A comment said that the most common way to encode q information in photons is using their internal degrees of freedom, not using a "there/not there" encoding.
When using photons, quantum information can indeed be encoded into an internal degree of freedom; for instance the polarization of the photon.
However, there are plenty of other systems where the information is encoded into an internal degree of freedom; a very clear example is an electron confined to a quantum dot (also loosely known as a semiconductor qubit). Here, the information is encoded into the qubits spin, which is definitely 'internal'. Such a semiconducting qubit definitely needs to be cooled (although its temperature can be higher than that of superconducting qubits!).
Moreover, there exist many encodings of quantum information for photons, and not all of them are 'internal'. In QKD systems, (on of) the most used encoding is the time-bin encoding, which to me is absolutely (in your terms) a "there/not there" encoding.
So does that mean that optical quantum computers that use photons doesn't suffer or suffers less from decoherence?
With those two previous things in mind I would argue that the discerning property for the amount of decoherence is not the 'internal-vs-external' encoding nature of the qubit.
You might be able to argue that photonic (quantum) computers suffer from different kinds of decoherence though...
What sort of environmental noise causes decoherence?
Well, pretty much anything. Decoherence is quite a broad term and can be seen as loosing the coherent quantum information because the system couples (uncontrolled and unknowingly) with the environment. Thermal noise is a huge issue for many qubit architectures, and indeed it is less so a problem for photonic quantum computers.
What is a large source of decoherence of photonics quantum computers is photon loss. Dependent on the encoding of the qubit, you may treat this as leakage (for 'internal' encodings) or as decoherence (or even amplitude damping; for 'external' encodings). Whatever you call it, photons may exit the system through modes that you don't intend/expect it to.
Of course there are other types of environmental noise, including, but not limited to:
- Magnetic coupling
- Electric coupling
- Stray photons
- Mechanical (not an issue for most architectures)
What is the expected number of qubits that will allow us to build a universal quantum computer?
This is an entirely different questions, and we need to treat it carefully.
If we are talking about a universal fault-tolerant computer (loosely speaking the high-in-the-sky 'end goal' of all quantum computer manufacturing efforts) the answer is:
many, many many qubits. No I really mean a lot. This paper states the need of $20$ million noisy (by then state-of-the-art noise levels, if I recall correctly) qubits to factor a $2048$ RSA key in $8$ hours. This number can be brought down in four ways:
- Better noise characteristics for the physical qubits.
- Better error correction and fault-tolerance schemes.
- Better connectivity between the qubits to reduce overhead.
- Smarter compilation & algorithms.
With the current fault-tolerant methods it is actually quite hard to determine the actual total need of physical qubits for a general fault-tolerant universal quantum computation, so that's why this paper was actually worth publishing at all.
If we are talking about quantum supremacy (loosely speaking the moment that we have a quantum computer that can perform something that a classical computer cannot do sensibly; some people don't even put on the constraint that it has to be something useful) then I would say that at around $~100$ qubits we will have a definite answer. Something useful (if not very limited) coming out of the computation will impose the need for a multiplying factor of (I guess) $2$ or $3$.
