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An x y z coordinate system is shown, along with a point P and the vector r from the origin to P. In this figure, the point P has positive x, y, and z coordinates. The vector r is inclined by an angle theta from the positive z axis. Its projection on the x y plane makes an angle theta from the positive x axis toward the positive y axis.
The relationship between the spherical and rectangular coordinate systems.

The factor r sin θ is the magnitude of a vector formed by the projection of the polar vector onto the xy -plane. Also, the coordinates of x and y are obtained by projecting this vector onto the x - and y -axes, respectively. The inverse transformation gives

r = x 2 + y 2 + z 2 , θ = cos −1 ( z r ) , ϕ = cos −1 ( x x 2 + y 2 ) .

Schrödinger’s wave equation for the hydrogen atom in spherical coordinates is discussed in more advanced courses in modern physics, so we do not consider it in detail here. However, due to the spherical symmetry of U ( r ), this equation reduces to three simpler equations: one for each of the three coordinates ( r , θ , and ϕ ) . Solutions to the time-independent wave function are written as a product of three functions:

ψ ( r , θ , ϕ ) = R ( r ) Θ ( θ ) Φ ( ϕ ) ,

where R is the radial function dependent on the radial coordinate r only; Θ is the polar function dependent on the polar coordinate θ only; and Φ is the phi function of ϕ only. Valid solutions to Schrödinger’s equation ψ ( r , θ , ϕ ) are labeled by the quantum numbers n , l , and m .

n : principal quantum number l : angular momentum quantum number m : angular momentum projection quantum number

(The reasons for these names will be explained in the next section.) The radial function R depends only on n and l ; the polar function Θ depends only on l and m ; and the phi function Φ depends only on m . The dependence of each function on quantum numbers is indicated with subscripts:

ψ n l m ( r , θ , ϕ ) = R n l ( r ) Θ l m ( θ ) Φ m ( ϕ ) .

Not all sets of quantum numbers ( n , l , m ) are possible. For example, the orbital angular quantum number l can never be greater or equal to the principal quantum number n ( l < n ) . Specifically, we have

n = 1 , 2 , 3 , l = 0 , 1 , 2 , , ( n 1 ) m = l , ( l + 1 ) , , 0 , , ( + l 1 ) , + l

Notice that for the ground state, n = 1 , l = 0 , and m = 0 . In other words, there is only one quantum state with the wave function for n = 1 , and it is ψ 100 . However, for n = 2 , we have

l = 0 , m = 0 l = 1 , m = −1 , 0 , 1 .

Therefore, the allowed states for the n = 2 state are ψ 200 , ψ 21 1 , ψ 210 , and ψ 211 . Example wave functions for the hydrogen atom are given in [link] . Note that some of these expressions contain the letter i , which represents −1 . When probabilities are calculated, these complex numbers do not appear in the final answer.

Wave functions of the hydrogen atom
n = 1 , l = 0 , m l = 0 ψ 100 = 1 π 1 a 0 3 / 2 e r / a 0
n = 2 , l = 0 , m l = 0 ψ 200 = 1 4 2 π 1 a 0 3 / 2 ( 2 r a 0 ) e r / 2 a 0
n = 2 , l = 1 , m l = −1 ψ 21 1 = 1 8 π 1 a 0 3 / 2 r a 0 e r / 2 a 0 sin θ e i ϕ
n = 2 , l = 1 , m l = 0 ψ 210 = 1 4 2 π 1 a 0 3 / 2 r a 0 e r / 2 a 0 cos θ
n = 2 , l = 1 , m l = 1 ψ 211 = 1 8 π 1 a 0 3 / 2 r a 0 e r / 2 a 0 sin θ e i ϕ

Physical significance of the quantum numbers

Each of the three quantum numbers of the hydrogen atom ( n , l , m ) is associated with a different physical quantity. The principal quantum number     n is associated with the total energy of the electron, E n . According to Schrödinger’s equation:

E n = ( m e k 2 e 4 2 2 ) ( 1 n 2 ) = E 0 ( 1 n 2 ) ,

where E 0 = −13.6 eV . Notice that this expression is identical to that of Bohr’s model. As in the Bohr model, the electron in a particular state of energy does not radiate.

How many possible states?

For the hydrogen atom, how many possible quantum states correspond to the principal number n = 3 ? What are the energies of these states?

Strategy

For a hydrogen atom of a given energy, the number of allowed states depends on its orbital angular momentum. We can count these states for each value of the principal quantum number, n = 1 , 2 , 3 . However, the total energy depends on the principal quantum number only, which means that we can use [link] and the number of states counted.

Solution

If n = 3 , the allowed values of l are 0, 1, and 2. If l = 0 , m = 0 (1 state). If l = 1 , m = 1 , 0 , + 1 (3 states); and if l = 2 , m = 2 , 1 , 0 , + 1 , + 2 (5 states). In total, there are 1 + 3 + 5 = 9 allowed states. Because the total energy depends only on the principal quantum number, n = 3 , the energy of each of these states is

E n 3 = E 0 ( 1 n 2 ) = −13.6 eV 9 = −1.51 eV .

Significance

An electron in a hydrogen atom can occupy many different angular momentum states with the very same energy. As the orbital angular momentum increases, the number of the allowed states with the same energy increases.

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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