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For the canonical commutation relation $[\hat{x},\hat{p}]=i\hbar\hat{\mathbb{1}}$ to be true, the operators $\hat{x}$ and $\hat{p}$ are to be described by infinite dimensional matrices only, cf. e.g. this Phys.SE post. Consequently other operators representing physical quantities must also be of infinite dimension (when represented in discrete base). But the matrix for total angular momentum squared $J^2$ turns out to be finite-dimensional in some cases. I know this happens because $j$ is bounded and finite, so we need $2m+1$ dimensional matrices to describe $J$ (or if talking about spin, if it is a spin-1 particle, 3x3 matrix for $S^2$). And not only $J^2$ or $S^2$, I have seen problems where Hamiltonian $H$ is being represented by finite dimensional matrices (2 level systems or other similar restricted energy spectrum). Does the finite dimensional nature of matrix representation of these operators somehow violate the commutation relation? If not (I hope this is true), how are the dimensions of these operators consistent?

Qmechanic
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Manas Dogra
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  • which "these operators" are you referring to? The angular momentum operators must satisfy the correct commutation relations irrespective of the dimension of the representation, and in all but very specialized cases these representations are finite dimensional or the Hilbert space can be truncated so that only finite dimensional representations appear. – ZeroTheHero May 17 '20 at 17:01
  • S^2 is a 2x2 matrix for spin-1/2 particles,this is in contrary with the necessity of matrices being infinite dimensional in QM. – Manas Dogra May 17 '20 at 17:06
  • The three operators for angular momentum along the three axes, $J_x$, $J_y$, and $J_z$, do not commute. (And $J^2$ communtes with all three of these.) – Peter Shor May 17 '20 at 17:06
  • Why would you suggest the matrices need be infinite-dimensional? Or more appropriately: if the angular momentum operators do not satisfy the commutation relations irrespective of the dimension of the Hilbert, they are NOT legitimate angular momentum operators. – ZeroTheHero May 17 '20 at 17:07
  • @ZeroTheHero $xp-px=i\hbar I$ take trace on both sides, trace vanishes on LHS but doesn't on RHS.Trace taken in this way and definition of I are for finite dimensional matrices only.Inference-Matrices are infinite dimensional.

    This is done in a lot of books,including Exploring Quantum Mechanics by Galitski Karnakov,Kogan,Galitski, Jr.

     – Manas Dogra May 17 '20 at 17:14
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    @manasdogra That argument works for operators whose commutator is a nonzero constant. Does it also work for the commutation relations of the angular momentum operators? Their commutator is NOT proportional to the identity. Try taking the trace of both sides of their commutation relations. Do you get a contradiction? – Jahan Claes May 17 '20 at 17:22

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I think the confusion arises because, while the Hilbert space of states is often infinite dimensional, this space decomposes into a sum of $J$-invariant subspaces (cf. Peter-Weyl theorem). The commutation relations holds in every subspace.

There will be some $L$ subspaces that are infinite-dimensional, but those rarely rarely appear in practical problems.

For instance, the largest $L$-value of hydrogen atom eigenstates $L= n-1$ so that, unless one works with arbitrary large $n$'s and thus arbitrary small energy differences one need not consider very very very large values of $L$ (on which the commutation relations would hold anyways).

Another example would be the $3D$ harmonic oscillator, where the possible value of $L$ go as $L,L-2,L-4,\ldots$ until one reaches $0$ or $1$. Again, unless one is interested in states of arbitrary large energies, only finite values of $L$ will appear and the commutation relations will hold on all the subspaces.

ZeroTheHero
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  • Though not exactly what I was asking,this is a valid reasoning and a good one too...Accepting it with a +1! – Manas Dogra May 17 '20 at 17:26
  • @ManasDogra One effect of working with energy and angular momenta as a good quantum number is precisely that it usually allows one to truncate the Hilbert space from an infinite-dimensional one to a finite dimensional one (albeit sometimes large). The skill is in truncating to a physically meaningful finite dimensional subspace that will include all the physics of interest. Of course if you technically require the computation of the commutator of $x$ and $p$, you can so either symbolically or by using very very large matrices where the commutator will only approximately be $i\hbar$. – ZeroTheHero May 17 '20 at 17:43