Eigenvector orthogonality and unitary operator diagonalization

[Gary Roach]

Homework Statement

For reference: Problem 1.8.5 parts (3) , R. Shankar, Principles of Quantum Mechanics.
Given array $$\Omega$$, compute the eigenvalues ($e^i^\theta$ and $e^-^i^\theta$). Then (3) compute the eigenvectors and show that they are orthogonal.

Homework Equations

Eulers equation needed:
$$e^\pm^i^\theta = \cos\theta \pm i\sin\theta$$

The matrix;
$$\Omega = \left |\begin{array}{cc}\cos\theta& \sin\theta\\-\sin\theta& \sin\theta\end{array} \right|$$

The Attempt at a Solution

Computer eigenvectors (Some QM ket notation used)
$$|\omega = e^i^\theta > \Rightarrow \left|\begin{array}{cc}\cos\theta - (\cos\theta + i\sin\theta)& \sin\theta \\-\sin\theta& \cos\theta - (\cos\theta +i\sin\theta) \end{array} \right |$$

$$|\omega = e^i^\theta > \Rightarrow \sin\theta\left|\begin{array}{cc}-i& 1\\-1& -i \end{array}\right |$$
Since the array must equal the zero array, $$\sin\theta$$ can be eliminated and:
$$-ix_1 + x_2 = 0 \\<br /> -x_1 + -ix_2 = 0 \\<br /> \Rightarrow x_1 = -ix_2$$

So the eigenvector should be:
$$|\omega = e^i^\theta > = \left|\begin{array}{cc}-i\\1\end{array}\right| \\ letting& x_2=1$$

Similarly:
$$|\omega = e^-^i^\theta > \Rightarrow \sin\theta\left|\begin{array}{cc}i& 1\\-1& i \end{array}\right |$$
$$ix_1 + x_2 = 0 \\ <br /> -x_1 + ix_2 = 0 \\ <br /> \Rightarrow x_1 = ix_2$$
So the eigenvector should be:
$$|\omega = e^-^i^\theta > = \left|\begin{array}{cc}i\\1\end{array}\right|$$

I think I have all of this right but unfortunately these two vectors are supposed to be orthogonal implying that their inner product sould be zero. And:
$$\left|\begin{array}{cc}-i\\1\end{array}\right| \cdot \left|\begin{array}{cc}i\\1\end{array}\right| = 1+1 =2 \neq 0$$ ??

It's probably a stupid error but I can't find it. Any help will be sincerely appreciated. I hope this get cleaned up in the final version. The preview feature won't show corrections.

Gary R.

 

[Dick]

The inner product here is <a,b>=conjugate(a^(T))b. You forgot the complex conjugate on the first vector.

 

[Gary Roach]

Oops. There seems to be a hole here in my knowledge. Oh well, back to the books. Thanks for the prompt reply.

Gary R.