Unitary & Hermitian Operator
I have two questions to ask :
1)Let $|U|$ unitary operator. I proved that $ |U'| = |U|' $ by induction. Can you give me any hint to prove it by other means?
2) Let $ \alpha $ and $ \beta $ hermitian operators. What can I say about $i[\alpha,\beta] $? ($[\alpha,\beta]$ is the commutator of $ \alpha $ and $\beta $ defined as follows: $ \alpha\beta- \beta\alpha $).
I can't understad why the presence of $i$ since the feature of the hermitian operators is linked to the dagger operation wich has no sense for scalars.
1)Let $|U|$ unitary operator. I proved that $ |U'| = |U|' $ by induction. Can you give me any hint to prove it by other means?
2) Let $ \alpha $ and $ \beta $ hermitian operators. What can I say about $i[\alpha,\beta] $? ($[\alpha,\beta]$ is the commutator of $ \alpha $ and $\beta $ defined as follows: $ \alpha\beta- \beta\alpha $).
I can't understad why the presence of $i$ since the feature of the hermitian operators is linked to the dagger operation wich has no sense for scalars.
Risposte
"gugo82":
hence it remains to prove (or disprove) that \(\overline{\det U}=\frac{1}{\det U}\).
Let's consider a generic complex number $ a+ib = detU $.
$ 1/detU=1/(a+ib)=(a-ib)/(|detU|^2)= \bar\det U/(|detU|^2) $. Since we are dealing with unitary operators $|detU|=|detU|^2=1$. So the thesis is proved.
I would not like to be unpleasant but I need this demonstration to demonstrate that
"slashino":
Since we are dealing with unitary operators $|detU|=|detU|^2=1$.
in the following way:
[tex]1=|I|=|UU^\dagger|=|U||U^\dagger|=|U||{\overline {U^T}}|=|U|\overline{|U^T|}=|U| \overline{|U|}[/tex]
Mmmm... If I'm not wrong (for complex linear algebra is not my cup of tea), in the case of hermitian matrices one gets:
\[
\overline{U}^T = U^{-1}\quad \Rightarrow \quad \det \overline{U} = \det \overline{U}^T = \det U^{-1} = \frac{1}{\det U}\; ,
\]
hence it remains to prove (or disprove) that \(\overline{\det U}=\frac{1}{\det U}\).
\[
\overline{U}^T = U^{-1}\quad \Rightarrow \quad \det \overline{U} = \det \overline{U}^T = \det U^{-1} = \frac{1}{\det U}\; ,
\]
hence it remains to prove (or disprove) that \(\overline{\det U}=\frac{1}{\det U}\).
"gugo82":
So, your first question is proving \(\overline{\det U} =\det \overline{U}\) if \(U^\dagger =U^{-1}\) (where \(U^\dagger := \overline{U^T}=\overline{U}^T\))?
I aimed at proving that not only for hermitian matrices ( I even don't know if it's true ), but since I need this demostration to complete another one involving only hermitian operatos we can add the hypothesis [tex]U^\dagger=U^{-1}[/tex].
So, your first question is proving \(\overline{\det U} =\det \overline{U}\) if \(U^\dagger =U^{-1}\) (where \(U^\dagger := \overline{U^T}=\overline{U}^T\))?
"gugo82":
Is it the transpose-conjugate operator?
Yes, it is.
My book merely distinguishes the conceptual and the pratical sides. It says that the dagger operator acts on a matrix returning the transposed-conjugated matrix.
How does the "dagger operator" act?
Is it the transpose-conjugate operator?
Is it the transpose-conjugate operator?
"gugo82":
1) What do the prime (\(^\prime\)) and the bars (\(|\cdot |\)) mean in this context?
2) What have you tried so far?
1) "$|U|$" means " the determinant of $U$ and "$'$" is the symbol for the conjugation of a complex number.
2) $i[\alpha,\beta]$= $i(\alpha\beta-\beta\alpha) $. I make the dagger operator act on it ( I will indicate the dagger by "$^t$") : $ i(\alpha\beta-\beta\alpha)^t = i((\alpha\beta)^t-(\beta\alpha)^t)$. Since $\alpha $ and $\beta$ are Hermitian $-> i(\beta\alpha-\alpha\beta)=i[\beta,\alpha]=-i[\alpha,\beta]$. My doubt is about the presence of "$i$"...
1) What do the prime (\(^\prime\)) and the bars (\(|\cdot |\)) mean in this context?
2) What have you tried so far?
2) What have you tried so far?
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