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(I'm not a physicist.)

It seems from the little I've read that the "canonical" example (or maybe just easiest or first example everyone goes to) of a quantum system whose Hilbert space has 2 dimensions comes from thinking about the spin of a particle; the 2 dimensions comes from the spin-up and spin-down states being a basis.

My question is very simple: Is there a "canonical" (or easy or first) example of a quantum system whose complex Hilbert space is $n$ dimensional for arbitrary $n$?

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  • $\begingroup$ Spin networks are the basis for Penrose's twistor theory, which I always assumed was precisely because of their simplicity. $\endgroup$ – user167453 Sep 15 '17 at 15:36
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The set of $n$-dimensional harmonic oscillator states of the type $\{a_k^\dagger\vert 0\rangle, k=1,\ldots,n\}$ span the $n$-dimensional defining representation of $u(n)$. The elements of $u(n)$ are of the form $$ C_{ij}=a_i^\dagger a_j\, ,\quad i,j=1,\ldots, n\, . $$ For $n=2$, one recovers angular momentum by the identification $$ C_{12}\to L_+\, ,\qquad C_{21}\to L_-\, ,\qquad \frac{1}{2}\left(C_{11}-C_{22}\right)\to L_z $$ with $C_{11}+C_{22}$ commuting with all the others when acting on state with fixed total number of excitations.

This is about as simple as it gets.

Note that, when the total number of excitations is greater than one, v.g. the set of states of the type $a_k^\dagger a_m^\dagger \vert 0\rangle$, with $k,m=1,\ldots,n$, the Hilbert is larger and has the dimensionality of the $(2,0,\ldots,0)$ irrep of $su(n)$. This is the simple generalization of higher spin states living in a larger Hilbert space.

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  • $\begingroup$ I've been doing some more reading; is it not the case that if one considers a spin-(n/2) particle (ignoring position and momentum) that the resulting Hilbert space is (n+1)-dimensional? If so, this would answer my question and the original example would be retained by taking n=1. $\endgroup$ – Brian Klatt Sep 19 '17 at 14:23
  • $\begingroup$ @BrianKlatt The Hilbert space for both systems has the same dimension but the spin-(n/2) example is limited to $3$ observables, whereas the n-dimensional h.o. would have $n^2-1$ observables acting on the basis states. $\endgroup$ – ZeroTheHero Sep 19 '17 at 15:36
  • $\begingroup$ I don't think I understand the significance of your remark. $\endgroup$ – Brian Klatt Sep 19 '17 at 15:41
  • $\begingroup$ In the spin-$n/2$ case, you still have only three operators: $L_x,L_y,L_z$, Their matrix elements in the space of dim (2s+1) must be so that $[L_x,L_y]=-\hbar L_z$. The simplest $3d$ harmonic oscillator has $9$ observables: $C_{12},C_{23}$ etc and the commutation relations and matrix elements are different from those of $L_x,L_y,L_z$. $\endgroup$ – ZeroTheHero Sep 19 '17 at 15:45
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Not particularly.

The reason that spin is a good model is because any nontrivial hamiltonian (and indeed any hermitian operator) can essentially be mapped onto the canonical spin matrix $$ \sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix} $$ by adding and multiplying with constants. Moreover, you have a clean basis $\{\sigma_x,\sigma_y,\sigma_y\}$ for representing all nontrivial hermitian operators on that basis, with clear geometrical understanding underpinning all of them ─ but, more importantly, with all of them included in the angular-momentum algebra of the quantum spin. This is why e.g. the Jaynes-Cummings model pretends that its two-level system is a spin, when it's most(?) often an electronic transition between states with a different spatial wavefunction.

This is no longer the case in higher dimensions, because the spin-$j$ algebra is too constrained to match the full generality of hermitian operators on dimension $2j+1$, and indeed it struggles begin with diagonal operators and move on from there.

And as for non-spin systems, there's simply no likely candidates.

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