Vector Space, Two bases related by Unitary Matrix

[Zatman]

Homework Statement

The vector space V is equipped with a hermitian scalar product and an orthonormal basis e₁, ..., e_n. A second orthonormal basis, e₁^', ..., e_n^' is related to the first one by

$\mathbf{e}_j^{'}= \displaystyle\sum_{i=1}^n U_{ij}\mathbf{e}_i$

where U_ij are complex numbers. Show that U_ij = <e_i, e_j^'>, and that the matrix U with entries U_ij is unitary.

2. The attempt at a solution
I have done the first part, by simply taking the scalar product with some e_k of each side of the defining equation for e_j^', and using the fact that each term is zero unless i=k, which leads to the first result.

I'm not quite sure how to go about the second part. I've attempted to show that UU* = I; but all I can see is that the diagonal elements of UU* are:

$\displaystyle\sum_{j=1}^n \left\langle \mathbf{e}_i, \mathbf{e}_j^{'} \right\rangle ^2$

for i = 1:n. I think that this probably could be shown to be 1 for all i using the definition given.

I'm not sure how the remaining elements are going to be zero though, which is necessary for UU* = I.

Any nudges in the right direction would be greatly appreciated!

 

[vela]

I haven't worked this out, so this is just a suggestion. It seems you haven't used the fact that the second basis is orthonormal yet. What do you get if you use the fact that $\langle \mathbf{e}'_i, \mathbf{e}'_j \rangle = \delta_{ij}$?

 

[Zatman]

Sorry, but I'm not sure how that can be used. The elements of U are scalar products of the i^th basis vector and the j^th primed basis vector, and the only relationship between the two bases is U itself.

What you wrote was used in a previous part of the question which I didn't bother typing out (it's not relevant to this part). Thank you, anyway.

 

[PeroK]

U looks like a linear transformation that maps

$e_i \ to \ e'_i$, so it preserves scalar products.

Then show that <U*Ux.y> = <Ux.Uy> = <x.y>

 

[vela]

Sorry, but I'm not sure how that can be used. The elements of U are scalar products of the i^th basis vector and the j^th primed basis vector, and the only relationship between the two bases is U itself.

What you wrote was used in a previous part of the question which I didn't bother typing out (it's not relevant to this part). Thank you, anyway.

No, it wasn't. I wrote the inner product of $e'_i$ and $e'_j$, both primed, not one primed and one unprimed.