Presentation is loading. Please wait.

Presentation is loading. Please wait.

Quantum mechanics unit 2

Similar presentations


Presentation on theme: "Quantum mechanics unit 2"β€” Presentation transcript:

1 Quantum mechanics unit 2
The SchrΓΆdinger equation in 3D Infinite quantum box in 3D & 3D harmonic oscillator The Hydrogen atom SchrΓΆdinger 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

2 Last time Time independent SchrΓΆdinger equation in 3D
βˆ’ ℏ 2 2π‘š 𝛻 2 𝑒 𝐫 +𝑉 𝐫 𝑒 𝐫 =𝐸𝑒 𝐫 u 𝐫 must be normalised, u 𝐫 and its spatial derivatives must be finite, continuous and single valued If 𝑉 𝐫 = 𝑉 1 π‘₯ + 𝑉 2 𝑦 + 𝑉 3 𝑧 then 𝑒 𝐫 =𝑋 π‘₯ π‘Œ 𝑦 𝑍 (𝑧) and the 3D S.E. separates into three 1D SchrΓΆdinger equations - obtain 3 different quantum numbers, one for each degree of freedom Time independent wavefunctions also called stationary states

3 3D quantum box 𝑉 π‘₯,𝑦,𝑧 =0 if βˆ’π‘Žβ‰€π‘₯β‰€π‘Ž, βˆ’π‘β‰€π‘¦β‰€π‘ and βˆ’π‘β‰€π‘§β‰€π‘
𝐸 𝑛 1 , 𝑛 2 , 𝑛 3 = πœ‹ 2 ℏ 2 8π‘š 𝑛 π‘Ž 𝑛 𝑏 𝑛 𝑐 2 If π‘Ž=𝑏 then 𝐸 𝑛 1 , 𝑛 2 , 𝑛 3 = πœ‹ 2 ℏ 2 8π‘š 𝑛 𝑛 π‘Ž 𝑛 𝑐 2 Quantum numbers, 𝑛 1,2,3 =1,2,3…

4 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 𝑒 𝑒 ( 𝑒 𝑒 )/2 Degeneracy related to symmetry

5 3D Harmonic Oscillator 𝑽 𝒓 = 𝟏 𝟐 π’Ž 𝝎 𝟐 𝒓 𝟐
Calculate the energy and degeneracies of the two lowest energy levels 𝐸 𝑛 1 𝑛 2 𝑛 3 =β„πœ”( 𝑛 1 + 𝑛 2 + 𝑛 ) Ground state 𝐸 000 = 3 2 β„πœ” is undegenerate, or has degeneracy 1 1st excited state 𝐸 100 = 𝐸 010 = 𝐸 001 = 5 2 β„πœ” is 3-fold degenerate 2nd excited state 𝐸 110 = 𝐸 101 = 𝐸 011 = 𝐸 200 = 𝐸 020 = 𝐸 002 = 7 2 β„πœ” has degeneracy 6 - don’t forget 𝑛 1,2,3 =0,1,2,3… for a harmonic oscillator

6 3D Harmonic Oscillator 𝑽 𝒓 = 𝟏 𝟐 π’Ž 𝝎 𝟐 𝒓 𝟐
Show that the lowest three energy levels are spherically symmetric 𝑒 𝑒 𝑒 average

7 Hydrogenic atom Potential (due to nucleus) is spherically symmetric
𝑧 Use spherical polar coordinates π‘₯=π‘Ÿ sin πœƒ cos πœ™ 𝑦=π‘Ÿ sin πœƒ sin πœ™ 𝑧=π‘Ÿ cos πœƒ π‘Ÿ πœƒ nucleus 𝑦 πœ™ π‘₯

8 Hydrogenic atom 𝑉 π‘Ÿ,πœƒ,πœ™ =𝑉 π‘Ÿ so, can separate the wavefunction
𝑒 π‘Ÿ,πœƒ,πœ™ =𝑅 π‘Ÿ π‘Œ πœƒ,πœ™ =𝑅 π‘Ÿ Θ πœƒ Ξ¦(πœ™) Solve separately for 𝑅 π‘Ÿ , Θ πœƒ , Ξ¦(πœ™) 𝑅, Θ, Ξ¦ continuous, finite, single valued, 𝑒 π‘Ÿ,πœƒ,πœ™ 2 d𝜏 = 1 Expect 3 quantum numbers 𝑛,𝑙,π‘š - as 3 degrees of freedom Expect 𝑒 π‘Ÿ,πœƒ,πœ™ β†’0 as π‘Ÿβ†’βˆž because state is bound Expect 𝐸 𝑛 = βˆ’ π‘š 𝑒 𝑍 2 𝑒 πœ‹ πœ– ℏ 2 𝑛 2 (result from Bohr’s theory) Expect degenerate excited states

9 SchrΓΆdinger equation in spherical polars
βˆ’β„ 2 2 π‘š 𝑒 𝛻 2 𝑒 π‘Ÿ,πœƒ,πœ™ +𝑉 π‘Ÿ 𝑒 π‘Ÿ,πœƒ,πœ™ =𝐸𝑒 π‘Ÿ,πœƒ,πœ™ , where 𝛻 2 = 1 π‘Ÿ 2 πœ• πœ•π‘Ÿ π‘Ÿ 2 πœ• πœ•π‘Ÿ π‘Ÿ 2 sin πœƒ πœ• πœ•πœƒ sin πœƒ πœ• πœ•πœƒ π‘Ÿ 2 sin 2 πœƒ πœ• 2 πœ• πœ™ 2 , and 𝑒 π‘Ÿ,πœƒ,πœ™ =𝑅 π‘Ÿ π‘Œ πœƒ,πœ™ =𝑅 π‘Ÿ Θ πœƒ Ξ¦ πœ™ .

10 Separation of SchrΓΆdinger equation
Radial equation βˆ’ ℏ 2 2 π‘š 𝑒 1 R πœ• πœ•π‘Ÿ π‘Ÿ 2 πœ•π‘… πœ•π‘Ÿ + π‘Ÿ 2 𝑣 π‘Ÿ βˆ’πΈ =βˆ’πœ† πœƒ, πœ™ equation βˆ’ ℏ 2 2 π‘š 𝑒 1 sin πœƒ 1 π‘Œ πœ• πœ•πœƒ sin πœƒ πœ•π‘Œ πœ•πœƒ βˆ’ ℏ 2 2 π‘š 𝑒 sin 2 πœƒ 1 π‘Œ πœ• 2 π‘Œ πœ• πœ™ 2 =πœ† π‘Œ(πœƒ,πœ™) represents the angular dependence of the wavefunction in any spherically symmetric potential

11 π‘Œ π‘™π‘š πœƒ,πœ™ = 2𝑙+1 4πœ‹ π‘™βˆ’|π‘š| ! 𝑙+|π‘š| ! 1 2 βˆ’1 π‘š 𝑃 𝑙 π‘š cos πœƒ 𝑒 π‘–π‘šπœ™
π‘Œ 00 (πœƒ,πœ™)= 1 4πœ‹ 1/2 π‘Œ 10 (πœƒ,πœ™)= πœ‹ cos πœƒ π‘Œ 1βˆ’1 (πœƒ,πœ™)= πœ‹ sin πœƒ 𝑒 βˆ’π‘–πœ™ π‘Œ 11 (πœƒ,πœ™)= βˆ’ 3 8πœ‹ sin πœƒ 𝑒 π‘–πœ™


Download ppt "Quantum mechanics unit 2"

Similar presentations


Ads by Google