# Is the quantization of the harmonic oscillator unique?

To put it a little better:

Is there more than one quantum system, which ends up in the classical harmonic oscillator in the classial limit?

I'm specifically, but not only, interested in an elaboration in terms of deformation quantization.

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 By deformation quantization, do you mean a different kind (parametrized?) of homomorphism? – Antillar Maximus Mar 18 '12 at 14:17 @AntillarMaximus Compare my answer here: physics.stackexchange.com/a/7591/667 – student Mar 18 '12 at 14:22

1) There are many inequivalent quantum systems that have the same classical limit $\hbar\to 0$.

2) For instance, assume for simplicity that the quantum system is described by a single pair of creation and annilation operators, $$[\hat{a},\hat{a}^{\dagger}] ~=~\hbar {\bf 1}, \qquad\qquad \hat{a}~=~\sqrt{\frac{m\omega}{2}}(\hat{q}+\frac{\mathrm{i}\hat{p}}{m\omega}).$$ The standard quantum harmonic oscillator Hamiltonian read $$\hat{H}~=~\omega (\hat{a}^{\dagger}\hat{a}+\frac{\hbar}{2}).$$ We can e.g. add any term of the form $\hbar P(\hat{a}^{\dagger}\hat{a},\hbar)$ (where $P$ is a polynomial) to the Hamiltonian $\hat{H}$ without affecting the classical limit. In particular, we can shift the zero-point energy term.

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 1. Your formula doesn't only shift the zero-point energy of the system but produces a nonlinear transformation of the spectrum! - 2. One can even add any term of the form $ℏP(aˆ*aˆ,a+a^*,ℏ)$, hence producing arbitrary anharmonic oscillators, and they still reduce to the harmonic oscillator classically. Writing $\hbar$ as a commutator, this becomes essentially the statement in my own answer. – Arnold Neumaier Mar 18 '12 at 13:51 A constant shift (proportional to $\hbar$) in the zero-point energy term corresponds to a constant (zeroth-order) polynomial $P$. Indeed, one can add more general polynomials to the Hamiltonian than indicated in the answer(v2). – Qmechanic♦ Mar 18 '12 at 20:22

This is not a complete answer to your question but quantization of the harmonic oscillator means quantization of the complex projective space $CP^n$ (as reduced phase space of the system).

In the context of deformation quantization you may then have a look at this paper:

(available as preprint: http://arxiv.org/abs/math/9802078)

where inequivalent star products of $CP^n$ are discussed.

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Link also available at arxiv.org/abs/math/9802078 – Qmechanic Mar 18 '12 at 13:06
@Qmechanic Thanks, I included it into my answer. – student Mar 18 '12 at 13:28

For every classical Hamiltonian $H(p,q)$ there are infinitely many quantized systems that reduce to it in the classical limit.

The reason is that adding to $H(p,q)$ an arbitrary number of expressions in $p$ and $q$ where at least one factor is a commutator doesn't change the classical syatem but changes the quantum version. To be specific, take, for example, $H'=H(p,q)+A(p,q)^*[p,q]^2A(p,q)$ for an arbitrary expression $A(p,q)$.

This holds for systems in which $p$ and $q$ are canonically conjugate variables. It is easy to generalize the argument to arbitrary quantum systems.

In short:

Quantization of individual systems is an ill-defined process.

Geometric quantization is more well-defined, as it effectively quantizes not a particular system but a particular group of symmetries. In the above case, it shows how to go from the classical $p$ and $q$ (i.e., from the classical Heisenberg Lie algebra with the Poisson bracket as Lie product) to the quantum version, but not how to go from a particular classical Hamiltonian (and hence a particular classical system) to its quantization.

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 by what?? It yields a quantum observable for each classical expression in the generators of the group. But different expressions for the same classical function of p and q yield different quantizations - this is the ordering ambiguity that always plagued quantization of classical systems. – Arnold Neumaier Mar 18 '12 at 18:38 Sorry, I had problems with posting/editing. "but not how to go from a particular classical Hamiltonian (and hence a particular classical system) to its quantization." This is not exactly true, as geometric quantization yields a quantum operator for each classical observable $f$, which acts as $\psi \rightarrow -i \hbar\nabla_{X_f} \psi + f \psi$. In the particular example of the harmonic oscillator one obtains by this equation the operator $H = \hbar \omega a^\deger a$ - the quantum one without the zero-point energy (which is subject to the so called metaplectic correction). – altertoby Mar 18 '12 at 23:23 Of course, one can make up a recipe and call that ''the'' quantization. But it has canonical properties only for $f$ in the Lie algebra on which the construction is based. If you want further discussion, please give an online reference, so that we have common ground. – Arnold Neumaier Mar 19 '12 at 9:32 The standard reference is "Woodhouse: geometric quantization", wikipedia and nlab have some facts about it online at ncatlab.org/nlab/show/geometric+quantization. As you can see, the first step in geometric quantization (called prequantization) yields a quantum operator for each! classical observable. But the constructed Hilbertspace is too big (wavefunctions depend on $q$ and $p$ simultaneously), so that one has to introduce a "polarization" and later on a metaplectic correction. As this is done in a geometrical way (maniflolds,...), the coordinates $q$ and $p$ does not play any role. – altertoby Mar 19 '12 at 20:01 I don't have woodhouse, and nlab is too abstract for my taste. I guess the ordering ambiguity shows in the lack of a canonical polarization. – Arnold Neumaier Mar 19 '12 at 20:08