## Equality of operators

Imagine we have two operators A and B on a complex hilbert space H such that:
$[A,B] \psi = (AB-BA) \psi=c \psi \ \ \ \ \psi \epsilon H \mbox{ and } c \epsilon C$
Then can we say that [A,B] is the same as cI when I is the identity operator?Why?

Thanks

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Let $T,S:H\rightarrow H$ (if you're working with unbounded operators then things change). By definition, we have $$T=S$$ if and only if $$T\psi = S\psi$$ for all $\psi\in H$. So yes, in that case we can say $[A,B]=cI$.
 So if we have unbounded operators,we should deal with the domains too? Let D(I)=H and let one of A and B be unbounded. What you tell now? Thanks

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## Equality of operators

@micromass: Shyan's questions are motivated by this thread, where there's some discussion about what's wrong with calculations like this one:
 Quote by Fredrik Suppose that [A,B]=cI, where I is the identity operator and c is a complex number. \begin{align}1 =\frac 1 c \langle a| cI|a\rangle =\frac 1 c \langle a|[A,B]|a\rangle =\frac 1 c \big(\langle a|AB|a\rangle - \langle a|BA|a\rangle\big) =\frac 1 c\big(a^*\langle a|B|a\rangle-a\langle a|B|a\rangle\big)=0. \end{align} I used that a is real in the last step.
I don't understand the issues well enough (or just haven't thought it through enough) to write down a complete explanation of everything that's going on here, but it's clear that it has something to do with the domains of the operators.

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 Quote by Fredrik @micromass: Shyan's questions are motivated by this thread, where there's some discussion about what's wrong with calculations like this one: I don't understand the issues well enough (or just haven't thought it through enough) to write down a complete explanation of everything that's going on here, but it's clear that it has something to do with the domains of the operators.
I see. But in that case, [A,B] is only densely defined. So

$$[A,B]\psi$$

doesn't even make sense for all $\psi \in H$.

And since $I$ is everywhere defined, we can never have $[A,B]=cI$. We can only have $[A,B]\subset cI$.

 Yes,But we can define an identity operator with the same domain as [A,B] and do the calculations.Or we can choose a vector from [A,B]'s domain.So what you say doesn't solve the problem. P.S. Ok guys,looks like here's going to be the same discussion as the topic related to that paradox,so I guess you guys may want to lock this one.

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