Consider a Hamiltonian of a system with two variables where it is possible to separate the two variables (here x_{1} and x_{2}) in the following way:
H(x_{1}, x_{2}) = H_{1}(x_{1}) + H_{2}(x_{2}) (Eq. 1)
For example, H_{1}(x_{1}), H_{2}(x_{2}) are two different Hamiltonians of two different particles in two different boxes. Suppose that we know the eigenvalues and eigenfunctions of the individual Hamiltonians:
H_{1}(x_{1})ψ_{k}(x_{1}) = E_{1,k}ψ_{k}(x_{1}) (Eq. 2)
H_{2}(x_{2})ψ_{l}(x_{2}) = E_{2,l}ψ_{l}(x_{2})
It is simple to show that the solution of the total Schrödinger equation
H(x_{1}, x_{2})ψ(x_{1}, x_{2}) = E_{TOTAL}ψ(x_{1}, x_{2}) (Eq. 3)
is simply related to the solution of its two independent components.
The eigenvalue is just the sum of the separate eigenvalues, and the wavefunction is the product of the wavefunctions.

In formulae:
H(x_{1}, x_{2}){ψ_{k}(x_{1}, x_{2})ψ_{l}(x_{2})} = [E_{1,k} + E_{2,l}]{ψ_{k}(x_{1}, x_{2})} (Eq. 4)
H(x_{1}, x_{2})ψ_{k,l}(x_{1}, x_{2}) = E_{k,l}ψ_{k,l}(x_{1}, x_{2}) (Eq. 5)
The eigenvalues of the total Hamiltonian E_{k,l} = E_{1,k} + E_{2,l} have two quantum numbers (the energy depends on two indexes) and the same is true for the eigenfunctions of the total Hamiltonian ψ_{k,l}(x_{1}, x_{2}) = ψ_{k}(x_{1}) + ψ_{l}(x_{2}) (the wavefunction depends on two indexes).
This property is used quite a lot  it will be essential to understand the hydrogen atom and it is the basis for the concept of orbital in multielectron systems.
Particle in a two dimensional box[edit  edit source]
Consider a particle of mass μ in a plane x,y which is constrained in a rectangular area with 0<x<L_{x} and 0<y<L_{y}. This is the two dimensional version of the particle in a box and the Hamiltonian can be written as:
${\hat {H}}(x,y)={\frac {\hbar ^{2}}{2\mu }}\left({\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}\right)+V_{x}(x)+V_{y}(y)$ (Eq. 6)
V_{x}(x)=0 0<x<L_{x} ; V_{x}(x)=+∞ elsewhere
V_{y}(y)=0 0<y<L_{y} ; V_{y}(y)=+∞ elsewhere
The Hamiltonian is clearly separable into two independent bits, each of them corresponds to a particle in the box in one dimension
${\hat {H}}(x,y)={\hat {H}}_{x}(x)+{\hat {H}}_{y}(y)=\left({\frac {\hbar ^{2}}{2\mu }}{\frac {\partial ^{2}}{\partial x^{2}}}+V_{x}(x)\right)+\left({\frac {\hbar ^{2}}{2\mu }}{\frac {\partial ^{2}}{\partial y^{2}}}+V_{y}(y)\right)$ (Eq. 7)
The eigenvalues and eigenfunctions have two different quantum numbers (say n and m) and take the following form:
$E_{n,m}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2\mu L_{x}^{2}}}+{\frac {m^{2}\pi ^{2}\hbar ^{2}}{2\mu L_{y}^{2}}}$ (Eq. 8a)
$\Psi _{n,m}(x,y)=\psi _{n}(x)\psi _{m}(y)={\frac {2}{\sqrt {L_{x}L_{y}}}}\sin \left({\frac {n\pi x}{L_{x}}}\right)\sin \left({\frac {m\pi y}{L_{y}}}\right)$ (Eq. 8b)
Particle in a three dimensional box[edit  edit source]
The eigenvalues of the particle in the box in 3 dimensions (here the particle is constrained to values 0<x<L_{x}, 0<y<L_{y}, 0<z<L_{z}) are:
$E_{n,m,l}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2\mu L_{x}^{2}}}+{\frac {m^{2}\pi ^{2}\hbar ^{2}}{2\mu L_{y}^{2}}}+{\frac {l^{2}\pi ^{2}\hbar ^{2}}{2\mu L_{z}^{2}}}$ (Eq. 9)
$\Psi _{n,m,l}(x,y,z)=\psi _{n}(x)\psi _{m}(y)\psi _{l}(z)={\frac {8}{\sqrt {L_{x}L_{y}L_{z}}}}\sin \left({\frac {n\pi x}{L_{x}}}\right)\sin \left({\frac {m\pi y}{L_{y}}}\right)\sin \left({\frac {l\pi z}{L_{z}}}\right)$
The proof of this relation and the derivation of the eigenfunctions are in the Exercise.
Polar coordinates and different form of the kinetic energy operator[edit  edit source]
For many problems (including the hydrogen atom) Cartesian coordinates are not the best choice and it is best to use polar coordinates. The geometric meaning of polar coordinates and their relation with Cartesian coordinates in 2 and 3 dimensions is given in the scheme below:
2D
x = r cosθ
y = r sinθ
M = (x, y) in Cartesian
M = (r, θ) in polar


3D
x = r cosθ sinφ
y = r sinθ cosφ
z = r cosφ
M = (x, y, z) in Cartesian
M = (r, θ, φ) in polar


The kinetic energy operator in 2 and 3 dimensions can also be written in polar coordinates:
2D: ${\frac {\hbar ^{2}}{2m}}\left\{{\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}\right\}={\frac {\hbar ^{2}}{2m}}\left\{{\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial }{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}}{\partial \theta ^{2}}}\right\}$ (Eq. 10)
3D: ${\frac {\hbar ^{2}}{2m}}\left\{{\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}\right\}={\frac {\hbar ^{2}}{2m}}\left\{{\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right\}$
In three dimensions the polar coordinates take the following values: r ≥ 0, 0 ≤ θ ≤ π and 0 ≤ Φ ≤ 2π.
Consider a particle of mass μ constrained to be on a circumference of radius R. If the position of the particle is expressed in polar coordinates the Schrödinger equation will be
$\left\{{\frac {\hbar ^{2}}{2\mu R^{2}}}{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right\}\psi (\phi )=E\psi (\phi )$ (Eq. 11)
The part of the kinetic energy operator containing the partial derivative with respect to r is not present, because the distance from the origin is constant.
The solution is very simple (it resembles the particle in the box but this time the complex form of the solution is used):
${\frac {\partial ^{2}}{\partial \phi ^{2}}}\psi (\phi )={\frac {2\mu R^{2}}{\hbar ^{2}}}E\psi (\phi )$ (Eq. 12)
$\psi (\phi )=\exp \left(i{\sqrt {\frac {2\mu R^{2}E}{\hbar ^{2}}}}\phi \right)$ (Eq. 13)
The boundary condition here is ψ(2π) = ψ(0) and this is only satisfied if ${\sqrt {\frac {2\mu R^{2}E}{\hbar ^{2}}}}=0,\pm 1,\pm 2,\pm 3,\cdots =m$ any integer (Eq. 14)
The allowed values for the energy are therefore:
$E_{m}={\frac {\hbar ^{2}}{2\mu R^{2}}}m^{2}$ (Eq. 15)
m = 0, ±1, ±2, ±3,... is the quantum number and the eigenfunctions are:
$\psi _{m}(\phi )={\frac {1}{\sqrt {2m}}}e^{im\phi }$ (Eq. 16)
Classical particle on a ring and angular momentum[edit  edit source]
This section is just to remind you that the linear momentum (pertinent to linear motion) is the product of the mass times the velocity. The angular momentum L (pertinent to rotational motion) is the product of the moment of inertia I times the angular velocity ω. The angular momentum is a vector directed along the axis of the rotation and, for a single particle, it is related to the linear momentum as ${\overrightarrow {L}}={\overrightarrow {r}}\times {\overrightarrow {p}}$ where ${\overrightarrow {p}}$ is the linear momentum and ${\overrightarrow {r}}$ is the distance from the origin. According to the rules of vector products the three components of ${\overrightarrow {L}}$ can be expressed as:
${\overrightarrow {r}}=x{\overrightarrow {i}}+y{\overrightarrow {j}}+z{\overrightarrow {k}};{\overrightarrow {p}}=p_{x}{\overrightarrow {i}}+p_{y}{\overrightarrow {j}}+p_{z}{\overrightarrow {k}}$
${\overrightarrow {L}}=L_{x}{\overrightarrow {i}}+L_{y}{\overrightarrow {j}}+L_{z}{\overrightarrow {k}}=(yp_{z}zp_{y}){\overrightarrow {i}}+(zp_{x}xp_{z}){\overrightarrow {j}}+(xp_{y}yp_{x}){\overrightarrow {k}}$ (Eq. 17)
For a particle rotating at distance R from a fixed point in the xy plane, the angular momentum is directed along z (L_{z} = μvR). The total energy is L_{z}^{2} ÷ 2μR^{2}. By comparison with Equation 15 we can argue that, in quantum mechanics, the momentum of the particle in a ring is quantized as L_{z} = ħm. This is one of the first ideas of Bohr leading to hsi primordial model or hydrogen atom.
Exercise
 Prove Equation 4 from Equation 2, i.e. if the Hamiltonian is the sum of two different Hamiltonians, the eigenfunctions are the product of the eigenfunctions of the component Hamiltonians, and the eigenvalues are the sum of the eigenvalues of the component Hamiltonians.
 Calculate the lowest 6 energy levels of a particle in a 2D box of side L = L_{x} = L_{y}.
 Calculate the ground state energy of a system made by an electron in
 a 1D box of length 1Å
 a 3D box of side L_{x} = L_{y} = L_{z} = 1Å
 a ring of radius 1Å
 Check that ħ has the dimension of the angular momentum.
 What is the zero point energy of the particle in a ring?
 Plot the probability density ψ(Φ)² for a state with m=0,1 of the particle in a ring.
 Using the concept of separable Hamiltonians, calculate eigenvalues and eigenfunctions of a particle of mass μ on a cylinder of radius R and length L whose Hamiltonian can be written as ${\hat {H}}={\frac {\hbar ^{2}}{2\mu }}{\frac {\partial ^{2}}{\partial x^{2}}}{\frac {\hbar ^{2}}{2\mu R^{2}}}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}+V(x)$ where V(x)=0 for 0<x<L and V(x)=∞ for x>L; x<0.
Solutions

 $H(x_{1},x_{2})\left\{\psi _{k}(x_{1})\psi _{l}(x_{2})\right\}=\left\{H_{1}(x_{1})+H_{2}(x_{2})\right\}\left\{\psi _{k}(x_{1})\psi _{l}(x_{2})\right\}=\psi _{l}(x_{2})H_{1}(x_{1})\psi _{k}(x_{1})+\psi _{k}(x_{1})H_{2}(x_{2})\psi _{l}(x_{2})=\psi _{l}(x_{2})E_{l,k}\psi _{k}(x_{1})+\psi _{k}(x_{1})E_{2,l}\psi _{l}(x_{2})=\left[E_{1,k}+E_{2,l}\right]\psi _{k}(x_{1})\psi _{l}(x_{2})$
 $E_{n,m}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL_{x}^{2}}}+{\frac {m^{2}\pi ^{2}\hbar ^{2}}{2mL_{y}^{2}}}=(n^{2}+m^{2}){\frac {\pi ^{2}\hbar ^{2}}{2mL^{2}}}$. The lowest 6 levels are for n=1 m=1; n=1 m=2; n=2 m=1; n=2 m=2; n=1 m=3; n=3 m=1.
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 6
 The given Hamiltonian is the sum of two Hamiltonians ${\hat {H}}={\hat {H}}_{1}(x)+{\hat {H}}_{2}(\varphi )$. ${\hat {H}}_{1}(x)={\frac {\hbar ^{2}}{2\mu }}{\frac {\partial ^{2}}{\partial x^{2}}}+V(x)$ is the Hamiltonian of the particle in the box with eigenvalues $E_{n}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2\mu L^{2}}}$ and eigenfunctions $\xi _{n}(x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)$ (with n=1, 2, ...). ${\hat {H}}_{2}(x)={\frac {\hbar ^{2}}{2\mu R}}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}$ is the Hamiltonian of the particle in a ring with eigenvalues $E_{m}={\frac {\hbar ^{2}}{2\mu R^{2}}}m^{2}$ and eigenfunctions $\chi _{m}(\varphi )={\frac {1}{\sqrt {2\pi }}}e^{im\varphi }$. The total eigenfunction is the product of the eigenfunctions of the Hamiltonians $\psi _{n,m}(x,\varphi )=\xi _{n}(x)\chi _{m}(\varphi )={\sqrt {\frac {1}{\pi L}}}\sin \left({\frac {n\pi x}{L}}\right)e^{im\varphi }$ and the total eigenvalue is the sum of the individual eigenvalues $E_{n,m}=m^{2}{\frac {\hbar ^{2}}{2\mu R^{2}}}+n^{2}{\frac {\pi ^{2}\hbar ^{2}}{2\mu L^{2}}}$.


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