I was wondering if the good old quadratic potential was the only potential with equally spaced eigenvalues. Obviously you can construct others, such as a potential that is infinite in some places and quadratic in others, but that's only trivially different. I am not referring to equally spaced as a limiting behavior either, I mean truly integer spaced.

Any ideas? If not, is there a proof for its uniqueness?

If there are other potentials with equally spaced eigenvalues, can one use them as starting points for a free-field QFT? It would be interesting to know if there is a deeper mathematical relation between all of these potentials and whether they could be used to study interacting systems.

## Best Answer

I believe my answer here to the question "How does one determine ladder operators systematically?" gives at least a partial answer to your question. It is a partial answer because I assume a little more than your bare question, but then, as we see by looking carefully at yuggib's answer observes you can obviously write down a hamiltonian with equally spaced eigenvalues and then characterise the whole family of such hamiltonians. It becomes clear that you need to talk about more than simply the Hamiltonian to answer your question: we need to define other

observablesand how they behave to define something akin to a "potential".Let's look at yuggib's answer. You can obviously write down a hamiltonian with equally spaced eigenvalues. Then, as you counter, in your comment:

For this question, I choose energy eigenstates

bounded from below. Otherwise, you could get arbitrarily negative energy states and there would be no quantum ground state. This may or may not be more than what you want to assume, but I think it is physically reasonable. As I said, I'm giving apartialanswer. So now our index set $I$ becomes in fact the set of semi positive integers $\mathbb{N}$. So, in the notation of yuggib's answer, choose an orthonormal basis $\left\{Y_j\right\}_{j=0}^\infty$ for ourassumed separable(this is yet another assumption we must bring to bear) Hilbert space of quantum states with $P_j Y_k = \delta_{j\,k} Y_k$, where $\delta$ is clearly the Kronecker delta and $P_j$ are the projection operators onto the basis vectors. Then, the most general Hamiltonian with equispaced eigenvalues is as written in yuggib's answer with:$$\lambda_i = E_0 + \sigma(i) \Delta$$

where $E_0$ is the ground state energy, $\Delta$ the energy spacing and $\sigma:\mathbb{N}\to\mathbb{N}$ a bijection between $\mathbb{N}$ and itself. So there are an infinite number of Hamiltonians with equispaced energy levels. All members of each family of such Hamiltonians defined by the family's ground state energy $E_0$ and spacing $\Delta$ are unitarily equivelent to one another: two members $\hat{H}_1,\,\hat{H}_2$ are equivalent by $\hat{H}_1 = U\,\hat{H}_2\,U^\dagger$ with $U$ some unitary operator.

So now, how to work this into something like a "potential"? My solution is then to abstractly define position and momentum observables $\hat{X},\,\hat{P}$ and we make our final three assumptions:

They fulfill the canonical commutation relationship $\hat{X}\,\hat{P} - \hat{P}\,\hat{X}=i\,\hbar\,\mathrm{id}$;

Measurements by these observables vary

sinusoidallywith time;Our observables are Hermitian operators.

So now we Writing a general quantum state as:

$$\psi = \sum\limits_{j\in\mathbb{N}} \psi_j e^{-i\,\left(j\,\omega_0+\frac{E_0}{\hbar}\right)\,t}$$

so that the mean of a general observable $\hat{A}$ is:

$$\left<\left.\psi\right|\right.\hat{A}\left.\left|\psi\right.\right> = \sum\limits_{j=0}^\infty a_{j,j}|\psi_j|^2 + 2\,{\rm Re}\left(\sum\limits_{j=0}^\infty\sum\limits_{k=j+1}^\infty a_{j,k} \psi_j \psi_k^* \exp(i\,\omega_0\,(k-j)\,t)\,\right)$$

we can readily see that observables with sinusoidally varying measurement means must have two symmetrically placed, complex conjugate off-leading-diagonal diagonal stripes. Moreover the displacement from the leading diagonal must be the same for both $\hat{X},\,\hat{P}$ if they are to fulfill the CCR. The simplest case is when the two stripes are immediately above and below the leading diagonal. In this case, the means of the observables will vary like $\cos(\omega_0\, t + \phi_0)+const.$: if the two stripes are displaced $N$ steps either side of the leading diagonal, then we have variation varies like $\cos(N\,\omega_0\, t + \phi_0)+const.$. The case with the stripes displaced $N$ from the leading diagonal yield analyses that are essentially the same as the below as discussed in my other answer: they essentially pertain to a quantum oscillator with $N$ times the energy spacing of the one we talk about here.

So now, without loss of generality, in our assumed basis we can write the Hermitian observables as:

$$\begin{array}{ll}\hat{X} = \sqrt{\frac{\hbar}{2}}\left(\tilde{X} + \tilde{X}^\dagger\right) \\ \hat{P} = \sqrt{\frac{\hbar}{2}}\left(\tilde{P} + \tilde{P}^\dagger\right)\end{array}$$

where:

$$\tilde{X} = \left(\begin{array}{ccccc}0&x_1&0&0&\cdots\\0&0&x_2&0&\cdots\\0&0&0&x_3&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{array}\right)\quad\tilde{P}=\left(\begin{array}{ccccc}0&p_1&0&0&\cdots\\0&0&p_2&0&\cdots\\0&0&0&p_3&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{array}\right)$$

are both arbitrary lone-striped upper triangular matrices. You'll need to look at my other answer that I referenced above for full details, but by writing down the CCR we find that:

$$\begin{array}{ll}\hat{X} = \sqrt{\frac{\hbar}{2\,m\,\omega_0\,\cos\chi}}\left(a^\dagger\,e^{-i\,\xi} + a\,e^{i\,\xi}\right) \\ \hat{P} = i\,\sqrt{\frac{\hbar\,m\,\omega_0}{2\,\cos\chi}}\left(a^\dagger\,e^{-i\,(\xi+\chi)} - a\,e^{i\,(\xi+\chi)}\right) \end{array}$$

where we have defined the arbitrary complex constant $\alpha = -i\,m\,\omega_0\,e^{i\,\chi}$ by writing it in terms of a second real positive magnitude $m$ with the dimensions of mass and an arbitrary phase factor $\chi$ and where we have also defined:

$$a = \left(\begin{array}{ccccc}0&\sqrt{1}&0&0&\cdots\\0&0&\sqrt{2}&0&\cdots\\0&0&0&\sqrt{3}&\cdots\\\cdots&\cdots&\cdots&\cdots&\cdots\end{array}\right)$$

and its Hermitian conjugate as the wonted ladder operators.

One can quite straighforwardly show that our derived $\hat{X},\,\hat{P}$ must have continuous spectrums. So, if we change our co-ordinates so that we work in position co-ordinates,

i.e.where $\hat{X}$ becomes the diagonal (multiplication) operator $\hat{X}f(x) = x f(x)$ for $f(x)\in \mathcal{H} = \mathbf{L}^2(\mathbb{R}^3)$, then we can argue as I do in my answer here that there is needfully such a co-ordinate system wherein $\hat{X}f(x) = x f(x)$and$\hat{P} f(x) = -i\,\hbar\,\nabla f(x)$. So now the Hamiltonian in these co-ordinates is:$$\begin{array}{lcl}\hat{H} &=& \hbar\,\omega_0 \left(a^\dagger\,a + \frac{1}{2}I\right) + \left(E_0 - \frac{\hbar\,\omega_0}{2}\right)I\\ &=& \frac{1}{2\,m\,\cos\chi} \hat{P}^2 + \frac{1}{2\,\cos\chi}\,m\,\omega_0^2\,\hat{X}^2 - \frac{\omega_0\,\tan\chi}{2}(\hat{X}\hat{P} + \hat{P}\hat{X}) +\left(E_0 - \frac{\hbar\,\omega_0}{2}\right)I\end{array}$$

so that the SchrÃ¶dinger equation in these co-ordinates is:

$$i\hbar\partial_t \psi = -\frac{1}{2\,m\,\cos\chi} \nabla^2 \psi + \frac{1}{2\,\cos\chi} m\,\omega_0^2 |\vec{x}|^2 \psi + i\,\hbar\,\omega_0\,\tan\chi\,\vec{x}\cdot\nabla \psi + \left(E_0 +i\,\frac{\tan\chi\,\hbar\,\omega_0}{2} - \frac{\hbar\,\omega_0}{2}\right)\psi$$

and more "traditional" SchrÃ¶dinger equation is recovered when $\chi = 0$ and $E_0 = \hbar\,\omega_0/2$:

$$i\hbar\partial_t \psi = -\frac{1}{2\,m} \nabla^2 \psi + \frac{1}{2} m\,\omega_0^2 |\vec{x}|^2 \psi$$

and we see that we must have a quadratic potential. So, in summary, let's list the assumptions that lead to this conclusion:

I have not yet analysed the case where we relax the assumption 4. From the above, we see that equispaced energy levels implies periodic variations of measurements with time, and it seems to me that this relaxation would likely yield a much more general potential in position co-ordinates.