## #012: Product of observables

Category: Quantum mechanics - Tags: Linear operators, Hilbert spaces

Make a decisive argument as to whether the product of observables is well-defined (and what it is) or not.

Mathematical problem. Given that the product of two self-adjoint operators is not necessarily self-adjoint, the questions is whether there exists a canonical way to create the product of two quantum observables.

Note that the sum of two self-adjoint operators is itself self-adjoint, so the sum would seem to be allowed. Also, the function $$f$$ applied to an operator corresponds to the self-adjoint operator where the eigenvectors are the same while the eigenvalues are given by applying the function to the related eigenvalues. Therefore $$A+B$$ and $$f(A)$$ are allowed.

Note that $$\frac{(a+b)^2 - a^2 -b^2}{2}=a*b$$. The left side is in terms of operations allowed, therefore a product consistent with those rules must be in terms of the anti-commutator $$A*B = \frac{\{A, B\}}{2}$$.

The issue is that the operation is not associative: $$\{\{A, B\}, C\} - \{A, \{B, C\}\} = A [ B, C ] - [B, A] C$$. However, for position and momentum we have $$[Q, P] = \imath \hbar I$$ and therefore $$Q [Q, P] - [Q, P] Q = 0$$. By recursively applying the product, it would seem we are able to construct all polynomials of $$Q$$ and $$P$$ and therefore (through Taylor expansion) generic functions. Are there any problems?

Physical significance. This would seem like a basic problem that should have a settled question in the literature, yet we were not able to find an in depth treatment. If the operation is well-defined, it would also be interesting to understand what symmetry it would generate and what the conjugate variable would be.