So, I was trying to construct an operator for an orthonormal vector space $\mathbb{C}^2$. Suppose, I have two orthonormal vectors, $|x\rangle, |y\rangle$ such that $$\langle x|x\rangle=1,\langle y|y\rangle=1,\langle x|y\rangle=0.$$

Then, let's assume I have an operator $A$ such that $$A|x\rangle=\lambda_1|x\rangle,\quad A|y\rangle=\lambda_2 |y\rangle \\ \text{Then}, \langle x|A|y\rangle=\langle x|A|A|y\rangle = 0\\ \rightarrow\left[A\right]=\left[A\right]\left[A\right]$$

So, I should be looking for a 2X2 square matrix $A$ which is equal to it's square? From my knowledge, only an identity matrix or a zero matrix would work in this circumstance, but shouldn't this matrix be a Hermitian, since the operator necessarily needs to be a Hermitian, if it needs to have orthogonal eigenvectors? Am I going wrong somewhere?

  • $\begingroup$ Where did the extra $A$ in $ \langle x|A|y\rangle=\langle x|A|A|y\rangle = 0$ come from? $\endgroup$ – John Hughes Dec 25 '16 at 20:17
  • $\begingroup$ just dotting the $\langle x|A$ and the $A|y\rangle$ vectors $\endgroup$ – ubuntu_noob Dec 25 '16 at 20:51

As $x,y$ form a basis of $\mathbb{C}^2$ you can express every vector $v\in \mathbb{C}^2$ in a unique way as a linear combination of $x$ and $y$. I.e. we can write

$$ v= rx +sy $$

for some $r,s\in \mathbb{C}$. We can define our operator $A$ in the following way

$$ A: \mathbb{C}^2 \rightarrow \mathbb{C}^2, \ A(rx +sy) = \lambda_1r x + s \lambda_2 y.$$

If you express this operator as a matrix (in the basis {x,y}) you will just get the diagonal matrix

$$ \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}.$$

If you want to have your matrix with respect to the standard basis, just perform the respective change of basis.


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