# Unitary operator/matrix

1. Jul 9, 2009

### facenian

I have a very basic question. I'm confused because I've read in a text that the matrix representation of a unitary operator is a unitary matrix if the basis is orthogonal, however I believe that the matrix is unitary whatever basis one uses. I'd appretiate any comments on this.

2. Jul 10, 2009

### malawi_glenn

which text?

3. Jul 10, 2009

### facenian

Quantum Mechanics(third edition) E. Merzbacher-Chapter 17 Page 418,I quote :"Since the operators U_a were assumed to be unitary, the representation matrices are also unitary if the basis is orthonormal"

4. Jul 10, 2009

### George Jones

Staff Emeritus
Try a simple example. Take the identity matrix on a 2-dimensional space, which is clearly unitary. Use linearity to compute the matrix elements with respect to the basis $e_{1}' = e_1$ and $e_{2}' = e_1 + e_2$, where $e_1$ and $e_1$ make up an orthonormal basis.

Does this give a unitary matrix?

5. Jul 10, 2009

### Fredrik

Staff Emeritus
A unitary matrix U satisfies $\sum_j U^*_{ji}U_{jk}=\delta_{ik}$. Is this satisfied by the matrix representation of a unitary operator?

$$\sum_j U^*_{ji}U_{jk}=\sum_j\langle j|U|i\rangle^*\langle j|U|k\rangle=\sum_j\langle i|U^\dagger|j\rangle\langle j|U|k\rangle=\langle i|U^\dagger\Big(\sum_j|j\rangle\langle j|\Big)U|k\rangle$$

This reduces to $\delta_{ij}$ if the parenthesis is the identity operator. I can prove that it is, if I use that the basis is orthonormal, but not without that assumption. So it looks like your book is right. What makes you think it's wrong?

6. Jul 10, 2009

### facenian

thank you people. The orgin of my mistake goes like this : Let T be a untiary operator and |a_i> (i=1,...n) a basis then the matrix elements satisfy,
<a_i|T|a_k>=<a_k|T^{\dag}|a_i>*=<a_k|T^{-1}|a_i>*
what a did not realize was that the matriz elementes in the basis |a_i> are <a_i|T|a_k> only if the basis is orthonomal.

7. Jul 10, 2009

### Fredrik

Staff Emeritus
Oops, I didn't realize that myself.