A node that behaves like 0V without a wire to prove it, and two input pins that act shorted together while drawing no current at all.
Every op-amp circuit-analysis shortcut you were taught — "assume V+ equals V−", "assume no current flows into the inputs" — rests on two ideas that get mangled constantly: the virtual short and the virtual ground. Neither one describes an actual wire. Both describe a consequence of negative feedback acting on an amplifier with enormous open-loop gain. Get the mechanism right and the shortcuts stop feeling like magic and start feeling inevitable.
A virtual short is the general case: in an ideal op-amp with negative feedback, the voltage difference between the inverting (−) and non-inverting (+) inputs is forced to essentially zero — not because current flows between the two input pins to "equalize" them, but because negative feedback drives the output to whatever value makes that difference vanish. A virtual ground is one specific, extremely common case of that: when the (+) input happens to be tied directly to real ground, the virtual-short behavior forces the (−) input to sit at ≈0V too — even though nothing physically connects that node to ground. That fictional 0V is what makes inverting-amplifier and summing-amplifier math so fast.
"Assume V+ = V−" and "assume zero input current" are two independent idealizations that happen to combine beautifully. Zero input current comes from the op-amp's enormous input impedance — nothing to do with feedback at all. The near-equal input voltages come from negative feedback: with gain this high, the only output value that keeps the circuit in its linear (non-saturated) region is the one that drives ε toward zero. Put both together at an inverting input tied only to a resistor network, and that node behaves exactly like ground for the purpose of computing currents through R1 and Rf — even though not a single electron is flowing into the op-amp to make that true, and no wire ties that node to earth.
Both halves of that sentence are wrong. Virtual ground is a node that behaves like 0V as a consequence of feedback action — there is no physical connection to earth at all, which is exactly why it's called "virtual." And virtual short describes a forced voltage equality between the two input pins, not a current path between them: the ideal op-amp's input current stays essentially zero the entire time, at both pins, whether or not the voltages are equal. "Short" here means "same voltage," not "a wire current can flow through." Confusing the two leads people to expect current flowing between the inputs, or a real ground connection you could ohmmeter-check at the (−) pin — neither of which exists.
Explains why an ideal op-amp's inverting input can behave like 0V (virtual ground) or track the non-inverting input (virtual short) without any current actually flowing between the input terminals — the feedback loop, not a physical wire or current path, is what forces the two inputs toward equal voltage.
Students often absorb "assume V+ = V−, assume no input current" as one blended rule without separating where each half comes from. That leads to two common errors: believing current flows between the inputs to "balance" them (it doesn't — ideal input current is ~0 regardless of the voltages), and believing a virtual ground node is physically wired to earth (it isn't — that's the entire point of calling it "virtual"). Both idealizations are true simultaneously, but for different reasons.
Zero input current follows directly from the op-amp's very high (ideally infinite) input impedance — it holds whether or not the circuit has feedback. Voltage equality between the inputs follows from negative feedback combined with very high open-loop gain: any sustained difference ε between V+ and V− would be multiplied by a gain of 10^5–10^6, demanding an output far beyond the supply rails. Since that's physically impossible in the linear region, the feedback loop settles at the one condition that keeps the output in range — ε ≈ 0. Tie the (+) input to real ground and the (−) input is forced to ≈0V too, without any wire connecting it there: that's the virtual ground.
This pair of idealizations is what makes hand analysis of inverting amplifiers, summing amplifiers, integrators, and differentiators fast: you can treat the inverting input as a fixed-voltage node (0V for a grounded non-inverting input, or whatever voltage is on the non-inverting input otherwise) and apply Ohm's law and KCL to the surrounding resistors directly. It stops being valid the instant the op-amp saturates, the feedback path opens, or the circuit is operated open-loop — which is exactly the regime a comparator deliberately lives in.
Current flows through the resistors connected to that node (input resistor and feedback resistor), but essentially none of it flows into the op-amp's input pin itself — by KCL, whatever current arrives through the input resistor must leave through the feedback resistor, since the op-amp pin draws (ideally) zero current.
Virtual ground is a specific case of the virtual short. Virtual short is the general statement that V+ ≈ V− under negative feedback. Virtual ground is what you call the (−) input specifically when the (+) input is tied to real ground, making V− ≈ 0V as well.
Yes, in practice you will measure very close to 0V there — but that reading is a consequence of feedback action holding the node at that potential, not evidence of a physical ground connection. Break the feedback path and that same node can sit at almost any voltage.
A comparator has no negative feedback forcing its inputs toward equality — it's designed to do the opposite, slamming its output to one rail or the other based on which input is larger. Without feedback, there's no mechanism driving ε to zero, so V+ and V− can differ substantially the entire time the circuit operates.
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