Equality of Operators A & B on Hilbert Space H

[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

 

[micromass]

Let $T,S<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />\rightarrow H$ (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 $[A,B]=cI$.

 

[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

 

[Fredrik]

@micromass: Shyan's questions are motivated by this thread, where there's some discussion about what's wrong with calculations like this one:

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.

 

[micromass]

@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 $[A,B]\subset cI$.

 

[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.

 

[micromass]

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.

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.