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I noticed the following:

$$[L_{+},L^2]=0,\qquad [L_{+},L_3]\neq 0,\qquad [L^2,L_3]=0.$$

This would suggest, that $L^2,L_+$ have a common system of eigenfunctions, and so do $L^2,L_3$, but $L_+,L_3$ don't. How is that possible?

glS
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Xin Wang
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2 Answers2

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Commutativity is not a transitive relation: If operator $A$ commutes with $B$ and $C$,

$$AB=BA \quad\text{and}\quad AC=CA,$$

then there is no reason that $B$ and $C$ should commute.

Example: Take $A=y$, $B=x$, and $C=p_x$.

In particular, if commuting selfadjoint operators $A$ and $B$ have a common basis of orthonormal eigenvectors, and if commuting selfadjoint operators $A$ and $C$ have a common basis of orthonormal eigenvectors, then these two bases need not be the same if the spectrum of $A$ is degenerate.

Qmechanic
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    Yes, that's what I noticed above. But the question is the following: There is the interpretation that they commute iff they have a common system of eigenfunctions. If you look at it from that perspective, then this should hold. So where does this reasoning lack? – Xin Wang Jan 26 '14 at 13:50
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    The answer is that (in the mentioned examples) the set ${A,B,C}$ of all three operators don't have a common basis of orthonormal eigenvectors. Only a subset of two operators have in certain cases a common basis of orthonormal eigenvectors. – Qmechanic Jan 26 '14 at 13:53
  • This suggests that there are different ways to express the operator basis of $L^2$. I am only aware of the spherical harmonics. What is the other one called? – Xin Wang Jan 26 '14 at 14:05
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    Another problem (in the mentioned example) is that $L_{+}$ is not a normal operator, so there doesn't exist an orthonormal basis of eigenvectors for $L_{+}$. – Qmechanic Jan 26 '14 at 14:17
  • But there can exist an eigenvector basis even if the operator is not normal!Take the matrix with the first row being (0,1) and the second being (4,0). This one has an eigenbasis but is not normal. – Xin Wang Jan 26 '14 at 14:21
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    Yes, but the eigenvectors are not orthogonal. See also this and this Phys.SE posts. – Qmechanic Jan 26 '14 at 14:30
  • just one last thing: What does it then mean that $[L^2,L_+]$ commute physically? Your reasoning in your 1st thread suggests that this has no actual physical meaning? So what does it mean to look at the commutator between non Hermitian quantities? – Xin Wang Jan 26 '14 at 14:39
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    Since (i) $L_{\pm}= L_x\pm i L_y$ and (ii) $L^2$, $L_x$, $L_y$ are all Hermitian, one may view the statement $[L^2,L_{+}]=0$ as equivalent to $[L^2,L_{-}]=0$, and, in turn, equivalent to the pair of statements about Hermitian operators: $[L^2,L_x]=0$ and $[L^2,L_y]=0$. The only caveat is that $L_x$ and $L_y$ do not commute! – Qmechanic Jan 26 '14 at 16:21
  • If I can add a comment, $[A,L_+]=0$ implies that $[A, L_x]=[A,L_y]=0$ as correctly said by Qmechanic (I assume $A=A^\dagger$). But one can say more: It also implies that $[A,L_z]=0$, since $L_z$ is proportional to $[L_x,L_y]$ and Jacobi identity holds. Therefore $[A,L_+]=0$ implies that (in fact is equivalent to) $A$ is invariant under the action of $SO(3)$: $A$ is a scalar. – Valter Moretti Jan 26 '14 at 17:01
  • So, the physical meaning of $[A,L_+]=0$, for $A=A^\dagger$, is that $A$ is a scalar quantity under rotations. – Valter Moretti Jan 26 '14 at 17:08
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NOTE. Since $L_+$ is not normal (normal means $A^\dagger A = AA^\dagger$) it does not admit a basis of orthonormal eigenvectors. However your question can be safely restated replacing $L_+$ for $L_2$ and I will assume it henceforth.

The most elementary case of this phenomenon is given by a triple of normal matrices in $\mathbb C^n$:

$$cI,A,B$$

with $[A,B]\neq 0$ and where $c\in \mathbb C$ is an arbitrarily fixed number. $A$ has a common basis of eigenvectors with $cI$: Every basis of eigenvectors of $A$ is such basis. Similarly, every basis of eigenvectors of $B$ is also a basis of eigenvectors of $cI$. However, though it could happen for some vector, there cannot exist a whole basis of eigenvectors in common with $A$ and $B$, otherwise referring to that basis $A$ and $B$ would be in diagonal form and thus $[A,B]=0$, which is forbidden by hypotheses.

All that is possible thanks to the fact that the eigenspaces of $cI$ are (maximally) degenerate. Two vectors $u$ and $v$ with the same eigenvalue ($c$) of respect to $cI$ remain eigenvectors of $cI$ with the same eigenvalue even if linearly composed: $au+bv$. Nevertheless if $u$ and $v$ are eigenvectors of $A$, in general $au+bu$ is not, but it could be an eigenvector of $B$ (remaining, as said, an eigenvector of $cI$)

The situation is essentially the same when dealing with $L^2$ and $L_2,L_3$. The eigenspaces $\cal H_l$ of $L^2$ are degenerate and, in each eigenspace, $L^2$ is represented by $l(l+1)I$. Moreover, as $[L^2,L_i]=0$, each eigenspace $\cal H_l$ is invariant under the action of $L_i$. I mean $L_i({\cal H}_l)\subset \cal H_l$.

Restricting to $\cal H_l$, we find the situation I outlined above: $L$ is represented by $cI$ and $L_2, L_3$ are represented by non commuting operators $A$ and $B$.

Valter Moretti
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  • Nice answer. A few questions if you don't mind. You state that since the operator is not normal it can'tt have orthononormal basis of eigenvectors. One of the postulates of QM assumes that the observables are Hermitian and then from this we have a theorem which states that every Hermitian operator has a basis of orthonormal eigenvectors. But you are saying that the this is not enough? So does the postulate actually state that every observable operator is in fact normal (hence also Hermitian)? – Alex Sep 09 '16 at 10:44
  • Also, why is it important in your answer to note that each eigenspace $\mathcal{H_l}$ is invariant under the action of $L_i$ as you stated "$L_i(\mathcal{H}) \subset \mathcal{H}$"? Thanks a lot. – Alex Sep 09 '16 at 10:44
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    QM assumes that every observable is self-adjoint (not simply Hermitian) and thus it is also normal. In general, in non-finite dimensions, even normal operators may have no orthonormal basis of (proper) eigenvectors. What is true is that a normal operator has a spectral decomposition. However, at the beginning of my answer the point was another. If an operator is not normal, then it cannot have an orthonormal basis of eigenvectors just because otherwise it would be normal! Since $L_+$ is not normal, it cannot have an orthonormal basis of eigenvectors. – Valter Moretti Sep 09 '16 at 11:21
  • Regarding your second question, the answer appears in my last sentence: Restricting to $\cal H_l$, we find the situation I outlined above: $L$ is represented by $cI$ and $L_2, L_3$ are represented by non commuting operators $A$ and $B$. If $\cal H_l$ were not invariant I would not be allowed to restrict the discussion to $\cal H_l$ and use the already discussed theory. – Valter Moretti Sep 09 '16 at 11:24
  • Okay I see. You state " Moreover, as $[L_2,L_i]=0$, each eigenspace $\mathcal{H}l$ is invariant under the action of Li". As I understand it, the commutativity $[L^2,L_i]=0$ implies there exists a common basis of orthonormal eigenvectors, not that every eigenbasis is shared. You assume that $\mathcal{H}{l}$ is an eigenspace of operator $L^2$, why does it follow then that $\mathcal{H}{l}$ is invariant under $L{i}$, what if $\mathcal{H}_{l}$ is some eigenbasis which is not shared. – Alex Sep 09 '16 at 15:09
  • What result or reasoning are you using to state the invariance of $\mathcal{H}_l$ under $L_i$ from the fact that $[L_2,L_i]=0$? Thanks. – Alex Sep 09 '16 at 15:09
  • If $L^2 \psi = s\psi$ then $L^2L_i\psi = L_i L^2 \psi = L_i s \psi = s L_i \psi$. – Valter Moretti Sep 09 '16 at 15:16
  • Thanks this is quite helpful. What material (notes or book) would you recommend to give the difference between Hermitian and Self- Adjoint? You said that the observable is self-adjoint not simply Hermitian. But in the answer in this post this post it consludes by stating Hermitian implies self-adjiont...In the question it states self-adjoint implies Hermitian...and on wiki they give the definition as what they defined as symmetric in the above MSE post. – Alex Sep 09 '16 at 15:33