Quick conditions to see that $I - E_1 -E_2$ is positive for $E_1, E_2$ positives (example from Nielsen and Chuang)

Question

In the Nielsen and Chuang ("Quantum Computation and Quantum Information"), section 2.2.6 POVM measurements, they define these three operators:

$E_1 = \frac{\sqrt{2}}{1+\sqrt{2}} |1\rangle \langle 1|$

$E_2 = \frac{\sqrt{2}}{1+\sqrt{2}} \frac{(|0\rangle-|1\rangle)(\langle 0|-\langle 1|)}{2}$

$E_3 = I - E_1 - E_2$

They say that it is straightforward to verify that these are positive operators. I agree that it is direct for $E_1, E_2$ and that we already know that $E_3$ is hermitian as a real linear combination of hermitian operators.

But I think I am missing something to quickly see that $E_3$ is positive. Is there a quick way to see it, other than compute $\langle x| E_3(x) \rangle$ for a general $x$? The only quicker way I found is to compute the spectrum of $E_3$ in order to see that all the eigeinvalues are positives. But I think I am missing a quicker way.

Also, I am not sure to understand why the factor of $E_1$ and $E_2$ are exactly $\frac{\sqrt{2}}{1+\sqrt{2}}$ and $\frac{\sqrt{2}}{2(1+\sqrt{2})}$, but I suppose this is in link with my first question (I think there is no reason for these factor except the need of positivity of $E_3$).

Answer 1

You obtain the factor by defining \begin{gather} E_1 = \alpha |1\rangle \langle 1| \\ E_2 = \alpha \frac{(|0\rangle-|1\rangle)(\langle 0|-\langle 1|)}{2} \\ E_3 = I - E_1 - E_2 \\ \end{gather} and then maximizing $\alpha$ such that $E_3$ is positive semidefinite. To check that the most straightforward way is to compute the spectrum.

You can also do it via the determinant; it will be positive if and only if both eigenvalues are positive or negative. But if the trace of $E_3$ is positive it is not possible for both eigenvalues to be negative, so conditioned on a positive trace you just need to check that the determinant is positive.

Note that you want $E_3$ to be positive semidefinite, though, it is fine to have an eigenvalue equal to zero. So at the end of the day what you want is to solve for $\alpha$ such that the determinant is zero and the trace is non-negative.