Quantum mechanics unit 2 The Schrödinger equation in 3D Infinite quantum box in 3D & 3D harmonic oscillator The Hydrogen atom Schrödinger equation in spherical polar coordinates Solution by separation of variables Angular quantum numbers Radial equation and principal quantum numbers Hydrogen-like atoms Rae – Chapter 3
Last time Time independent Schrödinger equation in 3D − ℏ 2 2𝑚 𝛻 2 𝑢 𝐫 +𝑉 𝐫 𝑢 𝐫 =𝐸𝑢 𝐫 u 𝐫 must be normalised, u 𝐫 and its spatial derivatives must be finite, continuous and single valued If 𝑉 𝐫 = 𝑉 1 𝑥 + 𝑉 2 𝑦 + 𝑉 3 𝑧 then 𝑢 𝐫 =𝑋 𝑥 𝑌 𝑦 𝑍 (𝑧) and the 3D S.E. separates into three 1D Schrödinger equations - obtain 3 different quantum numbers, one for each degree of freedom Time independent wavefunctions also called stationary states
3D quantum box 𝑉 𝑥,𝑦,𝑧 =0 if −𝑎≤𝑥≤𝑎, −𝑏≤𝑦≤𝑏 and −𝑐≤𝑧≤𝑐 𝐸 𝑛 1 , 𝑛 2 , 𝑛 3 = 𝜋 2 ℏ 2 8𝑚 𝑛 1 2 𝑎 2 + 𝑛 2 2 𝑏 2 + 𝑛 3 2 𝑐 2 If 𝑎=𝑏 then 𝐸 𝑛 1 , 𝑛 2 , 𝑛 3 = 𝜋 2 ℏ 2 8𝑚 𝑛 1 2 + 𝑛 2 2 𝑎 2 + 𝑛 3 2 𝑐 2 Quantum numbers, 𝑛 1,2,3 =1,2,3…
Degeneracy States are degenerate if energies are equal, eg. 𝐸 1,2,1 = 𝐸 2,1,1 Degree of degeneracy is equal to the number of linearly independent states (wavefunctions) per energy level 𝑢 121 2 𝑢 211 2 ( 𝑢 121 2 + 𝑢 211 2 )/2 Degeneracy related to symmetry
3D Harmonic Oscillator 𝑽 𝒓 = 𝟏 𝟐 𝒎 𝝎 𝟐 𝒓 𝟐 Calculate the energy and degeneracies of the two lowest energy levels 𝐸 𝑛 1 𝑛 2 𝑛 3 =ℏ𝜔( 𝑛 1 + 𝑛 2 + 𝑛 3 + 3 2 ) Ground state 𝐸 000 = 3 2 ℏ𝜔 is undegenerate, or has degeneracy 1 1st excited state 𝐸 100 = 𝐸 010 = 𝐸 001 = 5 2 ℏ𝜔 is 3-fold degenerate 2nd excited state 𝐸 110 = 𝐸 101 = 𝐸 011 = 𝐸 200 = 𝐸 020 = 𝐸 002 = 7 2 ℏ𝜔 has degeneracy 6 - don’t forget 𝑛 1,2,3 =0,1,2,3… for a harmonic oscillator
3D Harmonic Oscillator 𝑽 𝒓 = 𝟏 𝟐 𝒎 𝝎 𝟐 𝒓 𝟐 Show that the lowest three energy levels are spherically symmetric 𝑢 110 2 𝑢 200 2 𝑢 020 2 average
Hydrogenic atom Potential (due to nucleus) is spherically symmetric 𝑧 Use spherical polar coordinates 𝑥=𝑟 sin 𝜃 cos 𝜙 𝑦=𝑟 sin 𝜃 sin 𝜙 𝑧=𝑟 cos 𝜃 𝑟 𝜃 nucleus 𝑦 𝜙 𝑥
Hydrogenic atom 𝑉 𝑟,𝜃,𝜙 =𝑉 𝑟 so, can separate the wavefunction 𝑢 𝑟,𝜃,𝜙 =𝑅 𝑟 𝑌 𝜃,𝜙 =𝑅 𝑟 Θ 𝜃 Φ(𝜙) Solve separately for 𝑅 𝑟 , Θ 𝜃 , Φ(𝜙) 𝑅, Θ, Φ continuous, finite, single valued, 𝑢 𝑟,𝜃,𝜙 2 d𝜏 = 1 Expect 3 quantum numbers 𝑛,𝑙,𝑚 - as 3 degrees of freedom Expect 𝑢 𝑟,𝜃,𝜙 →0 as 𝑟→∞ because state is bound Expect 𝐸 𝑛 = − 𝑚 𝑒 𝑍 2 𝑒 4 2 4𝜋 𝜖 0 2 ℏ 2 𝑛 2 (result from Bohr’s theory) Expect degenerate excited states
Schrödinger equation in spherical polars −ℏ 2 2 𝑚 𝑒 𝛻 2 𝑢 𝑟,𝜃,𝜙 +𝑉 𝑟 𝑢 𝑟,𝜃,𝜙 =𝐸𝑢 𝑟,𝜃,𝜙 , where 𝛻 2 = 1 𝑟 2 𝜕 𝜕𝑟 𝑟 2 𝜕 𝜕𝑟 + 1 𝑟 2 sin 𝜃 𝜕 𝜕𝜃 sin 𝜃 𝜕 𝜕𝜃 + 1 𝑟 2 sin 2 𝜃 𝜕 2 𝜕 𝜙 2 , and 𝑢 𝑟,𝜃,𝜙 =𝑅 𝑟 𝑌 𝜃,𝜙 =𝑅 𝑟 Θ 𝜃 Φ 𝜙 .
Separation of Schrödinger equation Radial equation − ℏ 2 2 𝑚 𝑒 1 R 𝜕 𝜕𝑟 𝑟 2 𝜕𝑅 𝜕𝑟 + 𝑟 2 𝑣 𝑟 −𝐸 =−𝜆 𝜃, 𝜙 equation − ℏ 2 2 𝑚 𝑒 1 sin 𝜃 1 𝑌 𝜕 𝜕𝜃 sin 𝜃 𝜕𝑌 𝜕𝜃 − ℏ 2 2 𝑚 𝑒 1 sin 2 𝜃 1 𝑌 𝜕 2 𝑌 𝜕 𝜙 2 =𝜆 𝑌(𝜃,𝜙) represents the angular dependence of the wavefunction in any spherically symmetric potential
𝑌 𝑙𝑚 𝜃,𝜙 = 2𝑙+1 4𝜋 𝑙−|𝑚| ! 𝑙+|𝑚| ! 1 2 −1 𝑚 𝑃 𝑙 𝑚 cos 𝜃 𝑒 𝑖𝑚𝜙 𝑌 00 (𝜃,𝜙)= 1 4𝜋 1/2 𝑌 10 (𝜃,𝜙)= 3 4𝜋 1 2 cos 𝜃 𝑌 1−1 (𝜃,𝜙)= 3 8𝜋 1 2 sin 𝜃 𝑒 −𝑖𝜙 𝑌 11 (𝜃,𝜙)= − 3 8𝜋 1 2 sin 𝜃 𝑒 𝑖𝜙