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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. Schroedinger Equation in three dimensions 2 2 2 2 2 2 2 2m x y z U ( x, y , z ) U ( x, y, z ) E 1 e2 40 x2 y2 z 2 The equation cannot be solved analytically in Cartesian coordinates. However, if we use spherical polar coordinates, the equation becomes more complicated, but solvable (although it’s still not an easy task!): Now, (r , , ), and 2 2 2 1 1 2 2 U (r , , ) E sin 2 2 2 2 2m r r r r sin r sin This equation has a separable solution t hat can be factored as : (r , , ) R(r ) ( ) ( ) 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 ε0 . In addition, three parameters emerge in a natural way as “indices” or “labels” for the solutions. They are listed below: n the principal quantum number, which can take values of n 1,2,3 . It is essentiall y the same as n in the Bohr model. l the angular momentum quantum number. Its allowed val ues are l 0, 1, 2, 3, , n 1. ml the magnetic quantum number. Its allowed values are ml 0, 1, 2, 3, , l. Complete with these quantum numbers, the separated solutions of the Schroedinger Equation can be wrtiien as: n,l ,m (r , , ) Rn,l (r )l ,m ( ) m ( ) l l l 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: me4 1 13.6 eV En 2 2 2 2 32 0 n n2 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, ml ). 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: -1.5 eV (3, 0, 0) (3, 1, 1) (3, 1, 0) (3, 1, -1) (2, 0, 0) (2, 1, 1) (2, 1, 0) (2, 1, -1) (3, 2, 2) (3, 2, 1) (3, 2, 0) (3, 2, -1) (3, 2, -2) -3.4 eV -13.6eV (1, 0, 0) 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: P( x)dx ( x) dx 2 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 probability 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: 2 P( x, y )dxdy ( x, y ) dxdy Again, the same rule is valid – the squared modulus of the wave function is the probability density. In order to find the probabilit y that the particle in a 1 - D well can be found in a region of finite width, say, between certain points x1 and x2 , one has to perform an integratio n : x1 P ( x1 x x2 ) ( x) dx 2 x1 In the case of a 2 - D quantum well, in order to find the probabilit y that the particle can be found in a finite size surface area region defined by points x1 and x2 , and y1 and y2 , one hat to perform a double integratio n : P ( x1 x x2 , y1 y y2 ) y 2 x1 ( x, y ) y1 x1 2 dxdy In the case of the hydrogen atom the “philosophy” is essentially the same. The squared modulus of the wavefun2 ction: (r , , ) gives the probability density of finding the electron at the location ( r , , ). To compute the actual probability of Finding the electron in a volume element dV located at ( r , , ), we multiply the probability density by this volume element. In spherical polar coordinates the volume element is: dV r 2 sin drdd 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: n,l ,m (r , , ) Rn,l (r )l ,m ( ) m ( ) l l l Repeated from the preceding page : n,l ,m (r , , ) Rn,l (r )l ,m ( ) m ( ) and dV r 2 sin drdd l l l Therefore, the probability is: 2 2 2 2 n,l ,m (r , , ) dV Rn,l (r ) l ,m ( ) m ( ) r 2 sin drdd l l l 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. 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. r Nucleus dr Repeated from the last slide: 2 2 2 2 n,l ,m (r , , ) dV Rn,l (r ) l ,m ( ) m ( ) r 2 sin drdd l l l In order to find the probability in question, we have to integrate over the other two variables, and : 2 P(r ) dr r Rn,l (r ) dr l ,ml ( ) sin d ml ( ) d 2 2 2 0 2 0 This formula looks scary, right? Fortunately, however, it is not scary at all, because in the separated wavefunction each function is individually normalized, so that both integrals are equal to unity! Hence, the radial probability density takes a simple form: P(r ) r Rn,l (r ) 2 2 and the probabilit y of finding the electron within the shell volune is : 2 P(r )dr r Rn ,l (r ) dr 2 First few radial wave functions Rnl (r ) of the hydrogen atom : The radial probabilit y density functions for the three lowest states of hydrogen : P(r ) Rnl (r ) 2 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 probability density functions 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 mathematical form of this energy is identical as in the Bohr model. But what do the other two quantum numbers, l and ml , 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: L l (l 1) (sorry, this is another mathematical equation I have to ask you to “accept without proof”). For instance, in a state with l 1, the electron' s angular momentum vector length is L 2 ; for l 3, L 3(3 1) 12 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 momenta. 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. Lower ang. momentun than for circular orbit Circular orbit: Maximum L 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. Example : L l (l 1) What' s the maximum possible value of ml ? ml L (projectio n cannot be longer tha n the vector itself! ) ml2 (l 2 l ), so ml l is OK. But could ml be equal to l 1? Let' s check : (l 1) 2 l 2 2l 1 l 2 l The projection would be longer than the vector, so such a situa tion is not possible.