Chapter 6. Hydrogen Atom. 6.1 Schrödinger Equation. The Hamiltonian for a hydrogen atom is. Recall that. 1 r 2 sin 2 θ + 1. and.

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1 Chapter 6 Hydrogen Atom 6. Schrödinger Equation The Hamiltonian for a hydrogen atom is Recall that Ĥ = h e m e 4πɛ o r = r ) + r r r r sin θ sin θ ) + θ θ r sin θ φ and [ ˆL = h sin θ ) + )] sin θ θ θ sin θ φ The Hamiltonian can then be written [ Ĥ = h r )] + ˆL m e r r r m e r e 4πɛ o r

2 Using the time-independent Schrödinger equation Multiply by m e r to simplify Ĥψ = Eψ Ĥψ Eψ = m e r Ĥψ Eψ) = [ [ h r )] )] + r r ˆL e m e r 4πɛ o r + E ψr, θ, φ) = Separating the angular variable r) from radial variables θ, φ) The Schrödinger equation becomes ψr, θ, φ) = Rr)Yθ, φ) [ h r R ) + m ) ] er e R r r h 4πɛ o r + E R + Y ˆL Y = Because r, θ, φ are independent variables [ h r R ) + m ) ] er e R r r h 4πɛ o r + E R = Y ˆL Y Thus we can write and [ h r R ) + m ) ] er e R r r h 4πɛ o r + E R = β [ Y ˆL Y = h Y sin θ where β is a constant sin θ Y ) + θ θ sin θ ] Y = β φ

3 6. Angular part of wavefunction Spherical Harmonics) Note that the angular part of the hydrogen wavefunction is a rigid rotor, and the solutions are spherical harmonics By separation of variables Solving the eigen equation Yθ, φ) = Θθ)Φφ) Separating Θ and Φ ˆL Y = βy = βθφ h Θ sin θ d sin θ dθ ) Θ + β sin θ + d Φ dθ dθ Φ dφ = Θ sin θ d dθ where m is a constant sin θ dθ ) + β sin θ = m dθ d Φ Φ dφ = m We can easily find the solutions of Φ Φφ) = A m e imφ Φφ) = A m e imφ Φφ) must be single valued, so Φφ + π) = Φφ) 3

4 And therefore A m E imφ+π) = A m e imφ A m E imφ+π) = A m e imφ Thus we see that e ±imπ) = for m =, ±, ±... Using the normalization condition π Φ mφ)φ m φ)dφ = And taking A ±m as our constant of normalization π A m dφ = A m π = We obtain the normalized functions A m = π Φφ) = π e imφ, m =, ±, ±,... The solutions of Θ are not so easily found Recall the differential equation for Θ Θ sin θ d sin θ dθ ) + β sin θ = m dθ dθ For convenience, we change variables Θθ) = P x) where x = cos θ 4

5 Note that Also note d dθ = d cos θ d dθ d cos θ = sin θ d dx sin θ + cos θ = and sinθ = ± cos θ) This change of variables leads to where β = ll + ) x ) dp dx xdp dx + [β m x ] P x) = This is Legendre s equation, and it has well-known solutions These solutions are the Legendre polynomials P l x) The orthonormal condition is P m l x) = x ) m d m dx P lx) m P m l x)p n m x)dx = We get a normalization constant of [ l + ) N lm = l + m )! l + ) l m )! δ ln l m )! l + m )! The complete spherical harmonic then becomes Yθ, φ) = Θ lm θ)φ m φ) ] Y = N lm P m cos θ)e imφ π l The first few spherical harmonics can be seen in table 6.3 in your book. 5

6 6.3 Commutation of Angular Momentum Recall the components of angular momentum ˆL x = y ˆP z z ˆP y = i h y z z ) y ˆL y = z ˆP x x ˆP z = i h z x x ) z ˆL z = x ˆP y y ˆP x = i h x y y ) x Converting to spherical coordinates ˆL z = i h φ A spherical harmonic is an eigenfunction of ˆL z ˆL z Y m l θ, φ) = ˆL z [Θ lm θ)φ m φ)] = Θ i h)im) e imφ π = Θ hm)φ = hmy m l θ, φ) A spherical harmonic is also an eigenfunction of ˆL ˆL Y m l θ, φ) = ll + ) h Y m l θ, φ) Because spherical harmonics are eigenfunctions of both ˆL z and ˆL, we can simultaneously determine values of ˆL z and ˆL. Therefore, ˆL z and ˆL commute. Note that the spherical harmonics are not eigenfuctions of ˆL x and ˆL y example 6.7 in your book), and therefore these values cannot be simultaneously determined. 6

7 6.4 Radial part of wavefunction h d r ddr ) m e r dr R + [ h ] ll + ) e m e r 4πɛ o r E R = The solution of this equation introduces a new quantum number n. However from the above equation, we will obtain a different solution depending on the value of l, hence R = R n lr). This solution is R nl r) = where α n = n a o [ ] n l )!α 3 n αn r) l e α nr n[n + l)!] 3 L l+ n+l α nr) a o Bohr radius) and L l+ n+l is the associated Laguerre polynomial Here we have the restriction that l n, and n =,, 3... The corresponding energy eigenstates in the absence of an electric or magnetic field) depend only on n They are given by E n = ) m e e h = ) e 4πɛ o n 4πɛ o a o n = where a o = ɛ oh πm e e The radial solutions satisfy e 8πɛ o a o n r R nlr)r n lr) = δ nn where R nl r) has n l nodes. See figure 6.3 in your book) 7

8 And the complete hydrogen wave function is given by ψ nlm r, θ, φ) = R nl r)y m l θ, φ) = R nl r)θ lm θ)φ m φ) The orthonormality condition can be written ψ nlmψ n l m dτ = π π r dr sin θdθ dφψnlmr, θ, φ)ψ n l m r, θ, φ) = π Rnlr)R n l r)r dr π sin θdθ dφylmθ, φ)y l m θ, φ) } {{ } δ ll δ mm = δ nn δ ll δ mm n: principal quantum number n =,, 3... l: angular momentum quantum number depends on n), l =,,...n, n values for each l m: magnetic quantum number depends on l), l m l, l + values for each l Hydrogen wavefunctions sometimes are called orbitals s p d f l = 3 s = ψ s = ψ p o = ψ 8

9 6.5 Some integration help Note: Integral tables can be found in the CRC or other handbooks. Thus I β; a, b) = b a I n β; a, b) = e βr dr = b a r n e βr dr ) e βr b a = β β e βa e βb ) Note that I o β;, ) = e βr dr = β ) di o dβ = And in general, b a re βr dr = β ) n dn dβ n I oβ; a, b) = ae βa be βb) + β e βa e βb) b Which in the specific case of Iβ;, ) gives a r n e βr dr = I n β; a, b) And I β;, b) = I n β;, ) = b r n e βr dr = n! β n+ r e βr dr = ) ] [ + βb + β b e βb β 3 9

10 6.6 Quantum Mechanical Virial Theorem Using the following Hamiltonian Ĥ = h m + V x, y, z) The commutator between Ĥ and some linear operator  is [ [Ĥ, Â] = i h x V x + y V y + z V ] i h z ˆK Then, x V x + y V y + z V = ˆK z Example: Harmonic Oscillator Potential The partial derivatives are V = kr = k x + y + z ) V x = kx, V y = ky, V z = kz Thus, we have kx + ky + kz = ˆK ˆV = ˆK Example: For any r n potential or ˆV = ˆK V = Cr n ) V x = ncrn x r

11 We now have x + y + z }{{} nc rn r r = ˆK n Cr n = ˆK n ˆV = ˆK Example3: Coulomb Potential V = e 4πɛ o r n = Therefore we have ˆV = ˆK ˆV = ˆK Recall Thus for this case ˆK = E ˆV E = ˆV