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Equality of operators

by Shyan
Tags: equality, operators
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Shyan
#1
Dec22-12, 02:09 PM
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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?

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micromass
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Dec22-12, 02:52 PM
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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
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Dec22-12, 02:59 PM
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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?
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Fredrik
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Dec22-12, 03:03 PM
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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 Quote by Fredrik View Post
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
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Dec22-12, 03:25 PM
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Quote Quote by Fredrik View Post
@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

[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
#6
Dec22-12, 10:17 PM
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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
#7
Dec22-12, 10:42 PM
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Quote Quote by Shyan View Post
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.


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