Why does the power supply into the op amp change from DC to AC?
You need significantly more capacitance from each rail to ground. Try 470uF or something like that in parallel with C3 and in parallel with C5. 

But if you're using a fancy high performance amplifier like that one, you're probably going to want to regulate the supply rails in reality. So that would mean a positive and a negative regulator, in addition to the aforementioned electrolytic caps and whatever additional parts the regulators require. 

An approximate formula for the p-p ripple on an unregulated supply is:

Vp-p = \$\frac{t_{hcyc} C}{I_{SUPPLY}}\$ where Isupply is the current drawn from the supply, C is the filter capacitance in farads and thcyc is the half-cycle time of the mains (0.01 seconds in the case of a 50Hz mains supply). 

When you add regulators, provided the valleys of your ripple stays well above the dropout voltage of the regulators, your ripple will be greatly attenuated by the regulator, perhaps down to tens of microvolts for a good regulator.