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I'm struggling in understanding the meaning of this condition that I found in an operator equation:

\begin{equation} [A,B]=A \end{equation} where both $A$ and $B$ are hermitian operators. What can I say about the operator $B$?

Gbp
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    I think that's for commutator i.e. $[A,B]=AB-BA$. – user296113 Feb 15 '18 at 18:53
  • what if the commutator of two operators A and B is equal to A? – Gbp Feb 15 '18 at 18:54
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    Then I suppose you can assume that $AB - BA = A$. – J126 Feb 15 '18 at 18:59
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    ...and you'll be able, for example, to conclude that tr.$,A=0;$ ... – DonAntonio Feb 15 '18 at 19:04
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    If you are asking what the equation «$[A,B]=A$» means, then it is simply the same as «$AB-BA=A$». If you are asking something else, you should explain... – Mariano Suárez-Álvarez Feb 15 '18 at 19:06
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    Ok let's say that this condition is imposing some requirements on B, right? so my question would be – Gbp Feb 15 '18 at 19:32
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    "what information can I get on B from [A,B]=A?" – Gbp Feb 15 '18 at 19:33
  • Please edit the question. Whenever you are asked for clarifications on your questions, the correct place to put them is in the questions themselves. – Mariano Suárez-Álvarez Feb 15 '18 at 19:35
  • Unless you give some context, you cannot say anything. In the generality you wrote this $A$ could be zero, for example, and then the answer to your question is "nothing". Also, we do not even know of these are operators on a finite dimensional vector space or not, and so on... – Mariano Suárez-Álvarez Feb 15 '18 at 19:45
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    Perhaps you meant $[A,B]=iA$ when you say that the operators are hermitian (as the commutator is antihermitian) – user8268 Feb 15 '18 at 19:55
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    There is a (unique, up to isomorphism) Lie algebra of dimension 2 which is non-abelian, corresponding to the group of affine motions of the line, and your question is essentially «what are the (hermitian) representations of that algebra?», which corresponds to «what are the unitary representations of the group?». This is a rather open ended question... – Mariano Suárez-Álvarez Feb 15 '18 at 19:56

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If A and B are hermitian, then you need, instead,
$$ [A,B]= i A , $$ for consistency under taking the adjoint. So A is neither hermitian nor antihermitian, even though, conventionally, B is taken to be hermitian in applications. Let's stick with $[A,B]= A$.

These are the prototypes for the shift and counter operators, used routinely in quantum mechanics.

Specifically, for an eigenvector v of B with real eigenvalue λ, $$ B v = \lambda v , $$ utilize your simplest nonabelian Lie algebra you specified, to observe $$ BAv = A(B-1)v, $$ and hence $$ B(Av)=(\lambda-1)(Av). $$ That is, A defines new eigenvectors with down-shifted eigenvalues of B, it "ladders" the state v down.

A simple realization of the algebra is $A=x$ and $B=-x\partial_x$. Consider what A does to the eigenvectors $x^{-\lambda}$ of B.

Formally, this is the simplest nonabelian Lie algebra there is, with just one commutator, that of the affine group in one dimension . The upper-triangular 2x2 matrix representation thereof is given in the WP link provided. You probably do not need to dwell on its delightful group structure. In QM, people use number operators for B and lowering operators a for A.

Cosmas Zachos
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