# How to find interesting operators for a quantum system?

How can we find "interesting" operators for a quantum mechanical system?

I can think of the following method: Given some system with an associated Hilbert space $V$ and Hamiltonian $H:V\rightarrow V$, we consider a linear action of a Lie group $G$ on $V$ which commutes with the Hamiltonian, i.e. we have a representation $\rho:G\rightarrow GL(V)$ such that $\rho(g)H\psi = H\rho(g)\psi$ for all $g\in G$ and $\psi\in V$. Then $\rho$ induces an action of the Lie algebra on $V$, defined as follows: let $g(t)\in G$ be a smooth path in $G$ with $g(0)=e$, $\left.\frac{d}{dt}\right|_{t=0}g(t) = x\in\mathfrak{g}$, then $x(v) = \left.\frac{d}{dt}\right|_{t=0}\rho(g(t))v$. This can be used to find operators for conserved quantities of the system (i.e. operators $A$ such that $[H,A]=0$).

An example: Let $H=-\frac{\hbar^2}{2m}\Delta + V(|\mathbf{x}|):L^2(\mathbb{R}^3)\rightarrow L^2(\mathbb{R}^3)$, i.e. the potential is rotational-symmetric. Then we ahve an action of $G=SO(3)$ on $L^2(\mathbb{R}^3)$ given by $g\cdot\psi(\mathbf{x}) = \psi(g\mathbf{x})$. The action induced on the Lie algebra is given by $$x(v) = \left.\frac{d}{dt}\right|_{t=0}g(t)\cdot \psi(\mathbf{x}) = \left.\frac{d}{dt}\right|_{t=0}\psi(g(t)\mathbf{x}) = \frac{\partial\psi}{\partial x^i}(\mathbf{x})\left.\frac{d}{dt}\right|_{t=0}(g(t)\mathbf{x})^i$$ Since $x\in \mathfrak{so}(3)$ acts on $\mathbf{x}\in\mathbb{R}^3$ as $\omega\wedge\mathbf{x}$ for some vector $\omega\in\mathbb{R}^3$, we have that the resulting operator is basically the angular momentum operator (up to some constant, probably $i\hbar$).

My questions:

1. Is what I have done above correct? How can it be explained better (I am a bit confused on some points, as you might have noticed)?
2. What interesting examples/applications of this fact are there? In what situations can this be useful?
3. Are there other methods to find interesting operators for a quantum system?

I am also interested to references to articles/books treating the subject, although I really won't have time to look at them any time soon.

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What objective qualities make an operator interesting? –  Henry T. Horton Aug 5 '13 at 20:43
@HenryT.Horton Commutativity with $H$. This is what replaces "integral of motion" and "symmetries" in the quantum case. One of the practical uses: since such operator (call it $L$) and $H$ can be diagonalized simultaneously, the study of spectrum of $L$ often simplifies the study of spectrum of $H$. If there are several such $L$'s and they do not commute between themselves, one can e.g. generate a lot of eigenstates of $H$ from a given one. –  O.L. Aug 5 '13 at 20:52

The most interesting operators are those which commute with the Hamiltonian $H$. To find them, it is useful to first consider the corresponding classical system and try to construct the integrals of motion (IM), i.e. the quantities which commute with the classical Hamiltonian w.r.t. Poisson bracket: $$\left\{H,L\right\}=0.$$ The existence of such IM is related to explicit or hidden symmetries by Noether theorem (e.g. translations $\rightarrow$ conserved momentum, rotational symmetry $\rightarrow$ angular momentum, $SO(4)$ symmetry of Coulomb potential $\rightarrow$ Runge-Lenz vector, etc). In the classical case, IM allow to reduce the number/order of the differential equations of motion.

Hence the first thing one usually does is the construction of quantum analogs of classical conserved quantities, i.e. the operators satisfying $$\left[H,L\right]=0.$$

Why are such operators useful?

Example 1. As an illustration, let us consider a somewhat pathological example: the Hamiltonian of a free particle in 2D: $$H=-\Delta=-\partial_{xx}-\partial_{yy}.$$ It has translational symmetry, i.e. commutes with the operators $$P_x=-i\partial_x,\qquad P_y=-i\partial_y,$$ which in addition commute with each other. That means we can diagonalize them simultaneously. And now note the first important thing: diagonalization of $P_{x,y}$ is "simpler" than that of $H$: the eigenfunctions are given by plane wave exponentials $e^{ik_x x}$, $e^{ik_y y}$, and we immediately obtain common eigenfunctions in the form $e^{ik_x x+ik_yy}$ without working directly with "complicated" operator $H$.

Example 2. $H$ from the previous example also commutes with the third component of angular momentum $L_{z}=-i\partial_{\varphi}$ and we can similarly attempt to diagonalize them simultaneously. This will lead to common eigenfunctions in the form $|m\rangle=f_m(r)e^{im\varphi}$, $m\in\mathbb{Z}$, where $f_m(r)$ can be expressed in terms of Bessel functions. But now recall that we also have IM $P_x,P_y$ which do not commute with $L_z$. Since, however, they do commute with $H$, their action on the eigenstates $|m\rangle$ will produce new eigenstates of $H$ corresponding to the same eigenvalue. If you work out the details of this exercise, you will discover that this reasoning automatically gives e.g. the differentiation formulas for Bessel functions. (Actually, one can obtain all basic properties of Bessel functions by playing with the symmetry algebra of this problem.)

In summary, the existence of quantum IM 1) simplifies the study of the spectrum of the Hamiltonian and 2) is largely responsible for its structure - in particular, the degeneracy of energy levels. Another rather famous example is the Runge-Lenz vector. In the classical case it is responsible for conservation of orbit orientation, whereas in the quantum case it manifests itself in the hydrogen atom miracle: the existence of eigenstates with the same energy but different orbital quantum number.

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This is a great answer, thank you. I have a couple further questions about it: where does the $SO(4)$ symmetry of Coulomb potential come from? And also if I'm not wrong there are symmetries in QM which have no classical correspondent (like the $SU(2)$ symmetry of the hydrogen atom accounting for spin, if I'm not mistaken). How does someone find them? –  Daniel Robert-Nicoud Aug 7 '13 at 0:35
@DanielRobert-Nicoud For any rotationally symmetric potential one has $SO(3)$ symmetry, which implies the conservation of the angular momentum vector. In the classical picture, this means that the orbit is planar, but it is not necessarily closed even for bounded motion. However, there are a few very special potentials (essentially $r^{-1}$ and, I think, $r^2$) for which the orbits are periodic. This is a signature of enhanced symmetry. It does not come from anywhere, it just secretly exists and makes some problems simpler than others. –  O.L. Aug 7 '13 at 0:51
@DanielRobert-Nicoud Concerning your 2nd question, it is true that there are exist symmetries without classical analog. However, I am not sure about what do you mean by finding them. When the hamiltonian is written down, I would say they are obvious. It seems to me that the problem is rather inverse: to find the hamiltonian which correctly accounts for some properties of the spectrum (and hence, for some symmetries). But maybe I misunderstand your point. –  O.L. Aug 7 '13 at 1:06
Yes, the "special potentials" are $r^{-1}$ and $r^2$ (Bertrand's theorem). If I have to be honest, I really can't see the $SU(2)$ symmetry for the Hamiltonian $H=-\frac{\hbar^2}{2m}\Delta - \frac{e^2}{4\pi r}$ of the hydrogen atom, but probably once you get the hang of it they're easier to find... –  Daniel Robert-Nicoud Aug 7 '13 at 1:13