Final Exam solutions This is the half of the double delta function in ...

We find the same solution:Third way:〈n|x 4 |n〉 =( ) 2 (6n 2 + 6n + 3 )2mω〈n|x 4 |n〉 =∫dx x 4 |ψ n (x)| 2=( ) 2 ∫ 1 ∞mω 2 n n! √ dy e −y2 y 4 H n (y) 2π −∞We know:yH n = 1 2 H n+1 + nH n−1then,y 2 H n = 1 2 yH n+1 + nyH n−1Now we need,= 1 2 (1 2 H n+2 + (n + 1)H n ) + n( 1 2 H n + (n − 1)H n−2 )= 1 4 H n+2 + 2n + 1 H n + n(n − 1)H n−22∫ ∞−∞dy e −y2 (y 2 H n (y)) 2 ==∫ ∞−∞∫ ∞−∞( 1dy e −y2 4 H n+2 + 2n + 1) 2H n + n(n − 1)H n−22( ( ) 21 2n + 1dy e −y2 16 H2 n+2 +Hn 2 + (n(n − 1)) 2 H 22n−2)where in the last line, we have used the orthogonality of Hermite functions.∫ ∞dy e −y2 (y 2 H n (y)) 2 = √ (( ) 212n + 1π−∞16 2(n+2) (n + 2)! +2 n n! + (n(n − 1)) 2 2 (n−2) (n − 2)!)2= √ ( )(n + 2)(n + 1)π2 n (2n + 1)2 n(n − 1)n!+ +44 4which gives us,〈n|x 4 |n〉 =( ) 2 (6n 2 + 6n + 3 )2mω2

Eigenstates:⎛ ⎞⎛ ⎞⎛ ⎞0u 1 = ⎝1⎠ u 2 = √ 1 1⎝0⎠ u 3 = 1 1√ ⎝ 0 ⎠0212−1or eigenfunctions:u 1 (θ, φ) = Y 10 (θ, φ)u 2 (θ, φ) = 1 √2(Y 11 (θ, φ) + Y 1,−1 (θ, φ))u 1 (θ, φ) = 1 √2(Y 11 (θ, φ) − Y 1,−1 (θ, φ))5.First we solve the three dimensional harmonic oscillator in cartesian coordinates:with the following energyψ kpq = C kpq H k (x)H p (y)H q (z)e − mω (x2 +y 2 +z 2 )= C kpq H k (x)H p (y)H q (z)e − mωr2E kpq = ω(k + p + q + 3 2 )Since l = 1 the θ and φ dependence should be a combination of Y 11 (θ, φ), Y 10 (θ, φ)and Y 1,−1 (θ, φ). These spherical harmonics can be obtained from the above wavefunction when the degree of the polynomial part of the wave function is one:H 1 (x) ∼ x = r cos φ sin θ = − 1 8π2 3 r(Y 11(θ, φ) + Y 1,−1 (θ, φ))H 1 (y) ∼ y = r sin φ sin θ = − 1 8π2i 3 r(Y 11(θ, φ) − Y 1,−1 (θ, φ))H 1 (z) ∼ z = r cos θ = 4π 3 rY 10(θ, φ)Thus we have three options: (kpq) = (100), (010), (001). So the degeneracy is 3 and,E 100 = E 010 = E 001 = ω(1 + 3 2 )= 5 2 ωWe can find the wave functions as well (unnecessary in the exam),ψ 100 ∼ (Y 11 (θ, φ) + Y 1,−1 (θ, φ)) r e − mωr2ψ 010 ∼ (Y 11 (θ, φ) − Y 1,−1 (θ, φ)) r e − mωr2ψ 001 ∼ Y 10 (θ, φ) r e − mωr24

6.Mean charge density:ρ(⃗r) = −e|ψ(⃗r)| 2where in the ground state of Hydrogen atom we have,|ψ(⃗r)| 2 = 4 a 3 0e −2r/a 0Gauss Law:∇ 2 ϕ(⃗r) = ρ(⃗r)which gives( ( 1 ∂r 2 ∂ ) )l(l + 1)+ ϕ(⃗r) = −e 4 r 2 ∂r ∂r r 2 a 3 0e −2r/a 0put l = 0, then the electric potential ϕ, can be computed by twice integration as,ϕ(r) = −e 4 ∫ r ∫1 rr 2 e −2r/aa 3 00 r 2= −e 4 ∫ r( )1 −1a 3 0 r 2 4 a 0e −2r/a 0(2r 2 + 2ra 0 + a 2 0) + C( )a20 e −2r/a 0= −e 4 a 3 04r(r + a 0 ) − C rwe take C = a 3 0/4 to have regular solution at origin:ϕ(r) = −e e−2r/a 0a 0 r(r + a 0) + e rThis is the potential by electron cloud. We may add the proton potential, ϕ p =(+e/r).5