# Equality of operators

by Shyan
Tags: equality, operators
 P: 842 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
 Mentor P: 18,290 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 $[A,B]=cI$.
 P: 842 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
Emeritus
PF Gold
P: 9,396
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.
Mentor
P: 18,290
 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 $[A,B]\subset cI$.
 P: 842 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.
Mentor
P: 18,290
 Quote by 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.
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.

 Related Discussions Quantum Physics 3 Calculus 4 Fun, Photos & Games 1 Advanced Physics Homework 22 Linear & Abstract Algebra 3