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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. |
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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. |
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