Equality of operators

Shyan

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

Thanks

micromass

Let [itex]T,S:H\rightarrow H[/itex] (if you're working with unbounded operators then things change). By definition, we have

[tex]T=S[/tex]

if and only if

[tex]T\psi = S\psi[/tex]

for all [itex]\psi\in H[/itex].

So yes, in that case we can say [itex][A,B]=cI[/itex].

Shyan

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

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.

micromass

I see. But in that case, [A,B] is only densely defined. So

[tex][A,B]\psi[/tex]

doesn't even make sense for all [itex]\psi \in H[/itex].

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

Shyan

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.

micromass

What problem??
It's probably not a good idea to have two threads on the same topic. So I'm going to lock this thread and ask that we discuss this further in the thread that Fredrik linked.