1

NMR and ESR Spectroscopy

K. R. Shamasundar and M. Nooijen

University of Waterloo Introduction

Nuclear Magnetic Resonance (NMR) spectroscopy and Electron Spin Resonance (ESR)

spectroscopy are two widely used spectroscopic techniques to infer structure and properties of

complex molecules (even bio-molecules such as proteins). Both these methods use angular

momentum, .i.e., pure “spin” or “total angular momentum” of the relevant particles to extract

molecular structural information.

In ESR spectroscopy, molecules in a state containing unpaired electrons, .i.e., with non-

zero spin-angular momentum (molecules in non-singlet states, 0S ≠ ) are placed in constant

magnetic field. The ESR spectrum resulting from transitions between molecular states with

different SM components contains chemically relevant information. For example, the ESR

spectrum of unpaired electrons in transition metal complexes contains information on how

ligands are arranged around the metal ion. Fine structure results from the interaction between

electrons and nuclear spin and this can serve as a fingerprint for molecular, nuclear and electron

spin density.

Likewise, in NMR spectroscopy, a composite system of nuclei in molecules with non-

zero nuclear spins ( 0I ≠ ) is placed in a constant magnetic field. Similarly, the NMR spectrum

resulting from transitions between different states of nuclear spin system contains a wealth of

information regarding chemical environment of the nuclei, and this is used to extract structural

information of molecules.

Despite their wide-spread use, the underlying principles of both these spectroscopic

techniques are fairly simple. In fact, the machinery needed for a quantum mechanical description

of the basic phenomena of these methods mainly involves the angular momentum theory we

have discussed in the class. Additionally, we would require basic first-order (and some times

second-order) perturbation theory, and variational method and related secular problem. The

Hamiltonians used are simple enough to enable manual solution of first-order perturbation theory

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and the secular equations. Finally, use of simple symmetry arguments are helpful to reduce the

problem to smaller problems.

Magnetic moment of charged particles with Intrinsic Spin.

Both NMR and ESR involve appropriate (nucleus or electrons) charged particles with

non-zero intrinsic spin. The spin of individual particles is fixed and typically 1/2 or 1 (in units of

h ), although higher integral values are possible (and complicate the spectrum somewhat). A

charged particle possessing angular momentum is related to a classical charged particle rotating

around its axis. This gives raise to a magnetic (dipole) moment µr .

2

g Jqm

µ β

β

=

=

rr

(1)

Where, q and m are charge and mass of the particle, and Jr

is the total angular

momentum (spin or orbital or their combination) for the particle. The quantity β is known as

“magneton”, and g is known as the “g-factor”. The g-factor is unitless, and is typically of the

order of 1-10. Both these quantities are characteristics of the particle.

For electrons, β is negative (since electron charge q is negative) and is usually referred

to as “Bohr-magneton”, and is denoted by eβ . Usually, the negative sign in β is explicitly

written into Eqn (1), and it is rewritten as.

for electrons

2

e e e

ee

g Jem

µ β

β

= −=

rr

(1-e)

Therefore, for electrons the magnetic moment is in reverse direction of total angular

momentum Jr

. Using relativistic theory of electron, the g-factor can be shown to be 2 for a free

electron. For electrons in chemical environment, g-factor will depend on the molecular state.

Such electrons have non-zero spin angular momentum ( S ) as well as orbital angular momentum

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( L ), giving raise to a total angular momentum ( J ). For this case, the g-factor is given by Lande

factor as,

( 1) ( 1) ( 1)12 ( 1)e

J J L L S SgJ J

+ − + + += +

+ (2)

For ground and low-lying states of molecules, usually the orbital angular momentum ( L )

is zero or small 0L ≈ . Therefore, total angular momentum is almost equal to the spin-angular

momentum ( J S≈ ). In such a case, the g-factor is close to 2. For molecules involving unpaired

electrons in d-orbitals of transition metals, the g-factor can be significantly different from 2.

For nuclei, β is positive (because nuclear charge is positive), and is usually referred to as

“nuclear-magneton”, and is denoted by Nβ . For nuclei, Eqn (1) can be rewritten as,

for nuclei

2

N N N

NN

g Jqm

µ β

β

==

rr

(1-N)

Therefore, magnetic moment is the same direction as Jr

. The g-factor for nuclei depends

on constituent particles (protons and neutrons), and is not easy to calculate. For protons

(Hydrogen atom nucleus), g-factor is 5.58. For practical purposes, g-factor for a particular type

of nucleus is a fixed value, and is experimentally determined.

In literature, h appearing in Jr

is often absorbed into definition of magneton β and is

expressed in units of Joules per Tesla (J/T). In this case, magnetic moment is expressed in units

of “magnetons”. The magnitude of β determines the energy gap between the states whose

transitions are recorded, and sets the energy range (in the electromagnetic spectrum) where the

spectroscopy is observable. For electrons, the Bohr-magneton 2e

e

em

β =h is 9.2*10-24 J/T. For

magnetic field strengths used in practice (1-10 Tesla), this leads to transitions of few cm-1 and

falls in the micro-wave (one to a few hundreds of GHz) region. Since proton (and other nuclei)

are about 1000 times heavier than electrons, the nuclear-magneton for proton

( )2N p

p

eprotonm

β β= =h is about 1000 times smaller than Bohr-magneton. As a result, this leads

to transitions observable in radio-wave region (one to a few hundred MHz).

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Unlike electron which has spin 1/2, the spin of nucleus depends on individual spin of

nucleons (protons and neutrons). Again, for all practical purposes, spin of a particular type of

nucleus can be considered as a fixed value (integer or half-integer) determined theoretically or

experimentally. Most common nucleus used in NMR, the proton, has spin 1/2.

For our purposes, and for the purpose of chemists who use ESR and NMR, we will

consider β and g-factor as being provided to us. The product of β (without the h ) and g-factor

is known as gyro-magnetic ratio and is denoted byγ .

g

J

γ β

µ γ

=

=rr (3)

Again for nuclei, gyro-magnetic ratio is positive and for electrons it is negative. In

literature, h appearing in Jr

is not absorbed into gyro-magnetic ratio. As we shall see, the

magnitude of gyro-magnetic ratio (measured in per second per Tesla) determines the amount of

splitting of energy levels. While β sets the energy scale for the spectroscopy, γ determines the

sensitivity of the given spin-system within this scale. Within the energy scale set by β , different

spin-systems with different gyro-magnetic ratio will resonate at significantly different

frequencies.

Furthermore, the higher the value of gyro-magnetic ratio, the smaller the external

magnetic field required for achieving resonance transitions and the easier it is to observe them.

For example, the gyro-magnetic Hydrogen nucleus (proton) is about one-order more than that of 13C nucleus. Therefore, it is easier to observe proton nuclear resonances compared to 13C. It also

implies that a group of hydrogen nuclei in a given chemical environment resonate at different

frequencies than a group of 13C nuclei in the same chemical environment! This means that a

NMR of a group of nuclei of a given type in a given environment can be safely observed

disregarding groups of other nuclei in the same environment, if these groups of nuclei have

different gyro-magnetic ratios.

Interaction of the Magnetic moment with External Magnetic Field.

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When placed in external magnetic field, a magnetic dipole µr will interact with the

applied magnetic field Br

. The interaction potential energy of is given by (refer to MS Problem

13-49),

E Bµ= − •rr (4)

As can be seen, the interaction energy will be minimum when µr is in the same direction

as Br

. Therefore, a magnetic dipole will tend to align parallel to the magnetic field. Usually, the

direction of applied magnetic field is taken to be along the z-axis, and hence the energy is given

by,

0z z zE B Bµ µ= − = − (5)

The quantum mechanical Hamiltonian for this interaction is obtained by expressing zµ in

above expression by ˆz zJµ γ= obtained from Eq (3).

0 0ˆ ˆ

zH B Jγ= − (6)

The eigenstates of the are just the angular momentum eigenfunctions corresponding to a

total angular momentum quantum number. In absence of external magnetic field, these (2 1)J +

states of the spin-system with different axial quantum numbers ( J− to J+ ) are all degenerate. In

presence of magnetic field, this degeneracy completely breaks (known as Zeeman effect) and

energy levels split. Considering spin-1/2 systems (consisting of α and β levels), the magnitude

of the energy level splitting is given by,

0E Bγ∆ = h (7)

Exercises:

1. Prove Eq (7) for a single spin 1/2 system.

2. Derive the energy level splitting for a single spin system with total angular

momentum quantum number J .

3. For the system above, consider the selection rule that only transitions with 1JM∆ = ±

will be allowed. How many transitions will be observed? Does Eq (7) require any

modification?

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Note that the magnitude of the energy splitting depends on applied field strength.

Depending on the sign of gyro-magnetic ratio, either α state or β state will be lower in energy.

For electrons, β state will be lower, and for nucleus α state will be lower.

When electro-magnetic radiation of appropriate angular frequency (given by 0Bω γ= ) is

applied perpendicular to the direction of the constant magnetic field, .i.e., in x-y plane, the

system undergoes transition. This can be detected and recorded as a spectrum.

In the following, we will mainly consider the case of NMR spectroscopy. We will mainly

study quantum mechanical treatment of NMR spectroscopy using a many-particle Hamiltonian

obtained by generalization of Eq (6) to a system consisting of a group of Hydrogen nuclei (spin

1/2) in a given chemical environment (as in a molecule or a crystal). We will first study the

effect of chemical environment immediate to a given nuclei on its resonance positions, and

define chemical shift. We will then consider the finer effects of spin-spin interactions between

these nuclei in different chemical environments, and show that it leads to multiplet structure in

NMR spectra. The material presented here may be read in conjunction with Chapter 14 of

McQuarrie and Simon.

It should be kept in mind that techniques presented here can also be used to study hyper-

fine effects in ESR spectroscopy.

Magnetic Shielding Effect and Chemical Shift:

According to Eq (7), all Hydrogen nuclei in a molecule absorb at the “same” frequency

given by,

0( )

bare spectrometer

E E EB β α α βω γ ω− ∆ →

= = = =h h

(8)

Here the subscript bare is included because this is the same frequency at which a free (or

bare) proton will absorb at the same magnetic field. This is the frequency at which the

spectrometer is said to be operating.

If all Hydrogen nuclei in a molecule were to absorb at exactly the same frequency as

above, then NMR would not be more any more useful than to test for the presence of Hydrogen

atoms in a molecule. In reality, different Hydrogen nuclei in different immediate chemical

environments in the molecule absorb at slightly different frequencies around the frequency given

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by Eq (8). This difference is what makes NMR spectroscopy a very useful tool for extracting

structural information about the molecule.

The reason for this difference is that Hydrogen nuclei in a molecule are surrounded by

clouds of electrons. This cloud of electrons is locally different around different nuclei. When the

molecule is placed in external magnetic field, these electron clouds interact with the magnetic

field. As a result of this interaction, a small additional magnetic field elecB develops at the

nucleus. For most molecules, this generated field is in opposite direction to the applied field, and

its magnitude is proportional to the applied field.

0elecB Bσ= − (9)

Here, σ is the proportionality constant (which is unit-less). The negative sign in Eq.(9)

shows that elecB is opposite to applied field. The constant σ is a measure of effective shielding

of the nucleus offered by electrons, and is known as shielding constant. Typically the generated

field is about 5 orders of magnitude smaller than applied field, .i.e., σ is usually of the order of

10-5. The local field at the nucleus localB and corresponding change in Eq (6) is therefore given

by,

0 0 0

0 0

(1 )ˆ ˆ ˆ(1 )local

local z z

B B B B

H B I B I

σ σ

γ γ σ

= − = −

= − = − − (10)

In the above equation, nuclear-spin denoted by zI replaces the general angular

momentum ˆzJ used in earlier equations. As opposed to Eq (8), the resonant frequency including

shielding effects is given by,

0 (1 )

bare

Bω γ σω σω= −

∆ = − (11)

The shift in frequency ω∆ depends on the applied magnetic field (higher magnetic fields

will give better resolutions). It is usually very small and is of the order of a few hundred Hz

depending on the applied field. Please refer to Figure. 1 where this splitting into α and β , and

the transition between them, has been illustrated by two levels of the first proton

The value of shielding constant for a given nucleus depends on the type of local

electronic environment the nucleus is in. Therefore, two protons in different chemical

environments experience different elecB , and absorb at slightly different frequencies. Observing

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this frequency difference between different protons in a molecule, we can elucidate molecular

structure.

The frequency difference depends (or is proportional to) applied magnetic field or

operating frequency of the spectrometer, comparing NMR spectra from different spectrometers

becomes difficult. To resolve this, a number called as chemical shift Hδ is assigned to each

proton in molecule as follows.

( ) 610H TMSH

spectrometer

ω ωδ

ω−

= × (12)

Here, TMSω is the resonance frequency of the protons of reference molecule known as

tetra methyl silane (TMS) Si(CH3)4, and spectrometerω is the operating frequency (which is related to

the applied field) of the spectrometer. The TMS is a relatively unreactive, and its 12 equivalent

protons are highly shielded, and give a clear signal even if it is present in small traces.

Chemical shift is related to the difference in absorption frequencies of proton in a given

molecule and the 12 equivalent protons in TMS molecule. It measures how much a proton in a

given molecule is shielded compared to protons in TMS. It is a dimensionless quantity, and is

expressed in parts per millions (ppm). Its value is indicative of the kind and amount of electron

cloud around the proton. Higher chemical shift for a proton usually implies depletion of electrons

around that proton. Bare proton has highest chemical shift. Different protons in a molecule have

different chemical shifts, and their difference will be independent of spectrometer characteristics.

Non-interacting two proton system

Consider two protons (labelled by 1 and 2) in two different chemical environments with

shielding constants 1σ and 2σ . If the protons are not interacting, the total 2-particle Hamiltonian

for this system is given by,

0 0 1 2ˆ ˆ ˆ(1 ) (1) (1 ) (2)z zH B I Iγ σ σ = − − + − (13)

Exercise:

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Prove that 0H commutes with z-component of the total nuclear spin angular momentum,

.i.e., with ˆ ˆ ˆ ˆ(1) (2)totalz z z zI I I I= = + . Use the known eigenfunctions { }(1), (1)α β of ˆ (1)zI and

{ }(2), (2)α β of ˆ (2)zI , to construct a set of simultaneous eigenfunctions of 0H and ˆtotalzI . What

are eigenvalues of Hamiltonian and corresponding eigenfunctions? List them in order of

increasing energies and their ˆtotalzI eigenvalues.

Answer:

Consider,

0 0 0

0 1 0 2

0 1 0 2

ˆ ˆ ˆ ˆ ˆ ˆ, , (1) , (2)

ˆ ˆ ˆ ˆ (1 ) (1), (1) (1 ) (2), (1)

ˆ ˆ ˆ ˆ (1 ) (1), (2) (1 ) (2), (2)

totalz z z

z z z z

z z z z

H I H I H I

B I I B I I

B I I B I I

γ σ γ σ

γ σ γ σ

= + = − − − − − − − −

0=

The first and fourth terms vanish because operator commutes with itself, and second and

third terms vanish because operators corresponding to co-ordinates of different particles

commute.

The above commutation means that eigenfunctions of Hamiltonian are also

eigenfunctions of ˆtotalzI operator, and therefore can be constructed as a linear combination of

eigenfunctions of ˆtotalzI operator.

The possible states of this 2-proton system are,

1 1

2 2

3 3

4 4

(1, 2) (1) (2)

(1, 2) (1) (2)

(1, 2) (1) (2)

(1, 2) (1) (2)

ψ ψ α α

ψ ψ β α

ψ ψ α β

ψ ψ β β

= =

= =

= =

= =

(14)

In the above equations, we have used what is known as “Dirac” notation to represent

these states. In Quantum Mechanics, Dirac notation is often used to represent quantum

mechanical quantities such as system states, overlap integrals between two system states, and

expectation value of an operator between two system states, completeness of system states etc.,.

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Dirac representation is simple, and powerful, and provides a co-ordinate independent way to

represent quantum mechanical quantities.

A half bracket around the wave-function (1,2)iψ is used to represent it as iψ . The

co-ordinates 1,2 are dropped, and are written explicitly when it is needed. The overlap integral

of state iψ with another state jψ is written as follows.

*| (1, 2) (1,2) 1 2j i j i d dψ ψ ψ ψ= ∫

The relation ( )** *(1, 2) (1,2) 1 2 (1,2) (1,2) 1 2j i i jd d d dψ ψ ψ ψ=∫ ∫ which is the property of

overlap integrals, is now simply represented in Dirac notation as, *

| |j i i jψ ψ ψ ψ=

Similarly, the expectation value of a quantum mechanical operator A , for example say

0H or zI , between a two states iψ and jψ is represented as,

*ˆ ˆ| | (1, 2) (1,2) 1 2j i j iA A d dψ ψ ψ ψ= ∫

Adjoint operator of an operator A is defined as another operator B which is related to A

for any two states iψ and jψ through the following relation.

*ˆˆ| | | |j i i jB Aψ ψ ψ ψ=

If the above relation holds, B is said to be adjoint of A , and is denoted as †A . An

operator A which is its own adjoint, .i.e., †ˆ ˆA A= , is known as self-adjoint or Hermitian

operator. As taught in introductory quantum mechanics courses, physical quantities such as

position and momentum are represented by Hermitian operators. As we know, when a specific

basis set { }, 1,i i Nψ = K is used, an operators A is represented by corresponding matrix Α

whose elements are ˆ| |j iAψ ψ=jiA . When an orthonormal basis is used, Hermitian operators

are represented as Hermitian matrices (self-complex conjugate transpose), and adjoint of an

operator is represented as complex-conjugate transpose matrices.

The { }, 1, , 4i iψ = K are eigenfunctions of ˆtotalzI . The function 1ψ has 1IM = + , 2ψ

and 3ψ have 0IM = , and 1ψ has 1IM = − . By acting 0H on each of these functions, you can

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verify that these are also eigenfunctions of 0H (this happens due to non-interacting nature of

0H ). Therefore, we can conclude that 0H is diagonal in { }, 1, , 4i iψ = K basis. The energies are

just expectation value of 0H in this basis.

[ ] [ ]1 1 0 1 0 1 21ˆ(1) (2) (1 ) (1 )2

E E H Bα α ψ ψ γ σ σ= = = − − + −h

[ ] [ ] [ ]2 2 0 2 0 1 2 0 1 21 1ˆ(1) (2) (1 ) (1 )2 2

E E H B Bβ α ψ ψ γ σ σ γ σ σ= = = − − − + − = − −h h

[ ] [ ] [ ]3 3 0 3 0 1 2 0 1 21 1ˆ(1) (2) (1 ) (1 )2 2

E E H B Bα β ψ ψ γ σ σ γ σ σ= = = − − − − = + −h h

[ ] [ ]4 4 0 4 0 1 21ˆ(1) (2) (1 ) (1 )2

E E H Bβ β ψ ψ γ σ σ= = = + − + −h

These energy levels of non interacting two proton system is illustrated in Figure. 1, where

the second proton is added to the system of first proton.

2. Later on, we will show that NMR transitions are governed by the selection rule

1IM∆ = ± . What does this selection rule mean? Use this to identify allowed transitions in the

two proton system we considered above. Find the transition frequencies in terms of two shielding

constants. How many transitions are possible? how many of them have different frequencies?

Answer:

For NMR transitions, the selection rule 1IM∆ = ± means that in a spin system, only

those transitions which “flip the spin of a single proton” are allowed. The rest are forbidden. You

may find it useful to solve the Problem 14-37 in Chapter 14 of McQuarrie and Simon, to

understand NMR selection rules. Using the IM values for the four states above, 1IM∆ = ± will

give following transitions.

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1 2 1 0 1

1 3 2 0 2

2 4 3 0 2

(1 ) spin flip of first proton with second proton in state.

(1 ) spin flip of second proton with first proton in state

(1 ) s

B

B

B

ψ ψ ω γ σ α β α

ψ ψ ω γ σ α β α

ψ ψ ω γ σ α β

→ = − →

→ = − →

→ = − →

3 4 4 0 1

pin flip of second proton with first proton in state

(1 ) spin flip of first proton with second proton in stateB

β

ψ ψ ω γ σ α β β→ = − →

These four transitions are shown in Figure. 1, where the second proton is added to the

system of first proton to make a non-interacting two proton system. Although there are four

transitions, only two are observed because 1 4ω ω= , and 2 3ω ω= . Since the two spins are not

interacting, it takes the same amount of energy to flip first spin first proton from alpha to beta

state irrespective of whether the second proton is in alpha state or beta state. However, flipping

energy for first and second proton is different. Therefore only two lines are observed in the NMR

spectrum, one for each protons.

3. Does 0H commute with 2,ˆ totalI ? If not, under what conditions does it commute? Give physical

explanation for the results.

Using 2, 2ˆ ˆ ˆ ˆ ˆtotalz zI I I I I− += + + h

2, 20 0 0

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , ,totalz zH I H I I H I I− +

= + + h

The second term vanishes, and the first term can be worked out using basic angular

commutation relations. The final result is,

( )2,0 1 2

ˆ ˆ ˆ ˆ ˆ ˆ, (1) (2) (2) (1)totalH I I I I Iσ σ − + − + = − −

Notice that the term in square brackets is not a commutator. There is no reason for the

term in second brackets to vanish, and hence 0H in general does not commute with 2,ˆ totalI .

However, notice that it commutes when 1 2σ σ= , which happens when the two-protons are

equivalent.

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The reason why 0H does not commute with 2,ˆ totalI in general is because when there are

non-equivalent protons in external magnetic field, there is no spherical symmetry in spin-space.

Due to shielding being different for each of them, 1 2σ σ≠ , the two protons interact with different

strengths with external magnetic field. However, when these protons are equivalent, they are

shielded in exactly the same way, and hence these protons interact with external magnetic field

with same strength. This leads to spherical symmetry in spin-space. This can be easily seen by

noting that, for the case of equivalent protons, 1 2σ σ σ= = , The Hamiltonian 0H reduces to,

0 0 0ˆ ˆ ˆ ˆ(1 ) (1) (2) (1 ) total

z z zH B I I B Iγ σ γ σ = − − + = − −

The above form of the Hamiltonian also shows that it is just proportional to ˆtotalzI , and its

eigenvalues, unlike the general non-equivalent proton case, are just eigenvalues of ˆtotalzI operator

multiplied by a constant number 0 (1 )Bγ σ− − . The special form also clearly shows that 0H

commute with 2,ˆ totalI , since we know that ˆtotalzI commutes with 2,ˆ totalI by the property of angular

momentum.

The intensity of a spectral line is proportional to the total area around a spectrometer line.

This is in turn proportional to the total number of protons absorbing at that frequency. For

example, the NMR spectrum of Acetaldehyde molecule (CH3CHO) contains two lines. One line

corresponds to proton of aldehyde group (CHO), and other line corresponds to three protons

(CH3) which are all equivalent and absorb at same frequency (in the assignment problem, it is

proven that all equivalent protons absorb at the same frequency, we will prove it here for two

proton case). In a sample of acetaldehyde, there are three times more protons absorbing at one

frequency and one proton at other frequency. Therefore, the area under the second line is three

times more than that of the first line.

Spin-Spin Coupling and Multiplets in NMR spectrum.

At low resolutions, the NMR spectrum of Acetaldehyde (CH3CHO) hows two broad lines

showing presence of two different hydrogen atoms in the molecule. The area count reveals the

ratio of number of protons of one type to the number of other type of protons. We calculated this

spectrum. This is known as zeroth-order spectrum.

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However, when high resolution NMR spectrum of Acetaldehyde is obtained, the broad

line corresponding to CHO proton shows further fine structure with four narrowly separated lines

of different intensity, the areas under these lines being in the ratio 1:3:3:1 (it is called quartet).

Similarly, the broad line corresponding to CH3 protons shows two separated lines of same

intensity, areas under them being in the ratio 1:1 (called doublet). Similarly, in ESR

spectroscopy, the broad line corresponding to transition between different JM levels of unpaired

electrons splits into narrowly separated lines of different intensity. These multiplets structures

are observed in both NMR and ESR spectroscopy, and are caused by interaction of magnetic

moments of the particle with that of neighbouring particles. In NMR spectroscopy, the

interaction of the magnetic moment of a proton with magnetic moments of neighbouring protons

produces this multiplet structure. In ESR, it is caused by the interaction of total magnetic

moment of unpaired electrons with magnetic moments of nearby nucleus. It is referred to as

hyperfine structure in ESR. In both cases, it contains structurally relevant information.

Qualitative explanation for spin-spin coupling is as follows. Consider the example of two

protons with different shielding constants we discussed earlier. In that example, the energy

required to flip first proton from alpha to beta state was shown to be independent of whether

second proton was in alpha or beta state. Therefore, state of the second proton did not influence

the transition of first proton.

However, in reality, the two protons are not independent and interact with each other.

The interaction which influences the states of individual protons and which is relevant to NMR,

is the mutual interaction of magnetic moments via the electronic environment in the molecule.

As a result of this interaction, the two transitions which looked like the spin-flipping of the first

proton, .i.e., 1 4,ω ω will differ from each other depending on the magnitude of this interaction.

Similarly, 2 3,ω ω transitions will also differ.

Usually, the strength of spin-spin interaction is very small compared to their individual

screening. Therefore, the transitions 1 4,ω ω will be close to each other, and 2 3,ω ω will be close to

each other. This qualitatively explains the (doublet) splitting of 1 4ω ω= transition line of first

proton, and same doublet splitting for second proton.

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In NMR spectroscopy, the effects of spin-spin coupling can be calculated by adding the

spin-spin coupling term into the NMR Hamiltonian as follows. The interaction between two

magnetic moments can be written as,

12 12 1 2V J µ µ= ⋅r r

Here, the subscripts 1 and 2 corresponds to that of first and second protons. 12J is

constant known as spin-spin coupling constant. It measures the strength of the interaction. It

depends on the separation between two protons. The also depends on the electronic structure of

the molecule, as the two protons interact through the cloud of electrons surrounding them. The

constant 12J is specific to a pair of protons in a specific chemical environment. Since, µr for

each proton is proportional to its spin, the above interaction can be written in quantum

mechanical operator form as,

12 12 1 2ˆ ˆ ˆV J I I= ⋅

In the above formula, we have absorbed all the factors into definition of spin-spin

coupling constant. We will add this spin-spin coupling constant to our zeroth-order NMR

Hamiltonian, and study how spin-spin coupling affects different transitions in our two-proton

system.

Effect of spin-spin coupling constant for two different protons.

In presence of spin-spin interaction, the NMR Hamiltonian for our two different proton

problem can be written as,

0 12 0 1 0 2 12 1 21ˆ ˆ ˆ ˆ ˆ ˆ ˆ(1 ) (1) (1 ) (2)z zH H V B I B I J I Iγ σ γ σ= + = − − − − + ⋅h

We have added 1h

to enable us to express 12J in HZ (1/sec or radians per second), just

like 0Bγ which is expressed in same units. Since there is only one spin-spin coupling constant

12J for our case, we will drop the subscripts and write it as J . I hope you will not confuse with

the symbol J used for total angular momentum. It will be clear from the context. Usually, the

magnitude of spin-spin coupling constant is very small compared to 0Bγ , or rather it is smaller

compared to the difference in resonating frequencies of two protons. Or more precisely,

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0 1 2J Bγ σ σ<< −

Therefore, we can treat 12V as a small perturbation to the zeroth-order Hamiltonian 0H ,

and apply first-order perturbation theory to calculate the splitting of lines observed. Such a

spectrum is known as first-order NMR spectrum.

We already know the eigenstates and energies of zeroth-order Hamiltonian. The first-

order perturbation correction to energy of zeroth order state iψ is given by,

(1) ˆi i iE Vψ ψ=

The total energy of the perturbed state within first order perturbation theory is obtained

by adding the above correction to the zeroth order energies. (1)

0

ˆ

ˆ ˆ ˆ

i i i i

i i i i

i i

E V

H V

H

ψ ψ

ψ ψ ψ ψ

ψ ψ

= +

= +

=

E

We can calculate the corrections (1) ˆi i iE Vψ ψ= , and final first-order energies (1)

iE as

follows.

(1)1 2

1 ˆ ˆi i iE J I Iψ ψ= ⋅

h

We can evaluate these matrix elements, by expanding 1 2ˆ ˆI I⋅ as,

1 2 1 2 1 2 1 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ

x x y y z zI I I I I I I I⋅ = + +

However, we will use another convenient expression for 1 2ˆ ˆI I⋅ as follows.

( )1 2 1 2 1 2 1 21ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 z zI I I I I I I I+ − − +⋅ = + +

You will derive this expression in the assignment problem 8.

( )

(1)1 1 2

1 2 1 2 1 2

1 2

1 ˆ ˆ(1) (2) (1) (2)

1 1 ˆ ˆ ˆ ˆ ˆ ˆ (1) (2) (1) (2)2

1 ˆ ˆ (1) (2) (1) (2)

1 4

z z

z z

E J I I

J I I I I I I

J I I

J

α α α α

α α α α

α α α α

+ − − +

= ⋅

= + +

=

= +

h

h

h

h

17

The first two terms vanished because either 1I + or 2I + could not act on (1) (2)α α state.

You can verify yourself that, in general, ( )1 2 1 21 ˆ ˆ ˆ ˆ2

I I I I+ − − ++ can not contribute to diagonal terms.

Therefore, we get,

(1)2 1 2

1 1ˆ ˆ(1) (2) (1) (2)4

E J I I Jβ α β α= ⋅ = − hh

(1)3 1 2

1 1ˆ ˆ(1) (2) (1) (2)4

E J I I Jα β α β= ⋅ = − hh

(1)4 1 2

1 1ˆ ˆ(1) (2) (1) (2)4

E J I I Jβ β β β= ⋅ = + hh

Therefore, the total first-order energies are,

[ ](1)1 0 1 2

1 1(1 ) (1 )2 4

B Jγ σ σ= − − + − +h hE

[ ](1)2 0 1 2

1 12 4

B Jγ σ σ= − − −h hE

[ ](1)3 0 1 2

1 12 4

B Jγ σ σ= + − −h hE

[ ](1)4 0 1 2

1 1(1 ) (1 )2 4

B Jγ σ σ= + − + − +h hE

The first-order spectrum of NMR is shown in Figure. 1, where the first-order energies are

shown as deviations from zeroth-order energies. You can see that first and fourth states go up by

amount 14

J∆ = h . Similarly, the second and third states go down by the same amount. This

results in Four lines as shown. Using the above energies, and selection rules 1IM∆ = ± the first-

order NMR spectra for our 2-proton system will have four lines at the following frequencies.

18

1 2 1 0 1

1 3 2 0 2

2 4 3 0 2

3 4 4 0 1

1 with (1 )21 with (1 )21 with (1 )21 with (1 )2

B J

B J

B J

B J

ψ ψ ω γ σ

ψ ψ ω γ σ

ψ ψ ω γ σ

ψ ψ ω γ σ

→ = − −

→ = − −

→ = − +

→ = − +

The above spectrum shows four lines. The two lines with frequencies 1ω and 4ω center

around 0 1(1 )Bγ σ− are separated by J . These two lines correspond to the two lines with

1 4ω ω= we associated with flipping of first proton in the zeroth-order (non-interacting)

spectrum. Similarly, the two lines with frequencies 2ω and 3ω correspond to the two lines with

2 3ω ω= associated with second proton flip.

It should be kept in mind that when we assign the spectral lines of interacting two-proton

system to flipping of first or second proton, it is a qualitative picture only to be used for our

intuitive understanding of what is happening within the system. In quantum mechanics, when

these two protons are interacting, the transitions we observe are of the system as a whole, and not

of the parts. When we assign a transition of the whole system to some parts of the system, such

as being a transition of first or second proton, it is only a loose approximate picture. It is valid

only when it is used along with a correct zeroth-order non-interacting description of the whole

system.

The ( 1)n + rule and Intensity pattern for first-order multiplet structure in NMR

Within the validity of first-order spectrum, the number of lines in a multiplet and their

intensity ratios can be neatly explained using a simple rule known as ( 1)n + rule. It is explained

in great detail with examples, in the chapter 14 of McQuarrie and Simons.

The ( 1)n + rule says that, if a proton (or a group of equivalent protons) has n equivalent

neighbouring protons, then the line in the zeroth-order spectrum corresponding to this proton (or

protons) will split into ( 1)n + closely spaced lines in the first-order spectrum.

19

Therefore, in case of Acetaldehyde, the three equivalent protons of CH3 group have one

neighbouring proton of CHO group. Therefore, the single line in the zeroth-order spectrum

corresponding to these three equivalent protons will split into (1 1) 2+ = closely spaced lines

(called doublet) in first-order spectrum. Similarly, the proton of CHO group has three equivalent

neighbouring protons of CH3 group, and hence it splits into (3 1) 4+ = closely spaced lines

(called quartet).

To see the origin of ( 1)n + rule in an intuitive way, consider the quartet splitting of CHO

proton line in Acetaldehyde as above. The three neighbouring equivalent protons of CH3 group

above can exist in 8 states grouped into the following groups.

Group-1 (1) (2) (3)

Group-2 (1) (2) (3) , (1) (2) (3) , (1) (2) (3)

Group-3 (1) (2) (3) , (1) (2) (3) , (1) (2) (3)

Group-4 (1) (2) (3)

α α α

β α α α β α α α β

β β α β α β α β β

β β β

When the proton of CHO group interacts with these three protons in any of these eight

states, the interaction will in general be dependent on the states of the neighbouring system.

However, due to equivalence of three neighbouring protons, the CHO proton will interact in the

same way with any of the states in a group. Therefore, the number of different types of

interactions depend on the number of group of states which are equivalent. As can be seen, for

three equivalent protons, there are four groups of equivalent states, and hence there would be

four lines.

The intensity ratios of these four lines can also be explained by considering that, in a

given sample of Acetaldehyde molecules, the three equivalent protons can exist in any one of

these states with equal probability. Therefore, the probability of the equivalent proton system

being in a group-2 state is three times than being in group-1 state. Therefore, the line resulting

from transition CHO proton when the equivalent proton system is in a group-2 state is three

times more intense than when it is in group-1 state. Therefore, the intensity ratio of quartet lines

is 1:3:3:1.

The origin of ( 1)n + rule and associated intensity ratio pattern can be generalized by

observing that it is related to the number of equivalent spin states of these n equivalent

20

neighbouring protons. By denoting the number of ways of choosing r objects from among n

objects as,

( )!

! !n

rnC

r n r=

−

Among all possible spin states of these n equivalent neighbouring protons, there would

be 0 1nC = state in which all protons are in alpha state and no proton is in beta state, there will be

1nC n= states in which one proton is in beta state, 2

nC states in which two protons are in beta

state and so on. Because these n protons are all equivalent, all possible the nrC states with a

fixed number r protons of beta state are indistinguishable, and contribute to the splitting in the

same way. Therefore, there would be one peak corresponding to flipping of the proton (or

protons) being considered with the state of neighbouring protons being all in alpha state, one

peak when neighbouring protons in any of the 1nC n= states, and so on. Therefore, there will be

( )1n + peaks in all with the intensity ratios given by 0 1 2 2 1: : : : : :n n n n n nn n nC C C C C C− −L .

It should be kept in mind that this pattern is valid only within the approximations for the

validity of first-order spectrum., .i.e., .i.e., 0 1 2J Bγ σ σ<< − ., only when the zeroth-order lines

are well separated and spin-spin coupling between these two groups is much smaller than this

separation. Please refer to McQuarrie and Simon for details and examples of illustration of range

of validity of ( )1n + rule and intensity rule.

Transitions within a System of Equivalent Protons

While calculating Intensity ratios and using the ( 1)n + rule, we have considered

equivalent protons differently. In reality, they themselves form a spin system and there can be

transitions among themselves. However, it turns out that all transitions within this system have

the same frequency. Therefore, equivalent proton system shows only one transition with the

intensity multiplied by the number of equivalent protons.

Let us see what happens when two protons are equivalent? You can verify that, setting

1 2σ σ σ= = in the first-order expressions for transition energies, leaves us with two lines

21

separated by J . However, in the experiments, only one line is observed for the case of two

equivalent protons.

Exercises:

1. Why are the first-order results for equivalent protons not agreeing with experiment? Is it that

the strength of interaction is large when two protons are equivalent, and hence perturbation

theory fails,… or are there other factors? Explain. How to fix this up?

The reason why first-order perturbation theory fails to predict correct spectrum for

equivalent protons is not that the strength of interaction becomes large. The reason is that 0H has

a degenerate energy levels for equivalent protons. To see this, let us find what happens to the

energy levels and eigenstates of 0H , when two protons are equivalent .i.e. 1 2σ σ σ= = . We get,

[ ]1 1 0 1 0ˆ(1) (2) (1 )E E H Bα α ψ ψ γ σ= = = − −h

[ ]2 2 0 2ˆ(1) (2) 0E E Hβ α ψ ψ= = =

[ ]3 3 0 3ˆ(1) (2) 0E E Hα β ψ ψ= = =

[ ]4 4 0 4 0ˆ(1) (2) (1 )E E H Bβ β ψ ψ γ σ= = = + −h

Therefore, the second state (1) (2)β α and third state (1) (2)α β states become

degenerate when two protons are equivalent. When you apply first-order perturbation theory on

second and third state, the perturbation shifts their energies by the same amount 14

J− h and

hence it does not break this degeneracy. The first state is shifted up by 14

Jh and fourth state is

shifted up by 14

Jh . Therefore, we get following first-order energies.

(1)1 0

1(1 )4

B Jγ σ= − − +h hE

22

(1)2

14

J= − hE

(1)3

14

J= − hE

(1)4 0

1(1 )4

B Jγ σ= + − +h hE

The final result is that the spectrum calculated using first-order perturbation theory shows

two different lines which is not what is observed in experiments. The problem with first-order

perturbation theory for degenerate states is that the states are left unchanged. In reality, the states

are greatly changed, and the perturbation breaks the degeneracy. These problems are better

treated using the variational method.

The full variation method for a given Hamiltonian, and solution of the resulting secular

problem provides exact energies and eigenstates of that Hamiltonian. We know that this method

works always. We will apply this method to the NMR Hamiltonian for a general non-equivalent

interacting two-proton system, and obtain its energies and states, and calculate the exact

spectrum. We will see that this approach leads to correct solutions for the case of equivalent

protons as well.

Solution of the Secular problem to calculate the Exact Spectrum.

The Secular problem for a Hamiltonian H involves construction of matrix elements of

the Hamiltonian ˆi jHχ χ=ijH in some orthogonal basis { }, 1, ,i i Nχ = K . This produces the

Hamiltonian operator in the form of a N -by- N Hermitian matrix. Let us denote this

Hamiltonian matrix by H . The exact energies and eigenstates are obtained by diagonalizing H

matrix. This leads to the following Secular equation for exact energies E .

det( ) 0− =H 1E

Here, 1 denotes the unit matrix of the same dimension as H . Once these energies are

obtained by solving the above equation, the eigenfunction which corresponds to a particular

eigenvalue E , written as a column vector c in the same basis, can be found by solving the

following linear equation.

( ) 0− =H 1 cE

23

The above linear equation is known as eigenvalue equation. Note that, since E used in

this equation satisfies secular equation, det( ) 0− =H 1E and therefore the matrix ( )−H 1E is not

invertible.

For Hermitian matrices, we know that its eigenvectors are orthogonal, and span the same

space as the original basis { }, 1, , 4i iχ = K . If we choose these eigenvectors as new orthogonal

basis to represent the Hamiltonian operator, it is represented as diagonal matrix. Therefore, the

process of finding eigenvalues and eigenvectors of Hamiltonian is also referred to as

“Diagonalization”. The Hamiltonian operator for our two-proton NMR problem is given as,

0 0 1 0 2 1 21ˆ ˆ ˆ ˆ ˆ ˆ ˆ(1 ) (1) (1 ) (2) z zH H V B I B I J I Iγ σ γ σ= + = − − − − + ⋅h

In principle, we can use any orthogonal basis { }, 1, ,i i Nχ = K for solution of Secular

problem. However, the steps involving construction of matrix elements of Hamiltonian, and

solution of Secular equations are greatly simplified by using symmetry properties of the

Hamiltonian. We will try to find the symmetries of our NMR Hamiltonian. In a previous section,

we proved that 0H commutes with ˆtotalzI . You can prove that the spin-spin interaction term also

commutes with ˆtotalzI . Consider,

( )

( )

1 2

1 2 1 2 1 2

1 2 1 2 1 2

1ˆ ˆ ˆ ˆ ˆ, ,

1 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ,2

1 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ , , ,2

total totalz z

totalz z z

total total totalz z z z z

I V J I I I

J I I I I I I I

J I I I I I I J I I I

+ − − +

+ − − +

= ⋅

= + +

= + +

h

h

h h

Using 1 2ˆ ˆ ˆ ˆ ˆ(1) (2)total

z z z z zI I I I I= + = + , we can see that the last term is zero. We can use the

commutator relation [ ] [ ] [ ], , ,A BC B A C A B C= + to simplify the first term as,

1 2 1 1 2 2 1 2 1 2

1 2 2 1 1 2

1 2 1 2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , ,

ˆ ˆ ˆ ˆ ˆ ˆ , ,

ˆ ˆ ˆ ˆ 0

totalz z z z z

z z

I I I I I I I I I I I

I I I I I I

I I I I

+ − + − + −

+ − + −

+ − + −

= + + + = +

= − +=

h h

24

Similarly, other term can be worked and found to be zero. Therefore, we can conclude

that that spin-spin coupling term V commutes with ˆtotalzI , and the entire NMR Hamiltonian

commutes with ˆtotalzI .

0ˆ ˆ ˆ ˆ ˆ ˆ, , , 0total total total

z z zH I H I V I = + =

Now, let us understand how this symmetry of Hamiltonian is useful in solving its secular

equations. Although, we illustrate with ˆtotalzI operator, the conclusions are very general and hold

for any operator which commutes with Hamiltonian operator. Suppose that we know the set of

eigenfunctions of ˆtotalzI , and we also know that they are orthogonal (because ˆtotal

zI is Hermitian).

Let us choose this set as our orthogonal basis functions for construction of Hamiltonian matrix.

Consider the construction of a matrix element of Hamiltonian ˆi jHψ ψ=ijH between two

basis functions iψ and jψ having different ˆtotalzI eigenvalues im , and jm respectively. Since,

ˆ ˆ, 0totalzH I = , we must have,

ˆ ˆ, 0totali z jH Iψ ψ =

ˆ ˆ ˆ ˆ 0total totali z z jHI I Hψ ψ− =

This leads to,

( )

ˆ ˆ ˆ ˆ 0

ˆ ˆ ˆ ˆ 0

ˆ 0

total totali z z j

total totali z j i z j

j i i j

HI I H

HI I H

m m H

ψ ψ

ψ ψ ψ ψ

ψ ψ

− =

− =

− =

In the above, we have made use of the fact that iψ (and iψ ), and jψ (and jψ ) are

eigenfunctions of ˆtotalzI eigenvalues im , and jm respectively. Since, we have assumed that these

eigenvalues are different, j im m≠ , or ( ) 0j im m− ≠ we get,

ˆ 0i jHψ ψ =

Therefore, the Hamiltonian matrix elements between basis functions having different

eigenvalues of ˆtotalzI are necessarily zero, and we do not have construct such elements. This is

solely the consequence of ˆtotalzI commuting with H . We can straightforward set these elements to

25

zero. With this, we can easily see the advantage of using the set of eigenfunctions of an operator

which commutes with Hamiltonian as a basis set for constructing the Hamiltonian matrix

elements. When working in such a basis, the only Hamiltonian matrix elements we need to

construct are those between two basis functions having the same ˆtotalzI eigenvalues. The rest are

simply zero by symmetry (or commutation), and this procedure is known as symmetry-

adaptation. Therefore, the Hamiltonian matrix entering the secular equations attains a block

diagonal form, which one block corresponding to an eigenvalue of ˆtotalzI operator. As a result of

this, we can see that the original secular problem det( ) 0− =H 1E splits into secular sub-

problems for these blocks. These can be solved independently to obtain corresponding

eigenvalues and eigenstates. Furthermore, we can see that the eigenstates of a Hamiltonian block

are just linear combinations of only those basis states which used to construct this particular

block. Therefore, each eigenfunction of Hamiltonian obtained as eigenfunction of a particular

block is also an eigenfunction of ˆtotalzI with a definite ˆtotal

zI eigenvalue, same as the ˆtotalzI

eigenvalue of the basis states of that block.

From angular momentum theory, we already know all the eigenfunctions ˆtotalzI operator,

and their eigenvalues. For this two-proton case, they are just the functions { }1 2 3 4, , ,ψ ψ ψ ψ

used in previous sections. We already know that 1ψ has ˆtotalzI eigenvalue of 1IM = + , both that

2ψ and 3ψ have ˆtotalzI eigenvalue of 0IM = , and 4ψ has ˆtotal

zI eigenvalue of 1IM = − .

Therefore, Hamiltonian matrix in this basis, splits into the following block diagonal form.

0 0 00 00 00 0 0

=

11

22 23

32 33

44

HH H

HH H

H

These six non-zero matrix elements can be evaluated by using the techniques already

presented in previous sections. The final results are,

1 1 11ˆ4

H E Jψ ψ= = +11H h

22 2 2 21ˆ4

H E Jψ ψ= = −H h

26

33 3 3 31ˆ4

H E Jψ ψ= = −H h

44 4 4 441ˆ4

H E Jψ ψ= = +H h

2 31ˆ2

H Jψ ψ= =23H h

3 21ˆ2

H Jψ ψ= = =32 23H H h

Where { }, 1, , 4iE i = K are the zeroth-order energies we already calculated in a previous

section. Therefore, the Hamiltonian matrix for full secular problem is,

1

2

3

4

1 0 0 04

1 10 04 2

1 10 02 4

10 0 04

E J

E J J

J E J

E J

+ −

= − +

H

h

h h

h h

h

Note that, the diagonal elements are just the first-order energies we obtained in the

previous section. As mentioned, the 4-by-4 secular problem can be solved by independently

solving the secular sub-problems of these blocks. Here, we have three sub-problems. There are

two 1-by-1 blocks, and one 2-by-2 block. The diagonal elements of the two 1-by-1 blocks are

just the energies themselves. The energies of the 2-by-2 block secular problem can be obtained

by solving,

2

3

1 14 2det 01 12 4

E J J

J E J

− − =

− −

h h

h h

E

E

The exact energies are obtained as,

27

[ ]

( )

( )

[ ]

1 1 0 1 2

22 2 22 0 1 2

22 2 23 0 1 2

4 4 0 1 2

1 1 1(1 ) (1 )4 2 4

1 14 21 14 2

1 1 1(1 ) (1 )4 2 4

E J B J

J B J

J B J

E J B J

γ σ σ

γ σ σ

γ σ σ

γ σ σ

= + = − − + − +

= − + − +

= − − − +

= + = + − + − +

h h h

h h

h h

h h h

E

E

E

E

Note that the exact energies 1E and 4E are same as first-order perturbation theory results

(1)1E and (1)

4E we obtained earlier. The first-order perturbation energies (1)2E and (1)

3E are

recovered from the corresponding exact energies for 2E and 3E , by neglecting the term 2J

within the square root, .i.e., by assuming ( )22 2 20 1 2B Jγ σ σ− >> . This is same as the conditions

required for the validity of first-order perturbation theory.

We will also find eigenfunctions { }, 1, , 4i iϕ = K of the Hamiltonian. The

eigenfunctions of two 1-by-1 blocks are just the corresponding ˆtotalzI eigenfunctions with

eigenvalues 1IM = + and 1IM = − respectively.

1 1ϕ ψ=

4 4ϕ ψ=

The two eigenfunctions of 2-by-2 block will be some general linear combinations of the

two ˆtotalzI eigenfunctions of with eigenvalue 0IM = .

2 1 2 2 3c cϕ ψ ψ= +

3 1 2 2 3d dϕ ψ ψ= +

The eigenfunction 2ϕ (or equivalently 1 2,c c ) corresponding to energy 2E , is found by

solving the corresponding eigenvalue equation.

2 21

23 2

1 14 2 01 12 4

E J J ccJ E J

− − =

− −

h h

h h

E

E

28

These are two simultaneous homogenous linear equations in 1c and 2c variables, and it

has non-zero solution only if the determinant of the matrix on left hand side vanishes. This is

precisely the condition (the secular equations) that was used to find 2E . The solution to above

equation can be found by choosing 1 1c = , and use the first equation (or second equation both

give same results within over all sign, you can verify it) to find 2c , and then normalizing the

obtained solution.

Defining,

( ) ( )22 2 20 1 2 0 1 2X B B Jγ σ σ γ σ σ= − + − +

The final normalized solution can be written in terms of X and J as follows,

1

2 22

1c Jc XJ X

= =

+ c

Similarly, the eigenfunction 3ϕ (or equivalently 1 2,d d ) corresponding to energy 3E , is

found by solving the corresponding eigenvalue equation.

2 31

23 3

1 14 2 01 12 4

E J J ddJ E J

− − =

− −

h h

h h

E

E

By substituting the value of 3E we found out, we can solve these equations. Final

normalized solution is as follows.

1

2 22

1d Xd JJ X

= = −+

d

You can verify that the vectors c and d vectors are orthonormal, as they should be. We

have completely solved the secular equations, and we now have the complete energies and

eigenstates for the general case of two interacting non-equivalent protons.

In previous section, we have seen that first-order perturbation theory did not predict the

correct spectrum for the case of two interacting equivalent proton system. Now that we have

obtained the exact solutions using the full variational method, let us substitute 1 2σ σ σ= = in the

these expressions and obtain energies and eigenstates.

29

1 0

2

3

4 0

1(1 )4

1434

1(1 )4

B J

J

J

B J

γ σ

γ σ

= − − +

= +

= −

= + − +

h h

h

h

h h

E

E

E

E

Note that the second and third states which were degenerate for the zeroth-order

Hamiltonian are no longer degenerate. Let us also see what are the eigenstates of two interacting

equivalent protons. Substituting 1 2σ σ σ= = in the expressions for eigenstates we obtained from

solving the Secular problem, we get,

( ) ( )22 2 20 1 2 0 1 2X B B J X Jγ σ σ γ σ σ= − + − + ⇒ =

1

2 22

11 112

c Jc XJ X

= = ⇒ =

+ c c

1

2 22

11 112

d Xd JJ X

= = ⇒ = − −+

d d

Therefore, the four eigenstates of two-equivalent protons are,

1 1 (1) (2)ϕ ψ α α= =

( ) ( )2 2 31 1 (1) (2) (1) (2)2 2

ϕ ψ ψ β α α β= + = +

( ) ( )3 2 31 1 (1) (2) (1) (2)2 2

ϕ ψ ψ β α α β= − = −

4 4 (1) (2)ϕ ψ β β= =

You will recognize that these states are precisely the normalized eigenstates of total spin

operator for two spin-1/2 particles. We constructed these functions in the class when we coupled

spin angular momentum of two spin-1/2 particles. We also proved in the class that

1 2 4, ,ϕ ϕ ϕ are eigenfunctions of 2,ˆ totalI with same eigenvalue 1I = and they form a triplet,

and 3ϕ is eigenfunction of 2,ˆ totalI with eigenvalue 0I = , that is it is a singlet. The factor 12

is

the normalization factor.

30

In an earlier section, we considered non-interacting two proton system. We showed that

the Hamiltonian 0H for this system commutes with ˆtotalzI , and that eigenfunctions of ˆtotal

zI are

also eigenfunctions of 0H . We further proved that 0H does not in general commute with 2,ˆ totalI ,

but it does so only if two protons are equivalent. In this case, we can find eigenfunctions of 0H

which would be simultaneous eigenfunctions of 2,ˆ totalI operator as well. Therefore, we could

construct a set of functions, which are simultaneous eigenfunctions of 0H , ˆtotalzI and 2,ˆ totalI .

The appearance of 2,ˆ totalI and ˆtotalzI eigenfunctions 1 2 3 4, , ,ϕ ϕ ϕ ϕ as eigenfunctions

even for the case of complete interacting Hamiltonian H , shows that H and 2,ˆ totalI also

commute. This is true in general. It is possible to prove in general that Hamiltonian for a system

of interacting equivalent interacting protons commutes with 2,ˆ totalI and ˆtotalzI . Using this, it is

possible to also prove that eigenstates of equivalent proton Hamiltonian are same as the

simultaneous eigenfunctions of 2,ˆ totalI and ˆtotalzI operators. We will return to this in the following

section.

Let us use the same selection rule 1IM∆ = ± as we used earlier, and find out the allowed

transitions and calculate the exact spectrum. We get,

1 2 1 0

1 3 2 0

2 4 3 0

3 4 4 0

with (1 )

with (1 )

with (1 )

with (1 )

B

B J

B

B J

ϕ ϕ ω γ σ

ϕ ϕ ω γ σ

ϕ ϕ ω γ σ

ϕ ϕ ω γ σ

→ = −

→ = − −

→ = −

→ = − +

The selection rule 1IM∆ = ± seems to predict three lines each separated by amount J .

However, since we have calculated the energies by full variation principle, they can not be

incorrect. Therefore, the selection rule 1IM∆ = ± must be inappropriate for the special

equivalent proton case. In the next section, we will work out appropriate selection rules for both

equivalent and non-equivalent cases. Using those selection rules we will then show that, for the

case of equivalent protons, the transitions 1 3ϕ ϕ→ and 3 4 ϕ ϕ→ are actually forbidden. In

this special case, the only allowed transitions are 1 2ϕ ϕ→ and 2 4ϕ ϕ→ allowed, and as

31

can be seen from above, they have the same frequency. This leads to only one line in the NMR

spectrum of equivalent protons.

NMR Selection Rules

Selection rules are widely used in spectroscopy to find out allowed and forbidden

transitions when exposed to electro-magnetic radiation. Transitions can take place in general

between any two quantum states of a system. Since the quantum states of a system are

characterized by quantum numbers, the selection rules can be formulated in terms of allowed

changes in quantum numbers between initial and final states. Note that, although generally not

explicitly stated, selection rules are usually meant for a particular mechanism of transition.

Common and important mechanisms are the electric and magnetic dipole transitions, electric

quadrupole transitions and so on.

In practice, transitions only take place from those states of the system where it is likely to

be found in. This depends on the nature of quantum states (electronic, vibrational, rotational

etc.,), as well as the temperature. Since the energy gaps of between various states of nuclear spin

systems used in NMR (and even ESR) are very small compared to average thermal energy at

experimental temperatures, all these states are almost equally occupied, and transitions can take

place between any of them.

To obtain NMR selection rules, let us note that transitions are induced by a small

oscillating magnetic field 1( ) cos( )t tω= 01B B , applied in addition to the strong constant magnetic

field 0 ˆB z along z-axis which splits the energy levels of NMR spin system. Here, 01B is the vector

specifying direction and amplitude of the oscillating magnetic field. As shown in Problem 14-37

of McQuarrie and Simon, NMR spin system with total magnetic moment ˆˆ totalIµ γ= interacts

with this oscillating magnetic field as,

( )ˆ ˆ( ) cos( )totalH t I tγ ω= − ⋅ 01B

Accordingly, a transition from system initially in eigenstate iΨ to the final state jΨ

can happen only if the integral is non-zero.

ˆ 0totalj iIΨ ⋅ Ψ ≠0

1B

32

As discussed in Problem 14-37, it can be shown that, if the constant magnetic field is in

z-direction, the component of 01B in along z-direction does not induce any transition. And

therefore, the transition is possible if the following holds.

ˆ 0totalj x iIΨ Ψ ≠ and/or ˆ 0total

j y iIΨ Ψ ≠

Using angular momentum relations, you can show that this is equivalent to,

ˆ 0totalj iI+Ψ Ψ ≠ and/or ˆ 0total

j iI−Ψ Ψ ≠

To find selection rules, note that proton system eigenstates iΨ and jΨ are in general

eigenfunctions of ˆtotalzI . Let us assume that iΨ and jΨ are characterized by quantum

numbers IM and JM . Since ˆtotalI+ increases the ˆtotalzI quantum number by one, the final state

jΨ must be eigenfunction of ˆtotalzI with eigenvalue 1IM + , if the above integrals were not to

vanish. Similarly, using properties of ˆtotalI− we can argue that jΨ must be eigenfunction of

ˆtotalzI with eigenvalue 1IM − . Therefore, transitions can take place only between two states

which differ by one in their ˆtotalzI eigenvalues. Therefore, for NMR, the selection rule in general

is 1IM∆ = ± .

However, we know from angular momentum theory that ˆtotalI+ and ˆtotalI− operators can not

change the 2,ˆ totalI eigenvalue of a function. Therefore, if iΨ and jΨ happen to be eigenstates

of 2,ˆ totalI , then an additional selection rule 0I∆ = must also be considered. We have seen that in

general cases, our NMR Hamiltonian does not commute with 2,ˆ totalI . There is no spherical

symmetry in the spin space, and hence I is not a good quantum number in general, and it is not

defined. Therefore, 0I∆ = selection rule does not apply in general.

However, for the case of equivalent protons, Hamiltonian commutes with 2,ˆ totalI , and I

becomes a good quantum number. The eigenstates of equivalent proton system are the

eigenstates of total angular momentum. Therefore, additional 0I∆ = selection rule applies in

this case.

33

This is the reason why, in equivalent two proton system, the transitions 1 3ϕ ϕ→ and

3 4 ϕ ϕ→ are forbidden, as they involve change of total spin quantum number .i.e., triplet and

singlet states. Therefore, there is only one line for group of equivalent protons.

Exercises:

Deduce NMR selection rules for equivalent and non-equivalent cases from the condition of non-

zero values for the above integrals. Use angular momentum theory to express totalxL and total

yL in

terms of totalL+ and totalL− .

A revisit to the interacting equivalent two-proton case.

In this section, we will revisit the case of interacting equivalent two proton case, and

reconsider it from the point of view of angular momentum theory. The techniques used would

apply in a similar way for ESR spectroscopy and spin-orbit coupling in atoms. The interacting

two equivalent proton Hamiltonian can be written as,

( )0 0 1 0 2 1 2

0 1 2

0 1 2

1ˆ ˆ ˆ ˆ ˆ ˆ ˆ(1 ) (1) (1 ) (2)

1ˆ ˆ ˆ ˆ (1 ) (1) (2)

1ˆ ˆ ˆ (1 )

z z

z z

totalz

H H V B I B I J I I

B I I J I I

B I J I I

γ σ γ σ

γ σ

γ σ

= + = − − − − + ⋅

= − − + + ⋅

= − − + ⋅

h

h

h

Earlier, we proved that the spin-spin interaction V commutes with ˆtotalzI . We will now

prove that V commutes with 2,ˆ totalI as well, and hence the whole Hamiltonian commutes with 2,ˆ totalI . To prove this, we will try to rewrite V which will give insight into structure of this

Hamiltonian.

( ) ( )2,1 2 1 2

1 1 2 2 1 2 2 1

ˆ ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ

totalI I I I I

I I I I I I I I

= + ⋅ +

= ⋅ + ⋅ + ⋅ + ⋅

Since angular momentum operators of different particles commute, we have,

1 2 2 1ˆ ˆ ˆ ˆI I I I⋅ = ⋅

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Using this, we get, 2,

1 1 2 2 1 2

2 21 2 1 2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ2ˆ ˆ ˆ ˆ 2

totalI I I I I I I

I I I I

= ⋅ + ⋅ + ⋅

= + + ⋅

Therefore, we can rewrite 1 2ˆ ˆI I⋅ as follows.

( )2, 2 21 2 1 2

1ˆ ˆ ˆ ˆ ˆ2

totalI I I I I ⋅ = − +

Using the above relation to rewrite V as,

( )2, 2 21 2

1 1ˆ ˆ ˆ ˆ 2 2

totalV J I J I I= − +h h

Now, using this form consider commutation of V with 2,ˆ totalI . The first term can be

easily seen to commute. From angular momentum theory, we know each term in ( )2 21 2ˆ ˆI I+

commutes with 2,ˆ totalI . Let us prove this. The second term involves commutators such as 2, 2

1ˆ ˆ,totalI I (consider first particle as example). We will expand 2,ˆ totalI , by using the following

formula. 2, 2,ˆ ˆ ˆ ˆ ˆtotal total total total total

z zI I I I I− += + + h

For the proof, we need to just consider terms in 2,ˆ totalI involving first-particle only,

because all others commute with 21I . Collecting all terms in 2,ˆ ˆ ˆ ˆtotal total total total

z zI I I I− + + + h which

contain the co-ordinates of first particle, we get, 2, 2 2

1 1 1 1 1 1 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, ,total total total total total

z z z z zI I I I I I I I I I I I− + − + = + + + + h

Using 21 1ˆ ˆ, 0zI I = and 2

1ˆ ˆ, 0total

zI I = , we get

2, 2 2 21 1 1 1 1

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , ,total total totalI I I I I I I I− + − + = +

Using 21 1ˆ ˆ, 0I I±

= we get,

2, 2 2 21 1 1 1 1

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, , ,total total totalI I I I I I I I− + − + = +

Using 2 21 1 1

ˆ ˆ ˆ ˆ, , 0totalI I I I± ± = = , we prove that

2, 21

ˆ ˆ, 0totalI I =

35

Therefore, the Hamiltonian for equivalent protons commutes with 2,ˆ totalI as well. Note

that in this process we have also proved that Hamiltonian commutes with 21I and 2

2I as well.

Using this new form for spin-spin coupling, the Hamiltonian can be rewritten as follows,

( )2, 2 20 1 2

1 1ˆ ˆ ˆ ˆ ˆ(1 ) 2 2

total totalzH B I J I J I Iγ σ= − − + − +

h h

Let us now summarize the full symmetries of this Hamiltonian.

ˆ ˆ, 0totalzH I =

2,ˆ ˆ, 0totalH I =

21

ˆ ˆ, 0H I =

22

ˆ ˆ, 0H I =

Therefore, the Hamiltonian has full spherical symmetry. Actually, you can see that

Hamiltonian contains purely angular momentum operators { }2, 2 21 2

ˆ ˆ ˆ ˆ, , ,total totalzI I I I . From this, we

can already conclude that the eigenstates of this Hamiltonian must be same as the total angular

momentum states (coupled states) obtained by adding the angular momentum of two particles. Its

eigenvalue must be completely expressible in terms of angular momentum eigenvalues.

However, let us try to prove it in general, using the theory of angular momentum we discussed in

the class. We know that the 2-particle angular momentum states are simultaneous eigenfunctions

of { }2, 2 21 2

ˆ ˆ ˆ ˆ, , ,total totalzI I I I operators. Let us denote these states as ( )1 2, , ,II M I IΦ .

As a little digression, for atoms, the total atomic Hamiltonian including the spin-orbit

interaction term involving ˆL S⋅ can be shown to commute with { }2, 2 2ˆˆ ˆ ˆ, , ,total totalzJ J L S operators.

This means that atomic states (eigenfunctions of atomic Hamiltonian) are also eigenfunctions of

these operators, and can be labelled by eigenvalues of operators with eigenvalues { }, , ,JJ M L S .

For a given { }, ,J L S , all atomic states with ,...,JM J J= − + are all degenerate, and hence they

are grouped together and referred to as “levels”, denoted by (2 1)SJL+ . All states in levels which

are obtained by adding a given total orbital and spin angular momentum quantum numbers L

and S , are grouped together are referred to as “terms”, denoted by (2 1)S L+ .

36

Therefore, the 2-particle “coupled” angular momentum eigenfunctions ( )1 2, , ,II M I IΦ

are defined as,

( ) ( )( ) ( )

( ) ( )( ) ( )

2, 21 2 1 2

1 2 1 2

2 21 1 2 1 1 1 2

2 22 1 2 2 2 1 2

ˆ , , , ( 1) , , ,ˆ , , , , , ,ˆ , , , ( 1) , , ,ˆ , , , ( 1) , , ,

totalI I

totalz I I I

I I

I I

I I M I I I I I M I I

I I M I I M I M I I

I I M I I I I I M I I

I I M I I I I I M I I

Φ = + Φ

Φ = Φ

Φ = + Φ

Φ = + Φ

h

h

h

h

These are for the very general case of addition of arbitrary angular momentum 1I and 2I .

For our case of two protons, we have 112

I = and 212

I = . Possible values of 0, 0II M= = and

1, 1,0, 1II M= = − + . Therefore, we have only four functions.

1 1 2 11 11, 1, ,2 2II M I Iϕ ψ = Φ = = + = = =

( )2 1 2 2 31 1 11, 0, ,2 2 2II M I Iϕ ψ ψ = Φ = = = = = +

( )3 1 2 2 31 1 10, 0, ,2 2 2II M I Iϕ ψ ψ = Φ = = = = = −

4 1 2 41 11, 1, ,2 2II M I Iϕ ψ = Φ = = − = = =

We will try to prove for a general case, that ( )1 2, , ,II M I IΦ are eigenstates of

Hamiltonian in general, and obtain its eigenvalues.

( ) ( )

( )

( ) ( )

1 2 0 1 2

2,1 2

2 21 2 1 2

ˆ ˆ, , , (1 ) , , ,

1 ˆ , , ,21 ˆ ˆ , , ,

2

totalI z I

totalI

I

H I M I I B I I M I I

J I I M I I

J I I I M I I

γ σΦ = − − Φ

+ Φ

− + Φ

h

h

Each term on right hand side involves only angular momentum operators. Using the

definition of these operators, we get,

( ) ( )1 2 1 2 1 2ˆ , , , ( , , , ) , , ,I I IH I M I I I M I I I M I IΦ = ΦE

37

[ ]1 2 0 1 1 2 21( , , , ) (1 ) ( 1) ( 1) ( 1)2I II M I I B M J I I I I I Iγ σ= − − + + − + − +h hE

This is beautiful. We have not only proved that ( )1 2, , ,II M I IΦ are eigenstates of

Hamiltonian, but also obtained their eigenvalues, by just using angular momentum theory. We

can now use specific values relevant for two-proton system to get,

1 1 2 01 1 1( 1, 1, , ) (1 )2 2 4II M I I B Jγ σ= = = + = = = − − +h hE E

2 1 21 1 1( 1, 0, , )2 2 4II M I I J= = = = = = + hE E

3 1 21 1 3( 0, 0, , )2 2 4II M I I J= = = = = = − hE E

4 1 2 01 1 1( 1, 1, , ) (1 )2 2 4II M I I B Jγ σ= = = − = = = + − +h hE E

These are the same energies as those we obtained by using the tedious procedure of

solving secular equations.

Exercise:

Using the general energy formula, and using selection rules appropriate for this case, .i.e.,

1IM∆ = ± and 0I∆ = , prove that all allowed transitions for a general case have same frequency,

and show that the frequency is independent of magnitude of spin-spin coupling constant.

38