1 Classical mechanics Position at Position at time t=0 Newton's time t>0 Quantum Mechanics 2nd Law Lecture notes by Assaf Tal Quantum mechanics (wave formulation) NMR is a semi-classical phenomenon. On the one Wavefunction Wavefunction hand, we treat the electromagnetic fields in matter at time t=0 Schroedinger's at time t>0 as classical, which is justified at the long Equation wavelengths encountered1. On the other hand, the Quantum mechanics (density matrix) basic unit we'll be dealing with - the nuclear spin - is strictly a quantum mechanical (QM) entity. Density matrix Density matrix at time t=0 Liouville's at time t>0 Although in certain cases we can take a classical Equation limit in which our system of spins can be described by classical equations, for the most part NMR is Most books begin by talking about "wave best thought of in quantum mechanical terms, and mechanics" using vectors to describe a pure cannot be explained well by classical analogies. quantum system, and generalize to statistical Luckily, QM is fairly benign in NMR since it ensembles using the density matrix formulation. deals with simple systems (mostly spin-1/2). We We will circumvent that step completely to review here some of the basic principles which will simplify the discussion, which means we will be be of use to us in explaining NMR phenomena in saying "just because" at a few spots where we will subsequent chapters. lack the necessary foundations to explain them. 1. Kinematics: First, I will explain how to We will conclude by generalizing our notions to describe a QM system. We will start with a more than one spin. single spin in a magnetic field. Instead of
taking the "usual" view adopted by many QM textbooks which talk about wave functions and vectors, we will immediately start with the Density matrix density matrix formalism, which is in fact a generalization of those concepts (and, Wave mechanics paradoxically, easier to understand!). 2. Dynamics: I will show you how to “solve” any problem in quantum mechanics. That is, given the state of the system at time t=0 and the electromagnetic field it's in, I'll explain how to solve - at least in theory - for its state at any time t>0. Wave Mechanics Our philosophy will not be to "understand" quantum mechanics, which is rather ambitious, Physical Quantities of The System Are but to lay down the basic rules by which we can Given By Square Matrices carry out meaningful calculations. Physical observable quantities of a system such as energy, angular momentum and magnetic moment are given by matrices. The simplest system in QM is possibly an isolated spin-1/2. Some atomic nuclei have non- zero angular momentum. It turns out that the magnitude of the angular momentum only appears 2 34 mkg in "steps" of h /2, where h 6.62 10 sec is Planck's constant. So we have spin-1/2 particles 1 In physics-speak we'd say we have many photons per (having intrinsic angular momentum h/2), spin-1 unit volume. particles (having intrinsic angular momentum h), It is highest when m and B are anti-parallel and spin-3/2 particles (having intrinsic angular lowest when they are parallel. momentum 3h/2) and so forth. Some examples The corresponding quantum mechanical from nature include: observable is called the Hamilatonian, and is obtained by swapping m out for the quantum "Particle" n p Spin (radMHz/T) mechanical magnetization observables: Electron 0 0 1/2 2 Proton (1H) 0 1 1/2 H MBˆ Neutron 1 0 1/2 MˆˆˆBMBMB Deuterium 1 1 1 xx yy zz (2H) ˆ ˆ Carbon 6 6 0 Substituting the full forms for the Mis MSii 12 ( C) and adding up all three matrices, we obtain: Carbon 7 6 1/2 13 ( C) * BBzxy Lithium 4 3 3/2 H . BB (7Li) xy z
For a spin-1/2 particle - such as the electron, or If the magnetic field is time dependent, so is H. proton - the spin angular-momentum components are given by: The "Quantum" in Quantum Mechanics ˆˆ01 0i ˆ 1 0 A fundamental property of many quantum systems SSxy,, S z 2210 i 0 2 0 1 is that many of their observables (physical quantities) are quantized. This means that if we For a spin-1 particle such as deuterium, the take a pure, well isolated system, and measure that angular momentum components are 33 matrices: quantity, we will always get an answer that belongs to a discrete set of values. This set coincides with 2 010 0i 0 the possible eigenvalues of the matrix . We remind that reader that (1.) an mm matrix has up to m SSiiˆˆ101, 0 xy distinct eigenvalues, and (2.) a diagonal matrix's 22 010 0i 0 eigenvalues are simply its entries on the diagonal. 10 0 For example, the z-component of the angular ˆ momentum for a spin-1/2 particle, Sz 00 0 00 1 ˆ 10 Sz 2 01 In general, for a spin m/2, the spin operators will be (m+1)(m+1) matrices. has two eigenvalues, 2 . This means that when The System's Energy is Called The Hamiltonian. It Is A Matrix
We know from classical electromagnetism that the 2 You will get to practice eigenvalues a bit more in energy of a magnetic moment m is an external the tutorial, but to refresh your memory, a vector v field B is given by is an eigenvalue of a matrix M is Mv=(some number)v. A simple example is the rotation matrix E mB . Rz() about the z-axis. It is clear that zˆ 0,0,1 is
an eigenvector with eigenvalue 1, since it is
colinear with the z-axis: Rz zzˆˆ . The State Of An Ensemble is Given By positive-semidefiniteness (probabilities are 0). A Density Matrix However, we will not be taking this route and will not be using this interpretation except at one point A density matrix, often denoted , is just a square down the road, when we discuss the thermal matrix that follows a few rules. First, it is equilibrium state of the system. hermitian, just like observables:
A Simple Example: Spin-1/2 Particle † . Let's take an example by looking at a general 2×2
matrix for a spin-½ particle: It is also positive semidefinite, meaning that for any vector x, 11 12 , † xx 0 . 21 22
Finally, it has a trace of unity: where the different elements can be complex. Let's carefully apply the conditions for to be a density tr 1 . matrix. Hermiticity means
** The density matrix has one fundamental physical 11 12 11 21 † . property from which its entire behavior can be ** 21 22 12 22 derived: if the system is described by a density matrix , then the average value of a measured We see that * and * , meaning observable A is given by 11 11 22 22 both quantities are real. Let's denote them by
A tr A . 11 a
d Our next order of business will be to generalize the 22 state vector formulation to handle statistical where a, d are both real numbers. Furthermore, ensembles of particles, such as the ~1025 spin-1/2 let's denote particles in a 200 mL glass of water.
bic The Entries of a Diagonal Density 12
Matrix Represent Probabilities with b, c real, such that In the "conventional" way of teaching QM, one first learns about wavefunctions and probabilities. abic It is mentioned that a spin-1/2 particle can be in . bicd either an "up" or "down" state, each having its own energy. Using these, the following interpretation is then given to the density matrix: Next, we impose the condition tr()=1, from when in a diagonal form, the elements along the which a+d=1, so: diagonal represents the probability of the system of being in the up or down state, respectively: abic . bic1 a Pr 0 . Before imposing the final condition of positive- 0Pr semidefiniteness, let's calculate the expectation
value for the x, y and z-components of the intrinsic This is also consistent with the properties of , magnetic moment, MSˆ ˆ (i=x,y,z): namely tr()=1 (probabilities sum up to 1) and ii If H is time independent, we can solve the abic01 Liouville formally: Mbx 2 tr bic110 a te iHt//†0 e iHt UU. abici0 Mcy 2 tr bic10 a i The quantity Ut eiHt / is called the abic10 propagator. To see this solves the Liouville Maz 2 tr 2 1 bic101 a 2 equation, note that3 diHUU and dt differentiate: These are just enough equations to solve for the coefficients a, b & c: 0 d ††† UUUUUU000 dt 1 M z a iH† † iH 2 UUUU00
M M bcx , y 11 Ht tH H, ii from which so it satisfies the Liouville equation and is a solution. MiM 1 M z xy 2 MiM Example: Time Evolution of a Magnetic xy1 M z 2 Moment In A Constant Magnetic Field 12 I MSˆˆˆ MS MS Let's use all of the machinery we've seen so far to 2 x xyyzz 2 calculate the time evolution of a spin-1/2 magnetic 12ˆ moment, starting along the x-axis, and placed in a I 2 MS 2 constant magnetic field along the z-axis. z or
B0 12ˆ I 2 MS 2 y The Time Evolution Of The System Is M Given By The Liouville Equation x
Much like Newton's 2nd law, F=ma, dictates the dynamics of a classical system, Liouville's equation dictates the dynamics of a quantum system: 3 d at at For regular functions and numbers, dt eae . d 1 ˆ This does not change if we discuss matrices; the H, . dt i only time we need to be careful when handling matrices is when we have two different matrices Here Hˆ is the hamiltonian, and by definition, for which don’t commute. If we only have one matrix, any two matrices, we can treat it as a number, so: d Atˆˆˆˆˆ At At dt eAeeA. The last line is true because, A,BABBA . again, we’re dealing with Aˆ and a function of Aˆ , which commute (a matrix commutes with any function of itself). To “solve” this problem, we will ask ourselves four questions in this order: 23 x xx ex 1 1. What is our density matrix at time t=0? 2! 3! 2. What is our Hamiltonian? 3. What is our propagator, U(t)? ˆ so for the matrix exponential eA , 4. What is our density matrix as a function of
time? (obtained by computing UU † ) ˆˆ23 Aˆ ˆ AA Our initial density matrix is obtained by putting eIA 2! 3! M=(M0,0,0) in our expression for : 10 A 0 2 11 1 A11 0 2 M0 01 0 A 1 22 2!0 A22 12 2 ˆ 3 I 2 MS . M 0 1 A 0 2 1 11 2 3 3! 0 A 22 The constant magnetic field is Adding up the matrices, we get
0 23 10A AA11 11 B 0 Aˆ 11 2! 3! e 23 AA22 22 B0 01A22 2! 3!
Our Hamiltonian is We see that, along the diagonal, we get Taylor expansions of e A11 and e A22 , so 0 0 ˆˆ ˆ 2 HSBMB z 0 . A11 0 0 Aˆ e 0 2 e . 0 e A22 with This result is not true in general. For a non- diagonal matrix, in general 00 B .
AA11 12 rad kHz AA11 12 For a proton 2 42.576 in a B0=3 AA21 22 ee mT e . eeAA21 22 Tesla magnetic field, 0 2 127 MHz . The propagator is easy to calculate because the Hamiltonian is diagonal. For any diagonal matrix, Using this, we can immediately write down the A, matrix multiplication with itself is very simple: propagator:
it0 /2 A 0 iHtˆ / e 0 Aˆ 11 Ut e . it0 /2 0 A22 0 e AA00A2 0 ˆ 2 11 11 11 A Finally, with all of this in place, we can compute 00AA0 A2 22 22 22 the form of the density matrix as a function of time: N ˆ N A11 0 A N 0 A22
We can use this property by expanding the matrix exponential with a Taylor expansion. I remind you that Taylor expansion for ex is tUU t sense of direction is obtained by putting your left hand along B, curling it and noting the direction 1 M 0 it00/2 it /2 ee002 in which your fingers curl. it/2M it /2 00ee000 1 2 z 1 it00/2M 0 it /2 it0 /2 2 ee e 0 M it /2 0 eeit00/21 it /2 0 e 0 2 M 1 0 it0 2 e y M 0 eit0 1 2
This looks simple, but we’d like to also x
“understand” it by recasting it in the general form 12 ˆ tI 2 2 MS t . Note that, simply by Magnetic Moments Precess About An inspection, External Field
There was nothing special about the directions t chosen in the previous section. We could go back 10 it0 1 M0 0 e and solve for a general initial density matrix, but 2 it 01 e 0 0 our physical intuition should tell us already that it shouldn't matter how we start out: if B is constant it0 2M 0 e 1 0 22I 2 in space, the magnetization will just precess it0 e 0 around it (note the left handed sense of rotation):
2M 0 cos00ti sin t 1 0 22I 2 costi sin t 0 B 00 12I Mcos tSˆˆ iM sin tS 2 2 00xy 00 M
From this we can immediately read off the components of the magnetic moment as a function of time: We won't go back and actually solve the tedious
Liouville equation with an arbitrary field and M x tM 00cos t initial condition, although - aside from extremely
M y tM 00sin t long-winded algebra - it is quite possible to do so. Since we know M(t), we can also immediately Mtx 0 write down the density matrix as a function of
time, using the formula We can also write this using matrix notation to make it a bit clearer: 12 ˆ tI 2 MS t. 2 Mtx cos000ttM sin 0 Mty sin00 t cos t 0 0 A Spin-1/2 System Can Be Understood 0010 Mtz Classically: Bloch's Equations LH M0 Mt Rtz 0 The above quantum mechanical derivation has a completely classical analogue. To see this, we begin
This describes a circular motion of M0 in the xy- by proving Ehrenfest's Theorem, which states that plane about the axis along which the external the time evolution of the expectation of value an constant field B0 points, which is the z-axis. The rotational motion is left handed, meaning the observable, Aˆˆ tr A , evolves in time Hˆˆ MB. according to:
Substituting in Ehrenfest's theorem, we get: ˆ dA 1 dAˆ AHˆ, ˆ . dt i dt ˆ dMi 1 ˆˆ ˆ ˆ MMBMBMBixxyyzz, dt i The proof is obtained simply by applying the 11 ˆˆ ˆˆ definition of a derivative and using the fact the Mix,,MB x MMB iy y ii trace is linear in the derivative: 1 ˆˆ Miz,,,.MB z ixyz i dAˆ dtr Aˆ I leave it to the reader to verify that dt dt
ddA ˆ tr Aˆ tr ˆˆ ˆ MMxy, i M z dt dt ˆˆ ˆ MMy , zx i M Hˆ , ˆ ˆ dA tr A tr ˆˆ ˆ idt MMzx, i M y
HAˆˆˆˆˆ HA dA from which (along with the fact that [A,B]=-[B,A] tr tr tr ii dt and [A,A]=0 for any A,B) follows that
Now, the first term on the last line can be changed ˆ dMx MBˆˆ MB to tr1 AHˆ ˆ since the trace is cyclic: zy yz i dt dMˆ tr ABC tr BCA tr CAB . Using the fact y ˆˆ MBxz MB zx that AtrAˆˆ for any operator Aˆ , we can dt dMˆ write this as z ˆˆ MBy xxy MB dt ˆ dA 11 dAˆ AHˆˆˆˆ HA Looking closely at these three equations, we see dt i i dt they can be succinctly written using vector notation as: which is precisely equivalent to the claim made. With Ehrenfest's theorem, we can take each of d Mˆ the magnetization operators and write down: MBˆ . dt
ˆ dMi 1 The same equation can be derived completely ˆˆ Mi ,H dt i classically as follows: A microscopic (point-like) magnetic moment m in a magnetic field B will be where the explicit time derivative of the operators affected in two ways: it will feel a torque (which is is zero since they are time independent4. the time derivative of the angular momentum): Remember what the Hamiltonian of the system looks like: dL mB. dt
4 Some observables can be time dependent, but that is a topic we will not touch upon in this course. A nucleus has an intrinsic angular momentum S This means the term in the exponential is very and proportional moment M=S. Its derivative is small. Going back to our Taylor expansion, for small x we can approximate exx 1 quite well, ddMSd S whence mB. dt dt dt 0 Hˆ e2kT 0 The solution of the Bloch equations, which we will e kT 0 not pursue here, predicts - you guessed it - that the 2kT 0 e moment m will precess around any constant magnetic field B. 10 0 2kT The Initial State Of A System is Given 0 01 By Boltzmann's Distribution 2kT 10 1 0 We now know how to solve for the time evolution 00 ˆ I Sz of a system given its initial state. But what should 01kT2 0 1 kT that initial state be? Well, that’s a question that should be answered by you: what physical state is and your system in initially? However, one initial condition presents itself repeatedly: thermal 00 22kT kT equilibrium. I’ll give you the “answer” for how Ze e should look like and then we’ll discuss its meaning: 00 11 2. 22kT kT HHˆˆ 1 TE eZekT,tr. kT . Simple! Dividing the two, Z
TE 1 IS0 ˆ Let’s assume our field is along the z-axis. The 22kT z 2 Hamiltonian is: 12I 0 Sˆ 24 2 kT z
0 TE 2 0 12ˆ ˆˆ ˆ Since is now in the form 2 I 2 MS , HSBMB z 0 . 0 0 2 we can immediately write down the z-component of the magnetization5: Because it is diagonal, we can immediately write down: 2 eq 0 M z 4kT 0 Hˆ 2kT e 0 e kT This tells us something that’s physically intuitive: 0 2kT 0 e when B0 is static and along z, the spins align along Hˆ B0 because there is an energetic preference for 00 ZetrkT ee22 kT kT . them to point along the field. However, the amount of this alignment is quite small because of the thermal effects that disperse the magnetization. We will be working at room temperature, and can The x & y components of the magnetization are thus simplify, by noting zero, which can also be confirmed by calculating
26 810 J 5 0 We could have equally derived the thermal equilibrium kT 4 1021 J components of the magnetization by first calculating ˆ and then taking the traces tr M i . ˆˆ to many observables. For an observable Aˆ , if trM x , tr M y , and this is because there is I then6 no energetic preference for the spins to point perpendicular to the main static field. This calculation could have been repeated with AAAtr ˆˆ tr any spin and yielded a similar result. For a spin S (S=1/2, 1, 3/2, 2, ... ), the equilibrium trAˆ 0 magnetization is Many observables satisfy . You’re free to ˆˆˆ check that trSSSxyz tr tr 0 , and that 2 eq 0 SS 1 M . for a spin in an external constant magnetic field, z 3kT trHˆ 0 as well. This is why many authors What happens if we have N spins? If they do not simply omit it. We will do that as well and simply interact with each other, the total magnetic not write it out. One consequence of this is that moment of the ensemble will (on physical the density matrix at thermal equilibrium assumes grounds) just be multiplied by a factor of N: a simple form:
2 TE Sˆ . eq NBSS 0 1 z M z 3kT Product Operators
(it’s written in a slightly different way – I took 0 We've seen that for a single spin-1/2, and wrote it explicitly as B0) 12ˆ I 2 MS. The Identity Element Is Often Omitted 2 From The Density Matrix Remember that for a single spin-1/2, we saw that This means we can write it as a linear sum of the the most general density matrix has the form form:
aI aIˆˆˆ aI aI 12ˆ 0 xx yy zz I 2 MS . 2 where we've defined SIˆ , and where7 ISˆ 1 ˆ 0 j j In other words, it looks like this: (j=x,y,z). This is like saying that the set of ˆˆˆ 1 operators I,,,IIIx yz forms a basis for the space 2 I . of density matrices of spin-1/2 particles: every Now, whatever our propagator is, density matrix can be written as a linear sum of these four operators. Now, ††1 UU U2 IUUU . ˆ trSI0 tr 2 The part with the identity matrix stays the same: trSSSˆˆˆ tr tr 0 123 11†† 1 UIU22 IUU 2 I 6 Of course this is not strictly a density matrix because because the propagator is unitary (so UU† I ). its trace is not one. However, this reasoning holds for 1 I This means it’s “boring”. Not only does it not any part of that is proportional to I, like the 2 term. change with time – for whatever interaction we 7 I’m dividing by to make all basis “elements” have! – but it also usually does not even contribute dimensionless (it would be weird to have one element of the basis have no units, while another have units of angular momentum). LH where Rnˆ is a left handed rotation matrix The demand tr 1 means about an axis defined by the unit vector nˆ , which points along the direction of the constant field B. 3 ˆ The angular velocity of the precession is B . tr tr nnS n0 This means that the density matrix changes from 3 ˆ nntr S 12 n0 ˆ 00 I 2 MS 1 2 2 002 12 I MI0 ˆ This is of course nothing new: it means we get the 2 1 2 I term. at time t=0, to Time Evolution of Product Operators 12 ˆ The interesting thing about this formulation is the 00 IRt MI . way it makes us think about propagation. Given a 2 propagator U, an initial density matrix evolves as: Now, another way to represent the dot product is † using matrices: UU .
abab ab ab Plugging in our density matrix expression, we get x xyyzz b x 3 aaab UaIUˆ † x yzy nn n0 b z 3 ˆ † abT aI0 aUIUnn n1 The transpose T has a nice property8, by which it So, in this completely equivalent picture, the switches the order of the matrices (just like for the coefficients are time independent and the basis T TT hermitian conjugate): AB B A . This is true vectors themselves “evolve” over time. If we know regardless of the sizes of the matrices, and they what the propagator does to each of the don’t even have to be square. So, for our propagators, we can immediately write the solution multiplication, down. This way of thinking will become extremely useful when trying to understand basic LH ˆ experiments down the line. Rtnˆ MI0 Let’s do an important example. For a spin-1/2 T LH ˆ system, we want to calculate Rtnˆ MI0 T LH ˆ MI0 Rnˆ t UIˆ U † n under the influence of a general and constant 8 magnetic field B. I’ll show you that you already This is easily proved. The transpose of A has elements know the answer. We’ve said that M(t) precesses AAT . For the product AB, ij ji about a constant magnetic field B: N T AB AB Ajk B ki ij ji MMtR LH t 0 k 1 nˆ NN BA BTT A BA TT ki jk ik kj ij kk11
ˆˆˆ is now transformed into In other words, the “basis matrices” I x ,,IIyz themselves perform a precession around the magnetic field. For example, if our field is along aa11 12 10 AIˆ the z-axis, aa21 22 01
aa1100 12 0 00aa 11 12 B 0 aa00 21 22 B0 00aa21 22 then our Hamiltonian is where I is the identity operator of system two. 2. Any operator A of system (2), ˆˆ ˆ HMBz 00 Iz aa ˆ 11 12 and A aa21 22
iHtˆ itI ˆ Ut e e0 z is now transformed into so 10 aa11 12 IAˆ 01 aa21 22 UtIU ˆ † t aa 00 x 11 12 itIˆˆ itI eIe00zzˆ aa21 22 00 x ˆˆ 00aa11 12 I xycos00tI sin t 00aa 21 22 where in the last line we just treated Iˆ as a x For example, the x-component of the angular “vector” along the x-axis and applied a left handed momentum for spin #1 is rotation to it with an angle t . 0 01 10 ˆˆ SSI1xx Multiple Spin Systems 2 10 01 0010 Multiple Spin Systems Are Described 0001 By Outer (Kronecker) Products 2 0100 By now we've seen how to describe the state of a 1000 system and calculate its time evolution if we know its Hamiltonian, using Liouville's equation (at least Another example: the y-component of angular in theory, or for the simple system of a single spin- momentum for spin #2 is: 1/2). This can be extended to multiple spin systmes by using the kronecker product of these 10 0i ˆˆ SIS2xy spin spaces. The rules are simple and are listed 012 i 0 below for two spin-1/2 systems; they can be 000i generalized in a fairly straightforward manner: i 000 1. Any operator A of system (1), 2 000i 00i 0 aa11 12 Aˆ aa 21 22 These ideas can be extended to any number of spins, by taking successive kronecker products. For example, for three spin-1/2s, the x-component of these form a basis for all density matrix of the the 2nd spin is combined system:
ˆˆ 3 SISI2 yy ˆˆ nmSS n m nm,0 10 0i 10 012 i 0 01 This makes sense in terms of number of constants: 000i is now a 4 4 matrix having 16 complex i 000 10 elements, or 32 real numbers. Hermiticity cuts this 2 00 0i 01 down to 16, precisely the number of coefficients in 00i 0 the expansion above (note the requirement tr()=1 fixes 1 , leaving us with 15 real coefficients). 00i 00000 00 4 00 0i 00 0 0 i 00 00000 0i 0 00000 2 00 0 0 00i 0 00 0 0 00 0 i 00 0 0i 0 0 0 00 0 0 0i 0 0
That is truly a frightening matrix! Down the road, we’re going to come up with other ways of dealing with multiple spin systems that avoids mostly dealing with direct matrix multiplication.
Product Operators For Multiple Spins Can this be extended to two spin-1/2 operators easily. There are a total of 6 components of angular momentum:
ˆˆˆ SISISIxyz,,
ˆˆˆ I SISISx ,,yz
We can also form kronecker products that do not correspond to any physical observable:
ˆˆ ˆˆ ˆˆ SSx xxyxz, SS SS ˆˆ ˆˆ ˆˆ SSyx, SS yy SS yz ˆˆ ˆˆ ˆˆ SSz xzyzz, SS SS
Along with the trivial identity operator, II , we ˆˆ get 16 combinations SSnm where, as before, ˆ ˆˆ ˆˆ n,m=0,1,2,3 and SI0 , SS1 x , SS2 y , ˆˆ SS3 z . We won't prove it mathematically, but