Hydrogen-like atom

In physics and chemistry, a hydrogen-like atom (or hydrogenic atom) is an atom with one electron. Except for the hydrogen atom itself (which is neutral) these atoms carry positive charge e(Z-1), where Z is the atomic number of the atom and e is the elementary charge. Because hydrogen-like atoms are two-particle systems with an interaction depending only on the distance between the two particles, their non-relativistic Schrödinger equation can be solved in analytic form. The solutions are one-electron functions and are referred to as hydrogen-like atomic orbitals. These orbitals differ from one another in one respect only: the nuclear charge eZ appears in the radial part of the wave function.

Hydrogen-like atoms per se do not play an important role in chemistry. The interest in these atoms is caused mainly by the fact that in quantum mechanics their Schrödinger equation can be solved as easily as for the hydrogen atom itself, because Z enters the problem in a trivial way.

Quantum numbers

Hydrogen-like atomic orbitals are eigenfunctions of a Hamiltonian (energy operator) with eigenvalues proportional to 1/n², where n is a positive integer. Further the orbitals are usually chosen such that they are also eigenfunctions of the square of the one-electron angular momentum vector operator

$\mathbf {l} \equiv -i\hbar \,(\mathbf {r} \times {\boldsymbol {\nabla }})\equiv (l_{x},\;l_{y},\;l_{z}),$

where $\hbar$ is Planck's constant divided by 2π. Since l ² ≡ l_x² + l_y² + l_z² commutes with the three angular momentum components, it is possible to require an orbital to be an eigenfunction of any of the three. It is convention to choose l_z, which has an eigenvalue proportional to an integer usually denoted by m. The square l ² has an eigenvalue proportional to l(l+1), where l is a non-negative integer.

It is important to note that the energies of the hydrogen-like orbitals do not depend on the angular momentum quantum numbers l and m, but solely on the principal quantum number n. The degeneracy (maximum number of linearly independent eigenfunctions of same energy) of energy level n is equal to n². This is the dimension of the irreducible representations of the symmetry group of hydrogen-like atoms, which is SO(4).

A hydrogen-like atomic orbital is uniquely identified by the values of the principal quantum number n, the angular momentum quantum number l, and the magnetic quantum number m. These quantum numbers are integers and we summarize their ranges:

${\begin{aligned}n&=1,2,3,4,\ldots ,\\l&=0,1,2,\ldots ,n-1,\\m&=-l,-l+1,\ldots ,l-1,l.\end{aligned}}$

This set must be augmented by the two-valued spin quantum number ms = ±½ in application of the exclusion principle. This principle restricts the allowed values of the four quantum numbers in electron configurations of more-electron atoms: it is forbidden that two electrons have the same four quantum numbers.

Schrödinger equation

The atomic orbitals of hydrogen-like atoms are solutions of the time-independent Schrödinger equation in a potential given by Coulomb's law:

$V(r)=-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Ze^{2}}{r}}$

where

• ε₀ is the permittivity of the vacuum,
• Z is the atomic number (charge of the nucleus in unit e),
• e is the elementary charge (charge of an electron),
• r is the distance of the electron from the nucleus.

The Schrödinger equation is the following eigenvalue equation of the Hamiltonian (the quantity in large square brackets):

$\left[-{\frac {\hbar ^{2}}{2\mu }}\nabla ^{2}+V(r)\right]\psi (\mathbf {r} )=E\psi (\mathbf {r} ),$

where $\hbar$ is Planck's constant divided by 2π, and where μ is, approximately, the mass of the electron. More accurately, it is the reduced mass of the system consisting of the electron and the nucleus. Because the electron mass is about 1836 smaller than the mass of the lightest nucleus (the proton), the value of μ is very close to the mass of the electron m_e for all hydrogenic atoms. In the remaining of the article we will often make the approximation μ = m_e. Since m_e will appear explicitly in the formulas it will be easy to correct for this approximation if necessary.

In this article (in which l ² is defined without Planck's constant and imaginary unit i) it is shown that the operator ∇² expressed in spherical polar coordinates, can be written as

$\hbar ^{2}\nabla ^{2}={\frac {\hbar ^{2}}{r}}{\frac {\partial ^{2}}{\partial r^{2}}}r-{\frac {l^{2}}{r^{2}}}.$

The wave function is written as a product of functions in the spirit of the method of separation of variables:

$\psi (r,\theta ,\phi )=R(r)\,Y_{lm}(\theta ,\phi )\,$

where Y_lm are spherical harmonics, which are eigenfunctions of l ² with eigenvalues $\hbar ^{2}l(l+1)$. Making this substitution, we arrive at the following one-dimensional Schrödinger equation:

$-{\frac {\hbar ^{2}}{2\mu }}\left[{\frac {1}{r}}{\frac {d^{2}}{dr^{2}}}rR(r)-{\frac {l(l+1)R(r)}{r^{2}}}\right]+V(r)R(r)=ER(r),$

Wave function and energy

In addition to l and m, there arises a third integer n > 0 from the boundary conditions imposed on R(r). The expression for the normalized wave function is:

$\psi _{nlm}=R_{nl}(r)\,Y_{lm}(\theta ,\phi ).$

Below it will be derived that

$R_{nl}(r)={\sqrt {{\left({\frac {2Z}{na_{\mu }}}\right)}^{3}{\frac {(n-l-1)!}{2n[(n+l)!]}}}}e^{-Zr/{na_{\mu }}}\left({\tfrac {2Zr}{na_{\mu }}}\right)^{l}L_{n-l-1}^{2l+1}\left({\tfrac {2Zr}{na_{\mu }}}\right)$

where:

• $L_{n-l-1}^{2l+1}$ are the generalized Laguerre polynomials in the definition given here.
• $a_{\mu }={{4\pi \varepsilon _{0}\hbar ^{2}} \over {\mu e^{2}}}$

Note that $a_{\mu }\,$ is approximately equal to $a_{0}\,$ (the Bohr radius). If the mass of the nucleus is infinite then $\mu =m_{e}\,$ and $a_{\mu }=a_{0}\,$.

• $E_{n}=-{\frac {\mu }{2}}\left({\frac {e^{2}}{4\pi \varepsilon _{0}\hbar n}}\right)^{2}$. (Energy eigenvalues. As we pointed out above they depend only on n, not on l or m).

• $Y_{lm}(\theta ,\phi )\,$ function is a spherical harmonic.

Derivation of radial function

As we just saw, we must solve the one-dimensional eigenvalue equation,

$\left\{-{\hbar ^{2} \over 2m_{e}r}{d^{2} \over dr^{2}}r+{\hbar ^{2}l(l+1) \over 2m_{e}r^{2}}+V(r)\right\}R(r)=ER(r),$

where we approximated μ by m _e. If the substitution $u(r)\ {\stackrel {\mathrm {def} }{=}}\ rR(r)$ is made, the radial equation becomes

$-{\hbar ^{2} \over 2m_{e}}{d^{2}u(r) \over dr^{2}}+V_{\mathrm {eff} }(r)u(r)=Eu(r)$

which is a Schrödinger equation for the function u(r) with an effective potential given by

$V_{\mathrm {eff} }(r)=V(r)+{\hbar ^{2}l(l+1) \over 2m_{e}r^{2}}.$

The correction to the potential V(r) is called the centrifugal barrier term.

In order to simplify the Schrödinger equation, we introduce the following constants that define the atomic unit of energy and length, respectively,

$E_{\textrm {h}}=m_{e}\left({\frac {e^{2}}{4\pi \varepsilon _{0}\hbar }}\right)^{2}\quad {\hbox{and}}\quad a_{0}={{4\pi \varepsilon _{0}\hbar ^{2}} \over {m_{e}e^{2}}}$.

Substitute $y=Zr/a_{0}\,$ and $W=E/(Z^{2}E_{\textrm {h}})\,$ into the radial Schrödinger equation given above. This gives an equation in which all natural constants are hidden,

$\left[-{\frac {1}{2}}{\frac {d^{2}}{dy^{2}}}+{\frac {1}{2}}{\frac {l(l+1)}{y^{2}}}-{\frac {1}{y}}\right]u_{l}=Wu_{l}.$

Two classes of solutions of this equation exist: (i) W is negative, the corresponding eigenfunctions are square integrable and the values of W are quantized (discrete spectrum). (ii) W is non-negative. Every real non-negative value of W is physically allowed (continuous spectrum), the corresponding eigenfunctions are non-square integrable. In the remaining part of this article only class (i) solutions will be considered. The wavefunctions are known as bound states, in contrast to the class (ii) solutions that are known as scattering states.

For negative W the quantity $\alpha \equiv 2{\sqrt {-2W}}$ is real and positive. The scaling of y, i.e., substitution of $x\equiv \alpha y$ gives the Schrödinger equation:

$\left[{\frac {d^{2}}{dx^{2}}}-{\frac {l(l+1)}{x^{2}}}+{\frac {2}{\alpha x}}-{\frac {1}{4}}\right]u_{l}=0,\quad {\hbox{with}}\quad x\geq 0.$

For $x\rightarrow \infty$ the inverse powers of x are negligible and a solution for large x is $\exp[-x/2]$. The other solution, $\exp[x/2]$, is physically non-acceptable. For $x\rightarrow 0$ the inverse square power dominates and a solution for small x is x^l+1. The other solution, x^-l, is physically non-acceptable. Hence, to obtain a full range solution we substitute

$u_{l}(x)=x^{l+1}e^{-x/2}f_{l}(x).\,$

The equation for f_l(x) becomes,

$\left[x{\frac {d^{2}}{dx^{2}}}+(2l+2-x){\frac {d}{dx}}+(\nu -l-1)\right]f_{l}(x)=0\quad {\hbox{with}}\quad \nu =(-2W)^{-{\frac {1}{2}}}.$

Provided $\nu -l-1$ is a non-negative integer, say k, this equation has polynomial solutions written as

$L_{k}^{(2l+1)}(x),\qquad k=0,1,\ldots ,$

which are generalized Laguerre polynomials of order k. We will take the convention for generalized Laguerre polynomials of Abramowitz and Stegun. Note that the Laguerre polynomials given in many quantum mechanical textbooks, for instance the book of Messiah, are those of Abramowitz and Stegun multiplied by a factor (2l+1+k)! The definition given in this article coincides with the one of Abramowitz and Stegun.

The energy becomes

$W=-{\frac {1}{2n^{2}}}\quad {\hbox{with}}\quad n\equiv k+l+1.$

The principal quantum number n satisfies $n\geq l+1$, or $l\leq n-1$. Since $\alpha =2/n$, the total radial wavefunction is

$R_{nl}(r)=N_{nl}\left({\tfrac {2Zr}{na_{0}}}\right)^{l}\;e^{-{\textstyle {\frac {Zr}{na_{0}}}}}\;L_{n-l-1}^{(2l+1)}\left({\tfrac {2Zr}{na_{0}}}\right),$

with normalization constant

$N_{nl}=\left[\left({\frac {2Z}{na_{0}}}\right)^{3}\cdot {\frac {(n-l-1)!}{2n[(n+l)!]}}\right]^{1 \over 2}$

which belongs to the energy

$E_{n}=-{\frac {Z^{2}}{2n^{2}}}E_{\textrm {h}},\qquad n=1,2,\ldots .$

In the computation of the normalization constant use was made of the integral

$\int _{0}^{\infty }x^{2l+2}e^{-x}\left[L_{n-l-1}^{(2l+1)}(x)\right]^{2}dx={\frac {2n(n+l)!}{(n-l-1)!}}.$

Completeness of hydrogen-like orbitals

In quantum chemical calculations hydrogen-like atomic orbitals cannot serve as an expansion basis, because they are not complete. The non-square-integrable continuum (E > 0) states must be included to obtain a complete set, i.e., to span all of one-electron Hilbert space.