diborane fragment orbitals energy of dihydrogen

I want to ask a question about the fragment orbitals of the $\ce{B2H6}$ system, specifically when I combine the $\ce{B2H4}$ with the $\ce{H2}$ fragments. 


I am halfway through my inorganic chemistry molecular orbitals course and today we were discussing the application of MO theory in figuring out the structure of $\ce{B2H6}$. 


So far, I have been able to combine the two $\ce{B2H4}$ FOs but I am struggling to justify the placement of the $\ce{H-H}$ FO to complete the task. 


So far, I've been given this diagram. 


[![enter image description here](https://i.stack.imgur.com/jQ0wf.png)](https://i.stack.imgur.com/jQ0wf.png)


And the professor said the following:



> 
> the $\ce{b2}$ orbitals are non-bonding FOs so it does has nothing to bond with in the $\ce{B2H6}$ molecule. We assume the H $\ce{1s}$ orbital is the reference and line up the hydrogen $\ce{H-H}$ FO with the $\ce{b2}$ FO, giving the symmetry groups $\ce{ag}$ and $\ce{b\_{3u}}$ respectively. 
> 
> 
> 


I understand the use of Hydrogen as a reference and the assignment of point groups in this situation, but I can't seem to work out why Hydrogen lines up with the $\ce{b2}$ FOs and not, for instance, with the $\ce{a1}$ FOs. 


I've searched across the internet for the past day and failed to find any conclusive answer. 


**Why is $\ce{H-H}$ given to have the same energy as the $\ce{b2}$ FOs of $\ce{B2H4}$?**


