Physical Chemistry
13. Nuclear Magnetic ResonanceRuifeng ZHOU

Faculty of Science &Institute for the Advancement of Higher Education
Hokkaido University

Table of Content

1. Magnetic resonance
2. Chemical shifts
3. Spin-spin coupling: perturbation theory
4. Spin-spin coupling: variational principle

11. Magnetic resonance• For a spin ½ particle, there are 2 eigenspinors, |±⟩,which are the eigenstates of 𝑆𝑆̂𝑧𝑧 .
ℏ̂𝑆𝑆𝑧𝑧 |±⟩ = ± |±⟩2
• The eigenspinors and observables can be
represented by matrices.
|+⟩ = 10|−⟩ = 01
ℏ 1 0𝑆𝑆̂𝑧𝑧 =2 0 −121. Magnetic resonance• In a magnetic field, 𝓑𝓑, the Hamiltonian of a spin is� = −�
𝐻𝐻𝝁𝝁 � 𝓑𝓑 = −𝛾𝛾𝑺𝑺� � 𝓑𝓑
• 𝛾𝛾 is the gyromagnetic ratio.
• Suppose 𝓑𝓑 is a uniform constant field in z direction,
𝛾𝛾𝛾𝛾 1 0� = −𝛾𝛾𝛾 𝑆𝑆̂𝑧𝑧 = −𝐻𝐻
2 0 −1� 𝑆𝑆̂𝑧𝑧 = 0. The eigenspinors are stationary states.
• 𝐻𝐻,• The energies of the eigenspinors are𝛾𝛾𝛾𝛾
𝐸𝐸± = ∓2
• The energy difference is ∆𝐸𝐸 = 𝛾𝛾𝛾𝛾.

341. Magnetic resonance𝜃𝜃 𝑖𝑖𝑖𝑖𝑖𝑖𝑖
cos 𝑒𝑒 22𝝌𝝌 𝑡𝑡 =
𝜃𝜃 −𝑖𝑖𝑖𝑖𝑖𝑖𝑖
sin 𝑒𝑒 22ℏ𝑆𝑆𝑧𝑧 = cos 𝜃𝜃2ℏ𝑆𝑆𝑥𝑥 = sin 𝜃𝜃 cos 𝛾𝛾ℬ𝑡𝑡
2ℏ
𝑆𝑆𝑦𝑦 = − sin 𝜃𝜃 sin 𝛾𝛾ℬ𝑡𝑡
2• 𝛾𝛾ℬ is the Larmor frequency(of Larmor presession).

BB’1. Magnetic resonance• Suppose there is an EM radiation with magnetic
field polarized in x direction, amplitude ℬ′ andangular frequency 𝜔𝜔.
• The time-dependent perturbation is′𝛾𝛾𝛾ℬ0 1
� = −𝛾𝛾𝑆𝑆̂𝑥𝑥 ℬ′ cos 𝜔𝜔𝜔𝜔 = −cos 𝜔𝜔𝜔𝜔𝐻𝐻′
2 1 0𝛾𝛾𝛾ℬ′ 0 1
𝑉𝑉� = −2 1 0• Thus,′𝛾𝛾𝛾ℬ′𝛾𝛾𝛾ℬ0 1 0�
𝑉𝑉+− = + 𝑉𝑉 − = −=−1 0
1 0 122≠05

1. Magnetic resonance In a constant magnetic field, a spin absorbs EMradiation to be excited.
• The conservation of energy in the absorption isℏ𝜔𝜔 = ∆𝐸𝐸 = 𝛾𝛾𝛾ℏ The driving frequency should be equal to theLarmor frequency.
• If the spins are from nuclei, the absorption is called
nuclear magnetic resonance (NMR).
• If the spins are from electrons, it is called electron
spin resonance (ESR).

61. Magnetic resonance• For instance, the gyromagnetic ratio of proton is𝛾𝛾H = 267.513 × 106 s−1 � T −1
• If the magnetic flux density is ℬ = 1T.
𝜔𝜔 = 𝛾𝛾H ℬ = 267.513 × 106 rad � s −1
• In NMR, frequency is usually used instead ofangular frequency.
𝜔𝜔= 42.576MHz
𝜈𝜈 =2𝜋𝜋• In practice, proton NMR operates between 60MHz
and 750MHz (radio frequency).

7

2. Chemical shifts• In classical mechanics, electrons move circularly ina constant magnetic field. The current of electron
motion produces a magnetic field which offsets theexternal magnetic field.
• The result is the same as in quantum mechanics.
××××××××××××××××××××××××××××××8

2. Chemical shifts• The offset of magnetic flux is proportional to theexternal magnetic flux.
∆ℬ = −𝜎𝜎ℬ• 0 < 𝜎𝜎 ≪ 1 is the shielding constant.
• 𝜎𝜎 of a nucleus is determined by the electronic
density around it, which is called its chemical
environment.
• 𝜎𝜎 of a proton decreases with the electronegativityof the atom bonded to the hydrogen atom.
• For protons bonded to a same element of atoms, 𝜎𝜎depend on the charge of those atoms.

9

2. Chemical shifts

10• For example, 𝜎𝜎 of proton in H2O is less than CH4.
• 𝜎𝜎 of proton in CH3Cl is less than CH4.
• The modified Larmor frequency is𝜔𝜔 = 𝛾𝛾 1 − 𝜎𝜎 ℬ
• Usually, the driving frequency is fixed, and themagnetic field is tuned to find resonance.
𝜔𝜔𝜎𝜎 = 1 −𝛾𝛾𝛾
• Since 𝜎𝜎 ≪ 1, the accuracies of ℬ and 𝜔𝜔 are crucial.
• However, it is difficult to measure them very
accurately.

2. Chemical shifts

11• An internal reference is used with the sample toeliminate measurement errors.
• For NMR of proton, tetramethylsilane (TMS) isusually used.
• TMS has 12 chemically equivalent protons.

• The difference in Larmor frequencies from theproton in the reference is called the chemical shifts.

2. Chemical shifts12𝜔𝜔 − 𝜔𝜔TMS 𝛾𝛾 1 − 𝜎𝜎 ℬ − 𝛾𝛾 1 − 𝜎𝜎TMS ℬ≈ 𝜎𝜎TMS − 𝜎𝜎𝛿𝛿 ≡=𝜔𝜔TMS
𝛾𝛾 1 − 𝜎𝜎TMS ℬ• In TMS, the carbon atoms have (almost) the mostnegative charge of all the organic molecules.
• The protons of TMS have the highest 𝜎𝜎.
 All the other organic molecules have positive 𝛿𝛿.
• In other word, all the protons in other organicmolecules needs lower magnetic field to reachresonance at the same frequency. It is calleddownfield.

13

2. Chemical shifts• The unit of 𝛿𝛿 is ppm (parts per million, 10−6 )CH4CH3 ClCH2 Cl2CHCl3
𝛿𝛿 = 0.23𝛿𝛿 = 3.05 𝛿𝛿 = 5.33 δ = 7.26
CH3 I𝛿𝛿 = 2.16

CH3 Cl𝛿𝛿 = 3.05

CH3 Br𝛿𝛿 = 2.68

CH3 CH2 Cl
𝛿𝛿 = 1.42

CH3 Cl𝛿𝛿 = 3.05

CH3 Fδ = 4.26

CH3 CH2 CH2 Cl𝛿𝛿 = 1.04

2. Chemical shifts• Which peaks correspond to a and b, respectively?

14

3. Spin-spin coupling: perturbation theory• Sometimes, in the proton NMR spectra, multiple
peaks can be observed for a single type of proton.
• For example,• 2-pentanone• The split of peaks is caused by spin-spin coupling(J-coupling).

15

3. Spin-spin coupling: perturbation theory• Consider a 2-proton system.
• The total spin states are the combinations of theeigenspinors of spin 1 and spin 2.
• If the protons are chemically inequivalent, they aredistinguishable.
• The total spin states are𝜒𝜒1 =↑↑𝜒𝜒2 =↑↓𝜒𝜒3 =↓↑𝜒𝜒4 =↓↓• So far, the protons are independent.

16

3. Spin-spin coupling: perturbation theory• The energy of the system in a constant magnetic
field in z direction is𝛾𝛾 1 − 𝜎𝜎1 ℬℏ 𝛾𝛾 1 − 𝜎𝜎2 ℬℏ0𝐸𝐸1 = −−22𝜎𝜎1 + 𝜎𝜎2= −𝛾𝛾𝛾𝛾 1 −2𝛾𝛾𝛾𝛾0𝐸𝐸2 =𝜎𝜎1 − 𝜎𝜎22𝛾𝛾𝛾𝛾00
𝐸𝐸3 =𝜎𝜎2 − 𝜎𝜎1 = −𝐸𝐸22𝜎𝜎1 + 𝜎𝜎200= −𝐸𝐸1
𝐸𝐸4 = 𝛾𝛾𝛾𝛾 1 −217183. Spin-spin coupling: perturbation theory
0𝐸𝐸30− 𝐸𝐸10= 𝐸𝐸4= 𝛾𝛾𝛾𝛾 1 − 𝜎𝜎10
𝐸𝐸20− 𝐸𝐸10= 𝐸𝐸4= 𝛾𝛾𝛾𝛾 1 − 𝜎𝜎20− 𝐸𝐸20− 𝐸𝐸30

𝐸𝐸40
𝐸𝐸30
𝐸𝐸20

𝐸𝐸1

↓↓
↓↑

↑↓
↑↑

3. Spin-spin coupling: perturbation theory• A proton has a magnetic moment due to its spin,
which produces a magnetic field in its
surroundings, which interacts with other protons.
S SN NS NN S• The spin-spin coupling can be regarded as a timeindependent perturbation.
• The spin operator of nucleus is 𝑰𝑰�.
𝑗𝑗𝑗𝑗
�ss = 𝑰𝑰� � 𝑰𝑰� = 𝐼𝐼̂𝑥𝑥 𝐼𝐼̂𝑥𝑥 + 𝐼𝐼̂𝑦𝑦 𝐼𝐼̂𝑦𝑦 + 𝐼𝐼̂𝑧𝑧 𝐼𝐼̂𝑧𝑧𝐻𝐻ℏℏ• 𝑗𝑗 > 0 is a constant indicating their couplingstrength.

19

3. Spin-spin coupling: perturbation theory• The first order correction of energy is�ss 𝜒𝜒
𝐸𝐸 1 = 𝜒𝜒 𝐻𝐻• Use the properties of spin ½ particles.
ℏ𝐼𝐼̂𝑧𝑧 |±⟩ = ± |±⟩2ℏ̂𝐼𝐼𝑥𝑥 |±⟩ = |∓⟩2𝑖𝑖ℏ𝐼𝐼̂𝑦𝑦 |±⟩ = ± |∓⟩2
• To smooth the calculation, a table showing the�ss 𝜒𝜒′ can be made.
�ss 𝜒𝜒 and 𝜒𝜒 𝐻𝐻
results of all 𝐻𝐻

20

21

3. Spin-spin coupling: perturbation theory𝐼𝐼̂𝑥𝑥 𝐼𝐼̂𝑥𝑥𝐼𝐼̂𝑦𝑦 𝐼𝐼̂𝑦𝑦𝐼𝐼̂𝑧𝑧 𝐼𝐼̂𝑧𝑧�ss𝐻𝐻𝑉𝑉�↑↑ℏ2↓↓4ℏ2− ↓↓
4
ℏ2
↑↑
4𝑗𝑗ℏ
↑↑
4−𝛾𝛾ℬ ′ ↓↑+↑↓↑↓

ℏ2
↓↑
4
ℏ2
↓↑
4
ℏ2− ↑↓4𝑗𝑗ℏ
2 ↓↑−↑↓
4−𝛾𝛾ℬ ′ ↓↓+↑↑

↓↑

ℏ2↑↓
4
ℏ2
↑↓
4
ℏ2− ↓↑4𝑗𝑗ℏ
2 ↑↓−↓↑
4−𝛾𝛾ℬ ′ ↑↑+↓↓↑ ↑ = ↓ ↓ =1
↑ ↓ = ↓ ↑ =0↓↓ℏ2↑↑4
ℏ2− ↑↑
4ℏ2↓↓4𝑗𝑗ℏ↓↓
4−𝛾𝛾ℬ ′ ↑↓+↓↑

22

3. Spin-spin coupling: perturbation theory�ss𝐻𝐻↑↓

↑↑𝑗𝑗ℏ
40

↓↓0𝑗𝑗ℏ−4𝑗𝑗ℏ
200−𝛾𝛾ℬ′−𝛾𝛾ℬ′

0−𝛾𝛾ℬ′−𝛾𝛾ℬ′

↑↑↓↑
𝑉𝑉�↑↑

↑↓

↓↑
↓↓

0

↑↑

−𝛾𝛾ℬ′−𝛾𝛾ℬ′

↑↓↓↑

↓↓

𝑗𝑗ℏ
2𝑗𝑗ℏ
−
4000

↑↓
000

↓↑
0000

𝑗𝑗ℏ
4
↓↓
0−𝛾𝛾ℬ′−𝛾𝛾ℬ′
03. Spin-spin coupling: perturbation theory• Thus, the first order energy corrections are𝑗𝑗ℏ
1�ss ↑↑ =�ss 𝜒𝜒1 = ↑↑ 𝐻𝐻
𝐸𝐸1 = 𝜒𝜒1 𝐻𝐻
4𝑗𝑗ℏ
1�ss ↑↓ = −�ss 𝜒𝜒2 = ↑↓ 𝐻𝐻
𝐸𝐸2 = 𝜒𝜒2 𝐻𝐻4𝑗𝑗ℏ
1�ss ↓↑ = −�ss 𝜒𝜒3 = ↓↑ 𝐻𝐻
𝐸𝐸3 = 𝜒𝜒3 𝐻𝐻4𝑗𝑗ℏ
1�ss ↓↓ =�ss 𝜒𝜒4 = ↓↓ 𝐻𝐻
𝐸𝐸4 = 𝜒𝜒4 𝐻𝐻
4S S• Energy rises when the spins are parallel.
N N• Energy lowers when the spins are antiparallel. NS NS

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3. Spin-spin coupling: perturbation theory• The time-dependent perturbation is� ′ = −𝛾𝛾ℬ′ cos 𝜔𝜔𝜔𝜔 𝐼𝐼̂𝑥𝑥 1 − 𝜎𝜎1 + 𝐼𝐼̂𝑥𝑥 1 − 𝜎𝜎2𝐻𝐻≈ −𝛾𝛾𝐵𝐵′ 𝐼𝐼̂𝑥𝑥 + 𝐼𝐼̂𝑥𝑥 cos 𝜔𝜔𝜔𝜔 = 𝑉𝑉� cos 𝜔𝜔𝜔𝜔� .
� and 𝜒𝜒 𝑉𝑉𝜒𝜒′• The tables also show all the 𝑉𝑉𝜒𝜒

� = 0. This is in accordance with our• First, 𝜒𝜒 𝑉𝑉𝜒𝜒
previous assumption.
 Only the transitions with one spin preserved areallowed.

24

25

3. Spin-spin coupling: perturbation theory

Without ss coupling
0

𝐸𝐸40
𝐸𝐸3

0
𝐸𝐸2
0

𝐸𝐸1

With ss coupling
↓↓↓↓
↓↑

↓↑

↑↓
↑↑

↑↓

4 energies↑↑

0
𝐸𝐸40𝑗𝑗ℏ+
4𝐸𝐸3 −𝑗𝑗ℏ
4𝑗𝑗ℏ
−
4𝑗𝑗ℏ
0
𝐸𝐸1 +
40
𝐸𝐸2

26

3. Spin-spin coupling: perturbation theory𝛿𝛿1𝑗𝑗
2𝛾𝛾ℬ4↑2

𝛿𝛿2𝑗𝑗
2𝛾𝛾ℬ

𝑗𝑗
2𝛾𝛾ℬ3↑
1Chemical shift4↑
3𝑗𝑗2𝛾𝛾ℬ2↑
1

3. Spin-spin coupling: perturbation theory• 2 chemically equivalent protons (𝜎𝜎1 = 𝜎𝜎2 = 𝜎𝜎) areindistinguishable (identical) particles.
• The eigenstates are𝜒𝜒1 =↑↑1↑↓+↓↑𝜒𝜒2 =21↑↓−↓↑𝜒𝜒3 =2𝜒𝜒4 =↓↓• The unperturbed Hamiltonian is� = −𝛾𝛾 1 − 𝜎𝜎 ℬ 𝐼𝐼̂𝑧𝑧 + 𝐼𝐼̂𝑧𝑧𝐻𝐻27

3. Spin-spin coupling: perturbation theory• The energies and first order corrections of ↑↑ and ↓↓ are the same as the distinguishable protons.
0𝐸𝐸1 = −𝛾𝛾𝛾𝛾 1 − 𝜎𝜎00𝐸𝐸4 = 𝛾𝛾𝛾𝛾 1 − 𝜎𝜎 = −𝐸𝐸1𝑗𝑗ℏ
11𝐸𝐸1 = 𝐸𝐸4 =4• On the other hand,� ↑↓= 𝐻𝐻� ↓↑= 0𝐻𝐻
• Therefore,00
𝐸𝐸2 = 𝐸𝐸3 = 028

3. Spin-spin coupling: perturbation theory

11𝑗𝑗ℏ 𝑗𝑗𝑗 𝑗𝑗𝑗 𝑗𝑗𝑗�ss ↑↓+↓↑ == ↑↓+↓↑ 𝐻𝐻− + + −224224𝑗𝑗𝑗=
411
�ss ↑↓−↓↑
𝐸𝐸3 = ↑↓−↓↑ 𝐻𝐻
21𝑗𝑗ℏ 𝑗𝑗𝑗 𝑗𝑗𝑗 𝑗𝑗𝑗
3𝑗𝑗𝑗=−=− − − −2422441𝐸𝐸229

3. Spin-spin coupling: perturbation theory• The transition probabilities are↑↑ 𝑉𝑉� ↑↓+↓↑ = −𝛾𝛾ℬ′ − 𝛾𝛾ℬ′ = −2𝛾𝛾ℬ′↑↑ 𝑉𝑉� ↑↓−↓↑ = −𝛾𝛾ℬ′ + 𝛾𝛾ℬ′ = 0↓↓ 𝑉𝑉� ↑↓+↓↑ = −𝛾𝛾ℬ ′ − 𝛾𝛾ℬ′ = −2𝛾𝛾ℬ′↓↓ 𝑉𝑉� ↑↓−↓↑ = −𝛾𝛾ℬ′ + 𝛾𝛾ℬ′ = 0↑↓+↓↑ 𝑉𝑉� ↑↓−↓↑ = 0 + 0 + 0 + 0 = 0↑↑ 𝑉𝑉� ↓↓ = 030

3. Spin-spin coupling: perturbation theory� 2 = 𝜒𝜒2 𝑉𝑉𝜒𝜒
� 4 = −2𝛾𝛾ℬ′𝜒𝜒1 𝑉𝑉𝜒𝜒� 3 = 𝜒𝜒3 𝑉𝑉𝜒𝜒
� 4 =0𝜒𝜒1 𝑉𝑉𝜒𝜒31• 𝜒𝜒1 , 𝜒𝜒2 and 𝜒𝜒4 are symmetric states (triplet states).
• 𝜒𝜒3 is antisymmetric state (singlet state).
 Only the transitions preserving the symmetry of thesystem are allowed.
 In NMR of 2 chemically equivalent protons, only thetransitions between 2 of the triplet states areallowed.
• The singlet state is not accessible.

32

3. Spin-spin coupling: perturbation theory

Without ss coupling
0𝐸𝐸40

𝐸𝐸30= 𝐸𝐸2↓↓↓↓=0↑↓+↓↑
↑↓−↓↑

0↑↑

𝐸𝐸1

With ss coupling↑↓+↓↑↑↓−↓↑

1 energy↑↑0

𝐸𝐸4 +𝑗𝑗ℏ
4𝑗𝑗ℏ
4−3𝑗𝑗ℏ
40𝐸𝐸1𝑗𝑗ℏ+
43. Spin-spin coupling: perturbation theory33• 𝑛𝑛 chemically equivalent protons have 𝑛𝑛 + 1possible total spin values in z direction.
• For example, 1 proton close to 4 equivalent protons.

↑↑↑↑↑↑↑↑↓↑↑↓↓↑↓↓↓↓↓↓↓• Therefore, a proton close to 𝑛𝑛 chemicallyequivalent protons has 𝑛𝑛 + 1 peaks in NMR. It iscalled the 𝑛𝑛 + 1 rule.

3. Spin-spin coupling: perturbation theory• The 𝑘𝑘 th peak corresponds to 𝑘𝑘 − 1 spin ups.
• 𝑘𝑘 = 3:↑↑↓↓↑↑↓↑↓↑↓↓↑↓↑↑↓↓↑↓↑↓↓↑↑• The number of combination is𝑛𝑛!
𝑛𝑛=𝑘𝑘 − 1𝑘𝑘 − 1 ! 𝑛𝑛 − 𝑘𝑘 + 1 !
• The relative intensities of the peaks are binomial
coefficients, which form a pascal triangle.

34

3. Spin-spin coupling: perturbation theory

35

36

3. Spin-spin coupling: perturbation theory• 2-pentanone

0+1
2+1 3+1 ?
5 + 1?
2+12+1

4. Spin-spin coupling: variational principle

37• When the coupling is not small, perturbation theory
cannot provide accurate result.
• Perturbation theory cannot predict accurate
intensities either.
• The energies and intensities can be calculated
accurately using “variational method”.
• Any state of a 2-spin system can be expressed asa linear combination of all possibility states.
𝜒𝜒 = 𝑐𝑐1 ↑↑ +𝑐𝑐2 ↑↓ +𝑐𝑐3 ↓↑ +𝑐𝑐4 ↓↓4. Spin-spin coupling: variational principle• The Schrödinger equation of them in a magnetic
field including the spin-spin coupling is�ss 𝜒𝜒 = 𝐸𝐸𝜒𝜒
� + 𝐻𝐻𝐻𝐻�• ↑↑ is an eigenstate of 𝐻𝐻.
0�
𝐻𝐻 ↑↑= 𝐸𝐸1 ↑↑�ss .
• ↑↑ is also an eigenstate of 𝐻𝐻𝑗𝑗ℏ�ss ↑↑=↑↑𝐻𝐻4
� + 𝐻𝐻�ss .
• 𝜒𝜒1 =↑↑ is one of the eigenstates of 𝐻𝐻
𝑗𝑗ℏ
0𝐸𝐸1 = 𝐸𝐸1 +438

4. Spin-spin coupling: variational principle

� + 𝐻𝐻�ss .
• Similarly, 𝜒𝜒4 =↓↓ is another eigenstate of 𝐻𝐻𝑗𝑗ℏ0
𝐸𝐸4 = 𝐸𝐸4 +4• To find out other energies and eigenstates, takeinner product of the Schrödinger equation with ↑↓from left.
� + 𝐻𝐻�ss 𝜒𝜒 = ↑↓ 𝐸𝐸𝜒𝜒↑↓ 𝐻𝐻
• Use results from the previous table.
𝑗𝑗ℏ
𝑗𝑗ℏ
0+ 𝑐𝑐3 = 𝐸𝐸𝑐𝑐2𝑐𝑐2 𝐸𝐸2 −4239

4. Spin-spin coupling: variational principle• Similarly, take inner product with ↓↑ from left.
𝑗𝑗ℏ𝑗𝑗ℏ
0= 𝐸𝐸𝑐𝑐3𝑐𝑐2 + 𝑐𝑐3 𝐸𝐸3 −24• Those equations can be written as a matrixequation.
𝑗𝑗ℏ
𝑗𝑗ℏ
0𝐸𝐸2 − − 𝐸𝐸
𝑐𝑐242=0𝑐𝑐
𝑗𝑗ℏ
𝑗𝑗ℏ
30
𝐸𝐸3 − − 𝐸𝐸
2440

4. Spin-spin coupling: variational principle• To have nonzero solution, the secular equation is𝑗𝑗ℏ
𝑗𝑗ℏ
0𝐸𝐸2 − − 𝐸𝐸
42=0𝑗𝑗ℏ𝑗𝑗ℏ
0𝐸𝐸3 − − 𝐸𝐸24
• It is a quadratic equation of 𝐸𝐸.
𝛾𝛾ℬℏ
𝑗𝑗ℏ
𝛾𝛾ℬℏ
𝑗𝑗ℏ
𝜎𝜎1 − 𝜎𝜎2 − − 𝐸𝐸
𝜎𝜎2 − 𝜎𝜎1 − − 𝐸𝐸2424𝑗𝑗 2 ℏ2−=0441

4. Spin-spin coupling: variational principle• One of the solution is𝑗𝑗ℏ ℏ 2 2𝐸𝐸2 = − −𝛾𝛾 ℬ 𝜎𝜎1 − 𝜎𝜎2 2 + 𝑗𝑗 24 2
• Substitute𝛾𝛾ℬ 𝜎𝜎1 − 𝜎𝜎2𝑥𝑥 =𝑗𝑗
𝑗𝑗ℏ
𝑗𝑗ℏ
2
𝐸𝐸2 = −𝑥𝑥 + 1 −241𝑐𝑐3 = − 𝛾𝛾 2 ℬ2 𝜎𝜎1 − 𝜎𝜎2 2 + 𝑗𝑗 2 − 𝛾𝛾ℬ 𝜎𝜎1 − 𝜎𝜎2 𝑐𝑐2𝑗𝑗= − 𝑥𝑥 2 + 1 − 𝑥𝑥 𝑐𝑐2424. Spin-spin coupling: variational principle• For normalization,
𝑐𝑐2 2 + 𝑐𝑐3 2 = 𝑐𝑐2 2 1 + 𝑥𝑥 2 + 𝑥𝑥 2 + 1 + 2𝑥𝑥 𝑥𝑥 2 + 1 = 1
1𝑐𝑐2 =
2 + 2𝑥𝑥 2 + 2𝑥𝑥 𝑥𝑥 2 + 1− 𝑥𝑥 2 + 1 − 𝑥𝑥𝑐𝑐3 =2 + 2𝑥𝑥 2 + 2𝑥𝑥 𝑥𝑥 2 + 1

43

4. Spin-spin coupling: variational principle

44• Similarly,ℏ 2 2𝑗𝑗ℏ 𝑗𝑗ℏ
𝑗𝑗ℏ
222𝐸𝐸3 =𝛾𝛾 ℬ 𝜎𝜎1 − 𝜎𝜎2 + 𝑗𝑗 − =𝑥𝑥 + 1 −242
4
1𝑐𝑐2 ′ =2 + 2𝑥𝑥 2 − 2𝑥𝑥 𝑥𝑥 2 + 1𝑥𝑥 2 + 1 − 𝑥𝑥𝑐𝑐3 ′ =
2 + 2𝑥𝑥 2 − 2𝑥𝑥 𝑥𝑥 2 + 1• The transition probability can also be calculated.
22��𝜒𝜒1 𝑉𝑉𝜒𝜒2 = 𝜒𝜒4 𝑉𝑉𝜒𝜒2 = 𝛾𝛾𝐵𝐵′ 2 𝑐𝑐2 + 𝑐𝑐3 222
��𝜒𝜒1 𝑉𝑉𝜒𝜒3 = 𝜒𝜒4 𝑉𝑉𝜒𝜒3 = 𝛾𝛾𝐵𝐵′ 2 𝑐𝑐2 ′ + 𝑐𝑐3 ′ 2

45

4. Spin-spin coupling: variational principle

E (jħ)• Suppose
0𝐸𝐸1= −1 × 106 𝑗𝑗𝑗

E41M
54
32
10−1−2−3−4−5−1ME3E20246x8E1

10

4. Spin-spin coupling: variational principle𝜔𝜔1→2 − 𝜔𝜔TMS 𝐸𝐸2 − 𝐸𝐸1
𝛿𝛿1→2 =≈
−1=
𝜔𝜔TMS𝛾𝛾𝛾𝛾
𝑗𝑗𝜎𝜎+𝜎𝜎122=− 𝑥𝑥 + 1 − 1 −22𝛾𝛾𝛾
𝑗𝑗𝜎𝜎+𝜎𝜎122𝑥𝑥 + 1 − 1 −𝛿𝛿1→3 =22𝛾𝛾𝛾
𝑗𝑗
𝜎𝜎1 + 𝜎𝜎22𝑥𝑥 + 1 + 1 −𝛿𝛿2→4 =22𝛾𝛾𝛾
𝑗𝑗
𝜎𝜎1 + 𝜎𝜎22− 𝑥𝑥 + 1 + 1 −𝛿𝛿3→4 =22𝛾𝛾𝛾

46

47

4. Spin-spin coupling: variational principle.0
2.5
110y (a.u.)
Intensit.01Chem
ical shift (10 -6)

8.5
06.0
00

442

20

108x6

𝑗𝑗
= 10−6𝛾𝛾𝛾
𝜎𝜎1 + 𝜎𝜎2 = 10−5

4. Spin-spin coupling: variational principle

𝑥𝑥 = ∞𝑥𝑥 = 12𝑥𝑥 = 6𝑥𝑥 = 3𝑥𝑥 =3
2

1𝑥𝑥 =
2

𝑥𝑥 = 048