Irv-
Yes, they should experience the same deflections as either loaded with one-half the load but since there is no defined shear capacity at the interface, they will act as if you had two Nominal 2x4's side by side instead of a nominal 2x8. Actually treating them this way, I looked at current values for Douglas Fir No. 1 per 2005 NDS supporting DL = 15 psf and LL= 40 psf and found:
Flexure: 1927 psi calculated 1725 allowable (12% overstress)
Shear: 80 psi calculated 180 psi allowable (o.k.)
Deflection: Total: 1.37" calculated (L/110) 0.60" allowed (218% of allowed)
Live: 0.95" calculated (L/152) 0.40" allowed (238% of allowable)
Frequency loaded with DL = 5.9 Hz (See Dolan and Woeste for method). Result would be a bit bouncy but may be mitigated if you have walls attached to the floor producing additional damping.
You don't say what load is on it when you measured the deflection, but if you loaded one joist pair, the effect of the flooring in distributing the load would bring the deflections down to about what you measured.
As for visualizing why they can act as one for vertical load, the calculated shear stress at mid depth would be 80 psi (3/2 x average shear for rectangular section). If you don't have this capacity, the stress distribution would have to change. While it doesn't affect shear capacity (which is dependent on area only), the I is reduced from 61.62 in^4 (a 1-3/4 x 7-1/2 member) to 7.69 x2 = 15.38 in^4 (two 1-3/4 x 3-3/4 members) resulting in a potential increase of 401% in deflection. Likewise, the S is reduced from 16.41 in^3 to 8.20 in^3 resulting in increased bending stress increase of 200%.
Regards,
Bill Cain, SE
Berkeley CA