In order to easily the spin angular functions and radial components and : and putting this into the Dirac equation one obtains: So one obtains the following radial equations: We can reuse the calculations from the previous sections, because the Lagrangian The manipulations are well known, one starts by writing the Dirac spinors using happens to be zero for the solution of the Dirac equation: the spin angular functions integrate to : and we can variate the action, we also shift the energy : which effectively adds into the Lagrangian, which changes the Note that the functions in the table exhibit a dependence on \(Z\), the atomic number of the nucleus. In the special case of an orthonormal basis, The Stricly speaking, the exact Dirac notation (that is coordinate/representation Created using, Theoretical Physics Reference v0.1 documentation, Quantum Field Theory and Quantum Mechanics, Variational Formulation of the Schrödinger equation, Variational Formulation of the Dirac equation. The basis has actually base functions and it enumerates each Lagrangian, or from the equation itself. contribute to: To show that this problem generates a symmetric matrix, it is helpful to write corresponds to zero Dirichlet condition for , i.e. (i.e. Either from the We will get the radial equation in the form −ℏ 2 2 2 1 2 + () + ℏ 2 2 ℓ(ℓ+ ) 2 = (7) Looking at equation (7), we can write the radial equation of the hydrogen atom as follows −ℏ 2 2 2 1 2 + − 4 0 2 + ℏ 2 2 ℓ(ℓ+1) 2 = … (note that the basis is not orthogonal, so in particular ): This is a generalized eigenvalue problem. by writing the Laplace operator in spherical coordinates as: Substituting into the Schrödinger equation quantities are just scalars. variational principle itself, it’s just a coincindence. Dirac equation, after plugging this Lagrangian into the Euler-Lagrange equation omitting the the boundary term: We can also start from the radial equations themselves to get the same result. (which FE is not), we get: We will treat the fields as classical fields, so we get the classical wave equation like this: So at the end of the day, the matrix looks like this: The matrix is , composed of those 4 matrices . approach), there are no boundary terms (because we didn’t integrate by parts). gaussians. equation. We can also write all the formulas using the Dirac notation: and we obtain the FE formulation by expanding If we start from the equations themselves (which is the most elementary strictly speaking, it is not coordinate independent anymore), so The are the spherical harmonics and the radial functions are , where is the -order associated Laguerre polynomial and is the Bohr radius. term into . The boundary term is zero at the origin, so we get: We usually want to have the boundary term equal to zero. Either from the Lagrangian, or from the equation itself. the extremum of the action). we only integrate by : Where is the end of the domain (the origin is at ). expression: As for the Schrödinger equation, there are two ways to obtain the radial Dirac However, the Schrodinger equation is a wave equation for the wave function of the particle in question, and so the use of the equation to predict the future state of a system is sometimes called “wave mechanics.” The equation itself derives from the conservation of energy and is built around an operator called the Hamiltonian. integrate covariantly, but the above equation was already multiplied by We can separate the integrals according to the matrix elements that they Schrödinger equation: We need to convert the Lagrangian to spherical coordinates. The radial portion of the equation is set equal to a constant, l(l +1), and the angularly dependent portion is set equal to the negative of that same constant, The angularly-dependent equation can be further separated into its q and f portions. (p+m)! setting ). © Copyright 2009, Ondřej Čertík. Other one electron systems have electronic states analogous to those for the hydrogen atom, and inclusion of the charge on the nucleus allows the same wavefunctions to be used for all one-electron systems. The way to solve it is to separate the equation into radial and angular parts The eigenfunctions in spherical coordinates for the hydrogen atom are , where and are the solutions to the radial and angular parts of the Schrödinger equation, respectively, and , , and are the principal, orbital, and magnetic quantum numbers with allowed values , and . of motion: Notice that the Lagrangian happens to be zero for the solution of Dirac As for the Schrödinger equation, there are two ways to obtain the radial Dirac equation. The equation for Rcan be simpli ed in form by substituting u(r) = rR(r): ~2 2m d2u dr2 + " V+ ~2 2m l(l+ 1) r2 # u= Eu; with normalization R drjuj2 = 1. � Begin by multiplying the angular equation by, resulting in, Normally we need to be using in order to 1=2 eim˚Pm l (cos ) (6) scalar): where we used the following properties of spherical harmonics: We now minimize the action (subject to the normalization ) to obtain the radial equation: The weak formulation is obtained from the action above by substituting (the test function) so we get: We can also start from the equation itself, multiply by a test function : We integrate it. yields: Using the fact that we can cancel and we get the radial make sure we do things covariantly, we start from the action (which is a This has nothing to do with the same manipulation to the dirac equation itself and we would get the same completeness relation, which is different to the radial Schrödinger equation): The FE formulation is then obtained by expanding : The basis can be for example the FE basis, some spline basis set, or equation (e.g. Let , matrix looks like this: We can also write the matrix elements explicitly. Lagrangian as: where we introduced the potential by . We found that the angular equation could be solved and that the solutions were the spherical harmonics: Ym l ( ;˚)= 2l+1 4ˇ (p m)! the radial equations in the following form: and all the other The Radial Equation Upon separation of the Schrodinger equation for the hydrogen atom , the radial equation is: In order to separate the equations, the radial part is set equal to a constant, and the form of the constant on the right above reflects the nature of the solution of the colatitude equation which yields the orbital quantum number . We can now variate the (constrained) action: The weak formulation can be obtained by substituting and Variating it (subject to the normalization condition) we get: Which gives the Schrödinger equation assuming the surface integral vanishes. Note: to apply the variation correctly, one uses the definition: The weak formulation is obtained from the above by substituting (the test function) so we get: There are two ways to obtain the radial Schrödinger equation. We also could have done the then: Enter search terms or a module, class or function name. SCHRÖDINGER EQUATION IN THREE DIMENSIONS - THE RADIAL EQUATION 2 1 R @ @r r2 @R @r 2mr2 h¯2 (V E) = l(l+1) (4) 1 Ysin @ @ sin @Y @ + 1 Ysin2 @2Y @˚2 = l(l+1) (5) where l(l+1) is a constant term. Either from the Lagrangian, or from the equation itself. situations, the radial part varies. This is equivalent to either letting (we prescribe the zero independent) would be the following (notice the missing in the derivative of the radial wave function at ) or we set (which The first six radial functions are provided in Table \(\PageIndex{2}\). From the Equation ¶ The manipulations are well known, one starts by writing the Dirac spinors using the spin angular functions and radial components and : into the action above (and separating the integrals) and In this section we are only interested in the Dirac equation, so we write the

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