Mongodb query with fields in the same documents

You could use $where:

db.myCollection.find( { $where: "this.a1.a != this.a2.a" } )

However, be aware that this won't be very fast, because it will have to spin up the java script engine and iterate each and every document and check the condition for each.

If you need to do this query for large collections, or very often, it's best to introduce a denormalized flag, like areEqual. Still, such low-selectivity fields don't yield good index performance, because he candidate set is still large.

update

using the new $expr operator available as of mongo 3.6 you can use aggregate expressions in find query like this:

  db.myCollection.find({$expr: {$ne: ["$a1.a", "$a2.a"] } });

---

Although this comment solves the problem, I think a better match for this use case would be to use $addFields operator available as of version 3.4 instead of $project.

db.myCollection.aggregate([
     {"$match":{"a1":{"$exists":true},"a2":{"$exists":true}}},
     {"$addFields": {
           "aEq": {"$eq":["$a1.a","$a2.a"]}
         }
     },
     {"$match":{"aEq": false}} 
  ]);