A black hole will have a temperature that is a function of the mass, the angular momentum and the electric charge. For a fixed mass, Angular momentum and electric charge are bounded by the extremality condition
$$M^2 - a^2 - Q^2 \gt 0$$
Exactly at the extremality boundary, both entropy and temperature are zero.
Suppose i create a black hole with a spherically symmetric, incoming wavefront of electromagnetic radiation in a pure quantum state (that is, the density matrix satisfies the property $\rho^2 = \rho$). The wavefront is shaped in a way such that the whole energy of the packet will be inside the Schwarzschild radius, which will form an event horizon.
Since the wavepacket is as nearly as pure as it is physically possible to create, the quantum (Von Neumann) entropy is zero or nearly zero. But the formation of the black hole does not create nor destroy entropy, so the black hole must contain zero or nearly zero entropy as well. So the black hole seems to be extremal (it has zero temperature) but it nonetheless does not have any angular momentum (it is formed from a wavefront with zero net polarization over the whole sphere) and it does not have any charge (electromagnetic radiation is neutral).
Question: what "hair" does have a black hole formed from such a pure state, so that it can be extremal and still do not have angular momentum or electric charge (which are the classical hair that we come to expect from classical general relativity)
This question is a mutation of this question, but while that specific question tries to look what input black hole states create specific output (Hawking) radiation states that are far from thermal from a statistical point of view, this question is specific about extremality that is unrelated to angular momentum and charge