1 r 2 sin 2 θ. This must be the case as we can see by the following argument + L2

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1 PHYS 4 3. The momentum operator in three dimensions is p = i Therefore the momentum-squared operator is [ p 2 = 2 2 = 2 r 2 ) + r 2 r r r 2 sin θ We notice that this can be written as sin θ ) + θ θ r 2 sin 2 θ 2 ] φ 2 where p 2 r = 2 r 2 r L 2 = 2 [ sin θ p 2 = p 2 r + L2 r 2 r 2 r θ ) sin θ θ ) + 2 ] sin 2 θ φ 2 This must be the case as we can see by the following argument Therefore L 2 = r p) 2 = r 2 p 2 r p) 2 r 2 p 2 = r p) 2 + L 2 p 2 = ˆr p) 2 + L2 r 2 p 2 = p 2 r + L2 r 2 Putting this together, we can write the Shrödinger equation in three dimensions as { } 2 r 2 ) + L2 2m r 2 r r 2mr + V r) ψ r) = Eψ r) 2

2 PHYS 4 3. The potential for the hydrogen atom is V r) = e2 4πɛ r The potential is spherically symmetric and thus does not depend on θ and φ This will allow us to use the method of separation of variables. We also notice that that the angular momentum operator and the hamiltonian commute. [L 2, H] = Therefore we can find simultaneous eigenfunctions of L 2 and H. Fortunately we already know the eigenfunctions of L 2 They are the spherical harmonics, the Y lm So we write the solutions as ψr, θ, φ) = Rr)Y lm θ, φ) Substituting this form into the Schrödinger equation gives us the radial equation { } 2 r 2 ) + 2 ll + ) e2 Rr) = ERr) 2m r 2 r r 2mr 2 4πɛ r This equation simplifies if we define a new function ur) rrr) With this substitution d/dr)[r 2 dr/dr)] = rd 2 u/dr 2 and hence { } 2 d 2 2m dr + 2 ll + ) e2 ur) = Eur) 2 2mr 2 4πɛ r This looks like the one dimensional Shrödinger equation, except that the effective potential, V eff = e2 4πɛ r + 2 ll + ) 2mr 2 contains an extra piece, the so-called centrifugal barrier. 2

3 PHYS 4 3. As always we should find the natural distance scale in this problem. The constants in our problem are m, and e 2 /4πɛ This leads to The solution is [ ML [m] α [ ] β [e 2 /4πɛ ] γ = [M] α 2 ] β [ ML 3 ] γ = L T T 2 α + β + γ = β 2γ = 2β + 3γ = α =, β = 2, γ = Therefore the quantity with the dimensions of distance is 4πɛ 2 =.529 m me 2 This quantity is the so-called Bohr radius. We can also define the quantity 2mE κ and introduce ρ 2κr and n / κ The differential equation then becomes d 2 u [ dρ = 2 4 n ll + ) ] + u ρ ρ 2 Let s look at this equation at large and small ρ. As ρ, the constant term dominates and so we have The general solution is d 2 u dρ 2 = u 4 uρ) = Ae ρ/2 + Be ρ/2 but the second term cannot be normalized, so uρ) Ae ρ/2 On the other hand as ρ the centrifugal term dominates and we have 3

4 PHYS 4 3. The general solution is d 2 u ll + ) = u dρ2 ρ 2 uρ) = Cρ l+ + Dρ l Again we discard the second term because of singular behaviour near the origin. The next step is to peel off the asymptotic behaviour an write uρ) = ρ l+ e ρ/2 vρ) Substituting this form into the radial equation we get the following recursion relation and representing vρ) by a power series we find the recursion relation vρ) = k= a k ρ k l + k + n a k+ = a k k + )2l + k + 2) In order to obtain a satisfactory wave function, the series must terminate, so that the wave function can be normalized. If the polynomial is to terminate with the k th term, it is necessary that a k+ =. That is, we require that n = l + k + Hence n takes on integral values given by l +, l + 2, l + 3. polynomials are called the associated Laguerre polynomials d ) 2l+ [ d n+lρ ] L 2l+ n+l ρ) e ρ dρ dρ) n+l e ρ ) The Reassembling the radial wavefunctions for hydrogen we have the not yet normalized R nl = ρ l e ρ/2 L 2l+ n+l ρ) Normalization of the radial functions over ρ may be achieved by evaluating the integral 4

5 PHYS 4 3. e ρ ρ 2l [L 2l+ n+l ρ)]2 ρ 2 dρ = 2n[n + l)!]3 n l )! If we substitute this form into the Schrödinger equation we find that the energy eigenvalues are E n = E /n 2 where E = m e 2 ) 2 e 2 ) 2 = 2 2 4πɛ 2 mc2 = 4πɛ c 2 mc2 α 2 = 3.6 ev Here α = e2 4πɛ c 37 is the fine structure constant. It is a fundamental dimensionless constant of nature. The energy is of course negative as the electron is bound to the proton. It costs energy to ionize hydrogen. We also notice that the energy levels fall like n 2, the energy level spacing is closer and closer and the system becomes very weakly bound. One of the most interesting observations is that the energy levels only depend on n and not on l or m. The fact that they don t depend on m is easy to see. It s a consequence of rotational symmetry. This manifests itself in that the radial equation doesn t depend on m. It does however depend on l. In fact both the equation and the wavefunction has l dependence and yet the energy eigenvalue does not. This appears to be somewhat of a miracle! In turns out that there is a hidden symmetry in the problem which leads to this degeneracy. We list the first few radial wavefunctions explicitly The radial probability distribution is P r) = r 2 R 2 r) It is normalized to P r) dr = 5

6 PHYS 4 3. R nl Table : The first few radial wavefunctions for hydrogen ) 3/2 R 2 e r/ ) 3/2 ) R r ) 3/2 R 2 3 r 2 ) 3/2 R r 3 ) 3/2 R R 32 ) 3/ e r/2 e r/2 + 2 ) r r ) 2 e r/3 R nl r) 27 r/) 2 ) r e r/3 e r/3 The next few figures show the radial probability distributions. 6

7 PHYS

8 PHYS 4 3. Putting everything together we can finally write the complete hydrogen wavefunctions as ) 3 2 n l )! ψ nlm = 2 n n[n + l)!] 3 e r/n 2r ) l[ L 2l+ n+l na 2r/n) ] Y lm θ, φ) They are an orthonormal set satisfying ψ nlmψ n l m r2 sin θ dr dθ dφ = δ nn δ ll δ mm We can use these wavefunctions to calculate where the radial probability distribution has a maximum. For example for ψ, the radial distribution is P = r 2 R 2 = 4 a 3 r 2 e 2r/ Taking the derivative and setting it to zero, we find the maximum occurs at r = To find the expectation value of r we construct r = rp r) dr = r 3 4 a 3 ) e 2r/ dr = 4 x 3 e x dx = 4 3! = 3 2 which is not at the location of the maximum. Note that r = Also notice that ψ ψ has a maximum at the origin, but P = at r = 8