1. ## Same Eigenvalues

Quantum computation and quantum ... - Google Books

I am doing the problem above.

$\displaystyle A_{ij} = < v_i| T | v_j>$ then use the fact that $\displaystyle I = \sum_m |w_m><w_m|$ and $\displaystyle I = \sum_n |w_n><w_n|$

eventually you get

$\displaystyle A_{ij} = \sum_m \sum_n <v_i|w_m><w_n|v_j>B_{mn}$

how do I invoke the unitary matrix? We know that $\displaystyle U=|w_i><v_i|$

2. I assume you're doing Exercise 2.20? What is $\displaystyle B_{mn}?$

3. It's the same as in the question (B_{ij}) except your introducing new indices m and n because I already used i and j. Is that clear?

4. Is that clear?
Not exactly. Are you doing Exercise 2.20?

5. Yes. Sorry.

6. I don't see a matrix $\displaystyle B$ anywhere in the problem statement.

7. What does $\displaystyle |\rangle$ mean?

8. In this case B would be their A_{ij}^{''}

Is this the appropriate forum I should have posted in?

9. Drexel28: in quantum mechanics, "kets", or column vectors (at least in finite-dimensional Hilbert spaces) look like this: $\displaystyle |v\rangle,$ whereas "bras", or row vectors in the dual space look like this: $\displaystyle \langle w|.$ A "bracket" is an inner product of a bra with a ket: $\displaystyle \langle w|v\rangle.$ You can also form a matrix with the outer product thus:

$\displaystyle \displaystyle\sum_{i}|w_{i}\rangle\langle v_{i}|.$

10. Nusc:

This forum is just fine. You could have gone either here or in the Advanced Applied Math forum. The underlying math is definitely linear algebra, though, so you're fine.

So you're claiming that

$\displaystyle \displaystyle A_{ij}'=\sum_{m}\sum_{n}\langle v_{i}|w_{m}\rangle\langle w_{n}|v_{j}\rangle A_{ij}''?$

11. Originally Posted by Ackbeet
Drexel28: in quantum mechanics, "kets", or column vectors (at least in finite-dimensional Hilbert spaces) look like this: $\displaystyle |v\rangle,$ whereas "bras", or row vectors in the dual space look like this: $\displaystyle \langle w|.$ A "bracket" is an inner product of a bra with a ket: $\displaystyle \langle w|v\rangle.$ You can also form a matrix with the outer product thus:

$\displaystyle \displaystyle\sum_{i}|w_{i}\rangle\langle v_{i}|.$
Last question (there's a chance I can help if I understand this) if one has some finite dimensional Hilbert space $\displaystyle \mathcal{H}$ with inner product $\displaystyle \langle\cdot,\cdot\rangle$ and $\displaystyle \varphi\in\text{Hom}\left(\mathcal{H},F\right)$ (a covector if that notation is unfamiliar...I know the terminology covector is common in physics, right?) what does $\displaystyle \langle \varphi,v\rangle$ even mean for $\displaystyle v\in\mathcal{H}$?

12. That's correct. Then somehow I am supposed to come up with U_{im} and U_{nj}^{*} but I thought it was the other way around U_{mn} and U_{ij}^{*}.

Then I am supposed to show that they have the same eigenvalues. Similar matrices have the same eigenvalues but I am not sure how relevant that is based on what is given.

$\displaystyle A_{ij}'= U_{im} U_{nj}^{*} A_{ij}''$ which is unclear.

13. Reply to Drexel28: You're way over my head there. I do know this: a Hilbert space is a complete inner product space, by definition. So with any Hilbert space, there is an inner product, and it's defined for all the vectors in the Hilbert space.

Why not try this:

$\displaystyle A_{ij}'=\langle v_{i}|A|v_{j}\rangle$ implies

$\displaystyle \displaystyle\sum_{i}|v_{i}\rangle A_{ij}'=\sum_{i}|v_{i}\rangle\langle v_{i}|A|v_{j}\rangle=A|v_{j}\rangle,$ and hence

$\displaystyle \displaystyle\sum_{i,j}|v_{i}\rangle A_{ij}'\langle v_{j}|=\sum_{j}A|v_{j}\rangle\langle v_{j}|=A.$

Similarly,

$\displaystyle \displaystyle\sum_{i,j}|w_{i}\rangle A_{ij}''\langle w_{j}|=A.$

Therefore,

$\displaystyle \displaystyle\sum_{i,j}|w_{i}\rangle A_{ij}''\langle w_{j}| =\sum_{i,j}|v_{i}\rangle A_{ij}'\langle v_{j}|.$

15. That's much more elegant!