Hamiltonian operator hydrogen atom

Applications of quantum mechanics to chemistry invariably deal with systems (atoms and molecules) that contain more than one particle. Apart from the hydrogen atom, the stationary-state energies caimot be calculated exactly, and compromises must be made in order to estimate them. Perhaps the most useful and widely used approximation in chemistry is the independent-particle approximation, which can take several fomis. Conuiion to all of these is the assumption that the Hamiltonian operator for a system consisting of n particles is approximated by tlie sum... 

The reason a single equation = ( can describe all real or hypothetical mechanical systems is that the Hamiltonian operator H takes a different form for each new system. There is a limitation that accompanies the generality of the Hamiltonian and the Schroedinger equation We cannot find the exact location of any election, even in simple systems like the hydrogen atom. We must be satisfied with a probability distribution for the electron s whereabouts, governed by a function (1/ called the wave function. 

The sum of two operators is an operator. Thus the Hamiltonian operator for the hydrogen atom has — j as the kinetic energy part owing to its single election plus — 1/r as the electiostatic potential energy part, because the charge on the nucleus is Z = 1, the force is atrtactive, and there is one election at a distance r from the nucleus... 

The electronic Hamiltonian commutes with both the square of the angular momentum operator r and its z-component and so the three operators have simultaneous eigenfunctions. Solution of the electronic Schrddinger problem gives the well-known hydrogenic atomic orbitals... 

In order to apply quantum-mechanical theory to the hydrogen atom, we first need to find the appropriate Hamiltonian operator and Schrodinger equation. As preparation for establishing the Hamiltonian operator, we consider a classical system of two interacting point particles with masses mi and m2 and instantaneous positions ri and V2 as shown in Figure 6.1. In terms of their cartesian components, these position vectors are... 

If we replace the z-component of the classical angular momentum in equation (6.87) by its quantum-mechanical operator, then the Hamiltonian operator Hb for the hydrogen-like atom in a magnetic field B becomes... 

The spin magnetic moment Ms of an electron interacts with its orbital magnetic moment to produce an additional term in the Hamiltonian operator and, therefore, in the energy. In this section, we derive the mathematical expression for this spin-orbit interaction and apply it to the hydrogen atom. 

Thus, the total Hamiltonian operator H for a hydrogen atom including spin-orbit coupling is... 

While the Hamiltonian operator Hq for the hydrogen atom in the absence of the spin-orbit coupling term commutes with L and with S, the total Hamiltonian operator H in equation (7.33) does not commute with either L or S because of the presence of the scalar product L S. To illustrate this feature, we consider the commutators [L, L S] and [S, L S],... 

The Hamiltonian operator for a hydrogen atom in a uniform external electric field E along the z-coordinate axis is... 

The simplest atomic system that we can consider is the hydrogen atom. To obtain the Hamiltonian operator for this three-dimensional system, we must replace the operator d2/dx2 by the partial differential operator... 

The interaction between an electron and a nucleus in a hydrogen atom gives rise to a potential energy that can be described by the relationship -e2/r. Therefore, using the Hamiltonian operator and postulate IV, the wave equation can be written as... 

In order to solve the wave equation for the hydrogen atom, it is necessary to transform the Laplacian into polar coordinates. That transformation allows the distance of the electron from the nucleus to be expressed in terms of r, 9, and (p, which in turn allows the separation of variables technique to be used. Examination of Eq. (2.40) shows that the first and third terms in the Hamiltonian are exactly like the two terms in the operator for the hydrogen atom. Likewise, the second and fourth terms are also equivalent to those for a hydrogen atom. However, the last term, e2/r12, is the troublesome part of the Hamiltonian. In fact, even after polar coordinates are employed, that term prevents the separation of variables from being accomplished. Not being able to separate the variables to obtain three simpler equations prevents an exact solution of Eq. (2.40) from being carried out. 

Since Hj does not have spherical symmetry like the hydrogen atom the angular momentum operator L2 does not commute with the Hamiltonian, [L2,H] 7 0. However, Hj does have axial symmetry and therefore Lz commutes with H. The operator Lz = —ih(d/d) involves only the 0 coordinate and hence, in order to calculate the commutator, only that part of H that involves need be considered, i.e. 

Quantum numbers and shapes of atomic orbitals Let us denote the one-electron hydrogenic Hamiltonian operator by h, to distinguish it from the many-electron H used elsewhere in this book. This operator contains terms to represent the electronic kinetic energy ( e) and potential energy of attraction to the nucleus (vne),... 

The differential equation for <1> results from a replacement of each coordinate X in the Hamiltonian operator by iS/Sp. This approach is not at all simple because of the Coulomb potentials, which involve r . Nevertheless, Hylleraas did succeed in solving this equation for the hydrogen atom [35]. 

The hydrogenic atom energy expression has no 1-dependence the 2s and 2p orbitals have exactly the same energy, as do the 3s, 3p, and 3d orbitals. This degree of degeneracy is only present in one-electron atoms and is the result of an additional symmetry (i.e., an additional operator that commutes with the Hamiltonian) that is not present once the atom contains two or more electrons. This additional symmetry is discussed on p. 77 of Atkins. 

If the relativistic effects are sufficiently large and therefore cannot be accounted for as corrections, then as a rule one has to utilize relativistic wave functions and the relativistic Hamiltonian, usually in the form of the so-called relativistic Breit operator. In the case of an N-electron atom the latter may be written as follows (in atomic units, in which the absolute value of electron charge e, its mass m and Planck constant h are equal to one, whereas the unit of length is equal to the radius of the first Bohr orbit of the hydrogen atom) ... 

A complete treatment of this derivation can be found in Ref. [19]. The first three terms in the kinetic energy operator indicates the presence of a 3D harmonic oscillator, and the final two terms indicate the presence of a 2D rotator (as for the hydrogen atom). A similar conclusion was made by Auberbach et al. [22] where they use a semi-classical quantisation method and the molecule is said to undergo a unimodal distortion and then the semi-classical Hamiltonian is found to be separated into two parts - a harmonic oscillator part with three vibrational coordinates and a rotational part with two rotational coordinates. However, more progress in terms of specifying the wavefunctions of the system can be made by following a different approach. 

In this equation, H is the Hamiltonian operator, P is the wavefunction that represents the state of the electrons and nuclei, and E is the system s energy. Although in theory it can be solved for any system, exact solutions are impracticable for anything more complicated than a hydrogen atom. Therefore we must find approximate... 

By applying the Hamiltonian operator of the hydrogen atom on fu, we can readily obtain Eis ... 

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