5.61 Physical Chemistry Lecture #36 Page 1
NUCLEAR MAGNETIC RESONANCE
Just as IR spectroscopy is the simplest example of transitions being induced by light’s
oscillating electric field, so NMR is the simplest example of transitions induced by the
oscillating magnetic field. Because the strength of mattermagnetic field interactions are
typically two orders of magnitude smaller than the corresponding electric field interactions,
NMR is a much more delicate probe of molecular structure and properties. The NMR spin
Hamiltonians and wavefunctions are particularly simple, and permit us to demonstrate several
fundamental principles (about raising and lowering operators, energy levels, transition
probabilities, etc.) with a minimal amount of algebra. The principles and procedures are
applicable to other areas of spectroscopy electronic, vibrational, rotational, etc. – but for
these cases the algebra is more extensive.
Nuclear Spins in a Static Magnetic Field
For a single isolated spin in a static magnetic field, the contribution to the energy is:
ˆ iB 0 = −γ Iˆ iB 0
Ĥ 0 = −m
where γ is called the gyromagnetic ratio. If we choose our z axis to point in the direction of
the magnetic field then:
Ĥ 0 = −mˆ z B0 = −γ Iˆz B0
If we assume the nuclear spin is ½ (As it is for a proton) then we can easily work out the
energy levels of this Hamiltonian:
E± = ± 12 γ �B0 ≡ ± 12 �ω0
where ω0 = γB0 is called the nuclear Larmor frequency (rad/sec). Now, nuclei are never
isolated in chemistry – they are always surrounded by electrons. As we learned for the
hydrogen atom, the electrons near the nucleus shield the outer electrons from the bare
electric field produced by the nucleus. Similarly, the electrons shield the nucleus from the
bare electric field we apply in the laboratory. More specifically, the electron circulation
produces a field, B’ opposed to B0 and of
magnitude equal to σ B0 where σ is a constant. nucleus with
bare
Thus, the effective field, B, at the nucleus is nucleus electrons
B0 B0(1-σ)
B = (1 − σ )B0
Note that σ is different for each chemically
different nuclear spin – this is the famous
Energy
chemical shift – and permits resolution of lines in hω 0 hω 0 (1 − σ )
NMR spectra corresponding to chemically
different sites. The Hamiltonian is modified
accordingly
Hˆ 0 = −mˆ z B0 (1 − σ ) = −γ Iˆz B0 (1 − σ )
Zero Field High Field
Thus, instead of “seeing” a magnetic field of
magnitude B0, a proton in a molecule will see a
, 5.61 Physical Chemistry Lecture #36 Page 2
magnetic field of magnitude (1-σ)B0 and the associated Hamiltonian and spin state energies
will become:
E± = ± 12 γ �B0 (1 − σ ) ≡ ± 12 �ω0 (1 − σ )
This is illustrated in the figure above. Note the sign of the Hamiltonian is chosen so that the
α state (spin parallel to B0) is lower in energy than the β state ( spin antiparallel to B0).
Now, in the simplest NMR experiment, we probe this system with an oscillating magnetic field
perpendicular to the static field. By convention, we take this field to be along the x axis:
Ĥ1 ( t ) = − m
ˆ iB1 ( t ) = −γ Iˆ iB1 ( t ) = −γ Iˆx Bx cos (ωt )
We use Fermi’s Golden Rule to describe the spectrum of the spin in the oscillating field. The
selection rule is:
2 2 2
V fi = ∫ φ f * miB1φi dτ = γ Bx i ∫ φ f * Iˆxφi dτ
Now, we recall that Iˆx can be written in terms of the raising and lowering operators for
angular momentum:
(
Iˆx ∝ Iˆ+ + Iˆ− )
So that:
( )
2 2
V fi ∝ γ Bx i ∫ φ f * Iˆ+ + Iˆ− φi dτ
We immediately see that the integral is nonzero only if the initial and final spin states
differ by ±1 quantum of angular momentum (i.e. ΔM = ±1 ) , because the operator must either
raise or lower the eigenvalue of Iˆz . Thus, there are two possible transitions: α→β and β→α.
Futher, the energy conservation rule tells us that these transitions will only occur when the
photon energy matches the energy gap between the two states. As a result, we can
immediately draw the spectrum of a single shielded spin:
Intensity
(1−σ)ωo Frequency (ω)
This is perhaps not all that shocking: there is only one transition here, and so we might have
guessed that the spectrum would involve the frequency of that transition. However, we note
two generalizations of this result. First, we note that if we had chosen to apply the
oscillating field parallel to the static field, we would not have generated any transitions; we
NUCLEAR MAGNETIC RESONANCE
Just as IR spectroscopy is the simplest example of transitions being induced by light’s
oscillating electric field, so NMR is the simplest example of transitions induced by the
oscillating magnetic field. Because the strength of mattermagnetic field interactions are
typically two orders of magnitude smaller than the corresponding electric field interactions,
NMR is a much more delicate probe of molecular structure and properties. The NMR spin
Hamiltonians and wavefunctions are particularly simple, and permit us to demonstrate several
fundamental principles (about raising and lowering operators, energy levels, transition
probabilities, etc.) with a minimal amount of algebra. The principles and procedures are
applicable to other areas of spectroscopy electronic, vibrational, rotational, etc. – but for
these cases the algebra is more extensive.
Nuclear Spins in a Static Magnetic Field
For a single isolated spin in a static magnetic field, the contribution to the energy is:
ˆ iB 0 = −γ Iˆ iB 0
Ĥ 0 = −m
where γ is called the gyromagnetic ratio. If we choose our z axis to point in the direction of
the magnetic field then:
Ĥ 0 = −mˆ z B0 = −γ Iˆz B0
If we assume the nuclear spin is ½ (As it is for a proton) then we can easily work out the
energy levels of this Hamiltonian:
E± = ± 12 γ �B0 ≡ ± 12 �ω0
where ω0 = γB0 is called the nuclear Larmor frequency (rad/sec). Now, nuclei are never
isolated in chemistry – they are always surrounded by electrons. As we learned for the
hydrogen atom, the electrons near the nucleus shield the outer electrons from the bare
electric field produced by the nucleus. Similarly, the electrons shield the nucleus from the
bare electric field we apply in the laboratory. More specifically, the electron circulation
produces a field, B’ opposed to B0 and of
magnitude equal to σ B0 where σ is a constant. nucleus with
bare
Thus, the effective field, B, at the nucleus is nucleus electrons
B0 B0(1-σ)
B = (1 − σ )B0
Note that σ is different for each chemically
different nuclear spin – this is the famous
Energy
chemical shift – and permits resolution of lines in hω 0 hω 0 (1 − σ )
NMR spectra corresponding to chemically
different sites. The Hamiltonian is modified
accordingly
Hˆ 0 = −mˆ z B0 (1 − σ ) = −γ Iˆz B0 (1 − σ )
Zero Field High Field
Thus, instead of “seeing” a magnetic field of
magnitude B0, a proton in a molecule will see a
, 5.61 Physical Chemistry Lecture #36 Page 2
magnetic field of magnitude (1-σ)B0 and the associated Hamiltonian and spin state energies
will become:
E± = ± 12 γ �B0 (1 − σ ) ≡ ± 12 �ω0 (1 − σ )
This is illustrated in the figure above. Note the sign of the Hamiltonian is chosen so that the
α state (spin parallel to B0) is lower in energy than the β state ( spin antiparallel to B0).
Now, in the simplest NMR experiment, we probe this system with an oscillating magnetic field
perpendicular to the static field. By convention, we take this field to be along the x axis:
Ĥ1 ( t ) = − m
ˆ iB1 ( t ) = −γ Iˆ iB1 ( t ) = −γ Iˆx Bx cos (ωt )
We use Fermi’s Golden Rule to describe the spectrum of the spin in the oscillating field. The
selection rule is:
2 2 2
V fi = ∫ φ f * miB1φi dτ = γ Bx i ∫ φ f * Iˆxφi dτ
Now, we recall that Iˆx can be written in terms of the raising and lowering operators for
angular momentum:
(
Iˆx ∝ Iˆ+ + Iˆ− )
So that:
( )
2 2
V fi ∝ γ Bx i ∫ φ f * Iˆ+ + Iˆ− φi dτ
We immediately see that the integral is nonzero only if the initial and final spin states
differ by ±1 quantum of angular momentum (i.e. ΔM = ±1 ) , because the operator must either
raise or lower the eigenvalue of Iˆz . Thus, there are two possible transitions: α→β and β→α.
Futher, the energy conservation rule tells us that these transitions will only occur when the
photon energy matches the energy gap between the two states. As a result, we can
immediately draw the spectrum of a single shielded spin:
Intensity
(1−σ)ωo Frequency (ω)
This is perhaps not all that shocking: there is only one transition here, and so we might have
guessed that the spectrum would involve the frequency of that transition. However, we note
two generalizations of this result. First, we note that if we had chosen to apply the
oscillating field parallel to the static field, we would not have generated any transitions; we
