You can interpret it as $1\over \sqrt 2$ times as much as the Standard deviation of $a$ and $b$. According to definition $$\sigma=\sqrt{(a-m)^2+(a-m)^2}$$where $m={a+b\over 2}$ therefore $$\sigma=\sqrt{{(a-b)^2}\over 2}=\sqrt 2{\left|{a-b\over 2}\right|}$$and we obtain$$\left|{a-b\over 2}\right|={\sigma(a,b)\over \sqrt 2}$$but I don't know whether any other interpretation exists.
You can interpret it as $1\over \sqrt 2$ times as much as the Standard deviation of $a$ and $b$. According to definition $$\sigma=\sqrt{(a-m)^2+(a-m)^2}$$where $m={a+b\over 2}$ therefore $$\sigma=\sqrt{{(a-b)^2}\over 2}=\sqrt 2{\left|{a-b\over 2}\right|}$$and we obtain$$\left|{a-b\over 2}\right|={\sigma(a,b)\over \sqrt 2}$$but I don't know whether any other interpretation exists.