# Hydrogen Atom in Wave Mechanics The good news is that the Schroedinger equation for the hydrogen atom has an EXACT ANALYTICAL solution! (this is one of.

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Hydrogen Atom in Wave Mechanics The good news is that the Schroedinger equation for the hydrogen atom has an EXACT ANALYTICAL solution! (this is one of the few problems in Quantum Mechanics that does have such a solution – most problems in QM cannot be solved exactly). The bad news, however, is that the procedure of solving the equation is extremely complicated. In a standard QM textbook it usually spans over about forty pages loaded with math. In order to clearly understand the solution procedure, one has to be familiar with elements of highly advanced calculus, such as functions called the “spherical harmonics” and things called the “Laguerre polynomials”. This is much beyond the mathematical preparation level required for the Ph314 course. Therefore, we will not go through the details of the solution procedure – we will only review the general properties of the solutions, and you will be asked to “accept without proof” a number of things.

The equation cannot be solved analytically in Cartesian coordinates. However, if we use spherical polar coordinates, the equation beco- mes more complicated, but solvable (although it’s still not an easy task!): Schroedinger Equation in three dimensions

The solutions of the Schroedinger Equation for the hydrogen atom contains the variables r, , and , and fundamental constants, such as the the electron mass and charge, the Planck Constant, speed of light, and ε 0. In addition, three parameters emerge in a natural way as “indices” or “labels” for the solutions. They are listed below: Complete with these quantum numbers, the separated solutions of the Schroedinger Equation can be wrtiien as:

The principal quantum number n specifies the energy level, just as in the Bohr model. Te quantized energy levels obtained by solving the Schroedinger Equation are given by exactly the same formula as in the Bohr model: (1, 0, 0) (2, 0, 0) -13.6eV -3.4 eV (2, 1, 1) (2, 1, 0)(2, 1, -1) -1.5 eV (3, 0, 0) (3, 1, 1)(3, 1, 0)(3, 1, -1)(3, 2, 2)(3, 2, 1)(3, 2, 0)(3, 2, -1) (3, 2, -2) Each quantum state of the electron in the Wave Mechanics Model of the hydrogen atom can be described by a “triad” of integers: (n, l, m l ). The energy of the state depends only on the first of these numbers, but otherwise they are all different quantum states. All states for n=1, 2, and 3 are shown below: Remember the notion of DEGENERACY we talked about some time ago? What is the degeneracy of the n = 1 state? Of the states with n = 2? Of the states with n = 3?

Radial probability densities First, let’s shortly recapitulate what we told about probability density. Recently, we talked about the probability of finding a particle in the x region between x and x + dx in a 1-D quantum well. This probability Is related to the particle-in-the-well wavefunction as: In other words, the squared modulus of the wave function gives the probability density. In the case of a 2-D quantum well, we are interested in the proba- bility of finding the particle in a “box”, or surface area element defined by regions ( x, x + dx) and (y, y + dy). This probability is given by: Again, the same rule is valid – the squared modulus of the wave function is the probability density.

In the case of the hydrogen atom the “philosophy” is essentially the same. The squared modulus of the wavefun- ction: gives the probability density of finding the electron at the location To compute the actual probability of Finding the electron in a volume element dV located at we multiply the probability density by this volume element. In spherical polar coordinates the volume element is: A moment ago we said that the solution of the Schroedinger Equation For the hydrogen atom can be separated into three functions, each one depending only on a single variable – which can be written as:

Therefore, the probability is: Using this expression, one can calculate many “goodies”, or many features of the hydrogen atom. However, such calculations are not trivial, and we will use them to determine only a single function of interest – namely, the so-called radial probability density. r dr Nucleus Imagine that the nucleus (proton) is surrounded by a thin spherical “shell” of radius r and thickness dr. We want to determine the probability of finding the electron within the volume of this shell.

Repeated from the last slide: In order to find the probability in question, we have to integrate over the other two variables,  and  : This formula looks scary, right? Fortunately, however, it is not scary at all, because in the separated wavefunction each function is indivi- dually normalized, so that both integrals are equal to unity! Hence, the radial probability density takes a simple form:

Practical example: (from the final exam in the Fall 2006 Ph314 course).

Since the total wave function of hydrogen is a three-dimensional function, plotting the overall probability density function is a much more complicated task than plotting the radial probability. But the pictures shown here can give you a pretty good idea of how these Functions look like for various quantum states of the electron. Imagine each plot to be rotated by 360° about the vertical axis (z).

Another possible way of illustrating how the 3-D proba- bility density funct- ions look like is to plot the surfaces of constant |ψ(r, ,  )| 2

We already know what role the quantum number n plays: it describes the energy of a given state, and the mathe- matical form of this energy is identical as in the Bohr model. But what do the other two quantum numbers, l and m l, tell us? What is their physical meaning? The first of these numbers is related to the electron’s angular momentum. Namely, the length of the electron’s angular momentum vector is given by: (sorry, this is another mathematical equation I have to ask you to “accept without proof”).

As was said, for states with the quantum number n, l can take values of l = 0, 1, 2, … n-1. As follows from the above, there are states of the the same energy, but with different angular mo- menta. This is not a surprising situation in physics. There is a classical analog in the planetary motion: There are many orbits of the same energy, but with different ellipticity. The circular orbit corresponds to the largest angular momentum. With increasing ellipticity, the angular momentum decreases. Circular orbit: Maximum L Lower ang. momentun than for circular orbit Lowest ang. momentum of the three orbits shown

A comet: Very elongated orbit At this part of the orbit, the comet’s motion is very slow. Comets, which are celestial bodies with extremely elongated orbits, spend most time both close to the Sun, or far away from the Sun, in the peripheral region of the Solar System. Corresponding situation in hydrogen atom

The physical meaning of the ml quantum number: it’s the length of the projection of the angular momentum on the z axis. The projection can only take values that are integer multiples of ℏ.

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