1 Basic Principles of NMR 1.1.INTRODUCTION Energy states and population distribution are the fundamental subjects of any spectro scopic technique.The energy difference between energy states gives rise to the frequency ofthe spectra,whereas intensities of the spectral peaks are proportional to the population difference of the states.R her funda ntal phe n ne s es an ng these In principle,an NMR spectrometer is more or less like a radio.In a radio,audio signals in the frequency range of kilohertz are the signals of interest,which one can hear.However, the signals sent by broadcast stations are in the range of 100 MHz for FM and of up to 1GHz for AM broadcasting.The ansmis n requenci re they are t to spea ns hay me Ia h nals ofn are the chemical shifts generated by the electron density surrounding an individual proton which are in the kilohertz frequency range.Many of the protons in the molecule have different chemical environments which give different signals in the kilohertz range.One must find a egahertz Larmor frequency in order to observe the kilohertz chemical 1.2.NUCLEAR SPIN IN A STATIC MAGNETIC FIELD 1.2.1.Precession of Nuclear Spins in a Magnetic Field As mentioned above,energy and population associated with energy states are the basis of the frequency position and the intensity ofspectral signals.In order to understand the principles of NMR spectroscopy,it is necessary to know how the energy states of nuclei are generated What kind of nuclei will give N How do nuclear spins orent in the magnetic field?
1 Basic Principles of NMR 1.1. INTRODUCTION Energy states and population distribution are the fundamental subjects of any spectroscopic technique. The energy difference between energy states gives rise to the frequency of the spectra, whereas intensities of the spectral peaks are proportional to the population difference of the states. Relaxation is another fundamental phenomenon in nuclear magnetic resonance spectroscopy (NMR), which influences both line shapes and intensities of NMR signals. It provides information about structure and dynamics of molecules. Hence, understanding these aspects lays the foundation to understanding basic principles of NMR spectroscopy. In principle, an NMR spectrometer is more or less like a radio. In a radio, audio signals in the frequency range of kilohertz are the signals of interest, which one can hear. However, the signals sent by broadcast stations are in the range of 100 MHz for FM and of up to 1 GHz for AM broadcasting. The kilohertz audio signals must be separated from the megahertz transmission frequencies before they are sent to speakers. In NMR spectroscopy, nuclei have an intrinsic megahertz frequency which is known as the Larmor frequency. For instance, in a molecule, all protons have the same Larmor frequency. However, the signals of interest are the chemical shifts generated by the electron density surrounding an individual proton, which are in the kilohertz frequency range. Many of the protons in the molecule have different chemical environments which give different signals in the kilohertz range. One must find a way to eliminate the megahertz Larmor frequency in order to observe the kilohertz chemical shifts (more details to follow). 1.2. NUCLEAR SPIN IN A STATIC MAGNETIC FIELD 1.2.1. Precession of Nuclear Spins in a Magnetic Field As mentioned above, energy and population associated with energy states are the basis of the frequency position and the intensity of spectral signals. In order to understand the principles of NMR spectroscopy, it is necessary to know how the energy states of nuclei are generated and what are the energy and population associated with the energy states. Key questions to be addressed in this section include: 1. What causes nuclei to precess in the presence of a magnetic field? 2. What kind of nuclei will give NMR signals? 3. How do nuclear spins orient in the magnetic field? 1
2 Not any kind of nucleuswlve charge and mas ve a spin quantum nun f nucle will n to an N signa ctive th prith a no app eld Nue 1 nagnitud is determi angul P=hvI(I+1) er and his the Planck constant divided by2 ofl is or half as n integra er The z compone of the ntum P:is given by P:=hm 12) in which the magnetic quantum number m has possible values of./-1. -1-and a total of2/+1.This equation tells us that the pr oiection of nuclear angular momentum on the axis is quantized in space and has a total of2/+1possible values.The orientations ofnuclear angular momentum are defined by the allowed m values.For example,for spin nuclei,the an angle of 54 79 relative to the magnetic field (Figure 11) The nuclei with a nonzero spin quantum number will rotate about the magnetic field Bo due to the torque generated by the interaction of the nuclear angular momentum with the magnetic field.The magnetic moment(or nuclear moment),,is either parallel or antiparallel to their angular momentum: u=yP=yhvI(I.+万 .3) in which y is the nuclear gyromagnetic ratio,which has a specific value for a given isotope. Thus,y is a characteristic constant for a specific nucleus.The angular momentum P is the same for all nuclei with the same magnetic quantum number,whereas the angular moment m=- Figure 1.1.Orientation of lear angular mo and it comp the laboratory frame
2 Chapter 1 Not any kind of nucleus will give NMR signals. Nuclei with an even number of both charge and mass have a spin quantum number of zero, for example, 12C. These kinds of nuclei do not have nuclear angular momentum and will not give rise to an NMR signal; these are called NMR inactive nuclei. For nuclei with a nonzero spin quantum number, energy states are produced by the nuclear angular moment interacting with the applied magnetic field. Nuclei with a nonzero spin quantum number possess nuclear angular momentum whose magnitude is determined by: P = I (I + 1) (1.1) in which I is the nuclear spin quantum number and is the Planck constant divided by 2π. The value of I is dependent on the mass and charge of the nucleus and can be either an integral or half integral number. The z component of the angular momentum Pz is given by: Pz = m (1.2) in which the magnetic quantum number m has possible values of I , I −1, ... , −I +1, −I , and a total of 2I +1. This equation tells us that the projection of nuclear angular momentum on the z axis is quantized in space and has a total of 2I +1 possible values. The orientations of nuclear angular momentum are defined by the allowed m values. For example, for spin 1 2 nuclei, the allowed m are 1 2 and −1 2 . Thus, the angular momentum of spin 1 2 has two orientations; one is pointing up (pointing to the z axis) and the other pointing down (pointing to the −z axis) with an angle of 54.7◦ relative to the magnetic field (Figure 1.1). The nuclei with a nonzero spin quantum number will rotate about the magnetic field B0 due to the torque generated by the interaction of the nuclear angular momentum with the magnetic field. The magnetic moment (or nuclear moment), μ, is either parallel or antiparallel to their angular momentum: μ = γ P = γ I (I + 1) (1.3) in which γ is the nuclear gyromagnetic ratio, which has a specific value for a given isotope. Thus, γ is a characteristic constant for a specific nucleus. The angular momentum P is the same for all nuclei with the same magnetic quantum number, whereas the angular moment μ z x z 2 m 1 = 2 m 1 = − Figure 1.1. Orientation of nuclear angular moment μ with spin 1 2 and its z component, μz. The vectors represent the angular moment μ rotating about the magnetic field whose direction is along the z axis of the laboratory frame.
Basic Principles of NMR is different for different nuclei.For instance,3C and H have the same angular momentum P because they have same spin quantum number of,but have different angular momentsu because they are different isotopes with differenty.Therefore,the nuclear angular moment is used to characterize nuclear spins.The moment u is parallel to the angular momentum ify is positive or antiparallel ify is negative (e.g,5N).Similar to thecomponent of the angular momentum,P..the z component of angular moment u.is given by: H:=yP=yhm The equation indicates thathas a different value for diffe if they may have etic field.they the etic field rated hy the of the etic field Bo with the clear n nt The relativ to Bo is de Nuclei with called nucle spins because their ents make the in in the netic field in summar the nuelear ang lar mome tum is whateau ses the nucleus to rotate relative to the magnetic field.Different nuclei have a characteristic nuclear moment because the moment is dependent on the gyromagnetic ratio y.whereas nuclei with the same spin quantum number possess the same nuclear angular momentum.Nuclear moments have quantized orientations defined by the value of the magnetic quantum number,m.The interaction of nuclei with the magnetic field is utilized to generate an NMR signal.Because the energy and population of nuclei are proportional to the magnetic field strength(more details discussed below),the frequency and intensity of the NMR spectral signals are dependent on the field strength. 1.2.2.Energy States and Population It has been illustrated in the previous section that nuclei with nonzero spin quantum numbers orient along specific directions with respect to the magnetic field.They are rotating continuously about the field direction due to the nuclear moment u possessed by nuclei.For each orientation state,also known as the Zeeman state or spin state,there is energy associated with it,which is characterized by the frequency of the precession. Key questions to be addressed in this section include: 1.What is the energy and population distribution of the Zeeman states? 2.What are the nuclear precession frequencies of the Zeeman states and the frequency of the transition hetween the states and how are they different? 3.How are energy and population related to measurable spectral quantities? The intrinsic fred ency of the precession is the Lam the ze an state with magnetic quantum mhe m can be des armo frequency: E=-u:Bo =-mhy Bo=mhao 15 in which Bo is the magnetic field strength in the unit of tesla,T,and o=-yBo is the Larmor frequency.Therefore,the energy difference of the allowed transition (the selection rule is that
Basic Principles of NMR 3 is different for different nuclei. For instance, 13C and 1H have the same angular momentum P because they have same spin quantum number of 1 2 , but have different angular moments μ because they are different isotopes with different γ . Therefore, the nuclear angular moment μ is used to characterize nuclear spins. The moment μ is parallel to the angular momentum if γ is positive or antiparallel if γ is negative (e.g., 15N). Similar to the z component of the angular momentum, Pz, the z component of angular moment μz is given by: μz = γ Pz = γ m (1.4) The equation indicates that μz has a different value for different nuclei even if they may have the same magnetic quantum number m. When nuclei with a nonzero spin quantum number are placed in a magnetic field, they will precess about the magnetic field due to the torque generated by the interaction of the magnetic field B0 with the nuclear moment μ. The angle of μ relative to B0 is dependent on m. Nuclei with nonzero spin quantum numbers are also called nuclear spins because their angular moments make them spin in the magnetic field. In summary, the nuclear angular momentum is what causes the nucleus to rotate relative to the magnetic field. Different nuclei have a characteristic nuclear moment because the moment is dependent on the gyromagnetic ratio γ , whereas nuclei with the same spin quantum number possess the same nuclear angular momentum. Nuclear moments have quantized orientations defined by the value of the magnetic quantum number, m. The interaction of nuclei with the magnetic field is utilized to generate an NMR signal. Because the energy and population of nuclei are proportional to the magnetic field strength (more details discussed below), the frequency and intensity of the NMR spectral signals are dependent on the field strength. 1.2.2. Energy States and Population It has been illustrated in the previous section that nuclei with nonzero spin quantum numbers orient along specific directions with respect to the magnetic field. They are rotating continuously about the field direction due to the nuclear moment μ possessed by nuclei. For each orientation state, also known as the Zeeman state or spin state, there is energy associated with it, which is characterized by the frequency of the precession. Key questions to be addressed in this section include: 1. What is the energy and population distribution of the Zeeman states? 2. What are the nuclear precession frequencies of the Zeeman states and the frequency of the transition between the states, and how are they different? 3. How are energy and population related to measurable spectral quantities? The intrinsic frequency of the precession is the Larmor frequency ω0. The energy of the Zeeman state with magnetic quantum number m can be described in terms of the Larmor frequency: E = −μzB0 = −mγ B0 = mω0 (1.5) in which B0 is the magnetic field strength in the unit of tesla, T, and ω0 = −γ B0 is the Larmor frequency. Therefore, the energy difference of the allowed transition (the selection rule is that
4 △E=yB 1.6 Beca h,the frequency of the required electromagnetic radiation for the transition @=YBo 1) which ha on Larmo 23Twh amp agne 14 adian strength of ns per secc frequency can also be represent v-2 1.8 diffe ce be ween two transition states as doe ition The in lati on diff f th en d h the Rolt- such as H.or C.the lo rgy state (g round state)is defined as thestate form .whereas the higher energy state (excited state)is labeled a the B state for m .For I5 N,m =-is the lower energy a state because of its negativey The ratio ofthe populations in the states is quantitatively described by the Boltzmann equation: =e△ET=e-T= （1.9 N in which N and Ne pulation of the a and 8 states.respectively.T is the absolute stant Thee uation states that hath the e ergy difference of the transition states and the population difference of the states increase with the magnetic field strength.Furthermore,the opulation difference has a temperature dependence if the sample temperature reaches absolute zero,there is no population at the B state and all spins will lie in the a state,whereas both states will have equal population if the temperature is infinitely high.At T near room temperature,~300 K,hy BokT.As a consequence,a first-order Taylor expansion can be used to describe the population difference: ≈1、yB 1.10) Na 7 elver than that su small fra n will contribu sity due o the low energy diffe rence and ence NMR spectroscopy intrinsically isa very insensitive
4 Chapter 1 only a single-quantum transition, that is, m = ±1, is allowed), for instance, between the m = −1 2 and m = 1 2 Zeeman states, is given by: E = γ B0 (1.6) Because E = ω, the frequency of the required electromagnetic radiation for the transition has the form of: ω = γ B0 (1.7) which has a linear dependence on the magnetic field strength. Commonly, the magnetic field strength is described by the proton Larmor frequency at the specific field strength. A proton resonance frequency of 100 MHz corresponds to the field strength of 2.35 T. For example, a 600 MHz magnet has a field strength of 14.1 T. While the angular frequency ω has a unit of radians per second, the frequency can also be represented in hertz with the relationship of: ν = ω 2π (1.8) As the magnetic field strength increases, the energy difference between two transition states becomes larger, as does the frequency associated with the Zeeman transition. The intensity of the NMR signal comes from the population difference between two Zeeman states of the transition. The population of the energy state is governed by the Boltzmann distribution. For a spin 1 2 nucleus with a positive γ such as 1H, or 13C, the lower energy state (ground state) is defined as the α state for m = 1 2 , whereas the higher energy state (excited state) is labeled as the β state for m = −1 2 . For 15N, m = −1 2 is the lower energy α state because of its negative γ . The ratio of the populations in the states is quantitatively described by the Boltzmann equation: Nβ Nα = e−E/kT = e−γ B0/kT = 1 eγ B0/kT (1.9) in which Nα and Nβ are the population of the α and β states, respectively, T is the absolute temperature and k is the Boltzmann constant. The equation states that both the energy difference of the transition states and the population difference of the states increase with the magnetic field strength. Furthermore, the population difference has a temperature dependence. If the sample temperature reaches absolute zero, there is no population at the β state and all spins will lie in the α state, whereas both states will have equal population if the temperature is infinitely high. At T near room temperature, ∼300 K, γ B0 kT . As a consequence, a first-order Taylor expansion can be used to describe the population difference: Nβ Nα ≈ 1 − γ B0 kT (1.10) At room temperature, the population of the β state is slightly lower than that of the α state. For instance, the population ratio for protons at 800 MHz field strength is 0.99987. This indicates that only a small fraction of the spins will contribute to the signal intensity due to the low energy difference and hence NMR spectroscopy intrinsically is a very insensitive
Basic Principles of NMR spectroscopic technique.Therefore,a stronger magnetic field is necessary to obtain better sensitivity,in addition to other advantages such as higher resolution and the TROSY effect (transverse relaxation optimized spectroscopy). 1.2.3.Bulk Magnetization Questions to be addressed in this section include: re is it located? Why do no transverse components of bulk magnetization exist at equilibrium? The observable NMR sigr als ome from the sembly of nuclea spins in the presence the magne e (o mag gives in of the nlane onent of the bulk magnetization at the equilibrium state is av ged to ero and hen is not observable (Figure 1.2).The bulk magnetization Mo results from the small p lation difference hety een the a and B states.At equilibrium,this vector lies along h zaxis and is parallel to the magnetic field direction for nuclei with positive because the spin population in the a state is larger than that in the B state.Although the bulk magnetization is stationary along the zaxis,the individual spin moments rotate about the axis. B B Figure 1.2.Bulk magnetization of spin nuclei with positive v.r.y.and z are the axes of the labora frame.The thin arrows represent individual nuclear moments.The vector sum of the nuclear moments on the xy plane is zero because an individual nuclear moment has equal probability of being in any directionof th exy pla The bu arrow,is ge f they the smal
Basic Principles of NMR 5 spectroscopic technique. Therefore, a stronger magnetic field is necessary to obtain better sensitivity, in addition to other advantages such as higher resolution and the TROSY effect (transverse relaxation optimized spectroscopy). 1.2.3. Bulk Magnetization Questions to be addressed in this section include: 1. What is the bulk magnetization and where is it located? 2. Why do no transverse components of bulk magnetization exist at equilibrium? The observable NMR signals come from the assembly of nuclear spins in the presence of the magnetic field. It is the bulk magnetization of a sample (or macroscopic magnetization) that gives the observable magnetization, which is the vector sum of all spin moments (nuclear angular moments). Because nuclear spins precess about the magnetic field along the z axis of the laboratory frame, an individual nuclear moment has equal probability of being in any direction of the xy plane. Accordingly, the transverse component of the bulk magnetization at the equilibrium state is averaged to zero and hence is not observable (Figure 1.2). The bulk magnetization M0 results from the small population difference between the α and β states. At equilibrium, this vector lies along the z axis and is parallel to the magnetic field direction for nuclei with positive γ because the spin population in the α state is larger than that in the β state. Although the bulk magnetization is stationary along the z axis, the individual spin moments rotate about the axis. x y z M0 B0 Figure 1.2. Bulk magnetization of spin 1 2 nuclei with positive γ . x, y, and z are the axes of the laboratory frame. The thin arrows represent individual nuclear moments. The vector sum of the nuclear moments on the xy plane is zero because an individual nuclear moment has equal probability of being in any direction of the xy plane. The bulk magnetization M0, labeled as a thick arrow, is generated by the small population difference between the α and β states, and is parallel to the direction of the static magnetic field B0.
6 1.3.ROTATING FRAME The questions to be addressed in this section include 2 What is the rotating frame and why is it needed? lating electr netic feld? 3 the hulk t when a B field is lied to it? 4.What is the relationship between radio frequency (RF)pulse pow The larmor frequency ofa nuclear isotope is the resonance frequency ofthe isotope in the magnetic field.For example.H Larmor frequency will be 500 MHz for all protons of a sample in a magnetic field of 11.75 T.If the Larmor frequency were the only observed NMR signal. NMR spectroscopy would not be useful because there would be only one resonance signal for allH.In fact,chemical shifts are the NMR signals of interest(details in section 1.7),which have a frequency range of kilohertz,whereas the Larmor frequency of all nuclei is in the range of megahertz.For instance,the observed signals of protons are normally in the range of several kilohertz with a Larmor frequency of 600 MHz in a magnetic field of 14.1 T.How the Larmor frequency is removed before NMR data are acquired,what the rotating frame is,why we need it,and how the bulk magnetization changes upon applying an additional electromagnetic field are the topics of this section. not be present inany NMR spectrum,it is nec essary in the an e e xy pla the frequency with respect t tory frame of (a) I the ce simplifying on of the spi n the nuclei in the when the fre c of the of the le This Geld is turne ethe Larm edin NMR experiments,a new coordinate frame is introduced to eliminate the Larmor frequency from consideration.called the me In the otating frame.the xy plane of the laboratory frame is rotating at or near the Larmor frequency o with re ect to the z axis of the laboraton frame The transformation of the laboratory frame to the rotating frame can be illustrated by taking a merry-go-round as an example.The merry-go-round observed by one standing on the ground is rotating at a given speed.When one is riding on it,he is also rotating at the same speed.However,he is stationary relative to others on the merry-go-round.If the ground is considered as the laboratory frame,the merry-go-round is the rotating frame.When the person on the ground steps onto the merry-go-round,it is transformed from the laboratory frame to the rotating frame.The sole difference between the laboratory frame and the rotating frame is that the rotating frame is rotating in the xy plane about the z axis relative to the laboratory frame. By transforming from the laboratory frame to the rotating frame. e nuclear m ments ar The ter s that its abo rotating am a result. the bulk cy o rmor freq e quency e血he f thi rame results in a sta long an axis on the xy plan n the rotating frame fo
6 Chapter 1 1.3. ROTATING FRAME The questions to be addressed in this section include: 1. What is the rotating frame and why is it needed? 2. What is the B1 field and why must it be an oscillating electromagnetic field? 3. How does the bulk magnetization M0 react when a B1 field is applied to it? 4. What is the relationship between radio frequency (RF) pulse power and pulse length? The Larmor frequency of a nuclear isotope is the resonance frequency of the isotope in the magnetic field. For example, 1H Larmor frequency will be 500 MHz for all protons of a sample in a magnetic field of 11.75 T. If the Larmor frequency were the only observed NMR signal, NMR spectroscopy would not be useful because there would be only one resonance signal for all 1H. In fact, chemical shifts are the NMR signals of interest (details in section 1.7), which have a frequency range of kilohertz, whereas the Larmor frequency of all nuclei is in the range of megahertz. For instance, the observed signals of protons are normally in the range of several kilohertz with a Larmor frequency of 600 MHz in a magnetic field of 14.1 T. How the Larmor frequency is removed before NMR data are acquired, what the rotating frame is, why we need it, and how the bulk magnetization changes upon applying an additional electromagnetic field are the topics of this section. Since the Larmor frequency will not be present in any NMR spectrum, it is necessary to remove its effect when dealing with signals in the kilohertz frequency range. This can be done by applying an electromagnetic field B1 on the xy plane of the laboratory frame, which rotates at the Larmor frequency with respect to the z axis of the laboratory frame. This magnetic field is used for the purposes of (a) removing the effect of the Larmor frequency and hence simplifying the theoretical and practical consideration of the spin precession in NMR experiments and (b) inducing the nuclear transition between two energy states by its interaction with the nuclei in the sample according to the resonance condition that the transition occurs when the frequency of the field equals the resonance frequency of the nuclei. This magnetic field is turned on only when it is needed. Because the Larmor frequency is not observed in NMR experiments, a new coordinate frame is introduced to eliminate the Larmor frequency from consideration, called the rotating frame. In the rotating frame, the xy plane of the laboratory frame is rotating at or near the Larmor frequency ω0 with respect to the z axis of the laboratory frame. The transformation of the laboratory frame to the rotating frame can be illustrated by taking a merry-go-round as an example. The merry-go-round observed by one standing on the ground is rotating at a given speed. When one is riding on it, he is also rotating at the same speed. However, he is stationary relative to others on the merry-go-round. If the ground is considered as the laboratory frame, the merry-go-round is the rotating frame. When the person on the ground steps onto the merry-go-round, it is transformed from the laboratory frame to the rotating frame. The sole difference between the laboratory frame and the rotating frame is that the rotating frame is rotating in the xy plane about the z axis relative to the laboratory frame. By transforming from the laboratory frame to the rotating frame, the nuclear moments are no longer spinning about the z axis, that is, they are stationary in the rotating frame. The term “transforming” here means that everything in the laboratory frame will rotate at a frequency of −ω0 about the z axis in the rotating frame. As a result, the bulk magnetization does not have its Larmor frequency in the rotating frame. Since the applied B1 field is rotating at the Larmor frequency in the laboratory frame, the transformation of this magnetic field to the rotating frame results in a stationary B1 field along an axis on the xy plane in the rotating frame, for
Basic Principles of NMR 7 example the x axis.Therefore,when this B magnetic field is applied,its net effect on the bulk magnetization is to rotate the bulk magnetization away from the z axis clockwise about the axis of the applied field by the left-hand rule in the vector representation. In practice,the rotation of the B field with respect to the z axis of the laboratory frame is achieved by generating a linear oscillating electromagnetic field with the magnitude of 2B because it is easily produced by applying electric current through the probe coil (Figure 1.3). The oscillating magnetic field has a frequency equal to or near the Larmor frequency of the nuclei.As the current increases from zero to maximum,the field proportionally increases from zero to the maximum field along the coil axis(2BI in Figure 1.3).Reducing the current from the ma. 0 um creases the fiel Finally, the If the 2B13 人 one cycle. itude,B P9 two eq the ane at the se site dire o each other (thir When th field has the on the axis with an itude of B1.Th sum of the two is 2B[Fis e 1 4(a)l 八hen the current is ze which gives zero in the field magnitude.each c nt still has the same magnitude of B but aligns on the x and-x axes,respectively,which gives rise to a vector sum of zero [Figure 1.4(c)].As the current reduces,both components rotate into 000不 VVVV 2B Figure 1.3.The electromagnetic field generated by the current passing through the probe coil.The 000A000e000000①00000C a) (b) c) d Figure 1.4.Vector sum of the oscillating B field generated by passing current through aprobe coil.The vectors rotating in opp When R is said to he on resonance (a)When the cur reaches the ximum the twe vectors align on y axis.The sum of the two vectors is the same as the field produced in the coil.(b)As
Basic Principles of NMR 7 example the x axis. Therefore, when this B1 magnetic field is applied, its net effect on the bulk magnetization is to rotate the bulk magnetization away from the z axis clockwise about the axis of the applied field by the left-hand rule in the vector representation. In practice, the rotation of the B1 field with respect to the z axis of the laboratory frame is achieved by generating a linear oscillating electromagnetic field with the magnitude of 2B1 because it is easily produced by applying electric current through the probe coil (Figure 1.3). The oscillating magnetic field has a frequency equal to or near the Larmor frequency of the nuclei. As the current increases from zero to maximum, the field proportionally increases from zero to the maximum field along the coil axis (2B1 in Figure 1.3). Reducing the current from the maximum to zero and then to the minimum (negative maximum, −i) decreases the field from 2B1 to −2B1. Finally, the field is back to zero from −2B1 as the current is increased from the minimum to zero to finish one cycle. If the frequency of changing the current is νrf , we can describe the oscillating frequency as ωrf(ω = 2πν). Mathematically, this linear oscillating field (thick arrow in Figure 1.4) can be represented by two equal fields with half of the magnitude, B1, rotating in the xy plane at the same angular frequency in opposite directions to each other (thin arrows). When the field has the maximum strength at 2B1, each component aligns on the y axis with a magnitude of B1. The vector sum of the two is 2B1 [Figure 1.4(a)]. When the current is zero, which gives zero in the field magnitude, each component still has the same magnitude of B1 but aligns on the x and −x axes, respectively, which gives rise to a vector sum of zero [Figure 1.4(c)]. As the current reduces, both components rotate into i –i –2B1 2B1 Figure 1.3. The electromagnetic field generated by the current passing through the probe coil. The magnitude of the field is modulated by changing the current between −i and +i. The electromagnetic field is called the oscillating B1 field. i i -i x y -i (a) (b) –ω (c) rf ωrf (d) (e) Figure 1.4. Vector sum of the oscillating B1 field generated by passing current through a probe coil. The magnitude of the field can be represented by two equal amplitude vectors rotating in opposite directions. The angular frequency of the two vectors is the same as the oscillating frequency ωrf of the B1 field. When ωrf = ω0, B1 is said to be on resonance. (a) When the current reaches the maximum, the two vectors align on y axis. The sum of the two vectors is the same as the field produced in the coil. (b) As the B1 field reduces, its magnitude equals the sum of the projections of two vectors on the y axis. (c) When the two vectors are oppositely aligned on the x axis, the current in the coil is zero
8 Figure 1.5.B field in the laboratory frame.The bulk magnetization Mo is the vector sum of individual nuclear moments which are precessing about the static magnetic field Bo at the Larmor frequency Bfeld anguar lrequeney of ot he i hed is cquan to the Lamor rrequeney.hat s.ofh the-y region and the sum produces a negative magnitude [Figure 1.4(d)I.Finally.the two components meet at the-y axis,which represents a field magnitude of-2B [Figure 1.4(e)]. At any given time the two decomposed components have the same magnitude of B,the same frequency of f,and are mirror images of each other. If the frequency of the rotating frame is set to,which is close to the Larmor frequency @o,the component ofthe B field which has in the laboratory frame has null frequency in the rotating frame because of the transformation by.The other with in the laboratory frame now has an angular frequency of-2 after the transformation.Since the latter has a frequency far away from the Larmor frequency it will not interfere with the NMR signals which are心nhc8rof6c业therctor.心componnu freque8y2 ssion unless specifically mentioned.T ormer component wi rotating f ent the BI field en said to 15).Simce i t ncy o,the BI the r ting frame th ear spir T eld un whe crBstapliediedh will rotate about the ction u n its inter with the bulk m ag axis where B otating frame.The frequer ncy of the otati ined by: =-YB1 .1) This should not be misunderstood as f of the B field since f is the field oscillating frequency determined by changing the direction of the current passing through the coil,which is set to be the same as or the Larm frequency.Frequency f is often called the carrie mitter dete ed b f the B By modula the ng the am ere in t eld i when Mo moves from h eaxis to the xy plane,this is calleda9 pulse
8 Chapter 1 x y M0 B1 1 0 Figure 1.5. B1 field in the laboratory frame. The bulk magnetization M0 is the vector sum of individual nuclear moments which are precessing about the static magnetic field B0 at the Larmor frequency ω0. When the angular frequency ωrf of the B1 field is equal to the Larmor frequency, that is, ωrf = ω0, the B1 field is on resonance. the −y region and the sum produces a negative magnitude [Figure 1.4(d)]. Finally, the two components meet at the −y axis, which represents a field magnitude of −2B1 [Figure 1.4(e)]. At any given time the two decomposed components have the same magnitude of B1, the same frequency of ωrf , and are mirror images of each other. If the frequency of the rotating frame is set to ωrf , which is close to the Larmor frequency ω0, the component of theB1 field which has ω1 in the laboratory frame has null frequency in the rotating frame because of the transformation by −ωrf . The other with −ωrf in the laboratory frame now has an angular frequency of −2ωrf after the transformation. Since the latter has a frequency far away from the Larmor frequency it will not interfere with the NMR signals which are in the range of kilohertz. Therefore, this component will be ignored throughout the discussion unless specifically mentioned. The former component with null frequency in the rotating frame is used to represent the B1 field. If we regulate the frequency of the current oscillating into the coil as ω0, then setting ωrf equal to the Larmor frequency ω0, the B1 field is said to be on resonance (Figure 1.5). Since in the rotating frame the Larmor frequency is not present in the nuclei, the effect of B0 on nuclear spins is eliminated. The only field under consideration is the B1 field. From the earlier discussion we know that nuclear magnetization will rotate about the applied field direction upon its interaction with a magnetic field. Hence, whenever B1 is turned on, the bulk magnetization will be rotated about the axis where B1 is applied in the rotating frame. The frequency of the rotation is determined by: ω1 = −γ B1 (1.11) This should not be misunderstood as ωrf of the B1 field since ωrf is the field oscillating frequency determined by changing the direction of the current passing through the coil, which is set to be the same as or near the Larmor frequency. Frequency ωrf is often called the carrier frequency or the transmitter frequency. The frequency ω1 is determined by the amplitude of the B1 field, that is, the maximum strength of the B1 field. By modulating the amplitude and time during which B1 is turned on, the bulk magnetization can be rotated to anywhere in the plane perpendicular to the axis of the applied B1 field in the rotating frame. If B1 is turned on and then turned off when M0 moves from the z axis to the xy plane, this is called a 90◦ pulse
Basic Principles of NMR 9 receiver applyinga90°pulse anda180°pulsc pulse length (or the9 te puise power se a e nd .The puls sho 90 lei (all nuclei ptH.Be l ratios than r ons,they have longer 90 pulse lengths at a gi Thepulse length (w)is proportional to the Bfield stre gth: 1 1= 12 T 1 pw9o=2yB1-4v1 1.13) in which v is the field strength in the frequency unit of hertz.A higher B field produces a shorter 90 pulse.A90 pulse of 10 us corresponds to a 25 kHz B field.Nuclei with smaller gyromagnetic ratios will require a higher B to generate the same pwo as that with larger y.When a receiver is placed on the transverse plane of the rotating frame,NMR signals are observed from the transverse magnetization.The maximum signal is obtained when the bulk magnetization is in the xy plane of the rotating frame,which is done by applying a9 pulse. No signal is observed when a 180 pulse is applied(Figure 1.6). 1.4.BLOCH EQUATIONS As we now know,the nuclei inside the magnet produce nuclear moments which cause them to spin about the magnetic field.Inaddition,the interaction ofthe nuclei with the magnetic
Basic Principles of NMR 9 x y B0 M0 B1 90°x 180° x x y B0 M0 B1 x y B0 M0 B1 receiver receiver Figure 1.6. Vector representation of the bulk magnetization upon applying a 90◦ pulse and a 180◦ pulse by B1 along the x axis in the rotating frame. The corresponding time during which β1 is applied is called the 90◦ pulse length (or the 90◦ pulse width), and the field amplitude is called the pulse power. A 90◦ pulse length can be as short as a few microseconds and as long as a fraction of a second. The pulse power for a hard (short) 90◦ pulse is usually as high as half of a hundred watts for protons and several hundred watts for heteronuclei (all nuclei except 1H). Because heteronuclei have lower gyromagnetic ratios than protons, they have longer 90◦ pulse lengths at a given B1 field strength. The 90◦ pulse length (pw90) is proportional to the B1 field strength: ν1 = γ B1 2π = 1 4pw90 (1.12) pw90 = π 2γ B1 = 1 4ν1 (1.13) in which ν1 is the field strength in the frequency unit of hertz. A higher B1 field produces a shorter 90◦ pulse. A 90◦ pulse of 10 μs corresponds to a 25 kHz B1 field. Nuclei with smaller gyromagnetic ratios will require a higher B1 to generate the same pw90 as that with larger γ . When a receiver is placed on the transverse plane of the rotating frame, NMR signals are observed from the transverse magnetization. The maximum signal is obtained when the bulk magnetization is in the xy plane of the rotating frame, which is done by applying a 90◦ pulse. No signal is observed when a 180◦ pulse is applied (Figure 1.6). 1.4. BLOCH EQUATIONS As we now know, the nuclei inside the magnet produce nuclear moments which cause them to spin about the magnetic field. In addition, the interaction of the nuclei with the magnetic
10 Chapter 1 tra the the Bl ons to be add ed in the current section 1 What phenomena do the Bloch equations describe? 2 at is free induction de ay (FID)? 3.What are the limitations of the Bloch equations? In the presence of the magnetic field Bo.the torque produced by Bo on spins with the angular moment u causes precession of the nuclear spins.Felix Bloch derived simple semi- classical equations to describe the time-dependent phenomena of nuclear spins in the static magnetic field(Bloch,1946).The torque on the bulk magnetization,described by the change of the angular momentum as a function of time,is given by: T=dp -=M X B 1.14) dr n的 of magnetization with time is described by: dM =y(M x B) (1,15) dt When B is the static magnetic field Bo which is along the z axis of the laboratory frame the change of magnetization along the x,y,and z axes with time can be obtained from the determinant of the vector product: dM dt =业++k=yMM,M=iyM,-jM.a1 dt dt dt 00 Bo in whichi,j,k are the unit vectors along the,y,andaxes,respectively.Therefore, dM. dr =yMy Bo 17) dM dr =-yMx Bo (1.18 dM: =0 dt (1.19 The abo the n components agnetic field Boprodu n NMR sp ti pe and the rate of decay is depen nt on ield nuclear gyromagnetic ratio
10 Chapter 1 field will rotate the magnetization toward the transverse plane when the electromagnetic B1 field is applied along a transverse axis in the rotating frame. After the pulse is turned off, the magnetization is solely under the effect of the B0 field. How the magnetization changes with time can be described by the Bloch equations, which are based on a simple vector model. Questions to be addressed in the current section include: 1. What phenomena do the Bloch equations describe? 2. What is free induction decay (FID)? 3. What are the limitations of the Bloch equations? In the presence of the magnetic field B0, the torque produced by B0 on spins with the angular moment μ causes precession of the nuclear spins. Felix Bloch derived simple semiclassical equations to describe the time-dependent phenomena of nuclear spins in the static magnetic field (Bloch, 1946). The torque on the bulk magnetization, described by the change of the angular momentum as a function of time, is given by: T = dP dt = M × B (1.14) in which M × B is the vector product of the bulk magnetization M (the sum of μ) with the magnetic field B. Because M = γ P (or P = M/γ ) according to Equation (1.3), the change of magnetization with time is described by: dM dt = γ (M × B) (1.15) When B is the static magnetic field B0 which is along the z axis of the laboratory frame, the change of magnetization along the x, y, and z axes with time can be obtained from the determinant of the vector product: dM dt = i dMx dt + j dMy dt + k dMz dt = γ ijk Mx My Mz 0 0 B0 = iγMyB0 − jγMxB0 (1.16) in which i, j , k are the unit vectors along the x, y, and z axes, respectively. Therefore, dMx dt = γ MyB0 (1.17) dMy dt = −γ MxB0 (1.18) dMz dt = 0 (1.19) The above Bloch equations describe the time dependence of the magnetization components under the effect of the static magnetic fieldB0 produced by the magnet of an NMR spectrometer without considering any relaxation effects. The z component of the bulk magnetization Mz is independent of time, whereas the x and y components are decaying as a function of time and the rate of decay is dependent on the field strength and nuclear gyromagnetic ratio. The
