What does a nucleus look like?
I can't give you a real answer but I figured this may be better than nothing:
- I don't know of any better model, although I'm not an expert in nuclear physics so I can't be sure one doesn't exist.
- I suspect that any better model would not give a nice neat physical picture of the nucleus, the way we have a nice neat physical picture of the atom. In particular, I wouldn't expect to see a model that predicts concentric wavefunctions for nucleons, the way quantum mechanics describes electrons. Protons and neutrons are complex composite particles with many internal (and external) interactions, so their wavefunctions will tend to be much more localized than those of electrons. This is one case where the classical particle model actually is not too bad.
- From what I've heard (from people who know this business), the current state of the art in lattice QCD computations is calculating the mass of the proton with some accuracy. They're a long way away from being able to actually simulate a whole nucleus.
When you get down to these length scales, you have to be careful with words such as "look". In my mind, the "answer" for this question requires a roundabout in the philosophical area.
"Look" - what does it mean in physics? Even more relevant: "is". When a physicists uses the word "is", s/he is really using a shortcut for saying "behaves as it would be", that is, it behaves according to some model (this is a realization that took a couple of years for me to reach. Of course, I might change it in the future as I continue my physics intake). This is not only true for quantum physics, but in the macroscopic world, the distinction is often thought of as not as important. No one can deny the following though: a rod is a model for a certain object, and we can prescribe this model to real-life objects and calculate. In some cases, the physical object that we one day called a rod might be called a spring, depending on what we want to do. What "is" the object? In the physicists eyes, it is both a rod and a spring, and there is no incompatibility, as long as one knows what one is doing. What does it "look" like? It depends on the length scale and the application.
In the quantum world, de Broglie says that the constituents are both particles and waves. Contradiction? No.
How does a nucleus "look"? Which model is "right"? It depends. How I would intuitively imagine it in my mind is as a continuous domain of a charge (strong/weak/EM) cloud. Is this "correct"? It is one possible model, that works well in some applications. In other applications, it is just a hard sphere/wave/point/oscillator...
Sorry about the flaky amateur-philosophical mumbo-jumbo :-) Question to think about: what does "is" mean for a physicist (or a mathematician, or many other people).
In standard undergraduate nuclear physics course one learns about models such as Liquid Drop model and Shell model that explain some properties of nucleus.
But as I understand it these models are purely empirical and most importantly incompatible.
This is not really true.
They're not incompatible. The liquid drop model can be obtained as the classical limit of the shell model.
They're not purely empirical. The shell model is usually done using a residual interaction that is empirically determined, but that doesn't mean the model is purely empirical.
We can't define what a nucleus "looks like" in the optical sense because it doesn't interact with photons in ways that would allow that. But the mean field basically has the form of a sphere or prolate ellipsoid in almost all cases. This shape can fluctuate about the equilibrium, just like with any quantum-mechanical oscillator.
What is the current state of first-principles QCD computations of the nucleus?
Very small systems like the deuteron are somewhat doable. But in general QCD is neither necessary nor a sufficient for doing nuclear structure calculations. You're dealing with an n-body problem that is essentially intractable unless you can find tricks to make it tractable. For a nucleus with mass number $A$, we have $n=A$ if we treat it using nucleons as the particles, $n=3A$ if we use quarks (not even counting the gluons). There is no reason to make the problem even more intractable by tripling $n$. Low-energy nuclear structure is nonrelativistic, and there is simply no advantage to using a quark model.