Commutation Relation

**Mauritzvdworm** · September 27th 2010, 07:38 AM

This has to be something stupid which I'm just not seeing. Let $R_{\theta}(g)(z)=g(e^{2\pi i \theta}z)$ with $g\in C(\mathbb{T})$ and $z\in\mathbb{T}$
Further, let $M_f$ be the multiplication operator on $B(L^2(\mathbb{T}))$ with $f\in C(\mathbb{T})$

$R_{\theta}$ can be extended to a unitary operator on $B(L^2(\mathbb{T}))$

show that the following commutation relation holds
$R_{\theta}M_{id}=e^{2\pi i \theta}M_{id}R_{\theta}$

Where $id$ denotes the identity function

This is what I have:
$(R_{\theta}M_{id})(g)(z)=R_{\theta}(M_{id}(g)(z))= R_{\theta}(id(z)g(z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)$

$M_{id}R_{\theta}(g)(z)=M_{id}(R_{\theta}(g)(z))=M_ {id}(g(e^{2\pi i \theta}z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)$

but here the commutation relation does not hold. Am I using the definitions wrongly?

**Opalg** · September 27th 2010, 10:21 AM

Originally Posted by Mauritzvdworm

This has to be something stupid which I'm just not seeing. Let $R_{\theta}(g)(z)=g(e^{2\pi i \theta}z)$ with $g\in C(\mathbb{T})$ and $z\in\mathbb{T}$
Further, let $M_f$ be the multiplication operator on $B(L^2(\mathbb{T}))$ with $f\in C(\mathbb{T})$

$R_{\theta}$ can be extended to a unitary operator on $B(L^2(\mathbb{T}))$

show that the following commutation relation holds
$R_{\theta}M_{id}=e^{2\pi i \theta}M_{id}R_{\theta}$

Where $id$ denotes the identity function

This is what I have:
$(R_{\theta}M_{id})(g)(z)=R_{\theta}(M_{id}(g)(z))= R_{\theta}(id(z)g(z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)$

$M_{id}R_{\theta}(g)(z)=M_{id}(R_{\theta}(g)(z))=M_ {id}(g(e^{2\pi i \theta}z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)$

but here the commutation relation does not hold. Am I using the definitions wrongly?

The calculation for $(R_{\theta}M_{\text{id}})(g)(z)$ is correct. The other one is not. The operator $M_{\text{id}}$ takes a function and multiplies its value at z by the independent variable z. So $M_{\text{id}}R_{\theta}(g)(z)=M_{\text{id}}(R_{\th eta}(g)(z))=M_{\text{id}}(g(e^{2\pi i \theta}z))=zg(e^{2\pi i \theta}z).$

**Mauritzvdworm** · September 27th 2010, 11:41 AM

So then $(R_{\theta}M_{id})(g)(z)=e^{2\pi i \theta}zg(e^{2\pi i \theta})$
and
$(M_{id}R_{\theta})(g)(z)=zg(e^{2\pi i \theta}z)$
in which case the commutation relation holds

The only part that is bothering me is the multiplication operator $M_{id}$ or more generally $M_{f},f\in C(\mathbb{T})$

The way I have it is that it is defined in the following way
$M_{f}(g)(z)=f(z)g(z)$

now in the second case we had
$M_{f}(g)(e^{2\pi i \theta}z)$
so why do we not use the rotated coordinate but just $z$ ?

**Opalg** · September 27th 2010, 12:06 PM

Originally Posted by Mauritzvdworm

So then $(R_{\theta}M_{id})(g)(z)=e^{2\pi i \theta}zg(e^{2\pi i \theta})$
and
$(M_{id}R_{\theta})(g)(z)=zg(e^{2\pi i \theta}z)$
in which case the commutation relation holds

The only part that is bothering me is the multiplication operator $M_{id}$ or more generally $M_{f},f\in C(\mathbb{T})$

The way I have it is that it is defined in the following way
$M_{f}(g)(z)=f(z)g(z)$

now in the second case we had
$M_{f}(g)(e^{2\pi i \theta}z)$
so why do we not use the rotated coordinate but just $z$ ?

The definition of $M_f$ is $M_f g(z) = f(z)g(z)$ . Now suppose you have a function $h(z)$ defined by $h(z) = g(e^{2\pi i \theta}z)$ . It is still true that $M_f h(z) = f(z)h(z) = f(z)g(e^{2\pi i \theta}z)$ . In this case $h = R_\theta g$ so that $h(z) = R_\theta g(z)$ .

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