Solving the Schrödinger Equation for the 1 Electron Atom (Hydrogen-Like)

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1 Stockton Univeristy Chemistry Program, School of Natural Sciences an Mathematics 101 Vera King Farris Dr, Galloway, NJ CHEM 340: Physical Chemistry II Solving the Schröinger Equation for the 1 Electron Atom Hyrogen-Like) Now that we have tackle the particle in a box, we now want to apply the Schröinger equation to etermine the structure of the one-electron atom. We have to start here because application to more than one electron gets very tricky. The math here is going to get a bit messy, but the important things to unerstan are the strategy we use an the form of the solutions. You will finally get to see what all your previous chemistry instructors tol you woul come later. In this ocument, there are many instances where constants are not inclue in orer to make things clearer. If you were going to o calculations with the solutions, you woul want to ensure you have the exact solution with all the appropriate constants in front. However, for our purposes, we ll take some shortcuts an come up with the correct form of the solution an see what we can learn from it. So to remin you, the Schröinger Equation in three imensions is: where is the Laplacian operator x y z. h m e Ψ V Ψ = EΨ 1) For an electron in an atom, what is V the potential energy of the electron)? For a one-electron atom, this potential is simply the Coulomb potential of the interaction of the positive nucleus with Z protons an the single electron: V = Ze ) r where e is the elementary charge an r is the istance between the nucleus an electron. If you stare at this equation for a bit you ll notice it is written in terms of r, a raial istance. This tell us we shoul consier using a ifferent coorinate system when solving the Schröinger Equation for the hyrogen atom. For example, if we stuck to Cartesian coorinates x, y, an z) we woul nee to rewrite Equation using the following relationship between r an the Cartesian coorinates x, y, an z: r = x y z 3) giving us: V x, y, z) = Ze x y z 4) which woul be a mess to eal with when solving the Schröinger Equation. So instea we will switch to spherical coorinates. This makes a lot more sense when you think about the spherical symmetry of an atom. In spherical coorinates, the parameters are r, θ, an φ as shown in Figure 1. When we make the jump to spherical coorinates, the Laplacian operator takes on a ifferent form. It will look a bit frightening, but we will take it apart piece by piece as we move along: = 1 r r ) r r r sin θ ) sin θ θ θ r sin θ φ 5) We can now combine this with Equations 1 an to solve for the wave functions a.k.a. orbitals) of the hyrogen atom.

2 Figure 1: Spherical an cartesian coorinates. Well, first lets write the Schröinger Equation in spherical coorinates: r r Ψ ) r r r sin θ Ψ ) Ψ sin θ θ θ r sin θ φ m eze h Ψ me r h Ψ = 0 6) Remember our goals whenever we solve the Schröinger Equation : 1) fin Ψ, the wave functions an ) fin E, the allowe energies. We begin by using a familiar trick, separate the variables into three separate functions: Ψr, θ, φ) = Rr)Θθ)Φφ) 7) where each of the three separate function only epens on one of the position variables. Combining Equations 6 an 7 an pulling constants through the partial erivatives we get: ΘΦ r r R ) RΦ r r r sin θ Θ ) RΘ Φ sin θ θ θ r sin θ φ m eze h RΘΦ me RΘΦ = 0 8) r h Now that is quite a lot of math for one sitting. We can o some housekeeping by iviing by RΘΦ an multiplying by r sin θ: sin θ 1 r R ) sin θ sin θ Θ ) 1 Φ R r r Θ θ θ Φ φ me Ze h r m ) h r sin θ = 0 9) By oing this we have gotten to a point where most of the erivatives are now in terms with their associate functions. We can o even better by oing some more housekeeping an move all the φ terms to the right an factor out a sin θ: [ 1 R r r R r ) 1 Θ sin θ sin θ Θ ) m eze θ θ h r m ] h r sin θ = 1 Φ Φ φ = m 10) In Equation 10 we have introuce a constant m or m, much like we i when we solve the classical wave equation. If both sies are equal for all r, θ, an φ, then both sies must equal a constant. Calling it m will allow us to come up with solutions that fit into the framework of atomic structure that you have seen before. Well, we can solve the secon part of that equation for the φ epenence since we have hanle equations like that before. Φ φ = m Φ 11) Solutions to this equation have the form: Φφ) = e imφ e imφ...or... cos mφ sin mφ To align us with the text an other sources of quantum mechanical information we will use both the complex solutions an the trigonometric solutions. Remember, they are equivalent, but the trig solutions are more

3 useful when trying to visualize the solutions while the complex solutions work better for the abstract math. So we have solutions for the φ epenence of the wave function. This is a goo place to stop an take a eep breath. We on t have anything yet that tells us much about the final solution. However, it is important to note we have introuce one quantum number so far, m, to escribe the φ epenence. We will see shortly that m will be restricte to a set of integer values. Keep that in min as we go along. We now go back to Equation 10 to efine the r terms in one parameter we ll call λ which is just a parameter, no relation to wavelength); λ = 1 r R ) m eze R R r h r m h r 1) which allows us to rewrite Equation 10 as λ sin θ sin θ Θ sin θ Θ ) = m 13) θ θ We can move the m over to the left sie to rearrange things a bit: sin θ sin θ Θ ) λ sin θ m ) Θ = 0 14) θ θ Equation 14 is calle the Legenre Equation. If an equation has a name that means it has solutions! We can lookup solutions for Θ an fin they have the form: Θ lm θ) = Θ m l θ) = P m l cos θ) 15) where we can lookup solutions for Pl m x) an then insert cos θ to fin our solutions for Θθ). These functions are also calle spherical harmonic functions an Pl m x) has the form: Pl 0 x) = 1 l x 1) l l l! x l 16) P m l x) = 1 x ) m/ m x m P 0 l x) 17) where both m an l are integers an m l or l m l. Also, the parameter λ = ll 1). These solutions look a little messy, but let s look at a few specific cases to see how it works. For example, P1 0 x) with m = 0 an l = 1: P1 0 x) = 1 x 1) 1 = 1 x = x 18) 1! x But for the Θ solution we nee to set x = cos θ: Θ 10 θ) = P 0 1 cos θ) = cos θ 19) This is one solution for the Θ part of the wave function. Let s look at a couple more. For example, m = 0 an l = : P 0 x) = 1 x 1)! x = 1 8 P 0 x) = 1 8 x 4 8 = 3 x 1 = 3x 1 To get Θ 0 we set x = cos θ an obtain: x x4 x 1) = 1 8 x 4x3 4x) = 1 8 1x 4) 0) 1) Θ 0 θ) = 3 cos θ 1 ) One more example, P 1 1 x) with m = 1 an l = 1: P 1 1 x) = 1 x ) 1 x P 1 0 x) = 1 x ) 1 x x = 1 x ) 1 3)

4 Substituting in x = cos θ we get: Θ 1 1θ) = 1 cos θ) 1 = sin θ 4) We now have two-thirs of the complete solution for our wave functions. So far, we have the two angular parts Θ an Φ. We have foun that to specify these solutions we nee two integers, or quantum numbers, l an m. Let s look at a few of these solutions in tabular form in Table 1. Table 1: Solutions for the angular portions of the hyrogen electron wave functions l m Θ lm θ) Φ m φ) ΘΦ type s 1 0 cos θ 1 cos θ p z 1 1 sin θ e iφ sin θ cos φ p x 1-1 sin θ e iφ sin θ sin φ p y 0 3 cos θ cos θ 1 z 1 sin θ cos θ e iφ sin θ cos φ xz -1 sin θ cos θ e iφ sin θ sin φ yz sin θ e iφ sin θ cos φ x y - sin θ e iφ sin θ sin φ xy You will notice the last column in the table, type, contains the orbital names that you have seen before in general or inorganic chemistry. So the quantum numbers l an m escribe the shape of these ifferent wave functions, which we will eventually call orbitals see Figure ). You will also notice that solutions for m 0 have an imaginary component e iφ or e iφ ). In orer to remove the imaginary component in the angular part of the wave function ΘΦ) we take linear combinations of the two solutions. So for example, for l = 1 an m = 1 the solution takes the form: p x Θ 11 Φ 1 Φ 1 ) = sin θe iφ e iφ ) = sin θcos φ i sin φ cos φ i sin φ) sin θ cos φ 5) an for l = 1 an m = 1 the solutions look like: p y Θ 11 Φ 1 Φ 1 ) = sin θe iφ e iφ ) = sin θcos φ i sin φ cos φ i sin φ) sin θ sin φ 6) Figure : 3-D representations of the angular part of the hyrogen atom wave functions.

5 Now the only part we are missing is the raial part of the wave function Rr)). Going back to Equation 1 an rearranging things: r R ) me Ze r r h r m ) h r λ = 0 7) where λ = ll 1) an l 0. This is an orinary ifferential equation in r an it is somewhat complicate to solve. However, once solve we will fin that for the solutions to be acceptable wave functions, the energy must be quantize accoring to: E n m ee 4 Z h n 8) where we have left out a few physical constants as I warne you about earlier. The important thing to notice here is that the for the hyrogen-like atom the energy is only a function of n. If you compare this form to Bohr s solution, they are the same except now the electrons aren t confine to sharp orbits but are escribe by the wave function Ψ. Also notice that the energy is negative an gets smaller approaches zero) as n gets larger. When n = then E = 0 an we say that the electron is free. This is just an arbitrary efinition since there is no natural zero for energy. In the process of solving Equation 7 we fin that the quantum number n appears naturally an must satisfy the conition that n l 1 or 0 l n 1, n = 1,, ) The solutions to Equation 7 epen on both n an l an take the form: ) 1 ) n l 1)! l3/ ) r R nl r) = n[n l)!] 3 r l e r/nao L l1 n1 30) na o na o ) where a o is the Bohr raius an is equal to a o = ɛoh πm ee an L l1 r n1 na o are calle Laguerre polynomials. Since they are name, there are a set of solutions to these Laguerre polynomials that you can lookup. Equation 30 looks complicate by if you strip away the crazy looking math it is simply a polynomial multiplie by an exponential function. For example for n = 1 an l = 0: R 10 r) = 1 π Z a o ) 3/ e Zr ao 31) So, in summary what we have been able to o is solve for the wave functions of the hyrogen atom by incluing the Coulombic potential in the Schröinger Equation an solving for Ψ by breaking the solution into parts: with allowe energies of the form: Ψ nlm r, θ, φ) = R nl r) }{{} raial part Θ lm θ)φ m φ) }{{} angularparts E n m ee 4 Z h n 33) The three integers that escribe the solutions shoul look familiar: quantum number name restrictions escribes orbital n principal quantum number n > 0 size l angular momentum quantum number 0 l n shape m magnetic quantum number l m l orientation 3)