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Suppose that two matrices $A$ and $B$, representing real($\mathbb{R}$) physical quantity, can be multiplied commutatively with each other; i.e. $AB =BA$. However, each matrix cannot be multiplied commutatively with other matrices all the time (obviously). If two matrices represent physical quantity, can they be stated to have definite values togeether still? Or as two matrices cannot be multiplied commutatively with other matrices, is it not possible for two matrices to have definite values together?

Also, suppose $C=A+B$. If A and B can be multiplied commutatively with each other, does $C$ have definite values? If $A$ and $B$ cannot be multiplied commutatively with each other, is there any way $C$ can have definite values?

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anyone.........? –  mother Sep 27 '12 at 14:20
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If two matrices $A$ and $B$ commute with each other and are diagonalizable then they can be simultaneously diagonalized. Conversely if two matrices can be simultaneously diagonalized then they will commute. What is your question regarding $C$ ? –  user10001 Sep 27 '12 at 14:45
    
The first paragraph of the question(v1) is answered by @dushya's comment. The second paragraph is a duplicate of this Phys.SE post. –  Qmechanic Sep 27 '12 at 15:01
    
@Qmechanic: Only a partial duplicate, as the question seems to be whether $C$ has definite values together with $A$ and $B$. –  Arnold Neumaier Sep 27 '12 at 17:28
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1 Answer 1

If two real (i.e. self-adjoint) operators $A$ and $B$ do not commute on the intersection of their domain, they don't have a basis of common eigenvectors, hence they do not have a joint probability distribution. In this case, $C=A+B$ doesn't need to be selfadjoint, hence does not always represent an observable (though it is always Hermitian). One needs some domain compatibility condition for the selfadjointness of $C$. (For finite-dimensional matrices, Hermitian implies selfadjoint, though.)

But if two self-adjoint operators $A$ and $B$ commute on a common dense domain, they have a joint probability distribution, and in a state where $A$ and $B$ have definite values $a$ and $b$, any function $f(A,B)$ with finite $f(a,b)$ has the definite value $f(a,b)$ in such a state. In particular, this holds for $C=A+B$.

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Do you have any example (if possible, relevant to QM) where $A$, and $B$ are noncommuting self adjoint operators but $C=A+B$ is not self adjoint. –  user10001 Sep 27 '12 at 18:24
    
Harmonic Oscillator: Let $\hat{A}=\frac{\hat{x}}{\sqrt{2}}$, $\hat{B}=\frac{i\hat{p}}{\sqrt{2}}$ and $\hat{C}=\hat{A}+\hat{B}=\frac{1}{\sqrt{2}}(\hat{x}+i\hat{p})$ is the annihilation operator. $\hat{C}^{\dagger}\neq\hat{C}$ so $\hat{C}$ is not self adjoint –  Jorge Sep 27 '12 at 19:38
    
For clarity, it would probably be good to include in the answer the precise definition of commutativity for operators with domains different from the full Hilbert space. –  Qmechanic Sep 27 '12 at 20:58
    
@dushya: There must be some examples in vol. 1 of the Reed/Simon series on math physics. –  Arnold Neumaier Sep 28 '12 at 7:33
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@Burzum: But $B$ is not selfadjoint either, against the assumption. –  Arnold Neumaier Sep 28 '12 at 7:33
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