#### kmm

Gold Member

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Here is what I understand. The generalized uncertainty principle is: [itex] \sigma^{2}_{A} \sigma^{2}_{B} \geq ( \frac{1}{2i} \langle [ \hat{A}, \hat{B} ] \rangle )^2 [/itex]

So if [itex] \hat{A} [/itex] and [itex] \hat{B} [/itex] commute, then the commutator [itex] [ \hat{A}, \hat{B} ] = 0 [/itex] and the operators are compatible. What I don't understand about this is why two compatible operators must have shared eigenfunctions or that in other words, incompatible operators such as the position and momentum operators of the original Heisenberg uncertainty principle

Proof: If we have [itex] \hat{A} f_{n} = \lambda_{n} f_{n} [/itex] and [itex] \hat{B}f_{n} = \mu_{n} f_{n} [/itex] so that [itex] f_{n}(x) [/itex] are eigenfunctions of both operators and the set [itex] \{f_{n}\} [/itex] are complete so that any function [itex] f(x) [/itex] can be written as a linear combination of them [itex] f = \Sigma c_{n} f{n} [/itex]. Then: [tex] [ \hat{A}, \hat{B} ] f = (\hat{A} \hat{B} - \hat{B} \hat{A}) \Sigma c_{n} f{n} = \hat{A}( \Sigma c_{n} \mu_{n} f{n} ) - \hat{B} (\Sigma c_{n} \lambda_{n} f{n}) = \Sigma c_{n} \mu_{n} \lambda_{n} f{n} - \Sigma c_{n} \mu_{n} \lambda_{n} f{n} = 0 [/tex] Since this is true for any [itex] f(x) [/itex] then [itex] [ \hat{A}, \hat{B} ] = 0 [/itex]

I still don't quite see how this shows that incompatible operators don't share any eigenfunctions. I see how this means that if two operators share a complete set of eigenfunctions then they must commute. But I don't see how this rules out the possibility of two incompatible operators sharing some "incomplete" set of eigenfunctions. Maybe there is no such thing as an "incomplete" set of eigenfunctions? The way I understand it, the eigenfunctions of a Hermitian operator are complete, but does this mean that some subset of the eigenfunctions are also complete? If not, then it seems that this proof has not ruled out that two incompatible operators could share some set of eigenfunctions, although incomplete. I'm pretty certain that's wrong, but I'm not sure why or how I'm thinking wrong about this.

So if [itex] \hat{A} [/itex] and [itex] \hat{B} [/itex] commute, then the commutator [itex] [ \hat{A}, \hat{B} ] = 0 [/itex] and the operators are compatible. What I don't understand about this is why two compatible operators must have shared eigenfunctions or that in other words, incompatible operators such as the position and momentum operators of the original Heisenberg uncertainty principle

**cannot**share eigenfunctions. I'm having a hard time completely understanding the proof of this.Proof: If we have [itex] \hat{A} f_{n} = \lambda_{n} f_{n} [/itex] and [itex] \hat{B}f_{n} = \mu_{n} f_{n} [/itex] so that [itex] f_{n}(x) [/itex] are eigenfunctions of both operators and the set [itex] \{f_{n}\} [/itex] are complete so that any function [itex] f(x) [/itex] can be written as a linear combination of them [itex] f = \Sigma c_{n} f{n} [/itex]. Then: [tex] [ \hat{A}, \hat{B} ] f = (\hat{A} \hat{B} - \hat{B} \hat{A}) \Sigma c_{n} f{n} = \hat{A}( \Sigma c_{n} \mu_{n} f{n} ) - \hat{B} (\Sigma c_{n} \lambda_{n} f{n}) = \Sigma c_{n} \mu_{n} \lambda_{n} f{n} - \Sigma c_{n} \mu_{n} \lambda_{n} f{n} = 0 [/tex] Since this is true for any [itex] f(x) [/itex] then [itex] [ \hat{A}, \hat{B} ] = 0 [/itex]

I still don't quite see how this shows that incompatible operators don't share any eigenfunctions. I see how this means that if two operators share a complete set of eigenfunctions then they must commute. But I don't see how this rules out the possibility of two incompatible operators sharing some "incomplete" set of eigenfunctions. Maybe there is no such thing as an "incomplete" set of eigenfunctions? The way I understand it, the eigenfunctions of a Hermitian operator are complete, but does this mean that some subset of the eigenfunctions are also complete? If not, then it seems that this proof has not ruled out that two incompatible operators could share some set of eigenfunctions, although incomplete. I'm pretty certain that's wrong, but I'm not sure why or how I'm thinking wrong about this.

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