# Equality of operators

1. Dec 22, 2012

### ShayanJ

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

2. Dec 22, 2012

### micromass

Staff Emeritus
Let $T,S\rightarrow H$ (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 $[A,B]=cI$.

3. Dec 22, 2012

### ShayanJ

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. Dec 22, 2012

### Fredrik

Staff Emeritus
@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.

5. Dec 22, 2012

### micromass

Staff Emeritus
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 $[A,B]\subset cI$.

6. Dec 22, 2012

### ShayanJ

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.

Last edited: Dec 22, 2012
7. Dec 22, 2012

### micromass

Staff Emeritus
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.