A wave function is an infinite dimensional vector space, how can it "live" in $\mathbb{R}^3$?
Given the equation that is built like: $$\Psi (x,t) = \sum ^{\infty} _{n=1} c_n \psi _n (x) e^{-i E_n t / \hbar}$$
How does one "excite" a quantum particle in the lab? The excited states simply give a different probability distribution for the particle?
For the infinite square well, the wave function in excited state $n$ has exactly $n$ bases, all of which are non zero? If a particle is in its ground state, does that mean that all other bases must be zero? making the wave function dependent only on one basis?
Best Answer
Let's go step by step:
1) A particle (or a system in general) can be found in a given state $|\psi\rangle$, which mathematically is defined as a vector of an Hilbert state $\mathcal{H}$. This is to ensure that the concept of scalar product (necessary to define ortonormal basis) and norm (necessary to define the probability amplitude) are well defined. When you then decide to expand your wavefunction on the position basis you get something like: $$ |\psi\rangle = \mathbb{1}|\psi\rangle = \int dr |r\rangle\langle r|\psi\rangle = \int dr\, \Psi(r)|r\rangle $$
where I defined $\Psi(r): \mathbb{R}^3\rightarrow\mathbb{C}$ as the wavefunction of the state $|\psi\rangle$ represented at the position $r$. If you want your wavefunction at position $x$, all you have to do is multiply by $\langle x|$ the above expression: $$ \langle x|\psi\rangle = \int dr \,\Psi(r)\langle x|r\rangle = \int dr \,\Psi(r)\delta(x-r) = \Psi(x) $$
and there you go.
2) Suppose now that you want to perform a measurement to characterize your system. Mathematically an operator $\hat O$ is associated to the measurement process, which you suppose has a complete orthonormal set of eigenvalues $\{|\psi_n\rangle \}_{n=0}^\infty$. You can then expand the wavefunction associated to your system as in the formula you provided. Now, since you choose a set of basis functions, what specifies the wavefunction is the coefficients $c_n$ of the expansion. Therefore, if you perturbe the system, your wavefunction will have another set of coefficients $c'_n$ onto the same basis. Since the probability amplitude is defined as the square modulus of the wavefunction, you'll get (exercise!) $$ |\Psi(x,t)|^2 = \sum_n |c_n|^2 $$ therefore a different set of coefficients will give you a different probability distribution.
3) If you are in the ground state of the system, the result of the measurement will be $$ \hat O|\psi\rangle = |\psi_0\rangle $$ every time you perform the measurement (see the Von Neumann's postulate). This means that in the coefficient expansion only $|c_0|=1$, while or the other coefficients are zero. It's really wrong to say that the basis terms must be zero, the basis is just the basis and it's associated to the measurement operator, it's the coefficients that matter for the wavefunction. Also "making the wavefunction depend on one basis" doesn't make much sense. Of course the representation of the wavefunction depends on the basis of choice, but you choose the basis with respect to the measurement you want to perform on it, i.e. with respect to the operator you're considering.