Niles said:
Is this just that we write the wavefunction as a linear combination of each possible state, and the square of each constant in front of the possible state is the probability of a particle being in that state?
Yes
There's one caveat I have to add, however. When you say "each possible state", that should really be "eigenstate of a particular observable." You can't expand a wavefunction as a sum of eigenfunctions of different operators; you can't add a dirac delta function and a compex exponential, for example. Have you taken any linear algebra courses? The key mathematical idea behind advanced QM is that the one "state" of a system \Psi, as a point in a specific kind of vector space, can be represented in a basis of eigenfunctions \psi of one particular observable. (If you haven't taken any courses on linear algebra, don't worry if that sounds like double dutch!)
It's not homework - I am trying to get a better grasp on the infinite square well:
It didn't sound like a homework problem! The best place to discuss conceptual stuff like this isn't the homework forum, which is really for help in problem solving. If you take a problem like this to the Quantum Mechanics subforum, you'll find that 1)more people check it 2)The answers you get are much more straightforward as people will quite happily explain anything to you, wheras they don't want to do your homework for you.
1) This is easy. This just means that the probability of the particle being in this state is zero. But in my book (Griffiths, under the "Infinite Square Well") there are some graphs with position (x) along the x-axis and Psi up the y-axis. Then there is a node for Psi_2 (i.e. n = 2). Does this just mean that in the infinite square well, we particle cannot be found at that exact point?
If by "state" you mean "at that point" then that second sentence is correct. Actually, because |\Psi|^{2} is a probability
density function, the probability that you'll find a particle at
any single point is zero. (Gold has a huge mass density- but how much does zero cm^3 of gold weigh?) But a node means that the probability density associated with that point is zero, so that the chance of finding it in a small region centred around that point is zero.
2) This is just an indication of the energy of the particle?
Pretty much. The specific reason for it: in quantum mechanics the kinetic energy operator is obtained by applying the substitiution p\rightarrow -i\hbar\frac{d}{dx} (in one dimension) to the classical formula for kinetic energy \frac{p^2}{2m}. The momentum is linked to the wavefunction (think de Broglie) and hence so is the energy. The more nodes you have, the shorter your wavelength is. As you can only fit an integer number of half-wavelengths into an in infinite potential well, there's only certain allowed wavelengths, and hence allowed energies; this is the reason for the quantisation of energy in this problem, and the case of atomic energy levels is similar (only with much more horrible maths
)