresponding orthonormal basis. Commented Jun 8, 2016 at 6:40

The unbound free states are not plane waves. And, in the extended (rigged) Hilbert space that allows for un-normalizable states, they are orthogonal to all the bound states.

• Yes, I thought rigged Hilbert spaces might come into play. Can you expand on your answer? Commented Jun 8, 2016 at 0:43
• The "extended" rigged Hilbert space is not a Hilbert space, so there is no notion of orthogonality in there (usually it is the space of distributions $\mathscr{S}'(\mathbb{R}^d)$). Commented Jun 8, 2016 at 6:40