Muons are constantly being produced in the upper atmosphere by the interaction of cosmic rays with the atmospheric air. Due to their high energy, a large fraction of them reaches the ground and even penetrate deep into rocks [3]. Pions decay into muons. Pions are created in the following interactions: In 1953, using a NaI synchrocyclotron, two American scientists Fitch and Rainwater observed radiation from muonic atoms [4]. This work indicated that the size of atomic nuclei was smaller than what had been supposed. In 1964, the NaI scintillation spectrometer was replaced by Ge(Li) detector because of its high resolution to investigate the muonic X-ray spectra that made great advances in accuracy and precision. The production of muon atoms in laboratory is carried out with a particle accelerator. Protons are incident on atomic nuclei like lithium or carbon after being accelerated to high energies. This collision creates a whole bunch of particles including pions. They decay into a muon and a neutrino.

These muons are then selectively channeled into beam lines and transported to the spectrometer where the sample lies. The muon is slowed down and captured by an atom into an orbit forming muonic atom.

Muonic atoms are nuclear probes and are used to test principles from quantum electrodynamics to chemical physics. Prior to 1980, most research was in negative muonic atoms. During the last few years positive muonic atoms are also being studied. In hadronic atoms, due to strong interaction between the hadron and the nucleus, the energy levels are shifted and broadened as compared to muonic atoms with strong dependence on the nuclear charge. The proton and neutron distribution insidethe nucleus can be separately analyzed by means of combined muonic and hadronic atom data, as the muon interacts mainly with the protons and the hadrons with both types of nucleons.

In this section, we apply Bohr’s model of hydrogen atom [5] to muonic hydrogen and calculate the orbit size for various principal quantum number n.

The radius of the n^th orbit is given by where symbols have their usual meaning.

The reduced mass of muonic hydrogen is where $m_p$ and $\mu$ are rest masses of proton and muon respectively. Using 1.8835 × 10⁻²⁸kg, the reduced mass for muonic hydrogen is $m_{\mu }=1.6928\times -28\mathrm{kg}$. Table 1 lists the orbit sizes for various values of n for electronic and muonic hydrogen atoms.

The radius of the n^th orbit of muonic hydrogen atom (H${}_{\mu }$) is 186 times less than the electronic hydrogen atom (H${}_e$). This is shown in Table 1. The energy of muon in n^th orbit is The relation between the energy levels of H${}_e$ and H${}_{\mu }$ is i.e., the corresponding energy levels of H${}_{\mu }$ is 186 times greater than H${}_e$.

The hydrogen atom problem in quantum mechanics is a two-body problem for which Schrödinger’s equation should be solved. We present below a brief derivation of the stationary states of the hydrogen atom for Coulomb potential [6]. The Schrödinger equation is The Hamiltonian is where $(-{\hslash }^2/2m){\nabla }^2$ is the kinetic energy operator, and $V\left(r\right)=e^2/\left(4\pi {ϵ}_{0}r\right)$ is the potential energy term. If $|\psi \rangle$ represents spatial part of the wave function $|\mathsf{\Psi }\rangle$, then for stationary states, we can write where E represents the energy eigenvalue of the system. General solution is of the form The Laplacian operator can be expressed in spherical polar coordinates $(r,\theta ,\varphi )$ as One can write $\psi (r,\theta ,\varphi )$ as where $R\left(r\right)$ is the radial part and $Y(\theta ,\varphi )$ is the angular part of $\psi (r,\theta ,\varphi )$. Using Equation (1) in the Schrödinger’s equation, we get the radial equation as The angular part of the Schrödinger’s equation is where $l(l+1)$ is the separation constant. Equation (2) can be further simplified by setting $Y(\theta ,\varphi )=\mathsf{\Theta }\left(\theta \right)\mathsf{\Phi }\left(\varphi \right).$ Thus Equation (3) has two variables $\theta$ and $\varphi$ which can be separated as where $m^2$ is the separation constant. Equation (5) can be solved to get

Equation (4) can be solved as where $P_l^m\left(\cos \theta \right)$ is the associated Legendre polynomial. If $\left|m\right|>l$, then $P_l^m=0$. Therefore, we must have $\left|m\right|\le l$ for solution to exist; l is a non-negative integer; $m=-l,-l+1,\cdots ,-1,0,1,\cdots ,l-1,l$. Now, $Y(\theta ,\varphi )=C{P}_l^m\left(\cos \theta \right)e^{im\varphi }$ which represents spherical harmonics. Normalizing, we get where $ϵ={(-1)}^m$ for $m\ge 0$ and $ϵ=1$ for $m\le 0$. It can be shown that the stationary part of the wave function is where L is associated Laguerre polynomial, and  The allowed energies of the hydrogen atom are Equation (7) is identical to the energy expression derived using Bohr’s model. Hence quantum mechanically, the energy eigenvalues are the same.

We define

For $1s$ and $2s$ state of H${}_e$, A is of the order of ${10}^{15}$ and ${10}^{14}$ respectively. For $1s$ and $2s$ state of H${}_{\mu }$, A is of the order of ${10}^{19}$ and ${10}^{18}$ respectively. These normalization constants do not have any physical significance.

For H${}_e$, 5.290 × 10⁻¹¹ m and for H${}_{\mu }$, $a_{\mu }$ = 2.846 × 10⁻¹³ m from Equation (6). Clearly $a_e>a_{\mu }$. The most probable value of r in the ground state and for the first excited state for both H${}_e$ and H${}_{\mu }$ are calculated and compared. The most probable value of r means that the probability of finding an electron or muon is highest for a given stationary state $\psi$.

For $1s$ state, The probability of finding electron or muon within an elemental volume $4\pi r^{2}r$ is Using (8), where By definition, the slope of the curve is zero where the probability is a maximum [5]. Therefore, the most probable value of r in $1s$ state is the Bohr’s radius a. For electronic hydrogen, electron is more likely to be found at $r=5.290\times {10}^{-11}$ m. For muonic hydrogen, muon is more likely to be found at $r=2.846\times {10}^{-13}$ m. Thus one sees that the muon is closer to the nucleus than the electron in $1s$ state by a couple of orders.

Finite difference methods are a class of numerical techniques for solving differential equations by approximating derivatives with finite differences [7]. Suppose $y=f\left(x\right)$, its first derivative is By definition where $h>0$ is the change in the variable x. Therefore, one can write provided h is reasonably small. One can replace the derivative $\mathrm{d}y/\mathrm{d}x$ by a difference $\left(f(x+h)-f\left(x\right)\right)/h$ in the differential equation and solve it. Using index notation, one can write For 2^nd order derivative, we have Approximating using finite difference method we write, Using index notation, we write

While solving the Schrödinger’s equation analytically, we get an expression for radial part $R\left(r\right)$ of the stationary state $\psi (r,\theta ,\varphi )$ as where $u\left(r\right)=rR\left(r\right)$. We will be using Equation (9) to determine the energy eigenvalues for the hydrogen atom. Equation (9) can be written as Equation (10) is an eigenvalue equation. We need to express different terms on left hand side in matrix form. As previously mentioned, ${\mathrm{d}}^{2}u/\mathrm{d}{r}^2$ term can be replaced by a finite difference where the index i is iterated up to N. This can be written as a square matrix of order N as $l(l+1)/r^2$ can be written in a square matrix of order N as $e^2/\left(4\pi {ϵ}_{0}r\right)$ can be written as a square matrix of order N as

The sum of all the above matrices will be the square matrix of the eigenvalue equation for which eigenvalues and eigenvectors will be determined.

The +python+ program for energy computation is the following:

The +python+ program for radial density distribution computation is the following:

The equation used for energy computation is

For the case of H${}_{\mu }$, the electronic mass is replaced by the reduced mass of H${}_{\mu }$, and the range of r for calculating energy is taken from 4 ×10 ⁻¹¹ m to the point at nucleus, excluding it.

The number of iteration points is taken as $N=3000$ to get more accurate energies. Increasing N will increase the time to compute. Since the order of the square matrix is 3000, there will be 3000 eigenvalues. There are n stationary states for the hydrogen atom, where n is the principal quantum number. While solving numerically, a given eigenvector will have a corresponding energy eigenvalue. Therefore, while constructing the solution, many points will correspond to the same energy. Once, a stationary state is determined, the next stationary state will be found which has a different eigenvalue. In this way all eigenvectors and eigenvalues are determined.

The computed first ten eigenenergies of H${}_e$ and H${}_{\mu }$ are shown in Table 2. We observe that the eigenvalues nearly match the experimentally observed energies for H${}_e$.

The program also plots the probability density ($P_r$) as a function of r for the first four eigenstates of H${}_e$ and H${}_{\mu }$. Here by probability we mean ${\left|u\left(r\right)\right|}^2$. Figure 1 and Figure 2 show the plots for H${}_e$ and H${}_{\mu }$ respectively for different energy eigenstates.

A python program has been written to compute the spatial distribution of electron and muon in hydrogen for different values of n, l, and m quantum numbers.

Density distribution plots for $n=1,2,3,4$; $l=0,1,2,3$, and $m=0,1$ for H${}_e$ and H${}_{\mu }$ are shown in Figure 3, Figure 4, Figure 5, Figure 6, Figure 7, Figure 8, Figure 9, Figure 10, Figure 11, Figure 12, Figure 13, Figure 14, Figure 15, Figure 16.

The probability of finding the particle decreases from red to blue in the density plots.

Here, we will interpret the physical meaning of the values obtained. A comparative study of the energy levels of H${}_e$ and H${}_{\mu }$ is done, followed by a discussion about their spectra.

In the absence of magnetic and electric field, the hydrogen atom will be in a stationary state. Until and unless energy is absorbed or emitted, the atom’s energy will be the same. In quantum mechanics, a particle or a system cannot have any arbitrary energy. Only certain discrete energies are allowed. These are referred to as the Bohr energies. Bohr energy levels depend only on the principal quantum number n, and independent of other quantum numbers such as l, $m_l$, and $m_s$. Further, the Bohr energy comprises several closely spaced levels called the fine structure, and hyperfine structure. We restrict ourselves to only Bohr energy levels here.

Table 2 shows the computed Bohr energies for H${}_e$ and H${}_{\mu }$ atoms. The negative sign means that the particle is in a bound state; here it is bound to the nucleus. The particle in its lowest energy state (ground state) has $-13.597\mathrm{eV}$ energy in the case of electron and $-2528.004\mathrm{eV}$ energy for muon. The muonic energy levels have more energy than the corresponding electronic energy levels. One can say that the muons are in a deeper energy well than the electrons.

To excite a H${}_{\mu }$ it requires more energy than to excite an H${}_e$. The energy levels of H${}_e$ correspond to ultra-violet, visible and infrared region in electromagnetic spectrum, whereas those of H${}_{\mu }$ correspond to the X-ray region of the electromagnetic spectrum. The atom will make a transition from one energy level to a higher energy level when energy of the incident radiation is equal to the difference. H${}_e$ atoms can be excited by UV, visible, and infrared photons; muonic hydrogen atoms can be excited only by X-ray photons. During de-excitation, photons in the corresponding regions are emitted.

Let $E_1$ represent the energy of one stationary state and $E_2$ represent the energy of another stationary state. When a photon of frequency $\nu =(E_2-E_1)/h$ is absorbed by the atom, transition happens from $E_1\to E_2$. When the atom transitions from $E_2\to E_1$ then a photon of same frequency $\nu$ is emitted. From the calculated energy levels of H${}_e$ and H${}_{\mu }$ we can calculate the emission and/or absorption spectrum using the above formula. In literature, different series in the hydrogen spectrum are given different names. When an atom makes a transition from any higher level to first, second, third, fourth and fifth energy levels, the spectral series corresponding to these are called Lyman, Balmer, Paschen, Brackett, and Pfund series respectively (see Table 2).

Experimentally, when measuring the wavelengths of emission spectrum of muonic atoms, they have to be measured quickly because the mean lifetime of a muon is about $2.2\mathsf{\mu }\mathrm{s}$. Experimentalists have measured X-ray spectra of various muonic atoms to study the nuclear structure as the muonic probability density is higher in the vicinity of the nucleus.