# Lie algebra and Lie group about quantum harmonic oscillator

We know that in the quantum harmonic oscillator $H=a^\dagger a$, $a^\dagger$, $a$, $1$ will span a Lie algebra, where $a, a^\dagger$ are the annihilation and creation operators, and $H$ is the Hamiltonian operator.

$$[H,a^\dagger\ ]= a^\dagger$$ $$[H,a]=-a$$ $$[a,a^\dagger]=1$$ So these four operators, $H=a^\dagger a$, $a^\dagger$, $a$, $1$, can span a Lie algebra, because the commutator satisfies closure and Jacobi's identity.

We know that for any Lie algebra $\mathscr{G}$ there exists only one Lie group $G$ up to a difference in the topology, whose Lie algebra is $\mathscr{G}$.

So what is this Lie group whose Lie algebra is spaned by $\{H=a^\dagger a , a^\dagger ,a ,1\}$ ?

I apologize, this is my third correction to my answer. This question is very subtle indeed. I hope this answer is the ultimate one!

First of all, if you want to take advantage of Lie's theorem you mention (some time called third Lie theorem), the Lie algebra has to be real, as it must be the Lie algebra of a real Lie group. Then, if you are interested in quantum mechanics applications, I mean if you wish that the given generators are also generators of a unitary representation of a Lie group, the generators must be Hermitian at least and a, a† are not.

So you first have to pass to anti self-adjoint generators (*), for instance, introducing two constants $\omega, m >0$:

$$-iI,-iH,-iP, -iK:= -imX\qquad (1)$$

where, up to real factors (so without changing the real Lie algebra) X and P are given by $a+a^\dagger$ and $i(a−a^\dagger)$ as is well known.

$m$ has the physical meaning of mass of the particle and $H = \hbar \omega(a^\dagger a + \frac{1}{2}I)$ can be re-arranged to:

$$H = \frac{1}{2m}P^2 + \frac{m\omega^2}{2}X^2$$

Operators (1) are, in fact essentially self-adjoint on the dense set made of finite linear combinations of vectors $|n\rangle$, eigenstates of $H$.

Classically, the Galileo group in one dimension includes time translations, space translations along x, Galileian boosts along x. If we think of the point $(x,p)$ in the space of phases as the vector $(1,1,x,p) \in \mathbb R^4$, and the generic element of $G$ is denoted by a triple $(\tau,a,v)$ (time translation + space translation + boost) $G$ acts on the system as

$$(1,1, x, v)^t \mapsto A(\tau,a,v) (1,1, x, p)^t$$

where, for the harmonic oscillator system $A(\tau, a,v)$ is (barring errors in computations) the $4\times 4$ real matrix

$$A(t,a,v) = \begin{bmatrix} I & 0 \\ R_{t}T_{a,v}& R_t \end{bmatrix}$$

where $R_t$ and $T_{a,v}$ are $2\times 2$ respectively matrices defined as:

$$R_t = \begin{bmatrix} \cos \omega t & -\frac{\sin \omega t}{m\omega} \\ m\omega \sin \omega t & \cos \omega t \end{bmatrix}$$

and

$$T_{a,v} = \begin{bmatrix} a & 0 \\ 0 & mv \end{bmatrix}$$

In this way, we have $3$ generators $h, \pi, k$ obtaining by taking the derivative of $A(t,a,v)$ respectively in $t$, $a$ and $v$ at (0,0,0). The commutation relations of these generators are the same as for $$H,P,K$$ with the following exception: $$[\pi, k]=0\quad \mbox{instead of}\quad [\pi, k] = m$$ to be compared with: $$[-iP,-iK]= - m(-iI)$$ Notice that this commutator is just a number, so that, when you exponentiate the generators it gives rise to a phase which commutes with all operators. In other words, if you wish to construct an unitary representation of $G$ acting in the Hilbert space of the harmonic oscillator, you face a problem with the composition rule, as you find a so-called unitary-projective representation: $$U(g)U(g')= e^{i\alpha(g,g')}U_{gg'}\qquad (2)$$ The phase $e^{i\alpha(g,g')}$ arises when $g$ and $g'$ includes transformations generated by the momentum $P$ and the boost $K$. It is possible to compute $\alpha(g,g')$ using several procedures, e.g. Hausdorff-Campbell-ecc... identity. Notice that the mass $m$ explicitly shows up in $\alpha$ (which has just the form $\alpha(g,g')= m f(g,g')$) and this is related to Bargmann's superselection rule.

To obtain a true unitary representation of some Lie group one can deals with as follows. Start from the group $U(1) \times G$ (a so called central extension of $G$) with the composition rule:

$$(e^{ia}, g) \circ (e^{ia'}, g') = (e^{i(a+a'+ \alpha(g,g'))}, gg')$$

and define the map:

$$U(1)\times G \ni (e^{ia}, g) \mapsto V_{(e^{ia}, g)} := e^{ia}U_g\:.$$

Just in view of (2), this is a proper unitary representation of $U(1)\times G$.

Notice that $U(1)\times G$ has now a further generator commuting with all the other generators in view of the fact that we have added'' $U(1)$ to the initial group $G$. This generator, in the Hilbert space, is proportional to $-iI$. The anti-self-adjoint generators are just:

$$-iI,-iH,-iP, -iK\:.$$

So, we can conclude that the considered generators are a representation of the Lie algebra of a central extension of a group $G$, representing the action of Galileo group along the $x$ axis on the harmonic oscillator .

There are some open issues.

(1) $U(1) \times G$ is a Lie group. What is the differential structure but also the topology on it? This is a delicate problem solved by Wigner.

(2) In view of the commutation relations of $H$ and $P$, the latter is not a conserved quantity along time evolution. This is a consequence of the fact that the system, obviously, is not invariant under space translations (the location of the minimum value of the harmonic potential fixes a natural origin). Nevertheless the system admits a conserved quantity associated with the generator $P$.

Since $-iP$ belongs the the Lie algebra of the representation, $$e^{-itH} (-iP) e^{itH}$$ still belongs to that Lie algebra in view of the fact that $e^{-itH}$ is a one-parameter subgroup of the representation. As a matter of fact (barring trivial errors in computations) $$e^{-itH} P e^{itH} = -\frac{\omega\sin (\omega t)}{m} K + \cos(\omega t) P\:.$$ Therefore the explicitly depending on time observable in Schroedinger picture: $$P(t) := -\frac{\omega\sin (\omega t)}{m} K + \cos(\omega t) P$$ turns out to be a constant in Heisenberg picture: $$P(t)_H = e^{itH} P(t) e^{-itH} = P\:.$$

This is exactly the procedure exploited to associate a constant quantity (always in Heisenberg picture) to the boost generator, even in relativistic theories.

(*) When one unitarily represents Lie groups, the Lie algebra of the group is isomorphic to the corresponding Lie algebra of anti self-adjoint generators of the unitary representation. It is true when identifying the Lie algebra commutator with the operator commutator.

There is one more option. You can check that $aa$, $\{a,a^+\}$ and $a^+a^+$ form Lie algebra $sp(2)\sim sl(2)$. Then you can add $a^+$ and $a$ treating them as supergenerators. These are words that tell you to take anticommutators of $a$ and $a^+$ as I did in the first line. Then you get a $5$-dimensional superalgebra, which is $osp(1|2)$. There is a supergroup $OSP(1|2)$.

Another view point is just to take all generators mentioned above as they are and lift them to exponent and investigate the group law. For $exp(\alpha a+\beta a^++\gamma)$ it is fairy easy and you find the Heisenberg group, $H_2$ which is a semidirect product of two-vectors and numbers. If you add bilinears, whose algebra is $sp(2)$, then exponentiating them gives $SL(2)\sim SP(2)$ and the full five-dimensional group is the semidirect product of $SP(2)$ and $H_2$. $\{a,a^+\}$ is just one particular generator corresponding to the Cartan element.