The hydrogen atom stands as the simplest and most fundamentally significant system in quantum mechanics, embodying a pristine model for exploring atomic structures. It consists of a single electron orbiting a positively charged proton, an arrangement that echoes the simplicity and aspects of atomic physics. This system, in its essence, offers an invaluable starting point for exploring quantum mechanics, given its exact solvability — a trait that sets it apart from simpler quantum systems like the particle in a box or the harmonic oscillator.
The real power of studying the hydrogen atom lies in its direct applicability across a broad spectrum of scientific disciplines. From the realm of chemistry, where it helps elucidate the nature of chemical bonds, to the field of engineering, particularly in semiconductor physics, the hydrogen atom model is pivotal. The concept of Wannier excitons, for instance, relies on a scaled version of the hydrogen atom’s solutions to explain optical absorption phenomena in semiconductors, critical for the development of optoelectronic devices.
Moreover, the hydrogen atom introduces a sophisticated layer to quantum mechanics by involving a two-particle system: the electron and the proton. This complexity elevates the study, offering insights into the behavior of systems with multiple particles and the wave functions that describe them. Such an exploration is not just academic; it paves the way for understanding more complex atomic and molecular structures, underpinning much of modern physics and chemistry.
Through this atom, we approach the heart of quantum mechanics, tackling the complexities of particle interactions, and laying the groundwork for advancements in technology and theoretical physics.
For the hydrogen atom, a pivotal step in quantum mechanics is the generalization of the Schrödinger equation for systems involving more than one particle. This generalized form, especially relevant for time-independent problems, posits that the Hamiltonian operator acting on the wave function yields the product of the energy of the system and the wave function itself:
\hat{\mathbf H} \Psi = E \Psi
Here, \Psi represents the state of the entire system, which could encompass two or more particles. Consequently, the Hamiltonian (\hat{\mathbf H}) is now understood as the operator that encapsulates the energy of the entire system, irrespective of the number of particles it comprises.
In the context of the hydrogen atom, which includes an electron and a proton, this approach necessitates a wave function that is a function of both particles’ coordinates. Therefore, we have coordinates x_e, y_e, z_e for the electron and x_p, y_p, z_p for the proton. The wave function \Psi, thus, becomes a function of all six coordinates, capturing the state of both the electron and the proton:
\Psi = \Psi(x_e, y_e, z_e; x_p, y_p, z_p)
This comprehensive framework allows for the analysis of the hydrogen atom’s quantum behavior, taking into account the interactions between the electron and the proton, and setting the stage for exploring more complex quantum systems.
In advancing the understanding of the hydrogen atom within the quantum mechanical framework, it’s essential to recognize the multidimensional nature of the problem introduced by the Schrödinger equation’s generalization. This generalization extends the equation to encompass systems of multiple particles, integrating the Hamiltonian as the energy operator for the system and \Psi as the wave function representing the state of the entire system. For the hydrogen atom, this involves the electron and the proton, each contributing three spatial coordinates to the wave function, resulting in a function of six coordinates in total.
This comprehensive approach allows for a detailed description of the hydrogen atom’s behavior, accommodating scenarios where the spatial relationship between the electron and the proton is crucial, such as a hydrogen atom confined within a box. In such situations, while the atom as a whole could be located anywhere within the box, the proximity of the electron to the proton is significantly likely, a phenomenon that necessitates the wave function’s dependence on all six coordinates.
However, certain specific conditions permit simplifications that reduce the problem to more manageable forms. For instance, in the context of a hydrogen atom in free space, it becomes feasible to consider wave functions of three coordinates each, but not directly as separate entities for the electron and the proton. These simplified forms, while useful, diverge from representing the particles individually and instead, describe the system’s collective states.
This dimensional complexity underscores a fundamental challenge in quantum mechanics: as the number of particles increases, so does the number of required coordinates, escalating exponentially for systems with three or more particles. Thus, while quantum mechanics offers a profound understanding of atomic and subatomic systems, it also highlights the computational and conceptual complexities inherent in the full description of multi-particle systems, often rendering such problems intractable without further simplification or approximation techniques.
The Hamiltonian for the hydrogen atom, incorporating the masses of the electron (m_e) and the proton (m_p), and their electrostatic potential energy due to their charges, is given by:
\hat{\mathbf H} = -\frac{\hbar^2}{2m_e}\nabla^2_e - \frac{\hbar^2}{2m_p}\nabla^2_p + V(|\mathbf{r}_e - \mathbf{r}_p|)
where: - \hbar = 1.0545718 \times 10^{-34} \frac{kg \, m^2}{s} is the reduced Planck’s constant, - m_e and m_p are the masses of the electron and proton, respectively, - \nabla^2_e and \nabla^2_p are the Laplacian operators acting on the electron and proton, indicating their kinetic energy, - V - \mathbf{r}_e and \mathbf{r}_p are the position vectors of the electron and proton, respectively, - |\mathbf{r}_e - \mathbf{r}_p| is the distance between the electron and the proton, determining the electrostatic potential energy due to their interaction.
The he Coulomb potential energy, the potential energy resulting from the electrostatic attraction between the negative electron and the positive proton, only depends on the separation distance between the electron and the proton, the modulus of the vector \mathbf r_e minus the vector \mathbf r_p.
-\frac{e^2}{4\pi\epsilon_0|\mathbf{r}_e - \mathbf{r}_p|}
where: - 4\pi\epsilon_0 is the permittivity of free space 8.854187817 \times 10^{-12} \,\frac{F}{m} - e = 1.602176634 \times 10^{-19} \, C is the elementary charge,
The energy is negative because the closer the particle get, the more negative the energy became; the energy would be zero if the electron and the proton are infinitely far apart, and it would take energy to pull the electron away from the proton. It is important that it depends only from the relative distance and not from the absolute position of either, in order to simplify the solution.
Putting all together the Hamiltonian is:
\hat{\mathbf H} = -\frac{\hbar^2}{2m_e}\nabla^2_e - \frac{\hbar^2}{2m_p}\nabla^2_p - \frac{e^2}{4\pi\epsilon_0|\mathbf{r}_e - \mathbf{r}_p|}
and the Schrödinger equation can be written explicitly:
\left[-\frac{\hbar^2}{2m_e}\nabla^2_e - \frac{\hbar^2}{2m_p}\nabla^2_p - \frac{e^2}{4\pi\epsilon_0|\mathbf{r}_e - \mathbf{r}_p|} \right] \psi(x_e, y_e, z_e; x_p, y_p, z_p) = E\psi(x_e, y_e, z_e; x_p, y_p, z_p)
To tackle this complex problem effectively, we aim to simplify it by breaking it down into more manageable components. This approach involves making strategic choices in coordinate systems or variables, which allows us to separate the resulting differential equation into simpler parts. When dealing with a single particle, selecting its coordinates for analysis is straightforward. However, the presence of two particles introduces a choice: whether to analyze the individual coordinates of each particle separately. It turns out, the problem does not simplify well with this approach.
The second consideration involves deciding on the most appropriate mathematical axes for the problem at hand. Given the hydrogen atom’s spherical symmetry, it’s a reasoned deduction that spherical polar coordinates would simplify the problem more effectively than Cartesian coordinates. This choice is pivotal for tackling the problem with greater efficiency. Let’s explore how this approach is implemented.
Given that the potential energy in the hydrogen atom problem depends solely on the distance between the electron and the proton, it is beneficial to adopt a new set of six coordinates. Three of these coordinates will represent the relative positions of the electron and proton. Mathematically, this involves defining a new x-coordinate as the difference between the electron’s and proton’s x-coordinates, and similarly for the y and z-coordinates.
x = x_e - x_p, \quad y = y_e - y_p, \quad z = z_e - z_p
As a result, we establish a new set of coordinates (x, y, z) that depict the relative positioning of the electron to the proton, forming a relative position vector:
\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}
where \mathbf{i}, \mathbf{j}, \mathbf{k} are the standard unit vectors in the x, y, and z directions, respectively. Consequently, the distance between the electron and proton can be expressed as the square root of the sum of the squares of these new coordinates, aligning with the separation distance previously described:
r = \sqrt{x^2 + y^2 + z^2} = | \mathbf r_e - \mathbf r_p|
For the remaining three coordinates, after establishing the relative positions between the electron and proton, we turn our attention to the system’s center of mass. Drawing parallels from classical mechanics, such as the gravitational interaction between a moon and a planet, we observe that their center of mass remains stationary relative to their orbits. Despite the apparent orbit of the moon around the planet, both bodies actually orbit their shared center of mass, which stays fixed in space.
This concept becomes even more evident when considering bodies of comparable mass. If the moon and planet had equal masses, they would orbit around a central point located midway between them. As the mass disparity grows, the heavier body moves less, but the principle remains the same: both bodies orbit around their common center of mass.
Applying this to the hydrogen atom, we consider the center of mass (denoted by the vector \mathbf{R}) of the electron and proton system. This can be thought of as the balance point on a hypothetical beam with the electron and proton at opposite ends. The position of this balance point, or the center of mass, is determined by the weighted average of their positions, factoring in the masses of the electron and proton.
Thus, the other three coordinates we choose pertain to this center of mass vector \mathbf{R}, calculated as the weighted average of the electron and proton positions, weighted by their respective masses relative to the total mass of the system (M = m_e + m_p):
\mathbf R = \frac{m_e \mathbf r_e + m_p \mathbf r_p}{M} = X\mathbf{i} + Y\mathbf{j} + Z\mathbf{k}
This approach enables us to simplify the six-dimensional problem by separating the motion into the relative motion of the electron and proton and the motion of the system’s center of mass.
The new coordinates are (for the x direction):
\begin{aligned} & X = \frac{m_e x_e + m_p x_p}{M}, \quad x = x_e - x_p \\ & Y = \frac{m_e y_e + m_p y_p}{M}, \quad y = y_e - y_p \\ & Z = \frac{m_e z_e + m_p z_p}{M}, \quad z = z_e - z_p \end{aligned}
Applying the generic transformation of derivatives when changing from one coordinate system to another; specifically, from Cartesian coordinates to a set involving relative and center of mass coordinates,it is expressed as follows for the x direction:
\begin{aligned} & \frac{\partial}{\partial x_e} = \frac{\partial X}{\partial x_e} \frac{\partial}{\partial X} + \frac{\partial x}{\partial x_e} \frac{\partial}{\partial x} = \frac{m_e}{M}\frac{\partial}{\partial X} + \frac{\partial}{\partial x} \\ & \frac{\partial}{\partial x_p} = \frac{\partial X}{\partial x_p} \frac{\partial}{\partial X} + \frac{\partial x}{\partial x_p} \frac{\partial}{\partial x} = \frac{m_p}{M}\frac{\partial}{\partial X} - \frac{\partial}{\partial x} \end{aligned}
The second derivative can be computed in the same way as:
\frac{\partial^2}{\partial^2x_e} = \frac{\partial}{\partial x_e} \left(\frac{\partial}{\partial x_e} \right)
Using the expression previous calculated:
\begin{aligned} & \frac{\partial^2}{\partial^2 x_e} = \frac{m_e}{M}\frac{\partial}{\partial x_e} \frac{\partial}{\partial X} + \frac{\partial}{\partial x_e} \frac{\partial}{\partial x} = \left(\frac{m_e}{M}\right)^2 \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial x^2} +\frac{m_e}{M}\left(\frac{\partial}{\partial X}\frac{\partial}{\partial x} + \frac{\partial}{\partial x}\frac{\partial}{\partial X} \right)\\ & \frac{\partial^2}{\partial^2 x_p} = \frac{m_p}{M}\frac{\partial}{\partial x_p} \frac{\partial}{\partial X} - \frac{\partial}{\partial x_p} \frac{\partial}{\partial x} = \left(\frac{m_p}{M}\right)^2 \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial x^2} -\frac{m_p}{M}\left(\frac{\partial}{\partial X}\frac{\partial}{\partial x} + \frac{\partial}{\partial x}\frac{\partial}{\partial X} \right) \end{aligned}
Adding these together with a multiplicative factor:
\frac{1}{m_e}\frac{\partial^2}{\partial^2 x_e}+\frac{1}{m_p}\frac{\partial^2}{\partial^2 x_p} = \frac{m_e + m_p}{M}^2 \frac{\partial^2}{\partial X^2} + \left(\frac{1}{m_e} + \frac{1}{m_p}\right) \frac{\partial^2}{\partial x^2} = \frac{1}{M} \frac{\partial^2}{\partial X^2} + \frac{1}{\mu} \frac{\partial^2}{\partial x^2}
where:
\mu = \frac{m_e m_p}{m_e + m_p}
is called the reduced mass.Due to the proton being significantly heavier than the electron, this value closely approximates the electron mass. The exact ratio of the reduced mass to the electron mass turns out to be approximately 0.9995.
Given the Laplacian operators \nabla^2 for both the center of mass coordinates (\mathbf R) and the relative position coordinates (\mathbf r), we define them as follows:
\nabla^2_{\mathbf R} = \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial Y^2} + \frac{\partial^2}{\partial Z^2}
\nabla^2_{\mathbf r} = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}
With these definitions, the Hamiltonian in terms of the center of mass (R) and relative position (r) can be written as:
\hat{\mathbf H} = -\frac{\hbar^2}{2M} \nabla^2_{\mathbf R} - \frac{\hbar^2}{2\mu} \nabla^2_{\mathbf r} + V(\mathbf r)
This Hamiltonian form simplifies the analysis by separating the motion of the center of mass from the relative motion, focusing on the quantum mechanical behavior of the system’s internal structure.
This formulation allows the six-dimensional differential equation to be separated into two three-dimensional problems, presuming that the wave function can be written as the product of two three-dimensional wave functions, S(\mathbf R) for the center of mass motion and U(\mathbf r) for the relative motion, simplify the problem significantly:
\psi(\mathbf R, \mathbf r) = S(\mathbf R)U(\mathbf r)
Whenever a separation like this is suggested, it’s essentially an educated guess, and the method is tested to check its viability. If the approach yields a solution, the unique nature of solutions to differential equations confirms its correctness after the computation has been carried over. Substituting in the Schrödinger equation:
-U(\mathbf{r})\frac{\hbar^2}{2M}\nabla^2_{\mathbf{R}}S(\mathbf{R}) + S(\mathbf{R}) \left[-\frac{\hbar^2}{2\mu}\nabla^2_{\mathbf{r}} + V(\mathbf{r})\right]U(\mathbf{r}) = ES(\mathbf{R})U(\mathbf{r})
These manipulations where possible as each term of the Hamiltonian is depending only on one of the two variable, and therefore the other can be factored in front. Dividing by S(\mathbf R)U(\mathbf r) and separating the terms for each of the coordinates:
-\frac{1}{S(\mathbf{R})}\frac{\hbar^2}{2M}\nabla^2_{\mathbf{R}}S(\mathbf{R}) = E - \frac{1}{U(\mathbf{r})}\left[-\frac{\hbar^2}{2\mu}\nabla^2_{\mathbf{r}} + V(\mathbf{r})\right]U(\mathbf{r})
The equation’s left-hand side is solely influenced by the center of mass coordinates \mathbf{R}, while the right-hand side is exclusively affected by the relative position coordinates \mathbf{r}. Therefore, each side must be equivalent to a common separation constant due to their dependency on distinct variables. This separation constant, which we denote as the energy associated with the center of mass motion, E_{\text{CoM}}, serves as a bridge linking the two sides of the equation, each of which pertains to a different aspect of the system’s dynamics.
There are now two equations:
\begin{aligned} & -\frac{\hbar^2}{2M}\nabla^2_{\mathbf{R}}S(\mathbf{R}) = E_{CoM}S(\mathbf{R}) \\ & \left[-\frac{\hbar^2}{2\mu}\nabla^2_{\mathbf{r}} + V(\mathbf{r})\right]U(\mathbf{r}) = \left(E - E_{CoM}\right)U(\mathbf{r}) = E_H U(\mathbf{r}) \end{aligned}
one which describe the center of mass motion, the other which describe the relative motion, and these two can be solved separately.
The equation for the center of mass:
-\frac{\hbar^2}{2M}\nabla^2_{\mathbf{R}}S(\mathbf{R}) = E_{CoM}S(\mathbf{R})
It is very easy to solve, it is the Schrödinger equation for a free particle of mass M with solutions:
S(\mathbf R) = e^{i\mathbf K \cdot \mathbf R}
with energies:
E_{CoM} = \frac{\hbar^2K^2}{2M}
This term mirrors the equation previously encountered for an electron in free space, albeit now applied to a particle of mass M, representing the entire hydrogen atom. This forms the basis of the center of mass motion equation for the hydrogen atom, illustrating its capacity to navigate space as a singular entity. This equation delineates the atom’s translational movement.
In contrast, the equation for relative motion addresses the electron and proton’s internal dynamics within the hydrogen atom, delineating a distinction from the center of mass motion. It reveals the hydrogen atom’s internal states, which include the familiar orbitals and energy levels that define the atom’s intrinsic properties, as opposed to describing its external movement through space.
Having completing the solution for the center of mass, the next step is to solve the equation for the internal state of the atom, with orbitals and discrete energy levels; there is an additional separation that is possible for the internal equation, between an angular part and a radial part; the angular part is the same as the angular momentum with the same solution (spherical harmonics) so the focus will be on the radial equation.
The starting point is the equation previously derived:
\left[-\frac{\hbar^2}{2\mu}\nabla^2_{\mathbf{r}} + V(\mathbf{r})\right]U(\mathbf{r}) = \left(E - E_{CoM}\right)U(\mathbf{r}) = E_H U(\mathbf{r})
Since the problem is symmetric around the radius, it is possible to change in polar coordinates. The \nabla^2 is given by:
\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\left[\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial}{\partial \theta}\right) + \frac{\partial^2}{\partial \phi^2}\right]
The term in square brackets is the \nabla^2_{\theta,\phi} \equiv -\frac{L^2}{\hbar^2} that appears in the discussion of angular momentum, which solutions are spherical harmonics; therefore the assumption is to look for a solution of type:
U(\mathbf R) = R(r)Y(\theta,\phi)
Making a further assumption (just for making the mathematics easier) defining \chi(r)=rR(r) so that:
U(\mathbf R) = \frac{1}{r}\chi(r)Y(\theta,\phi)
The radial derivative:
\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R(r)}{\partial r}\right)\frac{\chi}{r} = \frac{1}{r}\frac{\partial^2\chi(r)}{\partial r^2}
Substituting into the Schrödinger equation:
-\frac{\hbar^2}{2\mu}Y(\theta,\phi)\frac{1}{r}\frac{\partial^2 \chi(R)}{\partial r^2} + \frac{\chi(r)}{r^3}\frac{1}{2\mu}\mathbf L^2Y(\theta,\phi) + Y(\theta,\phi) V(r) \frac{\chi}{r} = E_h \frac{1}{r}\chi(r)Y(\theta,\phi)
Dividing by the factor:
-\frac{\hbar\chi(r) Y(\theta,\phi)}{2\mu r^3}
gives:
\frac{r^2}{\chi(r)}\frac{\partial^2 \chi(R)}{\partial r^2} + r^2\frac{2\mu}{\hbar^2}\left(H_E - V(r)\right) = \frac{1}{\hbar^2}\frac{1}{Y(\theta,\phi)}\mathbf L^2 Y(\theta,\phi)
In this equation, the left side depends only on \mathbf r, and the right side only on \theta and \phi, and they must be equal to a constant, which is the same constant as the solution of the angular momentum equation so the constant is:
\ell(\ell+1)
So, in addition to the eigenequation for \mathbf L^2, there is a radial equation:
\frac{r^2}{\chi(r)}\frac{\partial^2 \chi(R)}{\partial r^2} + r^2\frac{2\mu}{\hbar^2}\left(H_E - V(r)\right) = \ell(\ell+1)
This is just a differential equation in \chi(r):
-\frac{\hbar^2}{2m}\frac{\mathrm d^2 \chi(r)}{\mathrm d r^2} + \left( V(r) + \frac{\hbar^2}{2\mu}\frac{\ell(\ell+1)}{r^2}\right)\chi(r) = E_H\chi(r)
Mathematically, this has the form of a Schrödinger wave equation with an additional potential energy term:
\frac{\hbar^2}{2\mu}\frac{\ell(\ell+1)}{r^2}
Considering the potential energy V(r), this separation of variable is valid for any potential that is only a function of the radial distance r, a central potential.
This assumption is important because it can be used in more complicated atoms as first approximation, and the solution will still be spherical harmonics.
In the particular case, using the Coulomb force, the equation becomes:
-\frac{\hbar^2}{2\mu}\frac{\mathrm d^2 \chi(r)}{\mathrm d r^2} + \left(\frac{e^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2\mu}\frac{\ell(\ell+1)}{r^2}\right)\chi(r) = E_H\chi
where l \in \mathbb N (from the spherical harmonics solution).
For mathematical convenience, setting (the minus is because E_H<0 to ensure \alpha is real:
E_H = -\frac{1}{n^2}\frac{\mu}{2}\left(\frac{e^2}{4\pi\epsilon_0 \hbar}\right)^2 \quad s =\alpha r = 2\sqrt{-\frac{2\mu}{\hbar^2}E_H}r
The equation simplify as:
\frac{\mathrm d^2 \chi(r)}{\mathrm d s^2} - \left(\frac{\ell(\ell+1)}{s^2} - \frac{n}{s} + \frac{1}{4}\right)\chi(r) = 0
Assuming that the solution has the form:
\chi(s) = s^{\ell+1}L(s)e^{-\frac{s}{2}}
Taking the derivatives:
\begin{aligned} & \frac{\mathrm d\chi}{\mathrm ds} = s^{\ell}e^{-\frac{s}{2}}\left[\left(-\frac{s}{2} + \ell + 1\right) L(s) + \frac{s}{2}\frac{\mathrm d L}{\mathrm ds}\right] \\ & \frac{\mathrm d^2\chi}{\mathrm ds^2} = s^{\ell}e^{-\frac{s}{2}} \left[ s \frac{\mathrm d^2 L}{\mathrm ds^2} + 2 \left( -\frac{s}{2} + \ell + 1 \right) \frac{\mathrm d L}{\mathrm ds} + \left(\frac{1}{4} s - \ell + 1 + \frac{\ell (\ell + 1)}{s}\right) L\right] \end{aligned}
Substituting back:
\begin{aligned} & s^{\ell}e^{-\frac{s}{2}} \left[ s \frac{\mathrm d^2 L}{\mathrm ds^2} + 2 \left( -\frac{s}{2} + \ell + 1 \right) \frac{\mathrm d L}{\mathrm ds} + \left(\frac{1}{4} s - \ell + 1 + \frac{\ell (\ell + 1)}{s}\right) L\right] - \left(\frac{\ell(\ell+1)}{s^2} - \frac{n}{s} + \frac{1}{4}\right)s^{\ell+1}L(s)e^{-\frac{s}{2}} = 0 \\ & s \frac{\mathrm d^2 L}{\mathrm ds^2} + \left[-s + 2\ell + 1 \right] \frac{\mathrm d L}{\mathrm ds} + \left(\frac{1}{4} s - \ell + 1 + \frac{\ell (\ell + 1)}{s}\right) - \left(\frac{\ell(\ell+1)}{s^2} - \frac{n}{s} + \frac{1}{4}\right)s L(s) = 0 \\ & s \frac{\mathrm d^2 L}{\mathrm ds^2} - \left[s - 2\ell + 1 \right] \frac{\mathrm d L}{\mathrm ds} + \left[n-(\ell+1)\right] L(s) = 0 \end{aligned}
The final assumption is that the equation:
s \frac{\mathrm d^2 L}{\mathrm ds^2} - \left[s - 2\ell + 1 \right] \frac{\mathrm d L}{\mathrm ds} + \left[n-(\ell+1)\right] L(s) = 0
have a solution that can be expressed as power series:
L(s) = \sum_{j=0}^\infty c_js^j
Differentiating (the first can start at either 0 or 1 as the derivative of c_0 constant is 0):
\begin{aligned} & \frac{\mathrm d L}{\mathrm ds} = \sum_{j=0}^\infty jc_js^{j-1} = \sum_{j=0}^\infty (j+1)c_{j+1}s^j \\ & \frac{\mathrm d^2 L}{\mathrm ds^2} = \sum_{j=0}^\infty j(j+1)c_{j+1}s^{j-1} \end{aligned}
Inserting into the equations:
\begin{aligned} & s \frac{\mathrm d^2 L}{\mathrm ds^2} - \left[s - 2\ell + 1 \right] \frac{\mathrm d L}{\mathrm ds} + \left[n-(\ell+1)\right] L(s) = 0\\ & \sum_{j=1}^\infty j(j+1)c_{j+1}s^{j} - \left[s - 2\ell + 1 \right] \sum_{j=0}^\infty (j+1)c_{j+1}s^j+ \left[n-(\ell+1)\right] \sum_{j=0}^\infty c_js^j = 0\\ & \sum_{j=1}^\infty j(j+1)c_{j+1}s^{j} - \sum_{j=0}^\infty j c_{j}s^{j} + 2\ell + 1 \sum_{j=0}^\infty (j+1)c_{j+1}s^j+ \left[n-(\ell+1)\right] \sum_{j=0}^\infty c_js^j =0 \end{aligned}
Equating the coefficients of like powers gives:
j(j+1)c_{j+1} + 2\ell + 1(j+1)c_{j+1} - jc_j + \left[n-(\ell+1)\right]c_j =0
or:
c_{j+1} = \frac{j+\ell+1-n}{(j+1)(j+2\ell+2)}
To ensure that this series is normalizable, it must terminate (otherwise adding infinite terms will explode for j\to \infty and the function won’t be normalizable), and therefore there must exist an N such as:
c_{n-1} \neq 0, \quad c_n = 0
and therefore:
c_{N} = \frac{N+\ell-n}{n(n+2\ell+1)}
So:
n = N + \ell
Using the formula previously defined:
Ry \equiv -\frac{1}{n^2}\left[\frac{\mu}{2}\left(\frac{e^2}{4\pi\epsilon_0 \hbar}\right)^2\right] = \frac{E_1}{n^2},\quad n \in \mathbb N
This is the Rydberg energy formula and from:
\alpha = 2\sqrt{-\frac{2\mu}{\hbar^2}E_H} = \frac{2}{n}\left(\frac{\mu e^2}{4\pi\epsilon_0 \hbar^2}\right) = \frac{2}{n}\frac{1}{a_0},\quad n \in \mathbb N
the Bohr radius is defined as:
a_0 \equiv \frac{4\pi\epsilon_0 \hbar^2}{\mu e^2} = 5.29 \times 10^{-11}\, m = 0.529 \, \text{\AA}
The solution to the equation and the recursion formulas are known functions (associated Laguerre polynomials) and can be computed as:
L_{n-\ell-1}^{2\ell+1}(x) = \sum_{m=0}^{n-\ell-1} (-1)^m \frac{(n+\ell)!}{(n-\ell-1-m)!(2\ell+1+m)!m!} x^m
The few can be written as:
\begin{aligned} L_0^k(x) &= 1 \\ L_1^k(x) &= -x + k + 1 \\ L_2^k(x) &= \frac{1}{2} \left[ x^2 - 2(k + 2)x + (k + 1)(k + 2) \right] \\ L_3^k(x) &= \frac{1}{6} \left[ -x^3 + 3(k + 3)x^2 - 3(k + 2)(k + 3)x + (k + 1)(k + 2)(k + 3) \right] \end{aligned}
The problem is now solved and the solution can be constructed.
\chi(s) = s^{\ell+1}L(s)e^{-\frac{s}{2}} = s^{\ell+1}L_{n-\ell-1}^{2\ell+1}(s)e^{-\frac{s}{2}}
using \chi(r)=rR(r) and since r\propto s:
R\left(\frac{a_0n s}{2}\right) \propto \frac{1}{r}s^{\ell+1}L_{n-\ell-1}^{2\ell+1}(s)e^{-\frac{s}{2}} = s^{\ell}L_{n-\ell-1}^{2\ell+1}(s)e^{-\frac{s}{2}}
Introducing a normalization factor, the solution will be:
R\left(\frac{a_0n s}{2}\right) =\frac{1}{A} s^{\ell}L_{n-\ell-1}^{2\ell+1}(s)e^{-\frac{s}{2}}
With the normalization integral of the wavefunction
U(\mathbf R) = R(r)Y(\theta,\phi)
to be:
\int_{r=0}^{\infty}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\left|R(r)Y(\theta,\phi)\right|^2 r^2\sin(\theta)\mathrm dr \mathrm d \theta \mathrm d\phi = 1
The spherical harmonics are already normalized, therefore the only integral remaining is the radial normalization:
\int_0^\infty R(r)^2r^2\mathrm dr = \int_0^\infty s^{2\ell}\left[L_{n-\ell-1}^{2\ell+1}(s)\right]e^{-s}s^2 \mathrm ds = \frac{2n(n+\ell)!}{(n-\ell-1)!}
So the normalized radial function is:
\sqrt{\left[\frac{(n-\ell-1)!}{2n(n+\ell)!} \left(\frac{2r}{na_0}\right)^3\right]}\left(\frac{2r}{na_0}\right)^{l}L_{n-\ell-1}^{2\ell+1}\left(\frac{2r}{na_0}\right)e^{-\frac{r}{na_0}}
This can be written as function of the Bohr radius with a dimensioness quantity:
\rho = \frac{r}{a_0}
As for the subscripts: