A ferromagnet has a nonvanishing total magnetization which results in a finite magnetic field induced outside the magnet. For an Heisenberg antiferromagnet instead, the total magnetization is zero and therefore the "macroscopic" magnetic field induced is also zero in first approximation.
However, since an Heisenberg antiferromagnet can be thought in first approximation as a discrete arrangement of antiparallel magnetic moments (e.g., the Néel state), it is reasonable to imagine that on a very short scale, very close to the system (assume a planar surface), the magnetic field induced is non zero.
Ferromagnet: $\quad\quad\quad\quad\quad\quad\quad\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow\,\uparrow$
Antiferromagnet (Néel state): $\quad\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,\uparrow \, \downarrow \,$
So my questions are:
1) Is this reasoning correct? It is safe to assume a Néel state to calculate the field outside the antiferromagnet?
2) Is it possible to have a rough estimate of how big is this field and how fast it decays with the distance from the surface?
Edit regarding 1) The Neel is an approximate groundstate for an AFM, not the exact (which can be obtained using the Bethe ansatz). Thus, if you calculate the AFM field from the exact groundstate (suppose you could do it), you will get an "exact" field which is probably different from the approximate field calculated using the Néel state. Is the difference between two solutions relevant in condensed matter systems? How good is this approximation?