# Lie algebra and lie group about quantum harmonic oscillator

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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 spanned by $\{H=a^\dagger a , a^\dagger ,a ,1\}$ ?

This post imported from StackExchange Physics at 2014-04-13 11:20 (UCT), posted by SE-user user34669

edited Apr 14, 2014

The same question was asked on MO, see http://mathoverflow.net/questions/163172/lie-group-about-the-quantum-harmonic-oscillator. It received two answers, but has now been put on hold and will probably disappear soon.

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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 $exp(\alpha a+\beta a^+)$ 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.

This post imported from StackExchange Physics at 2014-04-13 11:20 (UCT), posted by SE-user John
answered Apr 12, 2014 by (180 points)

True but it doesn't answer the question posed.

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This group is called the oscillator group. It is a solvable group whose commutator is the Heisenberg group (whose commutator is a cyclic group, whose commutator is trivial).

More explicitly, it is isomorphic to the group of complex upper triangular 3 by 3 matrices $M$ with $M_{11}=M_{33}=1$. The diagonal subgroup is generated by $H$, the subgroup of matrices with $M_{22}=1$ is the Heisenberg group generated by $a$ and $a^*$. Thus it is the semidirect product of a cyclic group $U(1)$and the Heisenberg group.

Its representation theory is given in the paper

R. F. Streater,
The Representations of the Oscillator Group,
Comm. Math. Phys. 4 (1967), 217-236.

answered Apr 14, 2014 by (15,488 points)
edited Apr 15, 2014

The 3 by 3 description in the revised answer is nicer than the 4 by 4 description given by Streater, mentioned in my first attempt to answer.

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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.

answered Apr 14, 2014 by (2,075 points)
edited Apr 14, 2014

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