ThehydrogenatomHamiltonianFor the hydrogen-like atom, then,aZeh*aa11lddH=T+Vr? sin0 a00r? sin? ao?2m.4元0r- ararWe can recognise in the angular part of the kinetic energy the expression for the square of theangular momentum, but in the case of the hydrogen atom it is conventional to use the symbol ifor the orbital angular momentum of the electron, reserving J (or J) for the total angularmomentum including spin.ThecompleteHamiltonianisthen12h?a Ze?1aH2m,r22m。 r2 rOr4元Note that the radial part can be written in several equivalent ways:1 a2adrorry2ararQut
The hydrogen atom Hamiltonian For the hydrogen-like atom, then, We can recognise in the angular part of the kinetic energy the expression for the square of the angular momentum, but in the case of the hydrogen atom it is conventional to use the symbol መ𝒍 for the orbital angular momentum of the electron, reserving 𝑱 (or 𝒋 Ƹ) for the total angular momentum including spin. The complete Hamiltonian is then Note that the radial part can be written in several equivalent ways: 2 2 2 2 2 2 2 2 2 0 1 1 1 sin 2 sin sin 4 e Z e H T V r m r r r r r r ò 2 2 2 2 2 2 0 ˆ 1 2 2 4 e e Ze H r m r r r m r r ò l 2 2 2 2 2 2 1 1 2 r r r r r r r r r r
Atomic unitsTheHatomHamiltonianisratherclutteredwithfundamental constants.Togetridoftheclutter, we use atomic units:UnitDefinitionSI valueSymbolNamebohrLength4元e.h2/m.e252.917721pma.Mass0.910938×10-30kgm.electronmassChargee1.6021765×10-19Cproton chargeEhEnergyHartreee?/4E.a.4.359744×10-18Angularh1.0545717×10-34JsmomentumWhen we work in atomic units, me, e, h and 4πe.all have a numerical value of 1, as do thequantities ao and E, derived from themThissimplifiestheequations,buthasthedisadvantagethatitbecomesdifficulttocheckthedimensions ofanexpression
The H atom Hamiltonian is rather cluttered with fundamental constants. To get rid of the clutter, we use atomic units: When we work in atomic units, me , e, ℏ and 4𝜋𝜖o all have a numerical value of 1, as do the quantities a0 and Eh derived from them. This simplifies the equations, but has the disadvantage that it becomes difficult to check the dimensions of an expression. Unit SymbolName Definition SI value Length a0 bohr 4𝜋𝜖oℏ 2/𝑚𝑒𝑒 2 52.917721 pm Mass me electron mass 0.910938 × 10-30 kg Charge e proton charge 1.6021765 × 10-19 C Energy Eh Hartree 𝑒 2/4𝜋𝜖o𝑎o 4.359744 × 10-18 J Angular momentum ℏ 1.0545717 × 10-34 Js Atomic units
We write r' = r /ao, and find thatZe?ZeV :4元蝌4元a0Similarly, writing I' = i/n, the angular kinetic energy becomesi2112112h?h?m,e?m,a 4元o,h2 2r'22m.l1122ma.In this way we find1a114a7H/E2r'aror, dropping the primes, and remembering that energies will be in Hartree,12a1aZH:2r22r2arOr
We write 𝑟 ′ = 𝑟/a0 , and find that Similarly, writing መ𝒍′ = መ𝒍/ℏ, the angular kinetic energy becomes In this way we find or, dropping the primes, and remembering that energies will be in Hartree, 2 2 0 0 0 4 4 h Ze Ze Z V E r a r' r' 蝌 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 0 ˆ ˆ ˆ ˆ 2 2 4 2 2 e h e e e ' ' ' m e E m r m a r' m a r' r' ò l l l l 2 2 2 2 ˆ 1 / 2 2 h ' Z H E r' r' r' r' r' r' l 2 2 2 2 ˆ 1 2 2 Z H r r r r r r l
5.3WavefunctionsforthehydrogenatomThe Hamiltonian contains the angular variables only as the operator i2; the potential energydepends only on r: Knowing that the eigenfunctions of 12 are the spherical harmonics, we lookfor solutions of the form (r, , ) = R(r)Y (m(, ). We get12a1RYm -ZRYm = ERYmHy:12lr? orarand since 12Yim =l(l +1)Ylm we can cancel out Yim to get the radial equation1 ,2ORI(I+1)R-ZR= ER2r? rOr2rFor each value of I (i.e., O, 1, 2, ...) there are infinitely many solutions of this equation. Theyare conventionally labelled by theprincipal guantum number n, which runs from l +1 to ooThe quantum number m has dropped out of this eq, so the radial wavefunctions don't dependon m. We label them Rnl, and the complete wavefunction is nlm = Rnl(r)Y m(, )
5.3 Wavefunctions for the hydrogen atom The Hamiltonian contains the angular variables only as the operator መ𝒍 2 ; the potential energy depends only on r. Knowing that the eigenfunctions of መ𝒍 2 are the spherical harmonics, we look for solutions of the form 𝜓(𝑟, 𝜃,𝜑) = 𝑅(𝑟)𝑌𝑙𝑚(𝜃,𝜑). We get and since መ𝒍 2𝑌𝑙𝑚 = 𝑙(𝑙 + 1)𝑌𝑙𝑚 we can cancel out 𝑌𝑙𝑚 to get the radial equation For each value of l (i.e., 0, 1, 2, .) there are infinitely many solutions of this equation. They are conventionally labelled by the principal quantum number n, which runs from 𝑙 + 1 to ∞. The quantum number m has dropped out of this eq, so the radial wavefunctions don't depend on m. We label them Rnl, and the complete wavefunction is 𝜓𝑛𝑙𝑚 = 𝑅𝑛𝑙(𝑟)𝑌𝑙𝑚(𝜃,𝜑). 2 2 2 2 ˆ 1 1 2 lm lm lm Z H r RY RY ERY r r r r r l 2 2 2 1 1 2 2 R Z l l r R R ER r r r r r