Unitary Operator
Hi guys, I can not demonstrate the following theorem about unitary operator:
-If one treats the columns of a $nXn$ unitary matrix as components of $n$ vectors, these vectors are orthonormal.
My book gives me two proofs, I'm particulary interested in the one that uses the property of unitary operatos of preserving inner products. It says the $j^(th)$ column of the matrix representing $U$ is the components of the $j^(th)$ basis vector after $U$ acts on it. I can not understad how to uses the property of the unitary operatos I mentioned before since a generic column of the matrix is in this form:
$((<1|U|1>),(<2|U|1>),(...),()$ ( I can't write the vector as a column, but It's the same )
...and a unitary operator preserves a scalar product only if it acts on both the factors of the product; so even if I know that the initial basis is orthormal, after U acts on a certain basis vector ( $|1> $ in this case ) how can I be sure that the inner products between $|1>$ and the other basis vectors doesn't change?
I try to better explain myself: A unitary operator is the generalization of rotation operators. Let consider a certain collection of vectors :$ |1>,|2>, |3> $. $U$ acts on the first, so : $U|1>$$=|1'> $ that is a vector having the same norm of $|1>$ but a different angle respect to $|2> $ e $ |3> $.
-If one treats the columns of a $nXn$ unitary matrix as components of $n$ vectors, these vectors are orthonormal.
My book gives me two proofs, I'm particulary interested in the one that uses the property of unitary operatos of preserving inner products. It says the $j^(th)$ column of the matrix representing $U$ is the components of the $j^(th)$ basis vector after $U$ acts on it. I can not understad how to uses the property of the unitary operatos I mentioned before since a generic column of the matrix is in this form:
$((<1|U|1>),(<2|U|1>),(...),(
...and a unitary operator preserves a scalar product only if it acts on both the factors of the product; so even if I know that the initial basis is orthormal, after U acts on a certain basis vector ( $|1> $ in this case ) how can I be sure that the inner products between $|1>$ and the other basis vectors doesn't change?
I try to better explain myself: A unitary operator is the generalization of rotation operators. Let consider a certain collection of vectors :$ |1>,|2>, |3> $. $U$ acts on the first, so : $U|1>$$=|1'> $ that is a vector having the same norm of $|1>$ but a different angle respect to $|2> $ e $ |3> $.
Risposte
While I was writing the post I understood!!! I get it now....Damn me.
Thanks for your helpfullness.
Thanks for your helpfullness.
Please, do the right computations using a concrete example.
"gugo82":
[quote="Slashino"][quote="gugo82"]
This is your claim, for \(U\mathbf{e}_k\) is the \(k\)-th column of the matrix representing \(U\) in the orthonormal basis \(\{\mathbf{e}_1,\ldots, \mathbf{e}_n\}\) (both in the domain and the codomain).
My problem is here. The $k^(th)$ column is not $ Ue_(k) $ but it's $
The point is : $
You're completely wrong, because \(\langle \mathbf{e}_i,U\mathbf{e}_k\rangle\) is a number, not a vector!
The \(k\)-th column of the matrix representing \(U\) is given by:
\[
U\mathbf{e}_k = \begin{pmatrix} \langle \mathbf{e}_1,U\mathbf{e}_k\rangle \\ \langle \mathbf{e}_2,U\mathbf{e}_k\rangle \\ \vdots \\ \langle \mathbf{e}_n,U\mathbf{e}_k\rangle
\end{pmatrix} \; ,
\]
or something like this.[/quote]
Wait, so is $
"Slashino":
[quote="gugo82"]
This is your claim, for \(U\mathbf{e}_k\) is the \(k\)-th column of the matrix representing \(U\) in the orthonormal basis \(\{\mathbf{e}_1,\ldots, \mathbf{e}_n\}\) (both in the domain and the codomain).
My problem is here. The $k^(th)$ column is not $ Ue_(k) $ but it's $
The point is : $
You're completely wrong, because \(\langle \mathbf{e}_i,U\mathbf{e}_k\rangle\) is a number, not a vector!
The \(k\)-th column of the matrix representing \(U\) is given by:
\[
U\mathbf{e}_k = \begin{pmatrix} \langle \mathbf{e}_1,U\mathbf{e}_k\rangle \\ \langle \mathbf{e}_2,U\mathbf{e}_k\rangle \\ \vdots \\ \langle \mathbf{e}_n,U\mathbf{e}_k\rangle
\end{pmatrix} \; ,
\]
or something like this.
"gugo82":
This is your claim, for \(U\mathbf{e}_k\) is the \(k\)-th column of the matrix representing \(U\) in the orthonormal basis \(\{\mathbf{e}_1,\ldots, \mathbf{e}_n\}\) (both in the domain and the codomain).
My problem is here. The $k^(th)$ column is not $ Ue_(k) $ but it's $
The point is : $
For sake of semplicity, let \(\mathbf{e}_k\) be the \(k\)-th vector of a fixed orthonormal basis of \(\mathbb{V}^n\), a real or complex vector space of dimension \(n\) with inner product \(\langle \cdot, \cdot \rangle\).
If \(U:\mathbb{V}^n\to \mathbb{V}^n\) is unitary, then:
\[
\langle U\mathbf{e}_k, U\mathbf{e}_h \rangle =\langle \mathbf{e}_k ,\mathbf{e}_h\rangle =\delta_k^h
\]
(here \(\delta_k^h\) is Kronecker's delta) hence the set \(\{U\mathbf{e}_1,\ldots, U\mathbf{e}_n\}\) is an orthonormal system of \(n\) vectors, i.e. an orthonormal basis of \(\mathbb{V}^n\).
This is your claim, for \(U\mathbf{e}_k\) is the \(k\)-th column of the matrix representing \(U\) in the orthonormal basis \(\{\mathbf{e}_1,\ldots, \mathbf{e}_n\}\) (both in the domain and the codomain).
If \(U:\mathbb{V}^n\to \mathbb{V}^n\) is unitary, then:
\[
\langle U\mathbf{e}_k, U\mathbf{e}_h \rangle =\langle \mathbf{e}_k ,\mathbf{e}_h\rangle =\delta_k^h
\]
(here \(\delta_k^h\) is Kronecker's delta) hence the set \(\{U\mathbf{e}_1,\ldots, U\mathbf{e}_n\}\) is an orthonormal system of \(n\) vectors, i.e. an orthonormal basis of \(\mathbb{V}^n\).
This is your claim, for \(U\mathbf{e}_k\) is the \(k\)-th column of the matrix representing \(U\) in the orthonormal basis \(\{\mathbf{e}_1,\ldots, \mathbf{e}_n\}\) (both in the domain and the codomain).
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