[JaredL]

Here's a quick shot. I'm only considering heads up limit poker here.

Folding always nets zero.
EV of calling is zero if and only if the odds against winning are the same as the odds offered by the pot.

Suppose the odds against winning are X:1 and the pot is offering X:1. In other words he is X times as likely to have a better hand than he is to have a worse hand than you. The EV of raising is the EV of calling plus the expected winnings from the action coming as a result of the raise. Since you are getting 1:1 on the raise and the odds against winning are X:1 it cannot be the same EV if he will always call.

The villain here must fold often enough and with the right hands so that the EV of the raise is also zero.

Suppose she calls with probability p when she is ahead of you and q when she is behind. Let Y be the probability that you are ahead when she bets and r be the probability she calls at all. Notice that Y = 1/(X+1) and r = pY + q(1-Y). The EV of raising is (1-r)X + (X+1)r(q(1-Y)/r) - 2r(pY/r). Replace r with pY+q(1-Y), set that bad boy equal to zero and solve for p with respect to q. If your opponent calls your raises with these proportions then they are all the same.

Jared