Physical Chemistry Week 10

Quantisation of energy (Cont.) At 𝑥=0, the wavefunction should be continuous: 𝑖𝑘−𝜅 𝑖𝑘+𝜅 𝑒 −𝑖𝑘𝐿 = 𝑖𝑘−𝜅 2𝑖𝑘 𝑒 −𝑖𝑘𝐿 + 𝑖𝑘+𝜅 2𝑖𝑘 𝑒 𝑖𝑘𝐿 i.e. 𝜅 2 − 𝑘 2 −2𝑖𝜅𝑘 cos 𝑘𝐿 −𝑖 sin 𝑘𝐿 = ( 𝜅 2 − 𝑘 2 +2𝑖𝜅𝑘) cos 𝑘𝐿 +𝑖 sin 𝑘𝐿 Real parts of LHS and RHS are identities: 𝜅 2 − 𝑘 2 cos 𝑘𝐿 −2𝜅𝑘 sin 𝑘𝐿 = 𝜅 2 − 𝑘 2 cos 𝑘𝐿 −2𝜅𝑘 sin 𝑘𝐿 Imaginary parts should satisfy: − 2𝜅𝑘 cos 𝑘𝐿 + 𝜅 2 − 𝑘 2 sin 𝑘𝐿 =2𝜅𝑘 cos 𝑘𝐿 + 𝜅 2 − 𝑘 2 sin 𝑘𝐿

Continued 4𝜅𝑘 cos 𝑘𝐿 =2 𝑘 2 − 𝜅 2 sin 𝑘𝐿 When cos 𝑘𝐿 ≠0 , we have tan 𝑘𝐿 = 2𝜅𝑘 𝑘 2 − 𝜅 2 , i.e. tan 2𝑚𝐸 𝐿 ℏ = 2 𝐸 𝑉−𝐸 2𝐸−𝑉 When cos 𝑘𝐿 =0 and 𝑘 2 = 𝜅 2 , we have 𝐸= 𝑉 2 and 𝐸= 𝑛+ 1 2 2 𝜋 2 ℏ 2 2𝑚 𝐿 2 where 𝑛=0,1,2,…. For a given 𝑉, if there is no such 𝐸 satisfy these two equations, then we discard this solution

One dimensional harmonic oscillator Hamiltonian 𝐻 =− ℏ 2 2𝑚 d 2 d 𝑥 2 + 1 2 𝑘 𝑓 𝑥 2 S.E.− ℏ 2 2𝑚 d 2 𝜓 𝑥 d 𝑥 2 + 1 2 𝑘 𝑓 𝑥 2 𝜓 𝑥 =𝐸𝜓 𝑥 Change variable from 𝑥 to 𝑦=𝑥/𝛼 where 𝛼= ℏ 2 𝑚 𝑘 𝑓 1 4 . Under this operation, 𝜓(𝑥) changes to 𝜙 𝑦 After some algebra, we have d 2 𝜙 𝑦 d 𝑦 2 + 𝜆− 𝑦 2 𝜙 𝑦 =0 where 𝜆= 2𝐸 ℏ𝜔 and 𝜔= 𝑘 𝑓 /𝑚

Solution of S.E. Asymptotic behaviour d 2 𝜙 d 𝑦 2 − 𝑦 2 𝜙=0 as 𝑦→±∞ 𝜙→ 𝑒 − 𝑦 2 2 Rewrite 𝜙 𝑦 =𝑁⋅𝐻 𝑦 𝑒 − 𝑦 2 2 , we have following Hermite eq. d 2 𝐻 d 𝑦 2 −2𝑦 d𝐻 d𝑦 + 𝜆−1 𝐻=0

Hermite polynomial Expand 𝐻 𝑦 as 𝐻 𝑦 = 𝑛=0 ∞ 𝑐 𝑛 𝑦 𝑛 . Plug in this expansion into H. eq. 𝑛=0 ∞ 𝑦 𝑛 𝑐 𝑛+2 𝑛+2 𝑛+1 − 2𝑛−𝜆+1 𝑐 𝑛 =0 𝑐 𝑛+2 = 2𝑛+1−𝜆 𝑛+2 𝑛+1 𝑐 𝑛 𝑛=0 ∞ 𝑐 𝑛 𝑦 𝑛 →∞ as fast as 𝑒 𝑦 2 when 𝑦→±∞. Thus 𝜙~ 𝑒 − 𝑦 2 2 𝑛=0 ∞ 𝑐 𝑛 𝑦 𝑛 →∞ as fast as 𝑒 𝑦 2 2

Boundary conditions To ensure 𝜙 ±∞ =0 the expansion of 𝐻 𝑦 must be truncated, i.e. 𝑐 𝜈 ≠0, 𝑐 𝜈+2 =0 then 2𝜈+1=𝜆= 2𝐸 ℏ𝜔 thus 𝐸 𝜈 =ℏ𝜔 𝜈+ 1 2 , 𝜈=0,1,2,… Ground state energy 𝐸 0 = 1 2 ℏ𝜔 – zero-point energy

Properties of Hermite polynomials Recursive relation 𝐻 𝜈+1 −2𝑦 𝐻 𝜈 +2𝜈 𝐻 𝜈−1 =0 Orthogonal relation −∞ +∞ d𝑦 𝐻 𝜈 ′ 𝐻 𝜈 𝑒 − 𝑦 2 = 0 & ,if 𝜈 ′ ≠𝜈 𝜋 2 𝜈 𝜈!& , if 𝜈 ′ =𝜈 First few Hermite polynomials 𝜈 1 2 𝐻 𝜈 2𝑦 4 𝑦 2 −2

Normalisation of wavefunction −∞ +∞ d𝑥 𝜓 𝜈 ∗ 𝑥 𝜓 𝜈 𝑥 &= 𝑁 𝜈 2 𝛼 −∞ +∞ d𝑦 𝐻 𝜈 2 𝑒 − 𝑦 2 &= 𝑁 𝜈 2 𝛼 𝜋 2 𝜈 𝜈! Thus 𝑁 𝜈 = 𝛼 𝜋 2 𝜈 𝜈! − 1 2 𝜓 𝜈 𝑥 = 𝛼 𝜋 2 𝜈 𝜈! − 1 2 𝐻 𝜈 𝑥 𝛼 𝑒 − 𝑥 2 2 𝛼 2

Mean value of 𝑥 𝑥 &= −∞ +∞ d𝑥 𝜓 𝜈 ∗ 𝑥 𝑥 𝜓 𝜈 𝑥 &= 𝑁 𝜈 2 𝛼 2 −∞ +∞ d𝑦 𝐻 𝜈 𝑦 𝑦 𝐻 𝜈 𝑦 𝑒 − 𝑦 2 &= 𝑁 𝜈 2 𝛼 2 −∞ +∞ d𝑦 𝐻 𝜈 𝐻 𝜈+1 +2𝜈 𝐻 𝜈−1 2 𝑒 − 𝑦 2 &=0

Mean value of 𝑥 2 𝑥 2 &= −∞ +∞ d𝑥 𝜓 𝜈 ∗ 𝑥 𝑥 2 𝜓 𝜈 𝑥 &= 𝑁 𝜈 2 𝛼 3 −∞ +∞ d𝑦 𝑦 𝐻 𝜈 𝑦 𝑦 𝐻 𝜈 𝑦 𝑒 − 𝑦 2 &= 𝑁 𝜈 2 𝛼 3 −∞ +∞ d𝑦 𝐻 𝜈+1 +2𝜈 𝐻 𝜈−1 2 2 𝑒 − 𝑦 2 &= 1 4 𝑁 𝜈 2 𝛼 3 −∞ +∞ d𝑦 𝐻 𝜈+1 2 +4𝜈 𝐻 𝜈+1 𝐻 𝜈−1 +4 𝜈 2 𝐻 𝜈−1 𝑒 − 𝑦 2 &= 𝛼 2 𝜋 2 𝜈+2 𝜈! 𝜋 2 𝜈+1 𝜈+1 !+𝜈 𝜋 2 𝜈+1 𝜈! &= 𝛼 2 𝜈+ 1 2

Mean value of potential energy 𝑉 &= 1 2 𝑘 𝑓 𝑥 2 &= 1 2 𝑘 𝑓 𝛼 2 𝜈+ 1 2 &= 1 2 ℏ𝜔 𝜈+ 1 2 &= 1 2 𝐸 𝜈

Tunnelling – classically forbidden region Classically forbidden region is where 𝑉 𝑥 > 𝐸 𝜈 Quantal oscillator can reach classically forbidden region with some tunnelling probability For ground state, 𝜓 0 = 𝑁 0 𝑒 − 𝑥 2 2 𝛼 2 , 𝐸 0 = 1 2 ℏ𝜔. Denote 𝑥 𝐿 and 𝑥 𝑅 as the negative and positive solutions of 𝑉 𝑥 = 𝐸 0 respectively. The tunnelling probability is then 𝑃 𝑥< 𝑥 𝐿 +𝑃 𝑥> 𝑥 𝑅 =2 𝑥 𝑅 +∞ d𝑥 𝜓 0 2 𝑥 ≈15.7%