The Hydrogen Atom Quantum Physics 2002 Recommended Reading: Harris Chapter 6, Sections 3,4 Spherical coordinate system The Coulomb Potential Angular Momentum Normalised Wavefunctions Energy Levels Degeneracy

The Hydrogen Atom The hydrogen atom is just an electron and a proton moving together under a mutual electrostatic (Coulomb) attraction. Bound system  total energy, E, is less than zero Mass of proton M p = 1836m e so to a first approximation we can assume that the proton is fixed and the electron moves around it in a circular orbit. Potential energy of electron is given by (1) Where r is the distance between the electron and the proton. r We now have a three-dimensional problem and we need three coordinates to specify the position of the electron relative to the proton.

The Coulomb Potential r U(r) Note that : U(r)  0 as r   and U(r)  -  as r  0 We want to find the wave functions and allowed energies for the electron when it is confined to this potential. (2)

Spherical Coordinates (r, ,  ) y x z r   r sin  y = r sin  sin  x = r sin  cos  r sin  Instead of the three coordinates (x,y,z), we can specify the position of the electron by three Spherical coordinates (r, ,  ). r = radius, range 0    = polar angle range 0    = azimuthal angle range 0  2 

Schrodinger Equation in Spherical Coordinates The time independent Schrodinger equation in three-dimensions is We must also express the Laplacian operator  2 in polar coordinates (see additional notes) the result is that therefore the Schrodinger equation in 3D polar coordinates is this looks terrible but it in not much different from a particle in a 3d box or the SHO equation. (3) (4) (5)

Schrodinger Equation in Spherical Coordinates We note that the potential energy of the electron only depends on its radial distance from the proton and does not depend on the angles  and , U = U(r). Because the force asociated with such a potential is always directed radially inward or outward, it is called a central force We try the same approach as before and use Separation of Variables, that is we assume we can write the wavefunction as a product of three one-dimensional wavefunctions We now look at the partial derivatives we need in the Schrodinger equation Now substitute these into the Schrodinger equation (just a lot of algebra) and we find (6)

Multiply across by r 2 sin 2  and divide across by R. .  and rearrange: On the left hand side we have only functions of r and , while the right hand side is only a function of , (Separation of Variables) This can only hold for all values of r,  and  if both sides are equal to a constant. We call this constant m L 2. Then taking the R.H.S. we have The the azimuthal equation describes how the wavefunction varies with angle  and  are the azimuthal wave functions. This is the same equation as that for a Particle on a Ring and Sphere which we solved previously. (7) (8) (9)

substituting equation (9) into equation (8) we get (10) divide across by sin 2  and rearrange again we have managed to separate the variables, the LHS only depends on the variable r while the RHS only depends on the variable . So each side must be equal to a constant, this time we call the constant l ( l +1). Setting the LHS of (10) equal to l ( l + 1) gives (11) equation 11 is called the Radial Equation, it describes how the wavefunction varies as the variable r changes R(r) are the Radial wavefunctions.

Setting the RHS of (10) equal to l ( l + 1) gives (12) Equation (12) is the Angular Equation we encountered previously, and it describes how the wave function varies with the polar angle . The wavefunction  (  ) are called the Angular or Polar wavefunctions. We now have three differential equations (9, 11 and 12) that provide solutions for the r,  and  components of the wavefunction. That is, we must solve these differential equations to find the wavefunctions R(r),  (  ) and  (  ). Once we have these solutions we can then obtain the total, three-dimensional wavefunction from We now look at each of these wavefunctions in turn starting with equation 9 (9)

This is the same equation we found for the particle confined to move on a ring and Sphere and we found that the solutions are: Recall our definition of the z-component of the angular momentum operator (13) (14) and using this definition equation (8) becomes (15) Equation (15) is an eigenvalue equation  the eigenvalues of the L z 2 operator are Thus the wavefunction  (  ) represents states for which the z- component of the angular momentum L z are quantised and can only take on the values (16)

sin(1.  ) cos(1.  ) 11 Visualisation of Azimuthal wavefunctions   1 (  ) 0 22 Angle  Imaginary part Real part

sin(1.  ) cos(1.  ) 11 Imaginary part Real part

sin(2.  ) cos(2.  )   2 (  ) 0 22 Angle  Imaginary part Real part

222 Imaginary part Real part

Solution to the  - Equation,  (  ) We now look at the solutions to the  -equation (eqn 16) Again this is a difficult equation to solve. It was first solved by the mathematician Adrien Marie Legendre ( ) and is named after him: The Associated Legendre Equation. Note that this equation depends on two quantum numbers l and m L, so we have found another quantization. Applying appropriate boundary conditions leads to the following restrictions on the quantum numbers l and m L. i.e. the quantum number l must be zero or a positive integer, and for a given value of l, m L can take on all integer values ranging from - l to + l. For example, if l = 3 then m L can only have the values -3, -2, -1, 0, 1, 2, 3 (16) (17)

What are the solutions of the The Associated Legendre Equation? The solutions are a called the Associated Legendre Functions, P l,mL (cos  ) they are polynomials that depend on the angle  and the two quantum numbers l and m l. For each allowed set of quantum numbers ( l, m L ) there is a solution  l,mL (  ). The first few are given in the following table the solutions to the angular equation are given by  l,m L (  ) = P l,m L (cos  ) What do these wavefunctions look like? Exactly the same as a particle on a sphere. Let us plot a few of them

Polar Plots to visualise the  (  ) wavefunctions  z x-y plane P 1,0 l = 1, m L = 0 x-y plane P 1,  1 l = 1, m L =  1

P 2,0 l = 2, m L = 0 P 2,  1 l = 2, m L =  1 P 2,  1 l = 2, m L =  2

P 3,0 l = 3, m L = 0 P 3,  1 l = 3, m L =  1 P 3,  2 l = 3, m L =  2 P 3,  3 l = 3, m L =  3

Solution to the Radial - Equation, R(r) We now look for solutions to the radial-equation (eqn 11) rearranged as (11) this equation is just expresses conservation of energy, the first two terms are the kinetic energies due to radial and rotational motion respectively, while the third term is the Potential energy. For the hydrogen atom we just put in the Coulomb potential to get (18) Again equation 18 is very difficult to solve, it is known as the Associated Laguerre equation after Edmund Nicolas Laguerre ( ) who was the first to solve it. The solutions are called the Associated Laguerre Functions R n,l (r), and they depend on two quantum numbers n and l. Let us look at the solutions in the special case where l = 0 (no angular momentum). In this case equation (18) becomes

we guess that the solutions to this equation have the form (20) (19) where A is a normalisation constant and a 0 is some constant, to be determined that has dimensions of length. Substitute (20) into (19) using the fact that and (21) this equation can only be satisfied for all values of r if each of the terms in brackets are equal to zero, (22)

rearrange equation 22 (23) substuting equation 23 into the first term in equation (21) gives (24) Amazingly, equation (27) is the same result we obtained for the ground state energy level of the hydrogen atom in the Bohr Model! While equation (23) is the Bohr radius of the atom! We have now obtained these parameters from the Schrodinger equation! What do the full solutions of the radial equation look like, (the assiciated Laguerre functions). Again they are polynomials in the variable r, and depend on two quantum numbers n and l. The allowed energies are given by where n is a positive integer, n = 1, 2, 3, 4,... (25)

for a given value of n, the quantum numbers l and m L can take on the values only one non- degenerate state four degenerate states since they all have the same values of n and therefore the same energy E n nine degenerate states

Radial wave functions R(r) Solutions to equation (18), the Associated Laguerre Polynomials

Radial wave functions R(r), n = 1 and 2 R 1,0 n = 1, l = 0 R 2,0 n = 2, l = 0 R 2,1 n = 2, l = 1 1s wavefunction 2s wavefunction 2p wavefunction

R 3,0 n = 3, l = 0 R 3,1 n = 3, l = 1 R 3,2 n = 3, l = 2 Radial wave functions R(r) for n = 3 3s wavefunction 3p wavefunction 3d wavefunction Since there is no upper limit to the value n can have, there are an infinite number of radial wavefunctions, but they all have finite energies E n