# "Equidistant" spectra in quantum mechanics [duplicate]

In one-dimensional quantum mechanics, it seems that the only kind of potential able to produce an "equidistant" spectrum, i.e. with $E_{n+1}-E_{n}=\text{constant}$, is the harmonic oscillator.

Why is that? And is there a way to prove it?

• An observation: it seems to be true iff the ladder operators $a$ obey $[H,a]=ka$ for a constant $k$. Commented Apr 26, 2015 at 18:21
• Possible duplicate: physics.stackexchange.com/q/132688/2451 Commented Apr 26, 2015 at 18:33

## 2 Answers

Any "harmonic oscillator", seen as the second quantization operator $$d\Gamma(1)=\int a^*(k)a(k)dk$$ of the symmetric Fock space $\Gamma_s(\mathscr{H})$ over a (separable) Hilbert space $\mathscr{H}$, has the natural numbers $\mathbb{N}$ as spectrum (i.e. evenly spaced spectrum). In addition, if e.g. $\mathscr{H}=\mathbb{C}$, the operator $aa^*$ has $\mathbb{N}^*=\mathbb{N}\setminus\{0\}$ as spectrum (again equidistant). So there are a lot of operators with equidistant spectrum, and all of them share a similar structure.

However I doubt that it is possible to prove that any operator with purely discrete equidistant spectrum can be written as either $d\Gamma(1)$ or $aa^*$ for some suitably defined creation and annihilation operators $a^{\#}$, because one may think at some quite artificial examples using projection operators (I am not sure about that, in addition you may also take a look at the answers, one is mine, to the question linked above by Qmechanic).

For sure there is a deep relationship between spectra of certain operators and interesting properties of number theory. For example, the prime numbers can be characterized as follows (this is a result of the Fields medal A.Connes):

• Let $\Gamma_s(\mathscr{H})=\bigoplus_{n=0}^{\infty}\mathscr{H}^{\otimes_s n}$ be the symmetric Fock space over $\mathscr{H}$; and let $A$ be a one-particle operator in $\mathscr{H}$. Then we define $\Gamma(A)$ to be the operator that acts on $n$-particle factorized functions $\psi(x_1,x_2,\dotsc, x_n)=\phi(x_1)\dotsm\phi(x_n)\in \mathscr{H}_n$ as $$\Gamma(A)\psi(x_1,x_2,\dotsc, x_n)= (A\phi)(x_1)\dotsm (A\phi)(x_n)\; ;$$ i.e. it acts at the same time on each particle. Furthermore let $\mathcal{P}$ be the set of prime numbers. Then the following result is true:

Let $T$ be a self-adjoint operator on $\mathscr{H}$; then, counting multiplicities (i.e. each value occurs with multiplicity one):

$\text{Spectrum } T=\mathcal{P}\Longleftrightarrow \text{Spectrum } \Gamma(T)=\mathbb{N}^*$.

• Wow... that's beautiful! How hard is this to prove? Commented Apr 26, 2015 at 18:39
• @CuriousOne Not so hard ;-) you can take a look at the short proof yourself in the Connes' book, page 529 Commented Apr 26, 2015 at 18:50
• @CuriousOne You're welcome! The result is indeed very interesting and, I think, not so well-known. I stumbled upon it by chance reading Connes' book, and thought it may be nice to share. Commented Apr 26, 2015 at 18:57
• I wasn't aware of it (which means nothing), but I am glad you showed it to me. Now I have to try to understand what it means. Commented Apr 26, 2015 at 19:10

In an one dimensional harmonic oscillator the energy observable is a complete set of observables of the system (we don't have degeneracy). If you have a list of possible energies you know everything about that system when we don't have degeneracy. If this list is an infinite, we have precise the list of energies of an harmonic oscillator.

You can construct a matrix representation of the hamiltonian in energy basis. Then you can construct a creation and annihilation operators in this basis too. Is quite simple, is a matrix with suitable factors in the tri-diagonalized. Then you can define operators as the position and momentum, and make sense of your harmonic oscillator