Question

A hydrogen atom electron is in a \(2 p\) state. If no experiment have been done to establish a \(z\) -component of angular momentum, the atom is equally likely to be found with any allowed value of \(L_{i}\). Show that if the probability densities for these different possible states are added (with equal weighting). the result is independent of both \(\phi\) and \(\theta\).

Short answer

By calculating the wavefunctions for each value of \(m_l\) in the \(2p\) state of a hydrogen atom and adding the probability densities together, we find that the result is independent of the azimuthal angle \(\phi\) and the polar angle \(\theta\). This confirms that if no experiment is conducted to determine the \(z\)-component of angular momentum, the atom is equally likely to be found with any possible value of angular momentum.

Step-by-step solution

Introduction to the problem

The \(2p\) state implies that the principle quantum number \(n = 2\) the angular momentum quantum number \(l = 1\), since \(p\) corresponds to \(l = 1\). The magnetic quantum number \(m_l\) can be from \(-l\) to \(+l\), so, in this case, \(m_l\) can be \(-1\), \(0\), \(+1\). Since there is no prior information about the \(z\)-component of angular momentum, all three states are equally probable.

Calculate the wavefunctions

For a hydrogen-like atom, the wavefunction can be expressed as \( \psi(r, \theta, \phi) = R(r)Y^{m_l}_l(\theta, \phi) \), where \(R(r)\) is the radial wavefunction and \(Y^{m_l}_l(\theta,\phi)\) is the spherical harmonics function. The radial part is the same for all three \(m_l\) states. The spherical part, for our case, will correspond to the Ylm-values with \(l = 1\) and \(m = -1, 0, +1\). The spherical part can be found from the standard quantum mechanics textbooks or online resources.

Add the probability densities

The probability density is given by \( |\psi|^2 \). To get the final outcome, we add the probability densities for the three \(m_l\) states. This is done by computing the absolute square of each \(Y^{m_l}_l\) and adding them up.

Conclusion

After performing the addition operation in the previous step, you will find that the outcome does not depend on the variables \(\phi\) and \(\theta\). This shows that without any experiment conducted to determine the \(z\)-component of angular momentum, the atom is indeed equally likely to be found with any possible value of angular momentum, regardless of \(\phi\) and \(\theta\).

Key concepts

Hydrogen Atom

The hydrogen atom, a fundamental building block of chemistry and physics, is a system consisting of a single electron bound to a proton. It is the simplest and most well-studied atomic system in quantum mechanics. The energy levels of the hydrogen atom are quantized, meaning the electron can only exist in certain discrete states characterized by quantum numbers.

In these quantum states, the electron orbits the nucleus in regions of space known as orbitals. Orbitals are not like the orbits of planets, but rather are probability clouds where the electron is likely to be found. Understanding the hydrogen atom's structure and behavior is critical to grasping the basics of atomic physics and quantum chemistry.

Angular Momentum

Angular momentum in quantum mechanics is a measure of an object's rotational motion. For particles like electrons in atoms, we can't talk about their rotation in the traditional sense, but they still possess angular momentum characterized by quantum numbers.

The orbital angular momentum is associated with the shape and orientation of an electron's orbital and is given by the quantum number 'l'. The hydrogen atom electron in the problem is in a '2p' state, meaning it has a principal quantum number 'n' of 2 and an angular momentum quantum number 'l' of 1. There are different orientations of this angular momentum defined by another quantum number, 'm', which can take values from '-l' to 'l'.

Spherical Harmonics

Spherical harmonics are mathematical functions that arise in the context of solving the Schrödinger equation in spherical coordinates. They are used to describe the angular part of the wavefunctions for particles in spherical potentials, such as electrons in atoms.

In the case of the hydrogen atom, spherical harmonics correspond to the different possible angular orientations of the electron's orbitals. These harmonics have distinct shapes and symmetries and are often denoted as 'Ylm', where 'l' is the orbital quantum number and 'm' is the magnetic quantum number. The importance of spherical harmonics goes beyond the hydrogen atom, as they find applications in many areas of physics and engineering for solving problems with spherical symmetry.

Wavefunction

A wavefunction is a mathematical function that describes the quantum state of a system. In the case of the hydrogen atom, the wavefunction \( \psi(r, \theta, \phi) \) represents the electron's state, which includes information about its position and energy. The square of the absolute value of the wavefunction, \( |\psi|^2 \) gives us the probability density, which tells us how likely we are to find the particle at a specific location in space.

The wavefunction itself is made up of two parts: the radial part, \( R(r) \) and the angular part, \( Y^{m_l}_l(\theta, \phi) \) - the spherical harmonics. It's the latter which dictates the directional distribution around the nucleus and it changes with the '2p' state considered in our exercise.

Probability Density

Probability density in quantum mechanics is a fundamental concept that quantifies the likelihood of finding a particle, such as an electron, in a particular region of space. It is derived from the wavefunction and represented by the square of the wavefunction's magnitude, \( |\psi|^2 \).

By considering the probability densities for different states of the electron in a hydrogen atom, we can understand where the electron is most likely to be found around the nucleus. In our exercise, when we sum the probability densities of different '2p' states, we get a distribution that is independent of the angles, showing that the electron's position is equally likely in all directions when there is no preferred orientation for its angular momentum.