Suppose the Earth is not rotating. As usual, the Moon follows its normal path around the Earth. Let's assume it's a circular motion and that there are no other gravitational influences.
A test particle on the surface of the Earth directed to the Moon on the line that connects the middle points of both will feel the Earth's gravity directed to its center and the Moon's gravity directed to the Moon.
A test particle on the opposite side of the Earth will also feel the Earth's gravity directed to its center (equal but opposite to the gravity experienced by the above-mentioned test particle) and the (smaller because of the greater distance to the Moon) gravity caused by the Moon.
So the surface water on Earth is attracted on both sides where the test particles reside with equal strength to the Earth. But the gravitational influence of the Moon on the water is bigger on the side directed to the Moon than it is on the opposite side. The difference is the greatest for the two test particles.
You would expect that because the water is pulled to the Moon by a little bigger gravitational force on the side of the Earth facing the Moon than on the opposite side (not facing the Moon), a bulge of water, directed to the Moon, will emerge that rotates around the Earth in sync with the Moon's motion.
But we must of course not forget that in this case there are also centrifugal forces that pull on the water due to the rotation of the Earth and the Moon around their CM (this is of course not the rotation of the Earth around its axis which I put zero). The CM between the mass and the Moon lies at 4600(km) from the center of the Earth in the direction of the Moon. The centrifugal force is bigger on the part of the Earth the furthest from the CM, where the gravitational effect of the Moon is the smallest (when the Earth is rotating also the centrifugal forces due to the rotation of the Earth itself come into play, which makes the situation more complicated).
So the question reduces to: What is the ratio between the centrifugal force plus the gravitational force caused by the Moon on the furthest part on Earth to the Moon and the centrifugal force plus the gravitational force caused by the Moon on the closest part of the Earth.
On the far side (from the Moon) of the Earth the Moon's gravitation is smaller but the centrifugal force bigger, while on the close side the Moon's gravitation is bigger and the centrifugal force smaller. Anybody who can do this quite straightforward calculation (for finding the ratio)? One thing is sure: two opposite bulges will develop which each go around the Earth in the same time as the Moon makes one complete cycle around the Earth. The ratio gives us information about the height of the bulbs.
I made an obvious error (to make my error clear I let the question as it is). The Earth isn't rotating (as can be read in my question), so no centrifugal forces are present. It's true that, in this case, the CM (4200(km) from the center of the Earth) rotates around the center of the Earth in sync with the rotation of the Moon, but the Earth doesn't rotate around the CM. So there are only tidal forces at work which causes the two bulges rotating around the Earth in one Moon cycle.