The Hydrogen Atom continued.. Quantum Physics 2002 Recommended Reading: Harris Chapter 6, Sections 3,4 Spherical coordinate system The Coulomb Potential.

Presentation on theme: "The Hydrogen Atom continued.. Quantum Physics 2002 Recommended Reading: Harris Chapter 6, Sections 3,4 Spherical coordinate system The Coulomb Potential."— Presentation transcript:

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

We now have all the components necessary to write down the solutions to the Schrodinger Equation for the Hydrogen Atom. We applied the method of Separation of Variables to wavefunction Put it all together and have found the following solutions for the radial and angular parts. And so the total wavefunction can be written as: where N n,l,mL is a normalisation factor which we still have to determine. Note that the wavefunction depends on three quantum numbers, n, the principal quantum number, l = the total angular momentum and m l the z-component of the angular momentum.

Example: the wavefunction corresponding to the state n = 1, l = 0, m l = 0, what is the explicit form of this wavefunction writing the explicit forms of the Laguerre and Legendre polynomials gives since P 0,0 (  ) = 1 and exp(  i.0.  ) = 1. Similary for n = 2, l = 2, m L = 1 we find

The angular part of the wavefunction Lets examine the angular part in more detail. We can write the angular part as as before we can combine these two terms together into a single function Y l,m L ( ,  ) the Spherical Harmonics, this function combines the  and  dependent part of the solution. The Spherical represent the solutions to the Schrodinger equation for a particle confined to move on the surface s a sphere of unit radius. The first few are tabulated on the next page. Note that the Spherical harmonics depends on two quantum numbers l and m l.

The Spherical Harmonics Y l,mL ( ,  )

continued... to see what these wavefunctions look like see the following websites http://www.quantum-physics.polytechnique.fr/en http://www.uniovi.es/~quimica.fisica/qcg/harmonics/harmonics.html http://wwwvis.informatik.uni-stuttgart.de/~kraus/LiveGraphics3D/ java_script/SphericalHarmonics.html

Y 0,0 Y 1,0 Y 1,+1 Y 1,-1

Y 2,0 Y 2,+2 Y 2,+1 Y 2,-1 Y 2,-2

z-Component of Angular momentum What happens if we operate on the angular part of the wavefunction with the L z operator ? Thus we see that the Spherical Harmonics are eigenfunctions of the L z - operator with eigenvalues this shows that the z-component of the angular momentum is quantised in units of What else can we learn from the angular part of the wavefunction ?

Total Angular Momentum Since the angular momentum is a vector we can write the total angular momentum as From this (after some math) we find that Let us operate on the angular part of the wavefunction and see what result it gives us, that is this shows that the the angular part of the wave function is an eigenfunction of the L 2 operator, with eigenvalue

Total Angular Momentum Since the square of a vector is equal to the square of its magnitude, this means that the magnitude of the angular momentum can take on only the values and recall that no physical solution exists unless  m l   l. So with this restriction we have The information we now have is the magnitude of the quantised angular momentum  L  and the magnitude of the z-component of the angular momentum. For example, if l = 2 then and the z-components of the allowed angular momenta are we can represent this graphically as follows

Space Quantisation The angular momentum vector can only point in certain directions in space  Space Quantisation

Example What is the minimum angle that the angular momentum vector may make with the z-axis in the case where (a) l = 3 and (b) for l = 1? When l = 3 the magnitude of the angular momentum is with 7 possible z-components, m L = -3, -2, -1, 0, 1, 2, 3. The angular momentum vector will make the minimum angle with the z-axis when the z-component is as large as possible, and then for l = 1:

Notation We can specify the angular momentum states of a single particle as follows so for example, if the particle has angular momentum l = 2, then it is called a d-state. In this state the magnitude of the angular momentum is and its z-component can have the following values, Note that each angular momentum state has a degeneracy of 2 l +1. Thus for example, a d-state has a degeneracy of 5. this means that there are 5 states with l = 2 which all have the same energy.

Degeneracy We have found that the energy levels are given by so each energy level has a degeneracy of note that the energy levels are degenerate since they do not depend explicitly on l and m l. Recall thatn = 1, 2, 3, 4, … l = 0, 1, 2, 3, … n-1 m l = - l, -( l -1), ….( l -1), l Example: the n = 2 state has a degeneracy of 4: a single 2-s state with l = 0, m l = 0 and a 3- fold degenerate 2-p state with l = 1, m l = -1, 0, 1

Normalisation Just as in the case of one dimensional problems, the total probability of finding the electron somewhere in space must be unity. Using the volume element in polar coordinates, the nomalisation condition becomes The last integral is just 2  so we are left with two integrals. and Using these two integrals we can find the correctly normalised Radial and Angular wavefunctions

Example Find the angular normalisation constant for P 1,+1 (  ). For l = 1, m L = +1 we have P 1,+1 (  ) = sin  (see table). then Carry out the integral, to find and the correctly normalised wavefunction is

Probability Density What is the probability of finding the electron within a small volume element dV at position r, ,  ? If we want to find the probability of finding the electron at a distance between r and r + dr (i.e. in a spherical shell of radius r and thickness dr), then we just integrate over  and . But, since the spherical harmonics are normalised (by definition), so the integral is equal to 1 and we then have or is the probability of finding the electron at a distance r from the nucleus at any angle. What do these distributions look like ?

1s State n = 1, l = 0

2s State n = 2, l = 0

2p State n = 2, l = 1

3s State n = 3, l = 0

3p State n = 3, l = 1

3d State n = 3, l = 2

Summary of Key Points 1) The wavefunctions of the hydrogen atom can be expressed as the product of three one dimensional functions in the variables r,  and  2) The mathematical form taken by the wavefunction depends upon three quantum numbers n, l, m l. 3) These three quantum numbers can only take integral values subject to the following restrictions: n = 1, 2, 3, … l < n, m l = 0,  1,  2 …  l orbitals with l = 0, 1 and 2 are known as s, p and d orbitals respectively. The radial solutions R n, l (r) are given by the Associated Laguerre Polynomials The angular solutions are given by the Spherical Harmonics Y l, m l ( ,  ) where Y l, m l ( ,  ) =P l, m l (  ).exp(  i m l  ), and P l, m l (  ) are the Associated Legendre Polynomials.

Download ppt "The Hydrogen Atom continued.. Quantum Physics 2002 Recommended Reading: Harris Chapter 6, Sections 3,4 Spherical coordinate system The Coulomb Potential."

Similar presentations