Schrodinger Equation in Three Dimensions:

The time-dependent and the time-independent Schrodinger equations in three dimensions for a particle of mass m and energy E are:

(-hbar²/2m)D² Y (r, t) + U(r) Y(r, t) = (ihbar) d Y(r,t)/dt,  Eq.(1)

(-hbar²/2m)D² Y(r) + U(r) Y(r) = E Y(r).   Eq. (2)

D² is called the Laplacian which is defined as -

D² = d²/dx² + d²/ dy² + d²/dz².

We will now study the time-independent Schrodinger equation [Eq. (2)] for some simple cases.

Particle In a Three-Dimensional Box:

The problem is similar to the one-dimensional particle in a box but in this case the particle is confined inside a cubic box (for simplicity we take the box to be cubic) of length L in each side. The potential energy U(x,y,z) = 0 in this case. The Schrodinger equation then reduces to

(-hbar²/2m)[ d²/dx² + d²/dy² + d²/dz²] Y(x,y,z) = E Y(x,y,z).  Eq. (3)

Using separation of variable technique the wave-function can be solved. The form of the wavefunction is:

Y(x,y,z) = A sin(k₁x) sin(k₂y) sin(k₃z),         Eq. (4)

where A, k₁, k₂, and k₃ are constants. From the boundary conditions that the wavefunction should vanish at x=0, x=L; y=0, y=L; and z=0, z=L it can be easily shown that  k₁, k₂, and k₃ can take only discrete values (so the motion is quantized)

k₁ = n₁ p/L, k₂ = n₂ p/L, k₃ = n₃ p/L.            Eq. (5)

n₁, n₂, and n₃ are quantum numbers which can take only positive integer values (n₁, n₂, n₃ = 1,2,3,...).

The total energy of the system is thus,

E_n1n2n3 = (p² hbar²/2mL²)[ n₁² + n₂² + n₃²].  Eq. (6)

Substituting the values of k₁, k₂, and k₃ in Eq. (4) we get -

Y_n1n2n3 (x,y,z) = A sin(n₁ p x/L)sin(n₂ p y/L)sin(n₃ p z/L).  Eq. (7)

Using the normalization condition that the total probability of finding the particle inside the box is 1, i.e. Int_{-infinity}^{infinity} dx dy dz | Y(x,y,z)|² =1 we obtain the value of A,

A = (2/L)^3/2.                     Eq. (8)

The ground state for which n₁=n₂=n₃=1 has energy,

E₁₁₁ = 3(p² hbar²/2mL²).

The first excited state (higher energy state) can correspond to either n₁=2, n₂=1, n₃=1, or n₁=1, n₂=2, n₃=1, or n₁=1, n₂=1, n₃=2. For all three cases the energy is the same

E₂₁₁ = E₁₂₁ = E₁₁₂ = 6 (p² hbar²/2mL²).

Whenever different states have the same energy, this energy level is said to be degenerate. As an example the first excited state is 3-fold degenerate.

Exercise: Write the wavefunctions for the second excited state. Calculate their energies and find out the degeneracy.

Central Forces, Angular Momentum, and the Hydrogen Atom:

The motion of an electron in a central field is one of the principal problems in the quantum mechanics of the atom. An atomic electron is attracted to the nucleus of the atom by the Coulomb force between opposite charges. This is an example of a central force, i.e., one directed toward a fixed point. The nucleus is the center of force. To analyze such central force problems it is convenient to rewrite the Schrodinger equation in spherical polar coordinates (r, q, j).

The Laplacian D² takes the form:

D² = d²/dr² + (2/r) d/dr + (1/r²)[d²/dq²+cot q d/dq + cosec²q d²/dj²].

For the case of hydrogen atom the orbiting electron of mass m and charge -e is bound to the nucleus of mass M and charge +Ze by the force of electrostatic interaction to the nucleus (we will assume it to be Coulombic in nature). Thus the potential is of the form

U(r) = -Ze²/r.

The time independent Schrodinger wavefunction in r, q, j coordinates is

Y(r) = Y(r, q, j) = R(r) Q (q) F(j).

The Schrodinger equation [Eq. (2)] can be separated in to three equations. The azimuthal part which only depends on the angle j satisfies a simple second order differential equation

d² F/dj² = -m_l² F(j),

which can be easily solved to be F(j) = exp(im_lj). The constant m_l known as the magnetic quantum number can take any positive or negative values upto a maximum value l (m_l = 0, +/-1,+/-2,+/-3, ......, +/-l).  The differential equation for the angular part (Q) can also be solved exactly and the general solution is in terms of special functions called spherical harmonics. The solution depends also on a constant l known as the orbital angular momentum which can only take positive integer values. It can be shown that the maximum value of l is (n-1) where n is another constant which comes from the solution of the radial (R) part. n is called the principal quantum number and an take only integer values n= 1,2,3,...... 

To see a three-dimesnional view of the hydrogen atom wavefunction (for different values of n, l, and m) please click on the link below:

Java Applet for 3D Hydrogen Atom Wave-function

Acceptable wavefunctions of the hydrogen atom can be found only if the energy E is restricted to be one of the following special values:

E_n = -(me⁴/2hbar²)(Z²/n²),    n=1,2,3,....

This result is in exact agreement with the Bohr's atomic theory discussed before. Note that the allowed energies depend only on the principal quantum number n. Keeping n fixed but allowing l [l can vary from 0 to (n-1)] and m_l [m_l varies from -l,-(l-1),...,0,...(l-1),l] to vary the energy doesnot change. This is another example of degeneracy. States with fixed n have same energies for different l and m_l.

Ground state of Hydrogen-like atoms:

The ground state of a one-electron atom or ion with atomic number Z, for which n=1, l=0, and m_l=0, has energy

E₁₀₀ = -(13.6 eV)Z².

The normalized wave-function for this state is,

Y_1,0,0 = (1/p)^1/2 (Z/a₀)^3/2 exp(-Zr/a₀).

The probability of finding an electron anywhere in a spherical shell of radius r and thickness dr, i.e. P(r) dr [ P(r) is called the radial probability distribution ] is

P(r) dr = |Y|² x (4 p r² dr).

The probability distribution for the ground state of hydrogen-like atoms doesnot depend on the angles q and j - so the distribution is spherically symmetric. The average distance of the electron from the nucleus can be obtained by evaluating the expression below:

<r> = Integrate_{0}^{infinity} r P(r) dr.

Exercise: For ground state of the hydrogen atom show that <r> is equal to the Bohr radius a₀.

In general, the average value of any function of distance f(r) is given by,

<f> = Integrate_{0}^{infinity} f(r) P(r) dr.

Angular Momentum:

It can be shown that the orbital quantum number l and the magnetic quantum number m_l are related to the observables |L| (the total angular momentum) and L_z (the z component of the angular momentum).

|L| = [l(l+1)]^1/2 hbar;    l=0,1,2,...

L_z = m_l hbar.         m_l = 0, +/-1, +/-2,....+/-l.

Together l and m_l specify the orientation of the angular momentum vector L. Let us now look at the possible orientations of L for a given value of orbital quantum number l. If q is the angle between L and the z-axis, we get

cos q = L_z/|L| = m_l/[l(l+1)]^1/2.

Classically q can take any values and so does the angular momentum. But quantum mechanically as m_l and l can take restricted values the angle q takes only special values. Thus we can say that the direction of L is quantized with respect to an arbitrary axis (the z axis) - this is known as space quantization.

Example: With l=2, |L| = 6^1/2 hbar, and m_l = 0, +/-1, +/-2. So, q can take only 5 special values to be determined from cos q = 0, +/-1/6^1/2, 2/6^1/2.

The Quantum Numbers:

Thus the three coordinates that come from Schrödinger's wave equations are the principal (n), angular (l), and magnetic (m_l) quantum numbers. These quantum numbers describe the size, shape, and orientation in space of the orbitals on an atom.

The principal quantum number (n) describes the size of the orbital. Orbitals for which n = 2 are larger than those for which n = 1, for example. Because they have opposite electrical charges, electrons are attracted to the nucleus of the atom. Energy must therefore be absorbed to excite an electron from an orbital in which the electron is close to the nucleus (n = 1) into an orbital in which it is further from the nucleus (n = 2). The principal quantum number therefore indirectly describes the energy of an orbital.

The angular quantum number (l) describes the shape of the orbital. Orbitals have shapes that are best described as spherical (l = 0), polar (l = 1), or cloverleaf (l = 2). They can even take on more complex shapes as the value of the angular quantum number becomes larger.

There is only one way in which a sphere (l = 0) can be oriented in space. Orbitals that have polar (l = 1) or cloverleaf (l = 2) shapes, however, can point in different directions. We therefore need a third quantum number, known as the magnetic quantum number (m), to describe the orientation in space of a particular orbital. (It is called the magnetic quantum number because the effect of different orientations of orbitals was first observed in the presence of a magnetic field.)

Rules Governing the Allowed Combinations of Quantum Numbers