Physical Chemistry Week 5 & 6

Uncertainty principle for energy and time From time-dependent S.E. 𝑖ℏ 𝜕 𝜕𝑡 𝜓= 𝐻 𝜓, we define 𝐻 =𝑖ℏ 𝜕 𝜕𝑡 ∵ 𝐻 ,𝑡 𝜙= 𝐻 𝑡𝜙 −𝑡 𝐻 𝜙 =𝑖ℏ𝜙 ∴ 𝐻 ,𝑡 =𝑖ℏ So that ∆𝐸⋅∆𝑡≥ℏ/2 If a quantum state has definite energy, i.e. ∆𝐸=0, then the life time of this state will be ∆𝑡→∞ In reality, energy level is broadened, and ∆𝑡~ℏ/∆𝐸 is regarded as life time of the energy level

Translation motion – 1D particle-in-a-box model One particle with mass 𝑚 confined in a box 0,𝐿 𝐻 =− ℏ 2 2𝑚 d 2 d 𝑥 2 +𝑉 𝑥 where 𝑉 𝑥 = 0,& 0<𝑥<𝐿 +∞,& 𝑥≤0 𝑜𝑟 𝑥≥𝐿 Within 0,𝐿 , the S.E. has solution 𝜓=𝐴 𝑒 𝑖𝑘𝑥 +𝐵 𝑒 −𝑖𝑘𝑥 , 𝑘= 2𝑚𝐸 /ℏ Outside 0,𝐿 , 𝜓=0

Continued Since wave function should be continuous, we impose boundary conditions 𝜓 0 =𝜓 𝐿 =0 𝜓 0 =𝐴+𝐵=0→𝐴=−𝐵 𝜓 𝐿 =−𝐵 𝑒 𝑖𝑘𝐿 +𝐵 𝑒 −𝑖𝑘𝐿 =−2𝑖𝐵 sin 𝑘𝐿 =0→𝑘𝐿=𝑛𝜋, 𝑛=1,2,… So that within 0,𝐿 , 𝜓 𝑥 =−2𝑖𝐵 sin 𝑛𝜋 𝐿 𝑥 After normalization, 𝜓 𝑛 𝑥 = 2 𝐿 sin 𝑛𝜋 𝐿 𝑥 , 𝑛=1,2,… Energy 𝐸 𝑛 = ℏ 2 𝑘 2 2𝑚 = 𝑛 2 𝜋 2 ℏ 2 2𝑚 𝐿 2 , 𝑛=1,2,…

Orthogonality For 𝑛≠𝑚, the following integral should be zero 0 𝐿 d𝑥 𝜓 𝑛 ∗ 𝜓 𝑚 &= 2 𝐿 0 𝐿 d𝑥 sin 𝑛𝜋𝑥 𝐿 sin 𝑚𝜋𝑥 𝐿 &=− 1 𝐿 0 𝐿 d𝑥 cos 𝑛+𝑚 𝜋𝑥 𝐿 − cos 𝑛−𝑚 𝜋𝑥 𝐿 &=− 1 𝐿 𝐿 𝑛+𝑚 𝜋 ⋅ sin 𝑛+𝑚 𝜋𝑥 𝐿 0 𝐿 − 𝐿 𝑛−𝑚 𝜋 ⋅ sin 𝑛−𝑚 𝜋𝑥 𝐿 0 𝐿 &=0

Check uncertainty principle for ground state We define uncertainty of an operator 𝐴 as ∆𝐴&= 𝐴 − 𝐴 2 &= 𝐴 2 −2 𝐴 𝐴 + 𝐴 2 &= 𝐴 2 −2 𝐴 𝐴 + 𝐴 2 &= 𝐴 2 − 𝐴 2

Continued 𝑥 &= 2 𝐿 0 𝐿 𝑥 sin 2 𝜋𝑥 𝐿 d𝑥 &= 1 𝐿 0 𝐿 𝑥 1− cos 2𝜋𝑥 𝐿 d𝑥 &= 𝐿 2 − 1 𝐿 0 𝐿 𝑥 cos 2𝜋𝑥 𝐿 d𝑥 &= 𝐿 2

Continued 𝑥 2 &= 2 𝐿 0 𝐿 𝑥 2 sin 2 𝜋𝑥 𝐿 d𝑥 &= 1 𝐿 0 𝐿 𝑥 2 1− cos 2𝜋𝑥 𝐿 d𝑥 &= 𝐿 2 3 − 1 𝐿 0 𝐿 𝑥 2 cos 2𝜋𝑥 𝐿 d𝑥 &= 𝐿 2 3 − 𝐿 2 2 𝜋 2

Continued 𝑝 &= 2ℏ 𝑖𝐿 ⋅ 𝜋 𝐿 0 𝐿 sin 𝜋𝑥 𝐿 cos 𝜋𝑥 𝐿 d𝑥 &= ℏ𝜋 𝑖 𝐿 2 0 𝐿 sin 2𝜋𝑥 𝐿 d𝑥 &=0

Continued 𝑝 2 &= 2 ℏ 2 𝐿 ⋅ 𝜋 2 𝐿 2 0 𝐿 sin 2 𝜋𝑥 𝐿 d𝑥 &= 𝜋 2 ℏ 2 𝐿 3 0 𝐿 1− cos 2𝜋𝑥 𝐿 d𝑥 &= 𝜋 2 ℏ 2 𝐿 2

Continued ∆𝑥&= 𝐿 2 3 − 𝐿 2 2 𝜋 2 − 𝐿 2 2 &= 𝐿 2𝜋 𝜋 2 3 −2 ∆𝑝&= 𝜋ℏ 𝐿 So ∆𝑥⋅∆𝑝= ℏ 2 ⋅ 𝜋 2 3 −2 ≈1.136⋅ ℏ 2 > ℏ 2

Appendix Let 𝛼 𝑘 = 0 𝐿 sin 𝑘𝑥 d𝑥 = 1− cos 𝑘𝐿 𝑘 d𝛼 𝑘 d𝑘 &= 0 𝐿 𝑥 cos 𝑘𝑥 d𝑥 &= 𝐿 sin 𝑘𝐿 𝑘 + cos 𝑘𝐿 −1 𝑘 2 So 0 𝐿 𝑥 cos 2𝜋𝑥 𝐿 d𝑥 = d𝛼 𝑘 d𝑘 𝑘=2𝜋/𝐿 =0

Continued Let 𝛽 𝑘 = 0 𝐿 cos 𝑘𝑥 d𝑥 = sin 𝑘𝐿 𝑘 d 2 𝛽 𝑘 d𝑘 2 &=− 0 𝐿 𝑥 2 cos 𝑘𝑥 d𝑥 &=− 𝐿 2 sin 𝑘𝐿 𝑘 − 2𝐿 cos 𝑘𝐿 𝑘 2 + 2 sin 𝑘𝐿 𝑘 3 So 0 𝐿 𝑥 2 cos 2𝜋𝑥 𝐿 d𝑥 =− d 2 𝛽 𝑘 d𝑘 2 𝑘=2𝜋/𝐿 = 𝐿 3 2 𝜋 2

Review One particle in a 1D box 𝜓 𝑛 𝑥 = 2 𝐿 sin 𝑛𝜋𝑥 𝐿 𝑛=1,2,3,… Wave Function: 𝜓 𝑛 𝑥 = 2 𝐿 sin 𝑛𝜋𝑥 𝐿 𝑛=1,2,3,… n=1 Energy Levels: x=L 𝑛=3 𝐸 3 = 9 𝜋 2 ℏ 2 2𝑚 𝐿 2 n=3 n=2 𝑛=2 𝐸 2 = 2𝜋 2 ℏ 2 𝑚 𝐿 2 𝑛=1 𝐸 1 = 𝜋 2 ℏ 2 2𝑚 𝐿 2 Ground State x

One particle in a 2D box Hamiltonian operator: y=L2 𝐻 =− ℏ 2 2𝑚 ( 𝜕 2 𝜕 𝑥 2 + 𝜕 2 𝜕 𝑦 2 )+𝑉(𝑥,𝑦) m 𝑉 𝑥,𝑦 = 0,& 0<𝑥< 𝐿 1 𝑎𝑛𝑑 0<𝑦< 𝐿 2 +∞,& 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 y=0 x=0 x=L1

Schrödinger Equation 𝐻 𝜓 𝑥,𝑦 =𝐸𝜓 𝑥,𝑦 Within 0<𝑥< 𝐿 1 and 0<𝑦< 𝐿 2 , − ℏ 2 2𝑚 𝜕 2 𝜓 𝑥,𝑦 𝜕 𝑥 2 + 𝜕 2 𝜓 𝑥,𝑦 𝜕 𝑦 2 =𝐸𝜓 𝑥,𝑦 Boundary conditions 𝜓 0,𝑦 =𝜓 𝐿 1 ,𝑦 =𝜓 𝑥,0 =𝜓 𝑥, 𝐿 2 =0

Separation of variables Let 𝜓 𝑥,𝑦 =𝑋 𝑥 𝑌 𝑦 and plug this equation into S.E., we get − ℏ 2 2𝑚 d 2 𝑋 d 𝑥 2 𝑌+𝑋 d 2 𝑌 d 𝑦 2 =𝐸𝑋𝑌 Divide both sides by 𝑋𝑌 − ℏ 2 2𝑚 1 𝑋 d 2 𝑋 d 𝑥 2 + 1 𝑌 d 2 𝑌 d 𝑦 2 =𝐸

Continued To ensure − ℏ 2 2𝑚 1 𝑋 d 2 𝑋 d 𝑥 2 + 1 𝑌 d 2 𝑌 d 𝑦 2 =𝐸, each term in the LHS should be some constant, viz. − ℏ 2 2𝑚 1 𝑋 d 2 𝑋 d 𝑥 2 = 𝐸 1 − ℏ 2 2𝑚 1 𝑌 d 2 𝑌 d 𝑦 2 = 𝐸 2 And 𝐸 1 + 𝐸 2 =𝐸

Continued The 2D S.E. of 𝜓 𝑥,𝑦 has been decomposed into two 1D S.E. − ℏ 2 2𝑚 d 2 𝑋 d 𝑥 2 = 𝐸 1 𝑋 − ℏ 2 2𝑚 d 2 𝑌 d 𝑦 2 = 𝐸 2 𝑌 With solution where 𝑛 1 , 𝑛 2 =1,2,3,… 𝜓 𝑛 1 , 𝑛 2 = 2 𝐿 1 𝐿 2 sin 𝑛 1 𝜋𝑥 𝐿 1 sin 𝑛 2 𝜋𝑦 𝐿 2 &, 𝑤𝑖𝑡ℎ𝑖𝑛 2𝐷 𝑏𝑜𝑥 0&, 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑏𝑜𝑥 𝐸 𝑛 1 , 𝑛 2 = 𝑛 1 2 𝜋 2 ℏ 2 2𝑚 𝐿 1 2 + 𝑛 2 2 𝜋 2 ℏ 2 2𝑚 𝐿 2 2

Mid-term summary Schrödinger Equation 𝐻 Ψ=𝐸Ψ 𝐻 = 𝐾 + 𝑉 Hamiltonian/Energy operator = Kinetic energy operator + potential energy operator For one-particle system: 3D 𝑇 =− ℏ 2 2𝑚 𝛻 2 +𝑉 𝑥,𝑦,𝑧 , where 𝛻 2 = 𝜕 2 𝜕 𝑥 2 + 𝜕 2 𝜕 𝑦 2 + 𝜕 2 𝜕 𝑧 2 is a Laplacian operator; 1D 𝑇 =− ℏ 2 2𝑚 d 2 d 𝑥 2 +𝑉 𝑥

Continued Probability density - Ψ 𝑥 2 The probability to find the particle between 𝑥 and 𝑥+d𝑥 is Ψ 𝑥 2 d𝑥 The probability to find the particle in whole space should be 1 Normalization of wave function Ψ/ d𝜏 Ψ 2 Hermitian operator d𝜏 𝜙 ∗ Ω 𝜓 = d𝜏 𝜓 ∗ Ω 𝜙 ∗ e.g. 𝑝 = ℏ 𝑖 ∇, 𝑝 𝑥 = ℏ 𝑖 d d𝑥

Continued Eigen value and eigen function ⅆ𝜏 𝜓 𝑖 ∗ 𝜓 𝑗 =0 if 𝐸 𝑖 ≠ 𝐸 𝑗 𝐻 𝜓 𝑖 = 𝐸 𝑖 𝜓 𝑖 ⅆ𝜏 𝜓 𝑖 ∗ 𝜓 𝑗 =0 if 𝐸 𝑖 ≠ 𝐸 𝑗 Uncertainty principle 𝑥 , 𝑝 𝑥 =𝑖ℏ, ∆𝑥⋅∆ 𝑝 𝑥 ≥ℏ/2 ∆𝐸⋅∆𝑡≥ℏ/2 − ℏ 2 2𝑚 d 2 𝜓 d 𝑥 2 +𝑉𝜓=𝐸𝜓, 𝐸>𝑉 𝜓=𝑁 𝑒 ±𝑖𝑘𝑥 , 𝑘= 2𝑚 𝐸−𝑉 /ℏ, 𝑁 is normalization factor