2006-7 Quantum Theory Slides Lecture 7

Quantum theory and atomic spectroscopy Lecture 7 The hydrogen atom

What use is quantum mechanics? • Replace the idea of particles with quantum objects which have a wavelength • Solve the Schrödinger equation for an appropriate situation to find the required wavefunctions • The square of the wavefunction is the probability density • Wave-particle duality necessary but not sufficient for quantization of energy • So how does this explain electron orbitals…..

Today’s question is…… • Can we solve the Schrödinger equation for hydrogen and will it demonstrate the four quantum numbers we need?

Wavefunctions must reflect reality… • A wavefunction must abide by its physical interpretation • It’s probability density integrated over all space must be 1 • It’s curvature reflects its kinetic energy

Basic requirements of wavefunctions • All wavefunctions must therefore obey the following: (1) Be finite everywhere (2) Be continuous (i.e. no breaks) (3) Have a continuous slope (4) Be singled value

Acceptable and unacceptable wavefunctions • Many wavefunctions fail because of physical constraints such as the Born Interpretation or the kinetic energy requirements

Boundary conditions

• These are mathematical restraints on wavefunctions, induced because of the physical system being studied • For example, at infinite potential walls the amplitude must be zero! • These boundary conditions can be expressed as mathematical constraints- source of QUANTIZATION!

The hydrogen atom • The Schrödinger equation for hydrogen is relatively simple to derive…. • But it is difficult to solve! 2 2 2 h h q 2 2 Hˆ = − 2 ∇ p − 2 ∇e − 4πε 0 r 8π m p 8π me

h2 ⎛ ∂2 h2 ∂2 ∂2 ⎞ − 2 ⎜⎜ 2 + 2 + 2 ⎟⎟ = − 2 ∇ 2 8π m ⎝ ∂x 8π m ∂y ∂z ⎠

Change of co-ordinate system • Replace cartesian coordinates with a spherical system • Final wavefunction separates into three functions, each only depending on one coordinate! Ψn ,l ,ml (r , θ , φ ) = Rn ,l ( r )Θ l , ml (θ )Φ ml (φ )

Particle on a ring • The Schrödinger equation for a particle on a ring is very similar to that of a free particle • Co-ordinate has changed from x to Φ d2 2 Φ = − a Φ 2 dφ

d2 2 Ψ = − a Ψ 2 dx

Φ (φ ) = N m e aφ = N m e imφ

But now here’s a change… • Unlike the free particle, we now have a boundary condition to take into account because of the periodic nature of the angular wavefunction

This leads to quantization! • We can now show that because of the boundary condition, m must take integer values only

e

e

imφ

=e

im (φ + 2 π )

im (φ + 2 π )

=e

imφ im 2 π

e

∴e

im 2 π

=1

• Only true if m is an integer: this is the origin of the ml quantum number!

Angular momentum wavefunctions • The combination of wavefunctions Θl,ml(θ) and Φml(φ) is called a spherical harmonic • Solving for Θl,ml(θ) gives us the l quantum no. Yl , ml (θ , φ ) = Θ l , ml (θ )Φ ml (φ )

These are the solutions to a particle sitting on the surface of a sphere

Spherical harmonics are orbitals! • The angular wavefunctions are in fact the orbitals you know and love

• Key here is that ANY angular momentum with integer values will look like the spherical harmonics i.e. nuclear rotation

The radial equation • This time, the boundary condition is the Coulomb potential well between the electron and the proton • This leads to discrete energy levels, using n as an index!

Radial wavefunctions • These distinguish between 1s and 2s orbitals etc. • For each value of l, the lowest n wavefunction is similar (except s) • The s-orbitals have a significant amplitude AT the nucleus • The number of radial nodes in the wavefunction is given by n – l - 1

Orthogonal wavefunctions • This is just a fancy way of saying that the wavefunctions do not overlap. ∞

∫ Ψ (x ) Ψ (x )dx = 0 ∗

i

−∞

j

when

i≠ j

Answers to the earlier questions.. • We can set up a Schrödinger equation for hydrogen and it can be solved • The presence of boundary conditions to the wavefunctions introducing quantization

But that’s a bit boring…… • Can we solve the Schrödinger equation for many electron atoms too? • In the last lecture, we will demonstrate the problems doing this and look at the wavefunctions for helium!