Question regarding Schrödinger's equation Here is my understanding of Schrödinger's equation and wave functions: (I'm considering the time independent equation.)

The equation takes in values of energy as an input and, being a differential equation, gives functions as outputs. These functions correspond to the different orbitals.
When we input coordinate values we get an answer that we can use to plot the probability distribution of an electron around the nucleus.

Can someone tell me if my understanding is correct, and correct me if I'm wrong?
 A: To solve the Schrodinger equation for a system like a hydrogen atom we need to know:


*

*the potential energy function

*the boundary conditions
The potential energy $V$ goes into the (time independant) Schrodinger equation:
$$ \left(-\frac{\hbar^2}{2m}\nabla^2 + V \right) \Psi = E\Psi $$
but this alone isn't enough to tell us what the wavefunction looks like. We also need the boundary conditions that the wavefunction must go to zero at infinity and be normalised to unity. It's only when we impose the boundary conditions that we restrict the set of solutions to the $1s$, $2s$, etc atomic orbitals that we know and love.
Once we have the solutions allowed by the boundary conditions then we can indeed plug our values of position into the equation for $\Psi$ and calculate the probability distribution of the electron.
A: The equation for a Hamiltonian $H$ which is time independent is given by,
$$ \left[\frac{-\hbar^2}{2m}\nabla^2 + V \right] \psi = E\psi$$
We can interpret the equation as simply stating the eigenvalues of the Hamiltonian correspond to the expectation value of the energy in a particular state. Once the equation is solved with appropriate boundary conditions, we obtain a wave function $\psi$ which is normalized such that,
$$\int_{-\infty}^{\infty} \mathrm{d}x_1 \, \dots \int_{-\infty}^{\infty} \mathrm{d}x_n \, \,  |\psi|^2=1$$
which is simply the demand that the probability we find the particle somewhere in the universe to be $100\%$ i.e. unity. If we want to know the probability within a volume $V$, it is given by,
$$P = \int_V \mathrm{d}^d x \, \, |\psi|^2$$
The orbitals we know from chemistry are indeed derived from $\psi$ which for the hydrogen-like ion carries three indices, known as quantum numbers, as they are required to fully label or index the states.

What if we want to know the average position in a state? For this example, we take the position operator $\hat{x}$ which when it acts upon a state, behaves as,
$$\hat{x}\lvert \psi \rangle  = x \lvert \psi \rangle$$
$$\langle \psi \lvert \hat{x}\lvert \psi \rangle = x \langle \psi \lvert \psi \rangle = x$$
because $x$ is only a constant, and we properly normalized our wave function. Hence,
$$x=\langle \psi \lvert \hat{x}\lvert \psi \rangle = \int \mathrm{d}^d x \, \, \psi^{\dagger}(x)\,  x \, \psi(x)$$
where on the l.h.s. we consider $\hat{x}$ an operator, and in the integral $x$ is treated as a variable which we integrate over. The 'dagger' arises because $\langle \psi \lvert = (\lvert \psi \rangle)^{\dagger}$.
