
#1
Dec2212, 02:09 PM

P: 740

Imagine we have two operators A and B on a complex hilbert space H such that:
[itex] [A,B] \psi = (ABBA) \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 



#2
Dec2212, 02:52 PM

Mentor
P: 16,703

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]. 



#3
Dec2212, 02:59 PM

P: 740

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 



#4
Dec2212, 03:03 PM

Emeritus
Sci Advisor
PF Gold
P: 9,018

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:




#5
Dec2212, 03:25 PM

Mentor
P: 16,703

[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]. 



#6
Dec2212, 10:17 PM

P: 740

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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