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We define a complete set of commuting observables as a set of observables $\{A_1,\ldots, A_n\}$ such that:

  1. $\left[A_i, A_j\right]=0$, for every $1\leq i,~j \leq n$;

  2. If $a_1,\ldots, a_n$ are eigenvalues of $A_1,\ldots, A_n$ respectively, there exists a unique state $\psi$ such that $A_i\psi=a_i\psi$.

I was wondering if there is a theorem or standard procedure to say if a set of observables is complete. In finite dimensional spaces it seems quite easy, but how do to it in infinite dimensional spaces, in particular, when the degenerescence is also infinite?

Some practical questions:

  1. How to prove $\{H,L^2,L_z\}$ is a complete set, where $L$ is the angular momentum, $L_z$ is the $z$-direction component of $L$, and $H=\frac{1}{2m}\nabla^2-\frac{e^2}{r}$ is the hamiltonian?

  2. How to prove $\{H,L_z\}$ is a complete set in the case of the Landau levels?

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It's complete if there is only one basis of common eigenvectors. That means, There is only one basis in which the matrices are diagonal matrices.

Let's start with only 2: operators $A$ and $B$. If $[A,B]=0$, there is at least one orthonormal basis of common eigenvectors.

If the eigenvalues of $A$ have no degenerancy, then the basis is unique (except for global phase factors), and hence the set is complete.

If $A$ has degenerate eigenvalues, then they form subspaces (the matrix has boxes along the diagonal). $B$ acts on each subspace without merging with others.

Inside every subspace, you can find a basis of $B$ which makes sub-sub-spaces (sub-boxes).

If those subspaces are more than 1-dimensional, then the system is not complete, but there's a third commuting observable $C$ that might make the matrices diagonal.

There might be more than 1 CSCO with different eigenvalues.

For a given CSCO, eigenvalues of all operators specify one only common eigenvector.


As for the practical question, you can show it in the particular case.

But one can show that any spherically-symmetric setup satisfies $[H,\vec{L}]=0$, and therefore there's $[H, L^2]=0, \ [H, L_z]=0$.

This is because any system which is invariant under rotations verifies that $H$ conmutes with rotations, and rotations are a function of $\vec{L}$, so $[H, \vec{L}]=0$. It's long to prove, but it is a very beautiful topic.

Note: invariant under rotations refers that you'll get the same result by a) Let it evolve in time and then rotate it. b) Rotate it first and then let it evolve.

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Generally, as a rule, if you know that one set of quantum numbers is complete, then you know that any other complete set must have the same number of quantum numbers - which is a handy way to check results. Additionally, the number of separable degrees of freedom also corresponds to the number of quantum numbers. In your example, the electron has 3 degrees of freedom around the atom (4 with spin) so 3 or 4 Quantum Numbers define the system. In Landau levels, it has only 2 degrees of freedom.

EDIT: Perhaps this is another important point worth noting: a set of commuting operators is complete if it represents the maximum number of linearly indepedent, commuting operators over the space. This is in general a challenging fact to prove given a candidate set of operators.

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  • $\begingroup$ It doesn't seems true. Since commuting operators can be diagonalized simultaneously, I will focus on diagonal operators to give an example. Consider a system in a four-dimensional Hilbert space and the following commuting positive operators $\endgroup$ – Ricardo Correa da Silva Feb 14 '18 at 1:02
  • $\begingroup$ $A=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}$, $B=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$, $C=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4 \\ \end{bmatrix}$. Then $\{A,B\}$ and $\{C\}$ are complete sets of commuting observables with different amount of quantum numbers. Is it right? $\endgroup$ – Ricardo Correa da Silva Feb 14 '18 at 1:03
  • $\begingroup$ @CorreadaSilva There are a couple of issues here. First, it's not clear what system you are modelling with these sets of matrices. Both sets must model the same system for us to even compare the QNs meaningfully. Second, these sets don't appear to be complete to me. Both sets still have other matrices which are linearly independent of them and commute with all members of the set. We need to have the maximum number of LI and commuting matrices to say that the set is complete. $\endgroup$ – Geoffrey Feb 14 '18 at 2:56
  • $\begingroup$ @RicardoCorreadaSilva Whoops, I messed up the tag the first time. Anyway, I though of something to add. Namely, this question is closely related to the problem of Clebsch-Gordan coefficients and a point of confusion some people have: the uncoupled basis (direct product of spin states) has the same number of basis vectors and commuting operators as the coupled basis (direct sum of spin states). For instance, $\frac{3}{2}\bigotimes\frac{3}{2}=3\bigoplus 2\bigoplus 1\bigoplus 0$, and in both cases the space is defined by 4 operators (and thus 4 QNs) and 16 basis vectors. $\endgroup$ – Geoffrey Feb 14 '18 at 6:08
  • $\begingroup$ In fact I don't want to specify the system, I was hoping for a general answer. Well, the sets are complete as you can see in the deffinition above and, as in Clebsch-Gordan coefficients, there is a correspondence beetween the basis: (i) in the set $\{A,B\}$ the vectors can be identified as $|1,1>=(1,0,0,0)$, $|1,2>=(0,1,0,0)$, $|1,3>=(0,0,1,0)$ and $|2,1>=(0,0,0,1)$; (ii) in the set $\{C\}$ the vectors can be identified as $|1>=(1,0,0,0)$, $|2>=(0,1,0,0)$, $|3>=(0,0,1,0)$ and $|4>=(0,0,0,1)$. But this relation are between vectors. $\endgroup$ – Ricardo Correa da Silva Feb 14 '18 at 12:47

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