Consider a quantum system with a hamiltonian $\hat{H}$, which is invariant under the action of a lie group $G$, meaning we have a unitary representation of $G$, $\hat{U}(g)$, in Hilbert space, and $\hat{H}$ commutes with every $\hat{U}(g)$.

As is well known, the generators of the Lie algebra of $G$ give rise to generators of the representation $\hat{U}(g)$. For example in the case of the 3-dimensional rotation group $SO(3)$ acting on a quantum system by rotating it, the 3 generators of the lie algebra of $SO(3)$ translate to the 3 angular momentum operators $\hat{J}_x, \hat{J}_y, \hat{J}_z$, and they are guaranteed to (separately) commute with $\hat{H}$. Furthermore we can also construct in Hilbert space the quadratic Casimir operator of $G$, call it $\hat{J}^2$, for example using the formula given in the question here. In the case of $SO(3)$, this is the total angular momentum operator $\hat{J}^2 = \hat{J}_x^2 + \hat{J}_y^2 + \hat{J}_z^2$. The quadratic Casimir operator is guaranteed to commute with $\hat{H}$, and is also guaranteed to commute (separately) with each of the generators of the group $\hat{J}_i$.

This means we can choose an eigenbasis of $\hat{H}$ which is also an eigenbasis of $\hat{J}^2$ and of one of the generators of $G$, typically $\hat{J}_z$.

We can also construct ladder operators $J_+, J_-$ that within any given eigenspace of $\hat{J}^2$, act on a state to raise or lower the eigenvalue of $\hat{J}_z$. Their construction is completely general and can be done for any group (not just $SO(3)$ in case the above assumptions are met. (For example, this is utilized in the algebraic solution of the hydrogen atom, where the $SO(4)$ symmetry of the hydrogen atom Hamiltonian is utilized in this way to construct ladder operators that change between the different eigenstates).

Now consider the simple harmonic oscillator, where as we all know, ladder operators $\hat{a}, \hat{a}^+$ can also be constructed that raise or lower the states. However, as far as I know, these operators are not induced from the action of any Lie group on the harmonic oscillator.

In mathematics, when one finds a similar phenomenon happening in seemingly different examples, one expects to find an abstract generalization that explains the similarities of the two examples. Is there a Lie group whose action on the harmonic oscillator explains the appearance of corresponding ladder operators? If not, is there a generalization of the structure that explains why the ladder operators can be constructed in both cases, and explains in which quantum systems we can expect to find ladder operators (that are somehow "naturally constructed" from the dynamical quantities)?


1 Answer 1


Ch 10 of the late W Miller's legendary book Symmetry Groups and their Applications tells you more than you'd ever want to know about this oscillator group, quite different than SO(3), with which it shares a similar ladder structure.

The oscillators provide the "defining" representation of it. For its four generators, $N=a^\dagger a$, $J^+=a^\dagger$, $J^-= a$, $E=\mathbb I$ (central) $[E,\bullet]=0$, so $$ [N, J^+ ]= J^+,\\ [N,J^-]=-J^-,\\ [J^+,J^-]=-E,$$ where the last commutation relation makes all the difference! It's really the Heisenberg algebra generating the eponymous group, and from its trace you see that all faithful irreducible representations of it are infinite-dimensional like the one you saw.

The Casimir is also radically different, $$ C= J^+ J^- - EN, $$ and vanishes for the oscillator representation above.

The corresponding (solvable, but not nilpotent) oscillator group is ${\cal G}(0,1)$; it can be written in the basis for its generic element $$ {\mathbb g}=e^{aE}e^{bJ^+}e^{cJ^-}e^{\tau N }, $$ which has simple, elegant, group element composition properties, just like its Heisenberg subgroup, τ=0.

Some like to realize it in 4×4 matrix notation, a reducible 4d representation with C=0, $$\begin{pmatrix} 1 &ce^\tau &a &\tau \\ 0&e^\tau& b& 0\\ 0&0&1&0\\ 0&0&0&1\\ \end{pmatrix}; $$ while some do the same for the algebra with the realization $$ N= \lambda+ z\frac{d}{dz}, ~~~~~~J^+=\mu z , ~~~ J^-={\xi\over z} +{d\over dz} , ~~~~ E=\mu~~\leadsto \\ C= \mu(\xi-\lambda). $$

Geeky Details

The above 4d representation follows directly from $$ \exp(a\begin{pmatrix} 0 & 0 &1 &0 \\ 0&0& 0& 0\\ 0&0&0&0\\ 0&0&0&0\\ \end{pmatrix})~ \exp(b\begin{pmatrix} 0 & 0 &0&0\\ 0&0& 1& 0\\ 0&0&0&0\\ 0&0&0&0\\ \end{pmatrix}) \times \\ \exp(c\begin{pmatrix} 0 &1 &0 &0 \\ 0&0& 0& 0\\ 0&0&0&0\\ 0&0&0&0\\ \end{pmatrix}) ~\exp(\tau \begin{pmatrix} 1 &0 &0 &1 \\ 0&1& 0& 0\\ 0&0&0&0\\ 0&0&0&0\\ \end{pmatrix}) ~. $$ This oscillator group is ${\cal G}(0,1)$.

Its partner, ${\cal G}(0,1)$, has the Lie algebra gl(2) which inspired you, and contains so(3). Contrast its realization, $$ J^3= \lambda+ z\frac{d}{dz}, ~~~~~~J^+=(2\lambda+\xi) z+ z^2 \frac{d}{dz} , ~~~ J^-={\xi\over z} -{d\over dz} , \\ E=\mu~ $$ to the above!


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