QM Lecture Notes - Students (Part II) PDF

KFT431 Physical Chemistry III Quantum Chemistry (Part II) Dr Ng Si Ling School of Chemical Sciences, USM PART II Quantum mechanics for simple systems The Free Particle Particle-in-a-box (1-D, 2-D and 3D) Harmonic oscillator Rigid rotor Quantum Mechanics for Simple Systems T...

KFT431 Physical Chemistry III Quantum Chemistry (Part II) Dr Ng Si Ling School of Chemical Sciences, USM PART II Quantum mechanics for simple systems The Free Particle Particle-in-a-box (1-D, 2-D and 3D) Harmonic oscillator Rigid rotor Quantum Mechanics for Simple Systems The Free Particle Consider a particle of mass m moving freely in one direction. Assume the potential energy is zero throughout the particle’s motion. The total energy is therefore the kinetic energy, 𝟐 𝟏 𝐩 𝐱 𝐄 = 𝐦𝐯𝐱𝟐 = 𝟐 𝟐𝐦 The quantum mechanical operator is ħ𝟐 𝐝𝟐 ෡ =− 𝐇 𝟐𝐦 𝐝𝐱 𝟐 The Schrödinger equation is ħ 𝟐 𝐝𝟐 𝛙 − 𝟐 = 𝐄𝛙 (25) 𝟐𝐦 𝐝𝐱 or 𝐝𝟐 𝛙 𝟐𝒎𝑬𝛙 (26) − 𝟐 + 𝟐 =𝟎 𝐝𝐱 ħ The solution of Eq. (26) are 𝛙+ = 𝐀+ 𝐞𝐢𝐤𝐱 and 𝛙− = 𝐀 − 𝐞−𝐢𝐤𝐱 𝟐𝐦𝐄 where k = , and 𝐀 + and 𝐀− are constant. ħ 𝛙+ 𝟐 = 𝛙∗+ 𝛙+ = 𝐀∗+ 𝐞−𝐢𝐤𝐱 𝐀 + 𝐞𝐢𝐤𝐱 = 𝐀∗+ 𝐀+ 𝟐 = 𝐀+ Similarly, 𝛙− 𝟐 = 𝐀− 𝟐 This are independent of x implying that there is an equal probability of finding particle at any distance along the x axis. The particle is therefore not localised and energy is not quantised. Quantum Mechanics for Simple Systems The Particle in a Box (one dimension) V Consider a particle at mass m constrained to move in one-dimensional box of length a. The potential energy V(x) is zero for 0 ≪ x ≪ a and infinite outside this region. Region I The Schrödinger equation can be written as ħ 𝟐 𝐝𝟐 𝛙 − = 𝐄𝛙 (27) 𝟐𝐦 𝐝𝐱 𝟐 𝐝𝟐 𝛙 𝟐𝒎𝑬𝛙 𝟐 or 𝟐 = − 𝟐 = −𝐤 𝛙 𝐝𝐱 ħ 𝟐𝐦𝐄 where k= ħ The general solution to this equation is 𝛙(𝐱) = 𝐀 𝐜𝐨𝐬𝐤𝐱 + 𝐁 𝐬𝐢𝐧𝐤𝐱 (28) Region II The Schrödinger equation can be written as ħ𝟐 𝐝𝟐 𝛙 − + 𝐕𝛙 = 𝐄𝛙 𝟐𝐦 𝐝𝐱 𝟐 ħ𝟐 𝐝𝟐 𝛙 or − 𝟐 = 𝐄−∞ 𝛙=𝟎 𝟐𝐦 𝐝𝐱 or 𝐝𝟐 𝛙 = ∞𝛙 𝐝𝐱 𝟐 The only solution to the above is 𝛙 = 𝟎. This implies that the probability of finding the particle outside the box is zero. Therefore, 𝛙 = 𝟎 at 𝐱 = 𝟎 and 𝐱 = 𝐚. By using the boundary condition in Eq. (28): at 𝐱 = 𝟎, 𝐀 = 𝟎 at 𝐱 = 𝐚, 𝛙 𝐚 = 𝟎 = 𝐁 𝐬𝐢𝐧𝐤𝐚 𝐁 ≠ 𝟎, ∴ 𝐬𝐢𝐧𝐤𝐚 = 𝟎 or 𝐤𝐚 = 𝐧𝛑 where n = 0,1,2,3 … 𝟐 𝟐 𝟐 𝐧 𝛑 𝟐𝐦𝐄 𝐤 = 𝟐 = 𝟐 (29) 𝐚 ħ Therefore, 𝐧𝟐 𝛑𝟐 ħ𝟐 𝐧𝟐 𝐡𝟐 𝐄𝐧 = 𝟐 = 𝟐 , 𝐧 = 𝟏, 𝟐, 𝟑 (30) 𝟐𝐦𝐚 𝟖𝐦𝐚 It is observed that a particle constrained between 𝐱 = 𝟎 and 𝐱 = 𝐚 has quantized energy levels [Eq. (30)]. The lowest value of 𝐧 is 1 because n = 0 will cause the wavefunction to be zero everywhere. Using Eq. (29), we can write the wavefunction as 𝐧𝛑𝐱 𝛙𝒏 𝐱 = 𝐁 𝐬𝐢𝐧 (31) 𝐚 To find 𝐁, we employed the normalisation condition, i.e. 𝐚 න 𝛙𝐧 ∗ 𝐱 𝛙𝐧 𝐱 𝒅𝒙 = 𝟏 (32) 𝟎 𝐚 𝟐 𝐧𝛑𝐱 𝟐 𝐁 න 𝐬𝐢𝐧 𝒅𝒙 = 𝟏 𝟎 𝐚 𝟐 𝐁= 𝒂 Thus, the normalised wavefunction for a particle in a 1-D box is 𝟐 𝐧𝛑𝐱 (33) 𝛙𝒏 = 𝐬𝐢𝐧 𝒂 𝐚 𝐄𝐧 𝐧=𝟑 𝐧=𝟑 Node of the wavefunction: the point where 𝛙 = 𝟎. The higher the energy, the more nodes in a wavefunction. 𝐧 =2 𝐧 =2 The number of nodes = (n − 1). 𝐧=𝟏 𝐧=𝟏 Fig. (1): The wavefunctions. Fig. (2): The probability density for a particle. 2a Wavelength = n Observation: 𝐧𝟐 𝐡𝟐 𝐄𝐧 = 𝟐 , 𝐧 = 𝟏, 𝟐, 𝟑 𝟖𝐦𝐚 (𝐧 + 𝟏)𝟐 𝐡𝟐 − 𝐧𝟐 𝐡𝟐 ∆𝐄= 𝟖𝐦𝐚𝟐 As 𝐚 gets larger (or 𝐦 gets larger), the energy levels become close together. In the limit, the energy levels are so close together that the quantisation may not be noticeable. In this case, the system behaves classically. The results is consistent with Bohr’s correspondence principle which states that at certain limits, quantum mechanical behaviour becomes classical behaviour. Examples: Calculation of ∆ 𝐄 Calculate the energy difference between n = 1 and n = 2 levels for an electron (m = 9.1 × 10−31 kg) confined to a 1-D box of length 4.0 × 10−10 m. What wavelength corresponds to a spectral transition between these levels? Solution: n2 h2 (n + 1)2 h2 − n2 h2 En = ; ∆E= 8ma2 8ma2 3 6.626 × 10−34 𝐽𝑠 2 ∆ E = E2 − E1 = 3E1 = = 1.13 × 10−18 J 8 9.1 × 10−31 𝑘𝑔 4.0 × 10−10 m 2 𝐜 𝐡𝐜 6.626 × 10−34 𝐽𝑠 (2.998 × 108 𝑚𝑠 −1 ) 𝝀 = = = = 1.76 × 10−7 𝑚 = 176 𝑛𝑚 𝐯 ∆𝐄 1.13 × 10−18 J (uv region) Examples: Calculation of ∆ 𝐄 Calculate the energy difference between n = 1 and n = 2 levels for a marble (m = 1 𝑔) confined to a 1-D box of length 0.10 𝑚. What wavelength corresponds to a spectral transition between these levels? Solution: n2 h2 (n + 1)2 h2 − n2 h2 En = ; ∆E= 8ma2 8ma2 3 6.626 × 10−34 𝐽𝑠 2 ∆ E = E2 − E1 = 3E1 = 2 = 5.48 × 10−62 J (Too small to be measurable) 8 0.001 𝑘𝑔 0.10 m 𝐜 𝐡𝐜 6.626 × 10−34 𝐽𝑠 (2.998 × 108 𝑚𝑠 −1 ) 𝝀 = = = = 1.20 × 1037 𝑚 𝐯 ∆𝐄 5.48 × 10−62 J (The wavelength is greater than the distance of any observed star! No quantization can be detected) Examples: Calculation: Orthogonality 2 πx 2 2πx Consider two wavefunctions for particle-in-a box system: ψ1 = sin ; ψ2 = sin , 𝑎 a 𝑎 a Show that ψ1 and ψ2 are orthogonal. Solution: a a a ∗ 2 πx 2 2πx Prove that ‫׬‬0 ψ1 ∗ ψ2 𝑑𝑥 =0 න ψ1 ψ2 dx = න a sin a a sin a dx 0 0 Formulae: 4 a 2 πx πx = න sin cos dx sin2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 a 0 a a 1 න𝑠𝑖𝑛2 𝑘𝑥 𝑐𝑜𝑠𝑘𝑥 𝑑𝑥 = 𝑠𝑖𝑛3 𝑘𝑥 a 3𝑘 4 a πx 3 = sin =0 a 3π a 0 This prove that ψ1 and ψ2 are orthogonal. Examples: Calculation: Probability A particle is in a linear box of 1 nm in length. What is the probability that it is between 0.45 and 0.55 nm in the ground state? 0.55 2 πx 2 πx P 0.45 ≤ x ≤ 0.55 = න sin sin dx Solution: 0.45 a a a a Calculate P (0.45 ≤ x ≤ 0.55). 2 0.55 2 πx 1 0.55 2πx = න sin dx = න 1 − cos dx a 0.45 a a 0.45 a 𝑎 ‫ 𝑎׬‬ψ1 ∗ ψ1 dx ; n = 1 2 1 0.55 1 𝑎 2πx = 𝑥− sin a 2π a 0.45 Formulae: cos 2𝜃 = 1 − 2𝑠𝑖𝑛2𝜃 1 1 = 0.55 − (−0.309) − 0.45 − (0.309) 2π 2π = 0.198 Examples: Calculation: Expected value Calculate the < px > for a particle in 1-D box. Solution: a px = න ψ∗ p ෞx ψ dx Operation for p ෞx : 0 𝜕 𝑎 2 nπx 𝜕 2 nπx − iħ =න sin − iħ sin dx 𝜕x a a 𝜕x a a 0 ෡ ψ dτ ‫ ׬‬ψ∗ R 𝑎 2nπ nπx nπx R = = − iħ න sin. cos dx ‫ ׬‬ψ∗ ψ dτ a 2 0 a a Formulae: nπ 𝑎 2nπx sin2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 = − iħ 2 න sin dx a 0 a =0 Examples: Calculation: Expected value a Calculate the < p2x > for a particle in 1-D box. ෢x2 ψ dx p2x = න ψ∗ p 0 a 2 2 nπx 𝜕 2 nπx Solution: =න sin − ħ2 2 sin dx 0 a a 𝜕x a a ෢x2 : Operator for p a 2 2ħ2 nπx n2 π2 nπx 𝜕 =− න sin. 2 −sin dx − ħ2 2 a 0 a a a 𝜕x 2ħ2 n2 π2 a nπx 2 = න sin dx a a2 0 a ෡ ψ dτ ‫ ׬‬ψ∗ R ħ 2 2 n π 2 a nπx R = = න 1 − cos dx ‫ ׬‬ψ∗ ψ dτ a a 2 0 a a ħ2 n2 π2 nπx Formulae: = x − sin a a2 a 0 sin2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 n2 ħ2 π2 = a2 Examples: Using eigenvalue equation, calculate the < p2x > for a particle in 1-D box. Solution: Operation for p෢x2 : 𝜕 2 − ħ2 2 𝜕x 𝜕 2 2 nπx ෢x2 ψ = − ħ2 p sin We obtain the eigenvalue: 𝜕x 2 a a n2 ħ2 π2 2 2 n π 2 nπx p2x = 2 a2 = −ħ − 2 sin a a a 𝑛ħπ px = ± n2 ħ2 π2 2 nπx a = sin a2 a a Examples: Calculation: Expected value Calculate the < x > for a particle in 1-D box. Solution: a x = න ψ∗ xො ψ dx Operation for xො : x 0 𝑎 2 nπx 2 nπx ‫׬‬ ෡ψ ψ∗ R dτ =න sin x sin dx 0 a a a a R = ‫ ׬‬ψ∗ ψ dτ 2 𝑎 2 2 nπx = න x. sin dx Formulae: a 0 a 2 𝑥 𝑥𝑠𝑖𝑛2𝑘𝑥 𝑐𝑜𝑠2𝑘𝑥 නx 2. 𝑠𝑖𝑛2 𝑘𝑥 𝑑𝑥 = − − a 4 4𝑘 8𝑘 2 = 2 Examples: Calculation: Expected value Calculate the < x 2 > for a particle in 1-D box. Solution: a x 2 = න ψ∗ x෢2 ψ dx ෢2 : x 2 Operation for x 0 𝑎 2 nπx 2 2 nπx ෡ ψ dτ =න sin x sin dx ‫ ׬‬ψ∗ R 0 a a a a R = ‫ ׬‬ψ∗ ψ dτ 2 𝑎 2 2 nπx Formulae: = න x. sin dx a 0 a න𝑥 2. 𝑠𝑖𝑛2 𝑘𝑥 𝑑𝑥 a 2 n2 π2 1 𝑥3 𝑥2 1 𝑥𝑐𝑜𝑠2𝑘𝑥 = − = − − 𝑠𝑖𝑛2𝑘𝑥 − nπ 3 2 4 4𝑘 8𝑘 3 4𝑘 2 Examples: Application: Electronic spectra of conjugated 𝛑-electron system The system may be represented by [(CH3)2CH=CH=(CH=CH)n-2C(CH3)2]+. The box length is taken to be the distance between two terminal carbons taking part in the delocalisation of the π electrons plus a penetration term, 𝑝. This term is included in recognition of the fact that the true potential does not change suddenly from zero to infinity at the terminal carbons. The polyenylic chains are depicted as follows: The box length is given by: 𝐚 = 𝒍𝒄 𝒄𝒐𝒔𝟑𝟎𝐨 𝒏𝒄 − 𝟏 + 𝟐𝒑 where 𝑛𝑐 = the number of carbons in the conjugated system 𝑙𝑐 = the C-C bond length The parameter 𝑝 is treated as an empirical parameter to be derived from experimental results. By making use of the particle-in-a-box treatment, the energy levels are n2 h2 En = 8m 𝑙𝑐 𝑐𝑜𝑠30o 𝑛𝑐 − 1 + 2𝑝 2 From the electronic transition from the nth to (n+1)th levels, the energy change is 𝑛 + 1 2 h2 − n2 h2 ∆En = 8m 𝑙𝑐 𝑐𝑜𝑠30o 𝑛𝑐 − 1 + 2𝑝 2 2𝑛 + 1 2 h2 = 8m 𝑙𝑐 𝑐𝑜𝑠30o 𝑛𝑐 − 1 + 2𝑝 2 𝑛𝑐 h2 = 8m 𝑙𝑐 𝑐𝑜𝑠30o 𝑛𝑐 − 1 + 2𝑝 2 Since 𝑛𝑐 − 1 n= 2 In terms of wavelength, hc 8mc 𝑙𝑐 𝑐𝑜𝑠30𝑜 𝑛𝑐 − 1 + 2𝑝 2 𝜆 = = ∆E 𝑛𝑐 h Using the values of 𝑝 = 0.14 𝑛𝑚, 𝑙𝑐 = 0.14 𝑛𝑚, the 𝜆max for the polyenylic ions are shown below: Quantum Mechanics for Simple Systems Particle in a Two-dimensional Box Consider a particle being confined to a rectangular box with sides of lengths 𝐚 and 𝐛. The potential is zero inside the box but infinite outside the box. The time-independent Schrödinger equation is: ෡ 𝛙(𝐱, 𝐲) = 𝐄𝛙(𝐱, 𝐲) 𝐇 (34) ෡ is given by where 𝐇 ħ𝟐 𝛛𝟐 𝛛𝟐 ෡ =− 𝐇 + 𝟐 𝟐𝐦 𝛛𝐱 𝟐 𝛛𝒚 Eq.(34) is a partial differential equation, and it can be solved by using the technique of separation of variables to obtain a set of ordinary differential equations. Assume that 𝛙 is a product of two functions: 𝛙(𝐱, 𝐲) = 𝛙𝒙 𝛙𝒚 (35) Substituting Eq. (35) in Eq. (34), we obtain ħ𝟐 𝛛𝟐 𝛛𝟐 − 𝟐 + 𝟐 𝛙𝒙 𝛙𝐲 = 𝐄𝛙𝐱 𝛙𝐲 (36) 𝟐𝐦 𝛛𝐱 𝛛𝒚 Dividing Eq. (36) by 𝛙𝒙 𝛙𝒚 , we obtain ħ𝟐 𝟏 𝐝𝟐 𝛙𝐱 𝟏 𝐝𝟐 𝛙 𝐲 − + =𝐄 𝟐𝐦 𝛙𝐱 𝐝𝐱 𝟐 𝛙𝐲 𝐝𝒚𝟐 Rearranging, 𝟏 𝐝𝟐 𝛙 𝐱 𝟏 𝐝𝟐 𝛙𝐲 𝟐𝐦𝐄 − 𝟐 − 𝟐 = 𝟐 (37) 𝛙𝐱 𝐝𝐱 𝛙𝐲 𝐝𝒚 ħ  𝟏 𝐝𝟐 𝛙𝐱 𝟐𝐦𝐄 𝟏 𝐝𝟐 𝛙 𝐲 + =− (37) 𝛙𝐱 𝐝𝐱 𝟐 ħ𝟐 𝛙𝐲 𝐝𝒚𝟐 Since both sides of Eq. (37) are each a function of a different independent variable and thus can be varied independently of each other. Each side must be equal to a constant. So, 𝟏 𝐝𝟐 𝛙𝐱 𝟐𝐦𝐄 𝟏 𝐝𝟐 𝛙𝐲 𝟐𝐦𝐄𝒚 𝟐 + 𝟐 =− 𝟐 = 𝛙𝐱 𝐝𝐱 ħ 𝛙𝐲 𝐝𝒚 ħ𝟐 Hence ħ𝟐 𝐝𝟐 𝛙𝐲 − 𝟐 = 𝐄𝒚 𝛙𝐲 (38) 𝟐𝐦 𝐝𝒚 and ħ𝟐 𝐝𝟐 𝛙𝐱 − = (𝐄 − 𝐄𝒚 )𝛙𝐱 = 𝐄𝐱 𝛙𝐱 (39) 𝟐𝐦 𝐝𝐱 𝟐 where 𝐄 = 𝐄𝐱 + 𝐄𝒚 Eqs. (38) and (39) are similar to Eq. (27). Hence the solutions are 𝟐 𝐧𝐱𝛑𝐱 𝐧𝐱 𝟐 𝐡𝟐 𝛙𝐱 = 𝐬𝐢𝐧 ; 𝐄𝒙 = 𝐚 𝐚 𝟖𝐦𝐚𝟐 𝟐 𝐧𝐲𝛑𝒚 𝐧𝐲 𝟐 𝐡𝟐 𝛙𝐲 = 𝐬𝐢𝐧 ; 𝐄𝐲 = 𝐛 𝐛 𝟖𝐦𝐛 𝟐 Thus, the normalised wavefunction for this system is 𝟒 𝐧𝐱𝛑𝐱 𝐧𝐲𝛑𝒚 𝛙= 𝐬𝐢𝐧 𝐬𝐢𝐧 (40) 𝐚𝐛 𝐚 𝐛 where 𝐧𝐱 = 𝟏, 𝟐, 𝟑 ….. and 𝐧𝐲 = 𝟏, 𝟐, 𝟑 ….. The allowed energy levels are 𝐡𝟐 𝐧𝐱 𝟐 𝐧𝐲 𝟐 𝐄= 𝟐 + 𝟐 𝟖𝐦 𝐚 𝐛 Degenerancy When the sides of the box are equal. i.e. 𝐚 = 𝐛, Eqs. (40) and (41) become 𝟐 𝐧𝐱𝛑𝐱 𝐧𝐲𝛑𝐲 𝛙 = 𝟐 𝐬𝐢𝐧 𝐬𝐢𝐧 (42) 𝐚 𝐚 𝐚 and 𝐡𝟐 𝐄= 𝐧𝐱 𝟐+𝐧 𝟐 𝐲 (43) 𝟖𝐦𝐚𝟐 𝐡𝟐 The state with the lowest energy (ground state) is 𝛙11 with 𝐄𝟏𝟏 =. 𝟒𝐦𝐚𝟐 𝟓𝐡𝟐 The state 𝛙12 and 𝛙21 each has the energy , ie. they have the same energy. 𝟖𝐦𝐚𝟐 The energy diagram for some of the stationary states is shown as follows: An energy level that corresponds to more than one state is said to be degenerate. The number of different states belonging to the level is the degree of degeneracy of the level. 5h2 Eg: the energy level of is twofold degenerate. 8ma2 Quantum Mechanics for Simple Systems Particle in a Three-dimensional Box Consider a particle being confined to a box with sides of lengths 𝐚, 𝐛 and 𝐜. The potential is zero inside the box but infinite outside the box. The time-independent Schrödinger equation is: ෡ 𝛙(𝐱, 𝐲, 𝐳) = 𝐄𝛙(𝐱, 𝐲, 𝐳) 𝐇 ෡ is given by where 𝐇 ħ𝟐 𝛛𝟐 𝛛𝟐 𝛛𝟐 ෡ =− 𝐇 + 𝟐+ 𝟐 𝟐𝐦 𝛛𝐱 𝟐 𝛛𝒚 𝛛𝒛 (1) Derive the expressions for 𝛙 and 𝐄. (2) In a table, write the quantum numbers, the energy levels and its degree of degeneracy, up to 9E1, where E1 is the ground state energy. Quantum Mechanics for Simple Systems Harmonic ocsillator The 1-D model harmonic oscillator is a useful model to treat the vibration of a diatomic molecule and also the vibrations of polyatomic molecules. The system possesses both kinetic energy and potential energy. Quantum Mechanics for Simple Systems Rigid Rotor This is a model treating rotation of a diatomic molecule. The rigid rotor consists of two masses m1 and m2 which are constrained to remain at a fixed distance R = r1 + r2 from each other. The system possesses kinetic energy only, and so V = 0.