PEPERIKSAAN IJAZAH SARJANA MUDA SAINS
EXAMINATION FOR THE DEGREE OF BACHELOR OF SCIENCE SESI AKADEMIK 2017/2018 : SEMESTER II
ACADEMIC SESSION 2017/2018 : SEMESTER II
SIC3019 : Kimia Kuantum Lanjutan Advanced Quantum Chemistry
Mei/Jun 2018 Masa: 2 jam
May/June 2018 Time: 2 hours
ARAHAN KEPADA CALON:
INSTRUCTIONS TO CANDIDATES:
Jawab SEMUA soalan.
Answer ALL questions.
(Kertas soalan ini mengandungi 11 soalan dalam 9 halaman yang bercetak) (This question paper consists of 11 questions on 9 printed pages)
2/9 1. (a) Dapatkan fungsi-fungsi eigen dan vektor-vektor eigen yang
dinormalkan bagi
Find the eigenvalues and normalized eigenvectors of 𝐀 = (2 2
2 −1)
(11 markah/marks) (b) Sahkan bahawa 𝐂−𝟏𝐀𝐂 bersamaan dengan matriks pepenjuru bagi
nilai-nilai eigen.
Verify that 𝐂−𝟏𝐀𝐂 equals the diagonal matrix of eigenvalues.
(4 markah/marks) 2. Operator peningkatan dan penurunan bagi pengayun harmonik satu dimensi
telah dibuktikan sebagai
The raising and lowering operators for the one-dimensional harmonic-oscillator were proved to be
𝐴̂+𝐴̂− = 𝐻̂ −1
2ℎ𝜈, 𝐴̂−𝐴̂+ = 𝐻̂ +1
2ℎ𝜈, [𝐴̂+, 𝐴̂−] = −ℎ𝜈 [𝐻̂, 𝐴̂+] = ℎ𝜈𝐴̂+, [𝐻̂, 𝐴̂−] = −ℎ𝜈𝐴̂−
Tunjukkan bahawa sesungguhnya 𝐴̂+ dan 𝐴̂− adalah operator tangga dan nilai-nilai eigen dijarakkan pada selang ℎ𝜈. Oleh kerana kedua-dua tenaga kinetik dan tenaga keupayaan adalah bukan negatif, kita jangkakan tenaga nilai-nilai eigen adalah bukan negatif. Demikian itu ianya mesti terdapat satu keadaan dengan tenaga minimum. Lakukan operasi ke atas fungsi gelombang bagi keadaan ini pada permulaannya dengan 𝐴̂− dan kemudian dengan 𝐴̂+ dan tunjukkan bahawa nilai eigen terendah adalah 1 2⁄ ℎ𝜈.
Akhirnya, simpulkan bahawa
Show that 𝐴̂+ and 𝐴̂− are indeed ladder operators and that the eigenvalues are spaced at intervals of ℎ𝜈. Since both the kinetic energy and the potential energy are nonnegative, we expect the energy eigenvalues to be nonnegative. Hence there must be a state of minimum energy. Operate on the wave function for this state first with 𝐴̂− and then with 𝐴̂+ and show that the lowest-energy eigenvalue is 1 2⁄ ℎ𝜈. Finally, conclude that
𝐸 = (𝑛 +1
2) ℎ𝜈, 𝑛 = 0,1,2, …
(11 markah/marks)
3/9 3. Operator 𝐋̂2 and 𝐿̂𝑧 adalah diagonal dalam asas |𝑙, 𝑚⟩:
The operators 𝑳̂2 and 𝐿̂𝑧 are diagonal in the basis |𝑙, 𝑚⟩:
⟨𝑙′, 𝑚′|𝐋̂2|𝑙, 𝑚⟩ = ℏ𝑙(𝑙 + 1)𝛿𝑙′,𝑙𝛿𝑚′,𝑚
⟨𝑙′, 𝑚′|𝐿̂𝑧|𝑙, 𝑚⟩ = ℏ𝑚𝛿𝑙′,𝑙𝛿𝑚′,𝑚 Pertimbangkan kes dimana 𝑙 = 1
Consider the case where 𝑙 = 1
(a) Dapatkan perwakilan matriks bagi operator 𝐋̂2 dan 𝐿̂𝑧. Find the matrices representing the operators 𝐋̂2 and 𝐿̂𝑧.
(5 markah/marks) (b) Dapatkan vektor eigen bersama bagi 𝐋̂2 dan 𝐿̂𝑧, dan sahkan bahawa
mereka membentuk satu asas yang ortonormal dan lengkap.
Find the joint eigenvectors of 𝐋̂2 and 𝐿̂𝑧, and verify that they form an orthonormal and complete basis.
(9 markah/marks) 4. Keadaan eigen |1 2⁄ , 1 2⁄ ⟩ dan |1 2⁄ , −1 2⁄ ⟩ bagi operator Hermitian 𝑆̂𝑧
membentuk satu set yang lengkap dan ortonormal, dan sebarang fungsi spin satu elektron boleh ditulis sebagai 𝑐1|1 2⁄ , 1 2⁄ ⟩ + 𝑐2|1 2⁄ , −1 2⁄ ⟩. Bagi perwakilan yang menggunakan |1 2⁄ , 1 2⁄ ⟩ dan |1 2⁄ , −1 2⁄ ⟩ sebagai asas,
The eigenstates |1 2⁄ , 1 2⁄ ⟩ and |1 2⁄ , −1 2⁄ ⟩ of the Hermitian operator 𝑆̂𝑧 form a complete, orthonormal set, and any one-electron spin function can be written as 𝑐1|1 2⁄ , 1 2⁄ ⟩ + 𝑐2|1 2⁄ , −1 2⁄ ⟩. For the representation that uses |1 2⁄ , 1 2⁄ ⟩ and
|1 2⁄ , −1 2⁄ ⟩ as the bases,
(a) Tulis vektor lajur (spinor) yang mewakili keadaan |1 2⁄ , 1 2⁄ ⟩ , |1 2⁄ , −1 2⁄ ⟩ dan 𝑐1|1 2⁄ , 1 2⁄ ⟩ + 𝑐2|1 2⁄ , −1 2⁄ ⟩.
Write down the column vectors (spinors) that correspond to the states
|1 2⁄ , 1 2⁄ ⟩ , |1 2⁄ , −1 2⁄ ⟩ and 𝑐1|1 2⁄ , 1 2⁄ ⟩ + 𝑐2|1 2⁄ , −1 2⁄ ⟩.
(4 markah/marks)
4/9 (b) Tunjukkan bahawa matriks bagi mewakli 𝑆̂𝑥, 𝑆̂𝑦 dan 𝑆̂𝑧 adalah
Show that the matrices that correspond to 𝑆̂𝑥, 𝑆̂𝑦 and 𝑆̂𝑧 are 𝑆̂𝑖 =ℏ
2𝜎𝑖
dimana 𝑖 = 𝑥, 𝑦, 𝑧 dan 𝜎𝑖 adalah spin matriks Pauli diberi oleh where 𝑖 = 𝑥, 𝑦, 𝑧 and 𝜎𝑖 are the Pauli spin matrices given by
𝜎𝑥 = (0 1
1 0), 𝜎𝑦 = (0 −𝑖
𝑖 0), 𝜎𝑥 = (1 0 0 −1)
(10 markah/marks) 5. Asas {|𝑗1, 𝑗2; 𝑚1, 𝑚2⟩} dan {|𝐽, 𝑀⟩} boleh dikaitkan satu sama lain dari
transformasi berikut
The {|𝑗1, 𝑗2; 𝑚1, 𝑚2⟩} and {|𝐽, 𝑀⟩} bases can be connected by means of a transformation as follows,
|𝐽, 𝑀⟩ = ∑ ⟨𝑗1, 𝑗2; 𝑚1, 𝑚2|𝐽, 𝑀⟩|𝑗1, 𝑗2; 𝑚1, 𝑚2⟩
𝑚1𝑚2
di mana perhubungan rekursi bagi pekali-pekali Clebsch–Gordan:
where the recursion relations for the Clebsch–Gordan coefficients:
√(𝐽 ∓ 𝑀)(𝐽 ± 𝑀 + 1)⟨𝑗1, 𝑗2; 𝑚1, 𝑚2|𝐽, 𝑀 ± 1⟩
= √(𝑗1±𝑚1)(𝑗1∓ 𝑚1+ 1)⟨𝑗1, 𝑗2; 𝑚1∓ 1, 𝑚2|𝐽, 𝑀⟩
+√(𝑗2± 𝑚2)(𝑗2∓ 𝑚2+ 1)⟨𝑗1, 𝑗2; 𝑚1, 𝑚2∓ 1|𝐽, 𝑀⟩
Dapatkan pekali-pekali Clebsch–Gordan berkaitan dengan nilai terendah 𝐽 dan 𝑀 bagi pengabungan spin elektron dan proton satu atom hidrogen dalam keadaan asasnya.
Find the Clebsch–Gordan coefficients associated with the lowest values of 𝐽 and 𝑀 for the coupling of the spins of the electron and the proton of a hydrogen atom in its ground state.
(8 markah/marks) 6. Satu operator 𝐵̂ mempunyai bentuk
An operator 𝐵̂ has the form
𝐵̂ = ∑ 𝑓̂𝑖+ ∑ ∑ 𝑔̂𝑖𝑗
𝑗>1 𝑛−1
𝑖=1 𝑛
𝑖=1
.
Jadual berikut memberi formula-formula Condon–Slater, The following table gives the Condon–Slater formulas,
5/9 dimana kamiran n-elektron ⟨𝐷′|𝐵̂|𝐷⟩ boleh diturunkan kepada hasil tambah sesetengah kamiran satu- dan dua-elektron dan 𝐷 adalah penentu Slater spin-orbital dinormalkan 𝑢𝑖. Guna peraturan-peraturan Condon–Slater bagi membuktikan pengortonormalan bagi dua penentu Slater n-elektron bagi spin-orbital ortonormal.
where the n-electron integral ⟨𝐷′|𝐵̂|𝐷⟩ can be reduced to sums of certain one- and two-electron integrals and 𝐷 is Slater determinant of orthonormal spin-orbitals 𝑢𝑖. Use the Condon–Slater rules to prove the orthonormality of two n-electron Slater determinants of orthonormal spin-orbitals.
(5 markah/marks) 7. Paras sifar bagi tenaga keupayaan adalah sebarangan dan kita sentiasa boleh menambah satu pemalar 𝐶 kepada fungsi tenaga keupayaan 𝑉. Jika kita menambah 𝐶 kepada 𝑉, nyatakan apa yang berlaku kepada setiap yang berikut bagi satu keadaan pegun: 〈𝑉〉, 〈𝑇〉, 𝐸. Adakah keputusan-keputusan ini bertentangan dengan teorem virial? Terangkan jawapan anda. [Petunjuk:
2〈𝑇〉 = 𝑛〈𝑉〉]
The zero level of potential energy is arbitrary and we can always add a constant 𝐶 to the potential energy function 𝑉. If we add 𝐶 to 𝑉, state what happens to each of the following for a stationary state: 〈𝑉〉, 〈𝑇〉, 𝐸. Do these results contradict the virial theorem? Explain your answer. [Hint: 2〈𝑇〉 = 𝑛〈𝑉〉]
(5 markah/marks)
6/9 8. Bezakan
Differentiate
𝐻̂ = 𝐻̂0+ 𝜆𝐻̂′
dan and
𝐸𝑛 = 𝐸𝑛(0)+ 𝜆𝐸𝑛(1)+ 𝜆2𝐸𝑛(2)+ ⋯ + 𝜆𝑘𝐸𝑛(𝑘)+ ⋯
terhadap 𝜆, gantikan hasil-hasilnya ke dalam teorem umum Hellmann–
Feynman, dan biarkan 𝜆 = 0 bagi menerbitkan persamaan teori pengusikan 𝐸𝑛(1) = ⟨𝜓𝑛(0)|𝐻̂′|𝜓𝑛(0)⟩.
with respect to 𝜆, substitute the results into the generalized Hellmann–Feynman theorem, and let 𝜆 = 0 to derive the perturbation-theory equation 𝐸𝑛(1)=
⟨𝜓𝑛(0)|𝐻̂′|𝜓𝑛(0)⟩.
[Petunjuk/𝐻𝑖𝑛𝑡: 𝜕𝐸𝑛
𝜕𝜆 = ∫ 𝜓𝑛∗𝜕𝐻̂
𝜕𝜆 𝜓𝑛𝑑𝜏]
(6 markah/marks) 9. Gambarajah 1 berikut menunjukkan sistem koordinat bagi satu molekul
diatomik dimana asalan adalah pada O.
The following Figure 1 shows the coordinate system for a diatomic molecule where the origin is at O.
Gambarajah 1 Figure 1
7/9 Gambarajah 2 berikut menunjukkan keratan rentas kawasan pengikatan dan antipengikatan dalam satu molekul homonuklear diatomik.
The following Figure 2 shows cross section of binding and antibinding regions in a homonuclear diatomic molecule.
Gambarajah 2 Figure 2
Nilai 𝑅𝑒 bagi keadaan asas elektronik HF adalah 0.92 Å. Permukaan yang menutupi kawasan antipengikatan “bersebelahan” proton dalam molekul ini bersilang dengan paksi internuklear pada dua titik, satu daripadanya adalah lokasi proton. Kira jarak antara dua titik persilangan ini bagi HF.
The 𝑅𝑒 value for the ground electronic states of HF is 0.92 Å. The surface enclosing the antibinding region “behind” the proton in this molecule intersects the internuclear axis at two points, one of which is the proton location. Calculate the distance between these two points of intersection for the HF.
[Petunjuk/Hint:
Menurut teorem elektrostatik Hellmann–Feynman, komponen 𝑧 bagi daya efektif ke atas nukleus 𝑎 disebabkan elemen cas elektronik dalam kawasan tak terhingga sekitar (𝑥, 𝑦, 𝑧) adalah
According to the Hellmann–Feynman electrostatic theorem, the 𝑧 component of the effective force on nucleus 𝑎 due to the element of electronic charge in the infinitesimal region about (𝑥, 𝑦, 𝑧) is
𝑍𝑎𝜌 (cos 𝜃𝑎
𝑟𝑎2 ) 𝑑𝑥𝑑𝑦𝑑𝑧.
Biarkan 𝑑 sebagai jarak dikehendaki antara dua titik persilangan ini. Pada titik kiri persilangan, Gambarajah 1 dan 2 memberikan 𝑟𝑎 = 𝑑, 𝑟𝑏 = 𝑑 + 𝑅𝑒, 𝜃𝑎 = 𝜋, dan 𝜃𝑏 = 0.
Let 𝑑 be the desired distance between these two intersection points. At the left intersection point, Figures 1 and 2 give 𝑟𝑎 = 𝑑, 𝑟𝑏= 𝑑 + 𝑅𝑒, 𝜃𝑎 = 𝜋, and 𝜃𝑏= 0.]
(7 markah/marks)
8/9 10. Operator Hartree-Fock 𝐹̂ adalah
Hartree-Fock operator 𝐹̂ is
𝐹̂(1) = 𝐻̂𝑐𝑜𝑟𝑒(1) + ∑[2𝐽̂𝑗(1) − 𝐾̂𝑗(1)]
𝑛/2
𝑗=1
𝐻̂𝑐𝑜𝑟𝑒(1) = −1
2∇12− ∑𝑍𝛼 𝑟1𝛼
𝛼
dimana operator Coulomb 𝐽̂𝑗 dan operator penukaran 𝐾̂𝑗 ditakrifkan sebagai where the Coulomb operator 𝐽̂𝑗 and the exchange operator 𝐾̂𝑗 are defined by
𝐽̂𝑗(1)𝑓(1) = 𝑓(1) ∫|𝜙𝑗(2)|2 1 𝑟12𝑑𝑣2 𝐾̂𝑗(1)𝑓(1) = 𝜙𝑗(1) ∫𝜙𝑗∗(2)𝑓(2)
𝑟12 𝑑𝑣2
dimana 𝑓 adalah fungsi sebarangan dan kamiran-kamiran adalah kamiran tentu terhadap keseluruhan ruang. Operator tenaga kinetik dan operator tenaga keupayaan dalam teras 𝐻̂𝑐𝑜𝑟𝑒 telah dibuktikan Hermitian. Buktikan bahawa operator Hartree–Fock satu elektron adalah Hermitian.
where 𝑓 is an arbitrary function and the integrals are definite integrals over all space.
The kinetic-energy operator and the potential-energy operator in core 𝐻̂𝑐𝑜𝑟𝑒 were proved to be Hermitian. Prove that the one-electron Hartree–Fock operator is Hermitian.
(8 markah/marks) 11. Sahkan bahawa Hamiltonian Kohn-Sham satu elektron ℎ̂KS(1) dalam
Verify that the one-electron Kohn–Sham Hamiltonian ℎ̂𝐾𝑆(1) in ℎ̂KS(1)𝜃𝑖KS(1) = 𝜀𝑖KS𝜃𝑖KS(1)
dimana where
ℎ̂KS(1) = −1
2∇12− ∑𝑍𝛼 𝑟1𝛼
𝛼
+ ∫𝜌(𝐫𝟐)
𝑟12 𝑑𝐫𝟐+ 𝑣𝑥𝑐(1), 𝜌 = ∑|𝜃𝑖KS|2
𝑛
𝑖=1
adalah sama dengan operator Fock is the same as the Fock operator
𝑓̂(1) = −1
2∇12 − ∑𝑍𝛼 𝑟1𝛼
𝛼
+ ∑[𝑗̂𝑗(1) − 𝑘̂𝑗(1)]
𝑛
𝑗=1
9/9 bagi elektron 1 dalam satu molekul 𝑛 elektron, kecuali operator-operator pertukaran dalam operator Fock ditukar oleh 𝑣𝑥𝑐.
for electron 1 in an 𝑛-electron molecule, except that the exchange operators in the Fock operator are replaced by 𝑣𝑥𝑐.
[Petunjuk: Guna persamaan dalam soalan 10]
[Hint: Use equation in question 10]
(7 markah/marks) TAMAT
END
