Stockton UniveristyChemistry Program, School of Natural Sciences and Mathematics

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CHEM 3420: Physical Chemistry II

Solving the Schrodinger Equationfor the 1 Electron Atom (Hydrogen-Like)

Now that we have tackled the particle in a box, we now want to apply the Schrodinger equation todetermine the structure of the one-electron atom. We have to start here because application to morethan one electron gets very tricky. The math here is going to get a bit messy, but the important thingsto understand are the strategy we use and the form of the solutions. You will finally get to see what allyour previous chemistry instructors told you would “come later”.

In this document, there are many instances where constants are not included in order to make thingsclearer. If you were going to do calculations with the solutions, you would want to ensure you have theexact solution with all the appropriate constants in front. However, for our purposes, we’ll take someshortcuts and come up with the correct form of the solution and see what we can learn from it.

So to remind you, the Schrodinger Equation in three dimensions is:

− h2

2me∇2Ψ + VΨ = EΨ (1)

where ∇2 is the Laplacian operator ∂2

∂x2 + ∂2

∂y2 + ∂2

∂z2 .

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 protonsand the single electron:

V =−Ze2

r(2)

where e is the elementary charge and r is the distance between the nucleus and electron.

If you stare at this equation for a bit you’ll notice it is written in terms of r, a radial distance. Thistell us we should consider using a different coordinate system when solving the Schrodinger Equation forthe hydrogen atom. For example, if we stuck to Cartesian coordinates (x, y, and z) we would need torewrite Equation 2 using the following relationship between r and the Cartesian coordinates x, y, and z:

r =√x2 + y2 + z2 (3)

giving us:

V (x, y, z) =−Ze2√

x2 + y2 + z2(4)

which would be a mess to deal with when solving the Schrodinger Equation . So instead we will switchto spherical coordinates. This makes a lot more sense when you think about the spherical symmetry ofan atom. In spherical coordinates, the parameters are r, θ, and φ as shown in Figure 1.

When we make the jump to spherical coordinates, the Laplacian operator takes on a different form. Itwill look a bit frightening, but we will take it apart piece by piece as we move along:

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2(5)

We can now combine this with Equations 1 and 2 to solve for the wave functions (a.k.a. orbitals) of thehydrogen atom.

Figure 1: Spherical and cartesian coordinates.

Well, first lets write the Schrodinger Equation in spherical coordinates:

1

r2∂

∂r

(r2∂Ψ

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂Ψ

∂θ

)+

1

r2 sin2 θ

∂2Ψ

∂φ2+

2meZe2

h2rΨ +

2mE

h2Ψ = 0 (6)

Remember our goals whenever we solve the Schrodinger Equation : (1) find Ψ, the wave functions and(2) find E, the allowed energies. We begin by using a familiar trick, separate the variables into threeseparate functions:

Ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ) (7)

where each of the three separate function only depends on one of the position variables.

Combining Equations 6 and 7 and pulling constants through the partial derivatives we get:

ΘΦ

r2d

dr

(r2dR

dr

)+

RΦ

r2 sin θ

d

dθ

(sin θ

dΘ

dθ

)+

RΘ

r2 sin2 θ

d2Φ

dφ2+

2meZe2

h2rRΘΦ +

2mE

h2RΘΦ = 0 (8)

Now that is quite a lot of math for one sitting. We can do some housekeeping by dividing by RΘΦ andmultiplying by r2 sin2 θ:

sin2 θ1

R

d

dr

(r2dR

dr

)+

sin θ

Θ

d

dθ

(sin θ

dΘ

dθ

)+

1

Φ

d2Φ

dφ2+

(2meZe

2

h2r +

2meE

h2r2)

sin2 θ = 0 (9)

By doing this we have gotten to a point where most of the derivatives are now in terms with theirassociated functions. We can do even better by doing some more housekeeping and move all the φ termsto the right and factor out a sin2 θ:[

1

R

d

dr

(r2dR

dr

)+

1

Θ sin θ

d

dθ

(sin θ

dΘ

dθ

)+

2meZe2

h2r +

2meE

h2r2]

sin2 θ = − 1

Φ

d2Φ

dφ2= m2 (10)

In Equation 10 we have introduced a constant m or m2, much like we did when we solved the classicalwave equation. If both sides are equal for all r, θ, and φ, then both sides must equal a constant. Callingit m will allow us to come up with solutions that fit into the framework of atomic structure that youhave seen before.

Well, we can solve the second part of that equation for the φ dependence since we have handled equationslike that before.

d2Φ

dφ2= −m2Φ (11)

Solutions to this equation have the form:

Φ(φ) =

eimφ

e−imφ

...or...cosmφsinmφ

To align us with the text and other sources of quantum mechanical information we will use both thecomplex solutions and the trigonometric solutions. Remember, they are equivalent, but the trig solutions

are more useful when trying to visualize the solutions while the complex solutions work better for theabstract math.

So we have solutions for the φ dependence of the wave function. This is a good place to stop and takea deep breath. We don’t have anything yet that tells us much about the final solution. However, it isimportant to note we have introduced one quantum number so far, m, to describe the φ dependence. Wewill see shortly that m will be restricted to a set of integer values. Keep that in mind as we go along.

We now go back to Equation 10 to define the r terms in one parameter we’ll call λ (which is just aparameter, no relation to wavelength);

λ =1

R

d

dR

(r2dR

dr

)+

2meZe2

h2r +

2meE

h2r2 (12)

which allows us to rewrite Equation 10 as

λ sin2 θ +sin θ

Θ

d

dθ

(sin θ

dΘ

dθ

)= m2 (13)

We can move the m2 over to the left side to rearrange things a bit:

sin θd

dθ

(sin θ

dΘ

dθ

)+(λ sin2 θ −m2

)Θ = 0 (14)

Equation 14 is called the Legendre Equation. If an equation has a name that means it has solutions! Wecan lookup solutions for Θ and find they have the form:

Θlm(θ) = Θml (θ) = Pml (cos θ) (15)

where we can lookup solutions for Pml (x) and then insert cos θ to find our solutions for Θ(θ). Thesefunctions are also called spherical harmonic functions and Pml (x) has the form:

P 0l (x) =

1

2ll!

dl(x2 − 1)l

dxl(16)

Pml (x) =(1− x2

)m/2 dm

dxmP 0l (x) (17)

where both m and l are integers and |m| ≤ l or −l ≤ m ≤ l. Also, the parameter λ = l(l + 1).

These solutions look a little messy, but let’s look at a few specific cases to see how it works. For example,P 01 (x) with m = 0 and l = 1:

P 01 (x) =

1

211!

d(x2 − 1)

dx=

1

22x = x (18)

But for the Θ solution we need to set x = cos θ:

Θ10(θ) = P 01 (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 and l = 2:

P 02 (x) =

1

222!

d2(x2 − 1)2

dx2=

1

8

d2

dx2(x4 − 2x2 + 1) =

1

8

d

dx(4x3 − 4x) =

1

8(12x2 − 4) (20)

P 02 (x) =

12

8x2 − 4

8=

3

2x2 − 1

2=

3x2 − 1

2(21)

To get Θ20 we set x = cos θ and obtain:

Θ20(θ) =3 cos2 θ − 1

2(22)

One more example, P 11 (x) with m = 1 and l = 1:

P 11 (x) = (1− x2)

12d

dxP 01 (x) = (1− x2)

12d

dxx = (1− x2)

12 (23)

Substituting in x = cos θ we get:Θ1

1(θ) = (1− cos2 θ)12 = sin θ (24)

We now have two-thirds of the complete solution for our wave functions. So far, we have the two angularparts Θ and Φ. We have found that to specify these solutions we need two integers, or quantum numbers,l and 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 hydrogen electron wave functions

l m Θlm(θ) Φm(φ) ΘΦ “type”0 0 1 1 1 s1 0 cos θ 1 cos θ pz1 1 sin θ eiφ sin θ cosφ px1 -1 sin θ e−iφ sin θ sinφ py2 0 3 cos2 θ − 1 1 3 cos2 θ − 1 dz2

2 1 sin θ cos θ eiφ sin 2θ cosφ dxz2 -1 sin θ cos θ e−iφ sin 2θ sinφ dyz2 2 sin2 θ e2iφ sin2 θ cos 2φ dx2−y2

2 -2 sin2 θ e−2iφ sin2 θ sin 2φ dxy

You will notice the last column in the table, “type”, contains the orbital names that you have seen beforein general or inorganic chemistry. So the quantum numbers l and m describe the shape of these differentwave functions, which we will eventually call orbitals (see Figure 2).

You will also notice that solutions for m 6= 0 have an imaginary component (eiφ or e−iφ). In orderto remove the imaginary component in the angular part of the wave function (ΘΦ) we take linearcombinations of the two solutions. So for example, for l = 1 and m = 1 the solution takes the form:

px ≡ Θ11(Φ1 + Φ−1) = sin θ(eiφ + e−iφ) = sin θ(cosφ+ i sinφ+ cosφ− i sinφ) ≈ sin θ cosφ (25)

and for l = 1 and m = −1 the solutions look like:

py ≡ Θ11(Φ1 − Φ−1) = sin θ(eiφ − e−iφ) = sin θ(cosφ+ i sinφ− cosφ+ i sinφ) ≈ sin θ sinφ (26)

Figure 2: 3-D representations of the angular part of the hydrogen atom wave functions.

Now the only part we are missing is the radial part of the wave function (R(r)). Going back to Equation12 and rearranging things:

d

dr

(r2dR

dr

)+

(2meZe

2

h2r +

2meE

h2r2 − λ

)= 0 (27)

where λ = l(l + 1) and l ≥ 0.

This is an ordinary differential equation in r and it is somewhat complicated to solve. However, oncesolved we will find that for the solutions to be acceptable wave functions, the energy must be quantizedaccording to:

En ∝ −mee

4Z2

2h2n2(28)

where we have left out a few physical constants as I warned you about earlier. The important thing tonotice here is that the for the hydrogen-like atom the energy is only a function of n. If you compare thisform to Bohr’s solution, they are the same except now the electrons aren’t confined to sharp orbits butare described by the wave function Ψ.

Also notice that the energy is negative and gets smaller (approaches zero) as n gets larger. When n =∞then E = 0 and we say that the electron is “free”. This is just an arbitrary definition since there is nonatural zero for energy.

In the process of solving Equation 27 we find that the quantum number n appears naturally and mustsatisfy the condition that n ≥ l + 1 or

0 ≤ l ≤ n− 1 , n = 1, 2, 3... (29)

The solutions to Equation 27 depend on both n and l and take the form:

Rnl(r) = −(

(n− l − 1)!

2n[(n+ l)!]3

) 12(

2

nao

)l+3/2

rle−r/naoL2l+1n+1

(2r

nao

)(30)

where ao is the Bohr radius and is equal to ao = εoh2

πmee2and L2l+1

n+1

(2rnao

)are called Laguerre polynomi-

als. Since they are named, there are a set of solutions to these Laguerre polynomials that you can lookup.

Equation 30 looks complicated by if you strip away the crazy looking math it is simply a polynomialmultiplied by an exponential function. For example for n = 1 and l = 0:

R10(r) =1√π

(Z

ao

)3/2

e−Zrao (31)

So, in summary what we have been able to do is solve for the wave functions of the hydrogen atomby including the Coulombic potential in the Schrodinger Equation and solving for Ψ by breaking thesolution into parts:

Ψnlm(r, θ, φ) = Rnl(r)︸ ︷︷ ︸radial part

Θlm(θ)Φm(φ)︸ ︷︷ ︸angularparts

(32)

with allowed energies of the form:

En ∝ −mee

4Z2

2h2n2(33)

The three integers that describe the solutions should look familiar:

quantum number name restrictions describes orbitaln principal quantum number n > 0 sizel angular momentum quantum number 0 ≤ l ≤ n shapem magnetic quantum number −l ≤ m ≤ +l orientation