Hydrogenlike atoms#

What you need to know

:class: note

  • The hydrogen atom is the simplest atom for which the Schrödinger equation can be solved exactly. In contrast, helium—though only having one additional electron—cannot be solved exactly due to the complexity introduced by electron-electron interactions.

  • Solving the Schrödinger equation for the hydrogen atom leverages the fact that the Coulomb potential from the nucleus is isotropic: it is radially symmetric and depends solely on the distance from the nucleus. This symmetry leads to degeneracies in the energy levels.

  • While the resulting energy eigenfunctions (atomic orbitals) are not necessarily isotropic, their dependence on angular coordinates arises fundamentally from the isotropy of the underlying potential.

  • Because angular momentum is conserved, energy eigenvalues can be classified by two angular momentum quantum numbers, \(l\) and \(m\) (both integers), which quantize the magnitude and projection of angular momentum.

  • Atomic orbitals derived for hydrogen are also foundational in describing the structure of multi-electron atoms and molecules.

Schrodinger equation for hydrogenlike atoms#

  • Consider one electron and one nucleus with charge \(Ze\) where \(e\) is the magnitude of the electron charge (1.6021773 \(\times\) \(10^{-19}\) C) and \(Z\) is the atomic number. Examples of such systems are: H, \(He^+\), \(Li^{2+}\)

  • We can reduce the two body problem to one body problem of electron with reduced mass moving with respect to fixed nucleus. Since electron is might lighter compared to the nucleus we can use electron mass instead of reduced mass \(\mu\approx m_e\)

  • We have a porblem of one particle moving in a symmetric potential field in 3D. Expect to get 3 quantum numbers, anticipate some degenreacies due to this raidal symmetry.

  • Kinetic energy operator in 3D is the same as in the particle in a box in 3D:

\[\frac{\hbar^2}{2m_e}\nabla^2\]

  • Potential energy operator operator consists of one coulomb term encoding the electrostatic attraction between nucleus and electron:

\[V= - \frac{Ze^2}{4\pi\epsilon_0 r}\]

H-atom in spherical coordinates system#

  • Because of the spherical symmetry of the Coulomb potential it is convenient to work in spherical coordinates:

\[{\left[ -\frac{\hbar^2}{2m_e}\Delta - \frac{Ze^2}{4\pi\epsilon_0 r}\right]\psi_i(r,\theta,\phi) = E_i\psi(r,\theta,\phi)}\]

  • Subscripts for \(\psi\) and \(E\) signify the fact that there are multiple (\(\psi_i\), \(E_i\)) combinations. The \(\Delta\) is expressed in spherical coordinates:

\[{\Delta\equiv\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2{sin}(\theta)} \frac{\partial}{\partial\theta}\left({sin}(\theta)\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2{sin}^2(\theta)}\frac{\partial^2}{\partial\phi^2}}\]

  • Note that the Coulomb potential term above depends only on \(r\) (and not on \(\theta\) or \(\phi\)). The Laplacian can be written in terms of the angular momentum operator \(\hat{L}\):

\[{\Delta = \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial}{\partial r}\right) - \frac{1}{r^2}\frac{\hat{L}^2}{\hbar^2} = \frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r} - \frac{1}{r^2}\frac{\hat{L}^2}{\hbar^2}}\]

By substituting this into and multiplying both sides by \(2m_er^2\), we get:

\[\left[\frac{-\hbar^2}{2m_e}\left(\frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r}\right) - \frac{Ze^2}{4\pi\epsilon_0r}+ \frac{\hat{L}^2}{2m_e r^2}\right]\psi_i(r,\theta,\phi) = \psi_i(r,\theta,\phi)\]

Since the operator can be split into \(r\) and angle dependent parts, the solution can be written as a product of the radial and angular parts

Separation of variables#

\[{\psi_i(r,\theta,\phi) = R_{nl}(r)Y_l^m(\theta,\phi)}\]

  • The \(R_{nl}\) is called the radial wavefunction which we need to obtain.

  • The \(Y_l^m\) are spherical harmonics which are eigenfunctions of angular momentum \(\hat{L}^2\) as discussed earlier. Hence this part is already solved.

\[\hat{L}^2 Y_{lm}=\hbar^2l(l+1)Y_{lm}\]

  • Plugging the \(R_{nl}(r)Y_l^m(\theta,\phi)\) into Schoringer euqation, applying angular momentum operator and calcelling spherical harmonics from both sides we end up with the radial part.

\[{\left[ -\frac{\hbar^2}{2m_e}\left(\frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r}\right) - \frac{Ze^2}{4\pi\epsilon_0r}+ \frac{l(l+1)\hbar^2}{2m_er^2}\right] R_{nl}(r) = E_{nl}R_{nl}(r)}\]

  • We find that electron is moving in an effective potential generated by attractive coloumb interaction and repulsive orbtial kinetic energy.

\[V_{eff} = - \frac{Ze^2}{4\pi\epsilon_0r}+ \frac{l(l+1)\hbar^2}{2m_er^2} \]

  • The first repuslive term more rapidly with \(r\) than the Coulomb potential, it dominates at small distances if \(l\neq0\). Both terms approach zero for large values of r. The resultant potential is repulsive at short distances for \(l>0\) and is more repulsive the greater the value of \(l\). The net result of this repulsive centrifugal potential is to force the electrons in orbitals with l \(l>0\) on average farther from the nucleus than l = 0.

Eigenvalues and Eigenfunctions#

  • The eigenvalues \(E_{nl}\) and and the radial eigenfunctions \(R_{nl}\) can be written as (derivations are lengthy but standard math):

\[{E_{nl} = -\frac{m_ee^4Z^2}{32\pi^2\epsilon_0^2\hbar^2n^2}{ \,\,\, }n = 1,2,3...{ (independent\, of\, }l,\,\,\,l<n{)}}\]
\[R_{nl}(r) = \rho^le^{-\rho/2}{L_{n-l-1}^{2l+1}(\rho)}\]

  • Bohr radius: \(a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_ee^2}\)

  • Dimensionless distance defined via ratio of Bohr radius: \(\rho = \frac{2Zr}{na_0}\)

  • Laguerre polynomials \(L_{n+l}^{2l+1}(\rho)\)

Examples of the radial wavefunctions for hydrogenlike atoms#

Orbital

\(n\)

\(l\)

\(R_{nl}\)

1s

1

0

\(2\left(\frac{Z}{a_0}\right)^{3/2}e^{-\rho/2}\)

2s

2

0

\(\frac{1}{2\sqrt{2}}\left(\frac{Z}{a_0}\right)^{3/2}(2 - \rho)e^{-\rho/2}\)

2p

2

1

\(\frac{1}{2\sqrt{6}}\left(\frac{Z}{a_0}\right)^{3/2}\rho e^{-\rho/2}\)

3s

3

0

\(\frac{1}{9\sqrt{3}}\left(\frac{Z}{a_0}\right)^{3/2}(6 - 6\rho - \rho^2)e^{-\rho/2}\)

3p

3

1

\(\frac{1}{9\sqrt{6}}\left(\frac{Z}{a_0}\right)^{3/2}(4 - \rho)\rho e^{-\rho/2}\)

3d

3

2

\(\frac{1}{9\sqrt{30}}\left(\frac{Z}{a_0}\right)^{3/2}\rho^2 e^{-\rho/2}\)

Spectrum of H atom#

Equation of hydroen atom energy can be expressed in units (\(m^{-1}\); usually \(cm^{-1}\) is used):

\[{\tilde{E}_n = \frac{E_n}{hc} = \frac{E_n}{2\pi\hbar c} = -\overbrace{\frac{m_ee^4}{4\pi c(4\pi\epsilon_0)^2\hbar^3}}^{\equiv R} \times\frac{Z^2}{n^2}{ }}\]

  • where \(R\) is the Rydberg constant and we have assumed that the nucleus has an infinite mass. To be exact, the Rydberg constant depends on the nuclear mass, but this difference is very small. For H atom \(R_H = 1.096 775 856 \times 10^7\) \(m^{-1}\).

  • The Eq of H atom energy can be used to calculate the differences in the energy levels:

\[\Delta\tilde{v}_{n_1,n_2} = \tilde{E}_{n_2} - \tilde{E}_{n_1} = -\frac{R_HZ^2}{n_2^2} + \frac{R_HZ^2}{n_1^2} = R_HZ^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]
\[{E_i = R_HZ^2\left(\frac{1}{1^2} - \frac{1}{\infty}\right)}\]

  • For a ground state hydrogen atom (i.e., \(n = 1\)), the above equation gives a value of 109678 cm\(^{-1}\) = 13.6057 eV. Note that the larger the nuclear charge \(Z\) is, the larger the binding energy is.

Quantum numbers \(n\), \(l\) and \(m\)#

The quantum numbers in hydrogenlike atoms take on the following values dicated by the solution of Schrodinger equation with boundary conditions imposed respective radial and anuglar parts

Quantum numbers of Hydrogen Atom

\[{n = 1, 2, 3, ...}\]
\[{l = 0, 1, 2, ..., n-1}\]
\[{m = 0, \pm 1, \pm 2,...,\pm l}\]
\[{m_s = \pm 1/2}\]

  • For historical reasons, the following letters are used to express the value of \(l\):

\[{\phantom{{symbo}}l = 0, 1, 2, 3, ...}{{symbol} = s, p, d, f, ...}\]

  • For the \(n=1\) we have only one wavefunction: \(R_{10}(r)Y_0^0(\theta,\phi)\). This state is usually labeled as \(1s\), where 1 indicates the shell number (\(n\)) and \(s\) corresponds to orbital angular momentum \(l\) being zero.

  • For \(n = 2\), we have several possibilities: \(l = 0\) or \(l = 1\). The former is labeled as \(2s\). The latter is \(2p\) state and consists of three degenerate states: (for example, \(2p_x\), \(2p_y\), \(2p_z\) or \(2p_{+1}\), \(2p_0\), \(2p_{-1}\)). In the latter notation the values for \(m\) have been indicated as subscripts.

  • There is one more quantum number that has not been discussed yet: Spin quantum number \(m_s=\pm 1/2\)

  • For one-electron systems this can have values \(\pm\frac{1}{2}\) (will be discussed in more detail later). In absence of magnetic fields the spin levels are degenerate and therefore the total degeneracy of the levels is \(2n^2\).

Total wave function#

The total wavefunction for a hydrogenlike atom is (\(m\) is usually denoted by \(m_l\)):

Where the normalization factor is:

\[{N_{nl} = \sqrt{\left(\frac{2Z}{na_0}\right)^3\frac{(n - l - 1)!}{2n\left[(n + l)!\right]}}}\]

Table of Wavefunctions in cartesian coordinates#

\(n\)

\(l\)

\(m\)

Wavefunction expressed in \(\sigma= \frac{zr}{a_0}\)

1

0

0

\(\psi_{1s} = \frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}e^{-\sigma}\)

2

0

0

\(\psi_{2s} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}(2 - \sigma)e^{-\sigma/2}\)

2

1

0

\(\psi_{2p_z} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-\sigma/2} {cos}(\theta)\)

2

1

\(\pm 1\)

\(\psi_{2p_x} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-\sigma/2}{sin}(\theta){cos}(\phi)\)

\(\psi_{2p_y} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-\sigma/2}{sin}(\theta){sin}(\phi)\)

3

0

0

\(\psi_{3s} = \frac{1}{81\sqrt{3\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\left(27 - 18\sigma + 2\sigma^2\right)e^{-\sigma/3}\)

3

1

0

\(\psi_{3p_z} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\left(6 - \sigma\right)\sigma e^{-\sigma/3} {cos}(\theta)\)

3

1

\(\pm 1\)

\(\psi_{3p_x} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\left(6 - \sigma\right)\sigma e^{-\sigma/3}{sin}(\theta){cos}(\phi)\)

\(\psi_{3p_y} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\left(6 - \sigma\right)\sigma \)

3

2

0

\(\psi_{3d_{z^2}} = \frac{1}{81\sqrt{6\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2e^{-\sigma/3}\left(3{cos}^2(\theta) - 1\right)\)

3

2

\(\pm 1\)

\(\psi_{3d_{xz}} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2 e^{-\sigma/3}{sin}(\theta){cos}(\theta){cos}(\phi)\)

\(\psi_{3d_{yz}} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2 \)

3

2

\(\pm 2\)

\(\psi_{3d_{x^2-y^2}} = \frac{1}{81\sqrt{3\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2 e^{-\sigma/3}{sin}^2(\theta){cos}(2\phi)\)

\(\psi_{3d_{xy}} = \frac{1}{81\sqrt{3\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2\)

Computationals with wavefunctions#

  • We can use the wavefunctions from the table above to calculate various averages by using the definition of average in quantum mechanics: \(\langle f(r) \rangle =\langle \psi |f(r)|\psi\rangle\):

\[ \langle r\rangle_{n,l,m} = \frac{a^2_0 n^4}{Z^2}\Big( 1+\frac{1}{2}\Big[1-\frac{l(l+1)}{n^2} \Big]) \]
\[ \langle r^2\rangle_{n,l,m} = \frac{a^2_0 n^4}{Z^2}\Big( 1+\frac{3}{2}\Big[1-\frac{l(l+1)-1/3}{n^2} \Big]) \]
\[ \langle r^{-1}\rangle_{n,l,m} = \frac{Z}{a_0 n^2} \]
\[ \langle r^{-2}\rangle_{n,l,m} = \frac{Z^2}{a^2_0 n^3 (l+1/2)} \]
\[ \langle r^{-3}\rangle_{n,l,m} = \frac{Z^3}{a^3_0 n^3 (l+1/2)(l+1)} \]