Let $A$ ne a $C^*$-algebra and $a,b\in A$ self-adjoint. a and b are orthogonal, iff $ab=0$.
Let A be nonunital and denote $A_1$ it's unitization, i.e., $A_1\cong A\oplus\mathbb{C}$ as vector spaces. Let $a+\alpha 1, b+\beta 1\in A_1$ be positiv elements which are orthogonal. This means $(a+\alpha 1)(b+\beta 1)=0$ and therefore at least one of $\alpha$ and $\beta$ has to be zero. Asumme $\beta =0$. This implies $b\ge 0$, because $b+\beta 1\ge 0$. Since $a+\alpha 1\ge 0$, it is $a^*=a$ and $\alpha\ge 0$. Let $(u_\lambda)_{\lambda\in \Lambda}$ an increasing approximate unit for $A$.
For every $\lambda\in\Lambda$ it is
$$a+\alpha 1=(a+\alpha 1)^{\frac{1}{2}}(1-u_\lambda)(a+\alpha 1)^{\frac{1}{2}}+(a+\alpha 1)^{\frac{1}{2}}u_\lambda (a+\alpha 1)^{\frac{1}{2}}.$$
It is $(a+\alpha 1)(b+01)=0$, but why follows, that $(a+\alpha 1)^{\frac{1}{2}}u_\lambda (a+\alpha 1)^{\frac{1}{2}}b=0$? Is this a hereditary property, because $0\le (a+\alpha 1)^{\frac{1}{2}}u_\lambda (a+\alpha 1)^{\frac{1}{2}} \le a+\alpha 1$? Regards
It is easier to see without the clutter notation. What you want to show is that if in some $C^*$-algebra $ab=0$ with $a\geq0$, then $a^{1/2}b=0$; there is no positivity requirement for $b$. Here are two proofs:
From $ab=0$, you get $$ (a^{1/2}b)^*a^{1/2}b=b^*ab=0,$$ so $a^{1/2}b=0$.
From $ab=0$, you get $a^nb=0$ for all $n\geq 1$, so $p(a)b=0$ for all polynomials $p$ with $p(0)=0$, and thus by taking limits $f(a)b=0$ for all continuous functions $f$ on the spectrum of $a$ that satisfy $f(0)=0$. The square root function is such a function.