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Transcript
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 nonzero spin-angular momentum (molecules in non-singlet states, S ≠ 0 ) are placed in constant
magnetic field. The ESR spectrum resulting from transitions between molecular states with
different M S 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 nonzero nuclear spins ( I ≠ 0 ) 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
1
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
r
around its axis. This gives raise to a magnetic (dipole) moment µ .
r
r
µ = gβ J
β=
(1)
q
2m
r
Where, q and m are charge and mass of the particle, and J 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 β e . Usually, the negative sign in β is explicitly
written into Eqn (1), and it is rewritten as.
r
r

µe = − g e β e J

e
 for electrons
βe =

2me

(1-e)
Therefore, for electrons the magnetic moment is in reverse direction of total angular
r
momentum J . 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
2
( L ), giving raise to a total angular momentum ( J ). For this case, the g-factor is given by Lande
factor as,
ge = 1 +
J ( J + 1) − L( L + 1) + S ( S + 1)
2 J ( J + 1)
(2)
For ground and low-lying states of molecules, usually the orbital angular momentum ( L )
is zero or small L ≈ 0 . 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 β N . For nuclei, Eqn (1) can be rewritten as,
r
r

µN = gN β N J

(1-N)
q
 for nuclei
βN =

2 mN

r
Therefore, magnetic moment is the same direction as J . 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.
r
In literature, h appearing in J 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 β e =
eh
is 9.2*10-24 J/T. For
2me
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
β N ( proton) = β p =
times
heavier
than
electrons,
the
nuclear-magneton
for
proton
eh
is about 1000 times smaller than Bohr-magneton. As a result, this leads
2m p
to transitions observable in radio-wave region (one to a few hundred MHz).
3
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 by γ .
γ = gβ
r
(3)
r
µ =γJ
Again for nuclei, gyro-magnetic ratio is positive and for electrons it is negative. In
r
literature, h appearing in J 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
13
C 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
13
C 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.
4
r
When placed in external magnetic field, a magnetic dipole µ will interact with the
r
applied magnetic field B . The interaction potential energy of is given by (refer to MS Problem
13-49),
r r
E = −µ • B
(4)
r
As can be seen, the interaction energy will be minimum when µ is in the same direction
r
as B . 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,
E = − µ z Bz = − µ z B0
(5)
The quantum mechanical Hamiltonian for this interaction is obtained by expressing µ z in
above expression by µ z = γ Jˆ z obtained from Eq (3).
Hˆ 0 = −γ B0 Jˆ z
(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 J + 1)
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,
∆E = hγ B0
(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 ∆M J = ±1
will be allowed. How many transitions will be observed? Does Eq (7) require any
modification?
5
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 ω = γ B0 ) 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 hyperfine 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,
ωbare = γ B0 =
Eβ − Eα
h
=
∆E (α → β )
= ωspectrometer
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
6
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 Belec 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.
Belec = −σ B0
(9)
Here, σ is the proportionality constant (which is unit-less). The negative sign in Eq.(9)
shows that Belec 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 Blocal and corresponding change in Eq (6) is therefore given
by,
Blocal = B0 − σ B0 = (1 − σ ) B0
Hˆ = −γ B Iˆ = −γ B (1 − σ ) Iˆ
0
local z
0
(10)
z
In the above equation, nuclear-spin denoted by Iˆz replaces the general angular
momentum Jˆ z used in earlier equations. As opposed to Eq (8), the resonant frequency including
shielding effects is given by,
ω = γ B0 (1 − σ )
∆ω = −σωbare
(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 Belec , and absorb at slightly different frequencies. Observing
7
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 δ H is assigned to each
proton in molecule as follows.
δH =
(ωH − ωTMS ) ×106
ωspectrometer
(12)
Here, ωTMS is the resonance frequency of the protons of reference molecule known as
tetra methyl silane (TMS) Si(CH3)4, and ωspectrometer 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 σ 1 and σ 2 . If the protons are not interacting, the total 2-particle Hamiltonian
for this system is given by,
Hˆ 0 = −γ B0 (1 − σ 1 ) Iˆz (1) + (1 − σ 2 ) Iˆz (2) 
(13)
Exercise:
8
Prove that Ĥ 0 commutes with z-component of the total nuclear spin angular momentum,
.i.e., with Iˆztotal = Iˆz = Iˆz (1) + Iˆz (2) . Use the known eigenfunctions {α (1), β (1)} of Iˆz (1) and
{α (2), β (2)}
of Iˆz (2) , to construct a set of simultaneous eigenfunctions of Ĥ 0 and Iˆztotal . What
are eigenvalues of Hamiltonian and corresponding eigenfunctions? List them in order of
increasing energies and their Iˆztotal eigenvalues.
Answer:
Consider,
 Hˆ 0 , Iˆztotal  =  Hˆ 0 , Iˆz (1)  +  Hˆ 0 , Iˆz (2) 

 
 

= −γ B0 (1 − σ 1 )  Iˆz (1), Iˆz (1)  − γ B0 (1 − σ 2 )  Iˆz (2), Iˆz (1) 
− γ B0 (1 − σ 1 )  Iˆz (1), Iˆz (2)  − γ B0 (1 − σ 2 )  Iˆz (2), Iˆz (2) 
=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 Iˆztotal operator, and therefore can be constructed as a linear combination of
eigenfunctions of Iˆztotal operator.
The possible states of this 2-proton system are,
ψ 1 = ψ 1 (1, 2) = α (1)α (2)
ψ 2 = ψ 2 (1, 2) = β (1)α (2)
ψ 3 = ψ 3 (1, 2) = α (1) β (2)
(14)
ψ 4 = ψ 4 (1, 2) = β (1) β (2)
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.,.
9
Dirac representation is simple, and powerful, and provides a co-ordinate independent way to
represent quantum mechanical quantities.
around the wave-function ψ i (1, 2) is used to represent it as ψ i . The
A half bracket
co-ordinates 1, 2 are dropped, and are written explicitly when it is needed. The overlap integral
of state ψ i with another state ψ j is written as follows.
ψ j | ψ i = ∫ψ *j (1, 2)ψ i (1, 2)d1d 2
The relation
∫ψ
*
j
(1, 2)ψ i (1, 2)d1d 2 =
( ∫ψ
*
i
(1, 2)ψ j (1, 2)d1d 2
)
*
which is the property of
overlap integrals, is now simply represented in Dirac notation as,
ψ j |ψ i = ψ i |ψ j
*
Similarly, the expectation value of a quantum mechanical operator  , for example say
Ĥ 0 or Iˆz , between a two states ψ i
and ψ j is represented as,
ψ j | Aˆ |ψ i = ∫ψ *j (1, 2) Aˆψ i (1, 2)d1d 2
Adjoint operator of an operator  is defined as another operator B̂ which is related to Â
for any two states ψ i
and ψ j through the following relation.
ψ j | Bˆ | ψ i = ψ i | Aˆ | ψ j
*
If the above relation holds, B̂ is said to be adjoint of  , and is denoted as † . An
operator  which is its own adjoint, .i.e., Aˆ = Aˆ † , 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
{ψ
i
, i = 1, K N } is used, an operators  is represented by corresponding matrix Α
whose elements are A ji = ψ j | Aˆ | ψ i . 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 { ψ i , i = 1,K, 4} are eigenfunctions of Iˆztotal . The function ψ 1 has M I = +1 , ψ 2
and ψ 3 have M I = 0 , and ψ 1 has M I = −1 . By acting Ĥ 0 on each of these functions, you can
10
verify that these are also eigenfunctions of Ĥ 0 (this happens due to non-interacting nature of
Ĥ 0 ). Therefore, we can conclude that Ĥ 0 is diagonal in { ψ i , i = 1,K, 4} basis. The energies are
just expectation value of Ĥ 0 in this basis.
1
E1 = E [α (1)α (2) ] = ψ 1 Hˆ 0 ψ 1 = − hγ B0 [ (1 − σ 1 ) + (1 − σ 2 ) ]
2
1
1
E2 = E [ β (1)α (2) ] = ψ 2 Hˆ 0 ψ 2 = − hγ B0 [ −(1 − σ 1 ) + (1 − σ 2 ) ] = − hγ B0 [σ 1 − σ 2 ]
2
2
1
1
E3 = E [α (1) β (2) ] = ψ 3 Hˆ 0 ψ 3 = − hγ B0 [ (1 − σ 1 ) − (1 − σ 2 ) ] = + hγ B0 [σ 1 − σ 2 ]
2
2
1
E4 = E [ β (1) β (2)] = ψ 4 Hˆ 0 ψ 4 = + hγ B0 [ (1 − σ 1 ) + (1 − σ 2 ) ]
2
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
∆M I = ±1 . 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 ∆M I = ±1 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 M I values for the four states above, ∆M I = ±1 will
give following transitions.
11
ψ1 → ψ 2
ω1 = γ B0 (1 − σ 1 )
α → β spin flip of first proton with second proton in α state.
ψ1 → ψ 3
ω2 = γ B0 (1 − σ 2 )
α → β spin flip of second proton with first proton in α state
ψ2 → ψ4
ω3 = γ B0 (1 − σ 2 )
α → β spin flip of second proton with first proton in β state
ψ3 → ψ4
ω4 = γ B0 (1 − σ 1 )
α → β spin flip of first proton with second proton in β state
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 ω1 = ω4 , and ω2 = ω3 . 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 Ĥ 0 commute with Iˆ 2,total ? If not, under what conditions does it commute? Give physical
explanation for the results.
Using Iˆ 2,total = Iˆ− Iˆ+ + Iˆz2 + hIˆz
 Hˆ 0 , Iˆ 2,total  =  Hˆ 0 , Iˆ− Iˆ+  +  Hˆ 0 , Iˆz2 + hIˆz 

 
 

The second term vanishes, and the first term can be worked out using basic angular
commutation relations. The final result is,
 Hˆ 0 , Iˆ 2,total  = (σ 1 − σ 2 )  Iˆ− (1) Iˆ+ (2) − Iˆ− (2) Iˆ+ (1) 




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 Ĥ 0 in general does not commute with Iˆ 2,total .
However, notice that it commutes when σ 1 = σ 2 , which happens when the two-protons are
equivalent.
12
The reason why Ĥ 0 does not commute with Iˆ 2,total 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, σ 1 ≠ σ 2 , 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, σ 1 = σ 2 = σ , The Hamiltonian Ĥ 0 reduces to,
Hˆ 0 = −γ B0 (1 − σ )  Iˆz (1) + Iˆz (2)  = −γ B0 (1 − σ ) Iˆztotal
The above form of the Hamiltonian also shows that it is just proportional to Iˆztotal , and its
eigenvalues, unlike the general non-equivalent proton case, are just eigenvalues of Iˆztotal operator
multiplied by a constant number −γ B0 (1 − σ ) . The special form also clearly shows that Ĥ 0
commute with Iˆ 2,total , since we know that Iˆztotal commutes with Iˆ 2,total 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.
13
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 M J 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., ω1 , ω4 will differ from each other depending on the magnitude of this interaction.
Similarly, ω2 , ω3 transitions will also differ.
Usually, the strength of spin-spin interaction is very small compared to their individual
screening. Therefore, the transitions ω1 , ω4 will be close to each other, and ω2 , ω3 will be close to
each other. This qualitatively explains the (doublet) splitting of ω1 = ω4 transition line of first
proton, and same doublet splitting for second proton.
14
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,
r r
V12 = J12 µ1 ⋅ µ2
Here, the subscripts 1 and 2 corresponds to that of first and second protons. J12 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
r
constant J12 is specific to a pair of protons in a specific chemical environment. Since, µ for
each proton is proportional to its spin, the above interaction can be written in quantum
mechanical operator form as,
Vˆ12 = J12 Iˆ1 ⋅ Iˆ2
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,
1
Hˆ = Hˆ 0 + Vˆ12 = −γ B0 (1 − σ 1 ) Iˆz (1) − γ B0 (1 − σ 2 ) Iˆz (2) + J12 Iˆ1 ⋅ Iˆ2
h
We have added
1
to enable us to express J12 in HZ (1/sec or radians per second), just
h
like γ B0 which is expressed in same units. Since there is only one spin-spin coupling constant
J12 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 γ B0 , or rather it is smaller
compared to the difference in resonating frequencies of two protons. Or more precisely,
15
J << γ B0 σ 1 − σ 2
Therefore, we can treat Vˆ12 as a small perturbation to the zeroth-order Hamiltonian Ĥ 0 ,
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 firstorder perturbation correction to energy of zeroth order state ψ i is given by,
Ei(1) = ψ i Vˆ ψ i
The total energy of the perturbed state within first order perturbation theory is obtained
by adding the above correction to the zeroth order energies.
Ei(1) = Ei + ψ i Vˆ ψ i
= ψ i Hˆ 0 ψ i + ψ i Vˆ ψ i
= ψ i Hˆ ψ i
We can calculate the corrections Ei(1) = ψ i Vˆ ψ i , and final first-order energies Ei(1) as
follows.
Ei(1) =
1
J ψ i Iˆ1 ⋅ Iˆ2 ψ i
h
We can evaluate these matrix elements, by expanding Iˆ1 ⋅ Iˆ2 as,
Iˆ1 ⋅ Iˆ2 = Iˆ1x Iˆ2 x + Iˆ1 y Iˆ2 y + Iˆ1z Iˆ2 z
However, we will use another convenient expression for Iˆ1 ⋅ Iˆ2 as follows.
(
)
1
Iˆ1 ⋅ Iˆ2 = Iˆ1+ Iˆ2− + Iˆ1− Iˆ2+ + Iˆ1z Iˆ2 z
2
You will derive this expression in the assignment problem 8.
1
J α (1)α (2)
h
1
= J α (1)α (2)
h
1
= J α (1)α (2)
h
1
= + Jh
4
E1(1) =
Iˆ1 ⋅ Iˆ2 α (1)α (2)
(
)
1 ˆ ˆ
I1+ I 2− + Iˆ1− Iˆ2+ + Iˆ1z Iˆ2 z α (1)α (2)
2
Iˆ1z Iˆ2 z α (1)α (2)
16
The first two terms vanished because either Iˆ1+ or Iˆ2+ could not act on α (1)α (2) state.
You can verify yourself that, in general,
(
)
1 ˆ ˆ
I1+ I 2− + Iˆ1− Iˆ2+ can not contribute to diagonal terms.
2
Therefore, we get,
E2(1) =
1
1
J β (1)α (2) Iˆ1 ⋅ Iˆ2 β (1)α (2) = − hJ
h
4
E3(1) =
1
1
J α (1) β (2) Iˆ1 ⋅ Iˆ2 α (1) β (2) = − hJ
h
4
E4(1) =
1
1
J β (1) β (2) Iˆ1 ⋅ Iˆ2 β (1) β (2) = + hJ
h
4
Therefore, the total first-order energies are,
1
1
E1(1) = − hγ B0 [ (1 − σ 1 ) + (1 − σ 2 ) ] + hJ
2
4
1
1
E2(1) = − hγ B0 [σ 1 − σ 2 ] − hJ
2
4
1
1
E3(1) = + hγ B0 [σ 1 − σ 2 ] − hJ
2
4
1
1
E4(1) = + hγ B0 [ (1 − σ 1 ) + (1 − σ 2 ) ] + hJ
2
4
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 ∆ =
1
hJ . Similarly, the second and third states go down by the same amount. This
4
results in Four lines as shown. Using the above energies, and selection rules ∆M I = ±1 the firstorder NMR spectra for our 2-proton system will have four lines at the following frequencies.
17
ψ1 → ψ 2
with
ψ1 → ψ 3
with
ψ2 → ψ4
with
ψ3 → ψ4
with
1
2
1
ω2 = γ B0 (1 − σ 2 ) − J
2
1
ω3 = γ B0 (1 − σ 2 ) + J
2
1
ω4 = γ B0 (1 − σ 1 ) + J
2
ω1 = γ B0 (1 − σ 1 ) − J
The above spectrum shows four lines. The two lines with frequencies ω1 and ω4 center
around γ B0 (1 − σ 1 ) are separated by J . These two lines correspond to the two lines with
ω1 = ω4 we associated with flipping of first proton in the zeroth-order (non-interacting)
spectrum. Similarly, the two lines with frequencies ω2 and ω3 correspond to the two lines with
ω2 = ω3 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 (n + 1) 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 (n + 1) rule. It is explained
in great detail with examples, in the chapter 14 of McQuarrie and Simons.
The (n + 1) 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 (n + 1) closely spaced lines in the first-order spectrum.
18
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 (n + 1) 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
α (1)α (2)α (3)
Group-2
β (1)α (2)α (3) , α (1) β (2)α (3) , α (1)α (2) β (3)
Group-3
β (1) β (2)α (3) , β (1)α (2) β (3) , α (1) β (2) β (3)
Group-4
β (1) β (2) β (3)
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 (n + 1) 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
19
neighbouring protons. By denoting the number of ways of choosing r objects from among n
objects as,
n
Cr =
n!
r !( n − r ) !
Among all possible spin states of these n equivalent neighbouring protons, there would
be nC0 = 1 state in which all protons are in alpha state and no proton is in beta state, there will be
n
C1 = n states in which one proton is in beta state, nC2 states in which two protons are in beta
state and so on. Because these n protons are all equivalent, all possible the nCr 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 nC1 = n states, and so on. Therefore, there will be
( n + 1)
peaks in all with the intensity ratios given by nC0 : nC1 : nC2 : L : nCn − 2 : nCn −1 : nCn .
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., J << γ B0 σ 1 − σ 2 ., 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 ( n + 1) rule and intensity rule.
Transitions within a System of Equivalent Protons
While calculating Intensity ratios and using the (n + 1) 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
σ 1 = σ 2 = σ in the first-order expressions for transition energies, leaves us with two lines
20
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 Ĥ 0 has
a degenerate energy levels for equivalent protons. To see this, let us find what happens to the
energy levels and eigenstates of Ĥ 0 , when two protons are equivalent .i.e. σ 1 = σ 2 = σ . We get,
E1 = E [α (1)α (2) ] = ψ 1 Hˆ 0 ψ 1 = −hγ B0 (1 − σ )
E2 = E [ β (1)α (2) ] = ψ 2 Hˆ 0 ψ 2 = 0
E3 = E [α (1) β (2) ] = ψ 3 Hˆ 0 ψ 3 = 0
E4 = E [ β (1) β (2) ] = ψ 4 Hˆ 0 ψ 4 = +hγ B0 (1 − σ )
Therefore, the second state
β (1)α (2)
and third state α (1) β (2)
states become
degenerate when two protons are equivalent. When you apply first-order perturbation theory on
1
second and third state, the perturbation shifts their energies by the same amount − hJ and
4
hence it does not break this degeneracy. The first state is shifted up by
shifted up by
1
hJ and fourth state is
4
1
hJ . Therefore, we get following first-order energies.
4
1
E1(1) = −hγ B0 (1 − σ ) + hJ
4
21
1
E2(1) = − hJ
4
1
E3(1) = − hJ
4
1
E4(1) = + hγ B0 (1 − σ ) + hJ
4
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 Ĥ involves construction of matrix elements of
the Hamiltonian H ij = χ i Hˆ χ j in some orthogonal basis
{χ
i
, i = 1,K, N } . 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(H − E 1) = 0
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.
( H − E 1) c = 0
22
The above linear equation is known as eigenvalue equation. Note that, since E used in
this equation satisfies secular equation, det(H − E 1) = 0 and therefore the matrix ( H − E 1) is not
invertible.
For Hermitian matrices, we know that its eigenvectors are orthogonal, and span the same
space as the original basis
{χ
, i = 1,K, 4} . If we choose these eigenvectors as new orthogonal
i
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,
1
Hˆ = Hˆ 0 + Vˆ = −γ B0 (1 − σ 1 ) Iˆz (1) − γ B0 (1 − σ 2 ) Iˆz (2) + J Iˆ1 ⋅ Iˆ2
h
In principle, we can use any orthogonal basis
{χ
i
, i = 1,K, N } 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 Ĥ 0 commutes with Iˆztotal . You can prove that the spin-spin interaction term also
commutes with Iˆztotal . Consider,
1
 Iˆztotal , Vˆ  = J  Iˆztotal , Iˆ1 ⋅ Iˆ2 

 h 

(
)
1  ˆtotal 1 ˆ ˆ

J  I z , I1+ I 2− + Iˆ1− Iˆ2+ + Iˆ1z Iˆ2 z 
h 
2

1
1
=
J  Iˆztotal , Iˆ1+ Iˆ2−  +  Iˆztotal , Iˆ1− Iˆ2+  + J  Iˆztotal , Iˆ1z Iˆ2 z 
h
2h
=
(
)
Using Iˆztotal = Iˆz (1) + Iˆz (2) = Iˆ1z + Iˆ2 z , 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,
 Iˆztotal , Iˆ1+ Iˆ2−  = Iˆ1+  Iˆ1z + Iˆ2 z , Iˆ2−  +  Iˆ1z + Iˆ2 z , Iˆ1+  Iˆ2−



 

= Iˆ1+  Iˆ2 z , Iˆ2−  +  Iˆ1z , Iˆ1+  Iˆ2−
= −hIˆ Iˆ + hIˆ Iˆ
1+ 2 −
1+ 2 −
=0
23
Similarly, other term can be worked and found to be zero. Therefore, we can conclude
that that spin-spin coupling term Vˆ commutes with Iˆztotal , and the entire NMR Hamiltonian
commutes with Iˆztotal .
 Hˆ , Iˆztotal  =  Hˆ 0 , Iˆztotal  + Vˆ , Iˆztotal  = 0

 
 

Now, let us understand how this symmetry of Hamiltonian is useful in solving its secular
equations. Although, we illustrate with Iˆztotal 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 Iˆztotal , and we also know that they are orthogonal (because Iˆztotal 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 H ij = ψ i Hˆ ψ j
between two
basis functions ψ i and ψ j having different Iˆztotal eigenvalues mi , and m j respectively. Since,
 Hˆ , Iˆztotal  = 0 , we must have,


ψ i  Hˆ , Iˆztotal  ψ j = 0
ˆ ˆtotal − Iˆtotal Hˆ ψ = 0
ψ i HI
z
z
j
This leads to,
ˆ ˆtotal − Iˆtotal Hˆ ψ = 0
ψ i HI
z
z
j
ˆ ˆtotal ψ − ψ Iˆtotal Hˆ ψ = 0
ψ i HI
z
j
i
z
j
(m
j
− mi ) ψ i Hˆ ψ j = 0
In the above, we have made use of the fact that ψ i (and ψ i ), and ψ j (and ψ j ) are
eigenfunctions of Iˆztotal eigenvalues mi , and m j respectively. Since, we have assumed that these
eigenvalues are different, m j ≠ mi , or ( m j − mi ) ≠ 0 we get,
ψ i Hˆ ψ j = 0
Therefore, the Hamiltonian matrix elements between basis functions having different
eigenvalues of Iˆztotal are necessarily zero, and we do not have construct such elements. This is
solely the consequence of Iˆztotal commuting with Ĥ . We can straightforward set these elements to
24
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 Iˆztotal eigenvalues. The rest are
simply zero by symmetry (or commutation), and this procedure is known as symmetryadaptation. Therefore, the Hamiltonian matrix entering the secular equations attains a block
diagonal form, which one block corresponding to an eigenvalue of Iˆztotal operator. As a result of
this, we can see that the original secular problem det(H − E 1) = 0 splits into secular subproblems 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 Iˆztotal with a definite Iˆztotal eigenvalue, same as the Iˆztotal
eigenvalue of the basis states of that block.
From angular momentum theory, we already know all the eigenfunctions Iˆztotal operator,
and their eigenvalues. For this two-proton case, they are just the functions { ψ 1 , ψ 2 , ψ 3 , ψ 4
}
used in previous sections. We already know that ψ 1 has Iˆztotal eigenvalue of M I = +1 , both that
ψ 2 and ψ 3 have Iˆztotal eigenvalue of M I = 0 , and ψ 4 has Iˆztotal eigenvalue of M I = −1 .
Therefore, Hamiltonian matrix in this basis, splits into the following block diagonal form.
 H11

0
H=
 0

 0
0
0
H 22
H 23
H 32
0
H 33
0
0 

0 
0 

H 44 
These six non-zero matrix elements can be evaluated by using the techniques already
presented in previous sections. The final results are,
1
H11 = ψ 1 Hˆ ψ 1 = E1 + hJ
4
1
H 22 = ψ 2 Hˆ ψ 2 = E2 − hJ
4
25
1
H 33 = ψ 3 Hˆ ψ 3 = E3 − hJ
4
1
H 44 = ψ 4 Hˆ ψ 4 = E44 + hJ
4
1
H 23 = ψ 2 Hˆ ψ 3 = hJ
2
1
H 32 = ψ 3 Hˆ ψ 2 = H 23 = hJ
2
Where { Ei , i = 1, K , 4} are the zeroth-order energies we already calculated in a previous
section. Therefore, the Hamiltonian matrix for full secular problem is,
1

 E1 + 4 hJ


0

H=

0



0

0
0
1
E2 − hJ
4
1
hJ
2
1
hJ
2
1
E3 − hJ
4
0
0




0



0

1 
E4 + hJ 
4 
0
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,
1

 E2 − 4 hJ − E
det 
1

hJ

2

1

hJ

2
=0
1
E3 − hJ − E 
4

The exact energies are obtained as,
26
1
1
1
E1 = E1 + hJ = − hγ B0 [ (1 − σ 1 ) + (1 − σ 2 ) ] + hJ
4
2
4
1
1
2
E2 = − hJ + h γ 2 B02 (σ 1 − σ 2 ) + J 2
4
2
1
1
2
E3 = − hJ − h γ 2 B02 (σ 1 − σ 2 ) + J 2
4
2
1
1
1
E4 = E4 + hJ = + hγ B0 [ (1 − σ 1 ) + (1 − σ 2 ) ] + hJ
4
2
4
Note that the exact energies E1 and E4 are same as first-order perturbation theory results
E1(1) and E4(1) we obtained earlier. The first-order perturbation energies E2(1) and E3(1) are
recovered from the corresponding exact energies for E2 and E3 , by neglecting the term J 2
within the square root, .i.e., by assuming γ 2 B02 (σ 1 − σ 2 ) >> J 2 . This is same as the conditions
2
required for the validity of first-order perturbation theory.
We will also find eigenfunctions
{ϕ
i
, i = 1,K , 4}
of the Hamiltonian. The
eigenfunctions of two 1-by-1 blocks are just the corresponding Iˆztotal eigenfunctions with
eigenvalues M I = +1 and M I = −1 respectively.
ϕ1 = ψ 1
ϕ4 = ψ 4
The two eigenfunctions of 2-by-2 block will be some general linear combinations of the
two Iˆztotal eigenfunctions of with eigenvalue M I = 0 .
ϕ2 = c1 ψ 2 + c2 ψ 3
ϕ3 = d1 ψ 2 + d 2 ψ 3
The eigenfunction ϕ2 (or equivalently c1 , c2 ) corresponding to energy E2 , is found by
solving the corresponding eigenvalue equation.
1

 E2 − 4 hJ − E2

1

hJ

2

1

hJ
  c1 
2
  = 0
c
1
E3 − hJ − E2   2 
4

27
These are two simultaneous homogenous linear equations in c1 and c2 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 E2 . The solution to above
equation can be found by choosing c1 = 1 , and use the first equation (or second equation both
give same results within over all sign, you can verify it) to find c2 , and then normalizing the
obtained solution.
Defining,
X = γ B0 (σ 1 − σ 2 ) + γ 2 B02 (σ 1 − σ 2 ) + J 2
2
The final normalized solution can be written in terms of X and J as follows,
c 
c= 1=
 c2 
J
 
J2 + X2  X 
1
Similarly, the eigenfunction ϕ3 (or equivalently d1 , d 2 ) corresponding to energy E3 , is
found by solving the corresponding eigenvalue equation.
1

 E2 − 4 hJ − E3

1

hJ

2

1

hJ
  d1 
2
  = 0
d
1
E3 − hJ − E3   2 
4

By substituting the value of E3 we found out, we can solve these equations. Final
normalized solution is as follows.
d 
d= 1=
 d2 
X 


J 2 + X 2  −J 
1
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 σ 1 = σ 2 = σ in the
these expressions and obtain energies and eigenstates.
28
1
E1 = −hγ B0 (1 − σ ) + hJ
4
1
E2 = + hJ
4
3
E3 = − hJ
4
1
E4 = + hγ B0 (1 − σ ) + hJ
4
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 σ 1 = σ 2 = σ in the expressions for eigenstates we obtained from
solving the Secular problem, we get,
X = γ B0 (σ 1 − σ 2 ) + γ 2 B02 (σ 1 − σ 2 ) + J 2 ⇒ X = J
2
c 
c= 1=
 c2 
J
1  1
 ⇒c=
 
2  1
J +X X
d 
d= 1=
 d2 
X 
1 1
⇒d=


 
2  −1
J 2 + X 2  −J 
1
2
2
1
Therefore, the four eigenstates of two-equivalent protons are,
ϕ1 = ψ 1 = α (1)α (2)
ϕ2 =
1
( ψ2 + ψ3
2
)=
1
( β (1)α (2) + α (1)β (2)
2
)
ϕ3 =
1
( ψ2 − ψ3
2
)=
1
( β (1)α (2) − α (1)β (2)
2
)
ϕ4 = ψ 4 = β (1) β (2)
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
ϕ1 , ϕ2 , ϕ4 are eigenfunctions of Iˆ 2,total with same eigenvalue I = 1 and they form a triplet,
1
is
and ϕ3 is eigenfunction of Iˆ 2,total with eigenvalue I = 0 , that is it is a singlet. The factor
2
the normalization factor.
29
In an earlier section, we considered non-interacting two proton system. We showed that
the Hamiltonian Ĥ 0 for this system commutes with Iˆztotal , and that eigenfunctions of Iˆztotal are
also eigenfunctions of Ĥ 0 . We further proved that Ĥ 0 does not in general commute with Iˆ 2,total ,
but it does so only if two protons are equivalent. In this case, we can find eigenfunctions of Ĥ 0
which would be simultaneous eigenfunctions of Iˆ 2,total operator as well. Therefore, we could
construct a set of functions, which are simultaneous eigenfunctions of Ĥ 0 , Iˆztotal and Iˆ 2,total .
The appearance of Iˆ 2,total and Iˆztotal eigenfunctions ϕ1 , ϕ 2 , ϕ3 , ϕ 4 as eigenfunctions
even for the case of complete interacting Hamiltonian Ĥ , shows that Ĥ and Iˆ 2,total 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 Iˆ 2,total and Iˆztotal . Using this, it is
possible to also prove that eigenstates of equivalent proton Hamiltonian are same as the
simultaneous eigenfunctions of Iˆ 2,total and Iˆztotal operators. We will return to this in the following
section.
Let us use the same selection rule ∆M I = ±1 as we used earlier, and find out the allowed
transitions and calculate the exact spectrum. We get,
ϕ1 → ϕ2
with
ω1 = γ B0 (1 − σ )
ϕ1 → ϕ3
with
ω2 = γ B0 (1 − σ ) − J
ϕ2 → ϕ4
with
ω3 = γ B0 (1 − σ )
ϕ3 → ϕ 4
with
ω4 = γ B0 (1 − σ ) + J
The selection rule ∆M I = ±1 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 ∆M I = ±1 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 ϕ1 → ϕ3 and ϕ3 → ϕ 4
this special case, the only allowed transitions are ϕ1 → ϕ2
are actually forbidden. In
and ϕ2 → ϕ 4 allowed, and as
30
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 B1 (t ) = B10 cos(ωt ) , applied in addition to the strong constant magnetic
field B0 zˆ along z-axis which splits the energy levels of NMR spin system. Here, B10 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 µˆ = γ Iˆtotal interacts
with this oscillating magnetic field as,
(
)
Hˆ (t ) = −γ Iˆtotal ⋅ B10 cos(ωt )
Accordingly, a transition from system initially in eigenstate Ψ i to the final state Ψ j
can happen only if the integral is non-zero.
Ψ j Iˆtotal ⋅ B10 Ψ i ≠ 0
31
As discussed in Problem 14-37, it can be shown that, if the constant magnetic field is in
z-direction, the component of B10 in along z-direction does not induce any transition. And
therefore, the transition is possible if the following holds.
Ψ j Iˆxtotal Ψ i ≠ 0 and/or Ψ j Iˆytotal Ψ i ≠ 0
Using angular momentum relations, you can show that this is equivalent to,
Ψ j Iˆ+total Ψ i ≠ 0 and/or Ψ j Iˆ−total Ψ i ≠ 0
To find selection rules, note that proton system eigenstates Ψ i and Ψ j are in general
eigenfunctions of Iˆztotal . Let us assume that Ψ i
and Ψ j
are characterized by quantum
numbers M I and M J . Since Iˆ+total increases the Iˆztotal quantum number by one, the final state
Ψ j must be eigenfunction of Iˆztotal with eigenvalue M I + 1 , if the above integrals were not to
vanish. Similarly, using properties of Iˆ−total we can argue that
Ψj
must be eigenfunction of
Iˆztotal with eigenvalue M I − 1 . Therefore, transitions can take place only between two states
which differ by one in their Iˆztotal eigenvalues. Therefore, for NMR, the selection rule in general
is ∆M I = ±1 .
However, we know from angular momentum theory that Iˆ+total and Iˆ−total operators can not
change the Iˆ 2,total eigenvalue of a function. Therefore, if Ψ i and Ψ j happen to be eigenstates
of Iˆ 2,total , then an additional selection rule ∆I = 0 must also be considered. We have seen that in
general cases, our NMR Hamiltonian does not commute with Iˆ 2,total . 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, ∆I = 0 selection rule does not apply in general.
However, for the case of equivalent protons, Hamiltonian commutes with Iˆ 2,total , and I
becomes a good quantum number. The eigenstates of equivalent proton system are the
eigenstates of total angular momentum. Therefore, additional ∆I = 0 selection rule applies in
this case.
32
This is the reason why, in equivalent two proton system, the transitions ϕ1 → ϕ3 and
ϕ3 → ϕ 4
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 nonzero values for the above integrals. Use angular momentum theory to express Lˆtotal
and Lˆtotal
in
x
y
and Lˆtotal
.
terms of Lˆtotal
+
−
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,
1
Hˆ = Hˆ 0 + Vˆ = −γ B0 (1 − σ 1 ) Iˆz (1) − γ B0 (1 − σ 2 ) Iˆz (2) + J Iˆ1 ⋅ Iˆ2
h
1
= −γ B0 (1 − σ ) Iˆz (1) + Iˆz (2) + J Iˆ1 ⋅ Iˆ2
h
1
= −γ B0 (1 − σ ) Iˆztotal + J Iˆ1 ⋅ Iˆ2
h
(
)
Earlier, we proved that the spin-spin interaction Vˆ commutes with Iˆztotal . We will now
prove that Vˆ commutes with Iˆ 2,total as well, and hence the whole Hamiltonian commutes with
Iˆ 2,total . To prove this, we will try to rewrite Vˆ which will give insight into structure of this
Hamiltonian.
(
)(
Iˆ 2,total = Iˆ1 + Iˆ2 ⋅ Iˆ1 + Iˆ2
)
= Iˆ1 ⋅ Iˆ1 + Iˆ2 ⋅ Iˆ2 + Iˆ1 ⋅ Iˆ2 + Iˆ2 ⋅ Iˆ1
Since angular momentum operators of different particles commute, we have,
Iˆ1 ⋅ Iˆ2 = Iˆ2 ⋅ Iˆ1
33
Using this, we get,
Iˆ 2,total = Iˆ1 ⋅ Iˆ1 + Iˆ2 ⋅ Iˆ2 + 2 Iˆ1 ⋅ Iˆ2
= Iˆ 2 + Iˆ 2 + 2 Iˆ ⋅ Iˆ
1
2
1
2
Therefore, we can rewrite Iˆ1 ⋅ Iˆ2 as follows.
(
)
1
Iˆ1 ⋅ Iˆ2 =  Iˆ 2,total − Iˆ12 + Iˆ22 

2
Using the above relation to rewrite Vˆ as,
(
1
1
Vˆ =
J Iˆ 2,total −
J Iˆ12 + Iˆ22
2h
2h
)
Now, using this form consider commutation of Vˆ with Iˆ 2,total . The first term can be
easily seen to commute. From angular momentum theory, we know each term in
( Iˆ
2
1
+ Iˆ22
)
commutes with Iˆ 2,total . Let us prove this. The second term involves commutators such as
 Iˆ 2,total , Iˆ12  (consider first particle as example). We will expand Iˆ 2,total , by using the following


formula.
Iˆ 2,total = Iˆ−total Iˆ+total + Iˆz2,total + hIˆztotal
For the proof, we need to just consider terms in Iˆ 2,total involving first-particle only,
because all others commute with Iˆ12 . Collecting all terms in Iˆ−total Iˆ+total + Iˆz2,total + hIˆztotal which
contain the co-ordinates of first particle, we get,
 Iˆ 2,total , Iˆ12  =  Iˆ1− Iˆ+total + Iˆ−total Iˆ1+ + Iˆ1z Iˆztotal + Iˆztotal Iˆ1z + hIˆ1z , Iˆ12 

 

Using  Iˆ1z , Iˆ12  = 0 and  Iˆztotal , Iˆ12  = 0 , we get
 Iˆ 2,total , Iˆ12  =  Iˆ1− Iˆ+total , Iˆ12  +  Iˆ−total Iˆ1+ , Iˆ12 

 
 

Using  Iˆ1± , Iˆ12  = 0 we get,
 Iˆ 2,total , Iˆ12  = Iˆ1−  Iˆ+total , Iˆ12  +  Iˆ−total , Iˆ12  Iˆ1+



 

Using  Iˆ±total , Iˆ12  =  Iˆ1± , Iˆ12  = 0 , we prove that
 Iˆ 2,total , Iˆ12  = 0


34
Therefore, the Hamiltonian for equivalent protons commutes with Iˆ 2,total as well. Note
that in this process we have also proved that Hamiltonian commutes with Iˆ12 and Iˆ22 as well.
Using this new form for spin-spin coupling, the Hamiltonian can be rewritten as follows,
(
1
1
Hˆ = −γ B0 (1 − σ ) Iˆztotal +
J Iˆ 2,total −
J Iˆ12 + Iˆ22
2h
2h
)
Let us now summarize the full symmetries of this Hamiltonian.
 Hˆ , Iˆztotal  = 0


 Hˆ , Iˆ 2,total  = 0


 Hˆ , Iˆ12  = 0


 Hˆ , Iˆ22  = 0


Therefore, the Hamiltonian has full spherical symmetry. Actually, you can see that
{
}
Hamiltonian contains purely angular momentum operators Iˆ 2,total , Iˆztotal , Iˆ12 , Iˆ22 . 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 Iˆ 2,total , Iˆztotal , Iˆ12 , Iˆ22 operators. Let us denote these states as Φ ( I , M I , I1 , I 2 ) .
As a little digression, for atoms, the total atomic Hamiltonian including the spin-orbit
{
}
interaction term involving L̂ ⋅ Sˆ can be shown to commute with Jˆ 2,total , Jˆ ztotal , Lˆ2 , Sˆ 2 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 { J , M J , L, S } .
For a given { J , L, S } , all atomic states with M J = − J ,..., + J are all degenerate, and hence they
are grouped together and referred to as “levels”, denoted by
(2 S +1)
LJ . 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 S +1)
L.
35
Therefore, the 2-particle “coupled” angular momentum eigenfunctions Φ ( I , M I , I1 , I 2 )
are defined as,
Iˆ 2,total Φ ( I , M I , I1 , I 2 ) = I ( I + 1)h 2 Φ ( I , M I , I1 , I 2 )
Iˆztotal Φ ( I , M I , I1 , I 2 ) = M I h Φ ( I , M I , I1 , I 2 )
Iˆ12 Φ ( I , M I , I1 , I 2 ) = I1 ( I1 + 1)h 2 Φ ( I , M I , I1 , I 2 )
Iˆ22 Φ ( I , M I , I1 , I 2 ) = I 2 ( I 2 + 1)h 2 Φ ( I , M I , I1 , I 2 )
These are for the very general case of addition of arbitrary angular momentum I1 and I 2 .
For our case of two protons, we have I1 =
1
1
and I 2 = . Possible values of I = 0, M I = 0 and
2
2
I = 1, M I = −1, 0, +1 . Therefore, we have only four functions.


1
1
ϕ1 = Φ  I = 1, M I = +1, I1 = , I 2 =  = ψ 1
2
2



1


1
1
ϕ2 = Φ  I = 1, M I = 0, I1 = , I 2 =  =
2
2

1
ϕ3 = Φ  I = 0, M I = 0, I1 = , I 2 =  =
2
2



1
1
( ψ2 + ψ3
2
)
1
( ψ2 − ψ3
2
)
1
ϕ4 = Φ  I = 1, M I = −1, I1 = , I 2 =  = ψ 4
2
2

We will try to prove for a general case, that Φ ( I , M I , I1 , I 2 )
are eigenstates of
Hamiltonian in general, and obtain its eigenvalues.
Hˆ Φ ( I , M I , I1 , I 2 ) = −γ B0 (1 − σ ) Iˆztotal Φ ( I , M I , I1 , I 2 )
1
J
2h
1
J
−
2h
Iˆ 2,total Φ ( I , M I , I1 , I 2 )
+
( Iˆ
2
1
)
+ Iˆ22 Φ ( I , M I , I1 , I 2 )
Each term on right hand side involves only angular momentum operators. Using the
definition of these operators, we get,
Hˆ Φ ( I , M I , I1 , I 2 ) = E ( I , M I , I1 , I 2 ) Φ ( I , M I , I1 , I 2 )
36
1
E ( I , M I , I1 , I 2 ) = −hγ B0 (1 − σ ) M I + hJ [ I ( I + 1) − I1 ( I1 + 1) − I 2 ( I 2 + 1) ]
2
This is beautiful. We have not only proved that Φ ( I , M I , I1 , I 2 ) 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
1
E1 = E ( I = 1, M I = +1, I1 = , I 2 = ) = −hγ B0 (1 − σ ) + hJ
2
2
4
1
1
1
E2 = E ( I = 1, M I = 0, I1 = , I 2 = ) = + hJ
2
2
4
1
1
3
E3 = E ( I = 0, M I = 0, I1 = , I 2 = ) = − hJ
2
2
4
1
1
1
E4 = E ( I = 1, M I = −1, I1 = , I 2 = ) = + hγ B0 (1 − σ ) + hJ
2
2
4
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.,
∆M I = ±1 and ∆I = 0 , 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.
37
38