The Schrödinger Equation provides a Probability Density map of the atom. In light of that, are either of the following possible:
I have taken two calculus based physics, and Modern Physics with the Schrödinger equation, Heisenberg Uncertainty Principle, Etc.
1.) No. All the calculations one does in elementary quantum mechanics courses are at zero temperature. If they were at a finite temperature, you could never reliably say what quantum mechanical state your system is in; it would always be in an ensemble of different states. Since the ground-state wavefunction and ground-state density is not a 2d surface, you don't get one at $T = 0$.
2.) No. At zero temperature, the probability density of your electron is given by the ground state wavefunction: $$\varrho(x) = \psi_0^*(x) \psi_0(x)$$ At finite temperature, your system is best described by an ensemble of states. Basically, you get $$\varrho(x) = \sum_i p_i \psi_i^*(x) \psi_i(x)$$ where $p_i$ is the ensemble-probability for your system to be in state $\psi_i(x)$. For a canonical ensemble, for example, you have $p_i \sim e^{-E_i/kT}$ if your $\psi_i(x)$ are the energy-eigenstates with eigenenergies $E_i$.
The same is true for any other expectation value: $$\langle \hat A \rangle = \sum_i p_i \langle \psi_i | \hat A | \psi_i \rangle$$ Note the two different expectation value here: One is $\langle \psi_i | \hat A | \psi_i \rangle$, the quantum mechanical expectation value of $\hat A$ when the system is in state $| \psi_i \rangle$. The sum over these, together with the $p_i$, then gives the thermodynamic expectation value
This framework is used everywhere in physics and has been proven to be mind-bogglingly exact.
