This question already has an answer here:
Let's consider any circuit which contains a differential pair, for example a telescopic cascode single stage OTA:
In every circuit like this one, my professor and every book compute the small signal differential gain by assuming a perfect differential input couple of signals. Thus, as far as regards our example, we have:
and the small signal differential gain, exploiting the symmetry (i.e. the ac grounds on the nodes on the axis of symmetry which allow me to consider just half circuit), becomes:
This procedure is general: every time we have a circuit with a differential input pair, under the hypothesis of differential input signals, we can simplify the computation of the differential gain by exploiting symmetry (half circuit).
The question now is: why do we always compute the differential gain in the particular case of differential signals as input? In the most general case, indeed, we don't have differential signals. Consider for example this basic circuit:
In every book I've read, we pretend that the gain of the op-amp is the differential gain calculated before (i.e. assuming differential input signals). In other words: in order to use the previous expression of the differential gain, we should have the following situation in which the inverting and the non-inverting terminals of the op-amp receive differential signals (in red in the following picture):
But actually those red signals are not present: indeed the non-inverting terminal is fixed to ground, whereas (assuming large differential gain) the inverting terminal is almost zero.
In conclusion: why do we always calculate, in circuits containing a differential pair, the small signal differential gain assuming perfect differential signals as input? Why do we wrongly assume that the gain found in this way is also the differential gain of op-amps used in negative feedback?