4 Quantum Mechanical Description of NMR

Transcription

1 4 Quantum Mechanical Description of NMR Up to this point, we have used a semi-classical description of NMR (semi-classical, because we obtained M 0 using the fact that the magnetic dipole moment is quantized). Though this approach is useful, in particular for understanding NMR in terms of vector diagrams, it cannot be used to describe all NMR phenomena. For instance, it is not possible to describe the evolution of the magnetization in coupled spins (e.g. via scalar coupling) using Bloch equations. Therefore, we need to be familiar with quantum mechanics in order to describe all NMR phenomena properly. 4.1 Mathematical Tools * Special thanks to Scotty for this section! Much of the maths used in QM uses complex numbers. Here are some useful definitions/relations to remember:

2 A complex number is defined as z = a + ib = r(cos φ + i sin φ), where i 2 = 1. The real and imaginary parts of z are: Re(z) = a, Im(z) = b. The norm and argument of z are: r and arg (z) = φ. The Euler representation of a complex number z: r(cos φ + i sin φ) = re iφ The complex conjugate of z = a + ib is z = a ib. The norm or modulus of z = a + ib = r(cos φ + i sin φ): z = zz = a 2 + b 2 = r Note that z 2 = zz = a 2 + b 2. Functions, Operators and Functionals A function maps numbers numbers. For example, if we define f(x) = sin(x), we can evaluate f(0) = sin(0).

3 NOTE: For the purposes of this discussion, we do not differentiate between scalar numbers, vectors, matrices or tensors. These are all numbers. An operator maps functions functions. For example, if we define ˆF {g(x)} = (g(x)) 2, then ˆF {sin(x)} = sin 2 (x) and ˆF {ln(x)} = ln 2 (x) This notation deviates from the standard notation: ˆF {g(x)} = ˆF g(x) Example: The ( nabla ) operator is defined as g( x N ) x 1 g( x {g( x N N ) x )} = 2... for a scalar function g( x N ). g( x N ) x N A functional maps functions numbers, e.g. if

4 we define F [g(x)] = 2π 0 g(x) dx, then F [sin(x)] = 2π 0 sin(x) dx = 0. We will use these definitions in this and subsequent chapters to describe our spin systems, where: the state of the system is given by the wavefunction ψ( q, Q, s e, s n, t) every observable has an operator  associated with it, where q refers to the electron coordinates, Q refers to the nuclear coordinates, s e refers to the electron spin, and s n refers to the nuclear spin. We can manipulate the state of the system using the postulates of quantum mechanics. 4.2 Postulates of Quantum Mechanics Wavefunction If a string is stretched between two fixed points, the displacement of the string from its equilibrium horizontal position is given classically by u(x, t). If a

5 perturbation moves along the string with velocity v, then we can write a one-dimensional wave equation: 2 u x 2 = 1 v 2 2 u t 2. (4.1) Separating u(x, t) into a temporal and a spatial part, we can rewrite it as u(x, t) = ψ(x)cos(ωt) (4.2) which in turn allows us to rewrite the wave equation as: 2 ψ(x) x 2 + ω2 ψ(x) = 0. (4.3) v2 which, we will see later, is the time-independent Schrödinger equation. Probability Given the wavefunction above, let us now consider a slightly more complex picture: a string stretched between two fixed points as before, with additional points (nodes) where it can be fixed.

6 psin(x) Three Wavefunctions of a 1-D Quantum Particle in a Box (L=10) sqrt(2.0/10.0)*sin(1*3.14*x/10) sqrt(2.0/10.0)*sin(2*3.14*x/10) sqrt(2.0/10.0)*sin(3*3.14*x/10) x In this case, we have many ψ n (x) with n = 1, 2,.... This is analogous to the particle in a one-dimensional box problem, where the wavefunction ψ n (x) describes the microstate of a quantum particle of mass m in a box of length L (x [0... L]) for every discrete energy level E n, n = 1, 2,... (the more nodes, the higher the energy - excited state) (Born Interpretation): the probability of finding the particle at a certain position x is ρ n (x) = ψ n (x) 2 = ψ n(x)ψ n (x) NOTE: We have used an operator to map ψ n (x)

7 onto ρ n (x). The wavefunction is normalised so that L 0 ρ n (x) dx = 1 psin^2(x) The Probability of Finding the Quantum Particle at Position x (2.0/10.0)*sin(1*3.14*x/10)**2 (2.0/10.0)*sin(2*3.14*x/10)**2 (2.0/10.0)*sin(3*3.14*x/10)** x NOTE: This is where the image of an electron cloud around a nucleus comes from: the cloud is dense where the probability of finding an electron is high. Operator postulate For any observable a there is an associated

8 Hermitian operator Â{ψ n(x)} such that Â{ψ n (x)} = a n ψ n (x) where a n is the observed value of a when the system is at energy E n. NOTE: The operator of an observable must be Hermitian because this guarantees that the observable is a real, rather than an imaginary number. An operator  is called Hermitian if ( ψ n(x)â{ψ n(x)} = ψ n(x)}) n(x)â{ψ (4.4) This makes sense when you consider the next postulate: Expectation value postulate The expectation value i.e. the microscopic experimentally measured value of any observable a is Exp a [ψ n (x)] = = L 0 L 0 ψ n(x)â{ψ n(x)} dx ψ n(x)a n ψ n (x) dx

9 NOTE: = L 0 a n ρ n (x) dx = a n (4.5) This is a functional. Example 1 - Let s assume we have some given potential V (x) that interacts with the particle at x. In order to calculate the potential energy, we need a potential energy operator, which is Ê pot {ψ(x)} = V (x)ψ(x) ; then the expectation value of the potential energy is Exp E pot [ψ n (x)] = = = L 0 L 0 L 0 ψ n(x)êpot{ψ(x)} dx ψ n(x)v (x)ψ(x) dx V (x)ρ(x) dx (4.6) A microscopic expectation value of any x dependent function f(x) is a weighted average Exp[f(x)] = L 0 f(x)ρ(x) dx

10 Example 2 - In order to calculate the kinetic energy of the particle, we need a kinetic energy operator, which is Ê kin {ψ(x)} = h2 2m 2 {ψ(x)}, where h = h 2π. Now we can calculate the expectation value for the kinetic energy of the system by evaluating the functional Exp E kin [ψ n (x)] = h2 2m h2 2m L 0 L 0 L 0 ψ n(x)êkin{ψ(x)} dx = ψ n(x) 2 {ψ(x)} dx = ψ n(x) d2 ψ n (x) d 2 x which we can solve if we know ψ n (x). dx, (4.7)

11 In general: To calculate a system property of a quantum system: 1 Start with the wavefunction 2 Choose the appropriate operator for the system property 3 Evaluate the expectation value functional A Quantum Particle in a Box As we saw above, a quantum particle of mass m in a one-dimensional box of length L (x [0... L]) can be likened to a string stretched between two fixed points, with nodes. Starting from equation 4.3, we can rewrite the term ω 2 /v 2 as ω 2 v 2 = (2πν)2 v 2 = 4π2 λ 2. (4.8) Using de Broglie s formula, which states that λ = h p (4.9) and using the definition of the total energy: E = p2 2m + V (x) (4.10)

12 where p is momentum, equation 4.3 can be written as the time-independent Schrödinger equation: h2 2m 2 {ψ(x)} + V (x)ψ(x) = Eψ(x), (4.11) where h = h 2π, and h is Planck s constant, ψ(x) is the complex wavefunction, (V (x) is the potential and E is the energy (a scalar, real value). As V (x) = 0 inside the box, and 2 {ψ(x)} = d2 ψ(x) d 2 x we have h2 d 2 ψ(x) 2m d 2 = Eψ(x), (4.12) x for which the general solution is known to be: ψ(x) = A sin (kx) + B cos (kx) E = k2 h 2 2m, (4.13) where A and B are complex numbers and k is real. We use boundary conditions to determine A, B and k. We choose ψ(0) = ψ(l) = 0. ψ(0) = 0: 0 = ψ(0) = A sin (0) + B cos (0) = A 0 + B 1

13 B = 0. Now set ψ(l) = 0 (4.14) 0 = ψ(l) = A sin (kl) (4.15) Either trivial, uninteresting solution A = 0, or for A 0 which has many solutions: sin (kl) = 0 k = nπ L (4.16) for n = 1, 2,.... NOTE: This means only complete waves are allowed in the box, which explains how only discrete energy levels are allowed for the quantum system. Analogy with sound waves: resonance. The probability of finding the particle at a point x is ψ(x) 2, so we know that L 0 ψ(x) 2 dx = 1

14 Lets use this to determine A: 1 = L We use the fact that [sin (ax)] 2 dx = 1 2 x 1 4a So then A 2 = and with a = nπ L 0 A 2 [sin ( nπ L x)]2 dx (4.17) ( 1 2 L 1 4a ( a we determine that 2 A = L sin (2ax) (4.18) ) sin (2aL) ) sin (2a0) (4.19) So: A is any complex number with an absolute value of 2 L ; E n = n2 h 2 π 2 2mL 2 = n2 h 2 n = 1, 2,... 8mL2 2 ( nπx ) ψ n = L sin L (4.20)

15 Hamiltonians Given the definition of an operator, we can see that equation 4.12 can be written in general form as Â{ψ n (x)} = a n ψ n (x) (4.21) or Ĥψ n (x) = Eψ n (x) (4.22) In other words, the classical energy function is related to the Hamilton operator or energy operator. Classical energy function, E H^ For example, recall from Chapter 1, the definition of the energy of a magnetic moment in an external magnetic field B 0 : E = µ B 0 = γ L B 0 (4.23) The corresponding quantum mechanical definition is

16 given by Ĥ = γ h ˆL B0 In general, one can write = γ h[ˆl x B x + ˆL y B y + ˆL z B z ] (4.24) Ĥ = γ Ĵ B 0 (4.25) where Ĵ can be the angular momentum associated with the electron spin, which we denote as h Ŝ or the angular momentum associated with the nuclear spin, which we denote as h Î. Quantum - Classical LINK: Recall from chapter 1, that we said that NMR involves measuring the total magnetic dipole moments in our samples. Translating this into QM lingo - since µ translates into Î for nuclear spins - NMR involves measuring the expectation value of Î (usually only one component of it actually, see below). 4.3 Angular Momentum in MR The angular momentum operators ( ˆL, Î) Ŝ, and

17 have the special property that their components do not commute with each other. The commutation relation for two operators  and ˆB is given by [Â, ˆB] =  ˆB ˆBÂ. (4.26) If  and ˆB commute with each other, then the equation above is zero. The physical implication of this is that the two observables can be measured at the same time - measuring one does not affect the outcome of the other, and vice versa. Applying this to the components of Î (the same applies for the other two), which are given by ( Î x = i y z z ) y ( Î y = i z x x ) z ( Î z = i x y y ) x we can obtain the following cyclic commutation (4.27)

18 relations: [Îx, Îy] [Îy, Îz] [Îz, Îx] = iîz = iîx = iîy (4.28) A good way to remember this is with the picture: ^x I I^ z ^ I y Another important property of momentum is the definition Î 2 = Î2 x + Î2 y + Î2 z (4.29) It can be shown that [Îx, Î2] = 0 [Îy, Î2] = 0 [Îz, Î2] = 0 (4.30)

19 i.e. Î 2 commutes with the components Îj where j = x, y, orz. 4.4 Matrix representation of operators It is possible to express the wavefunction ψ as a weighted sum of known, appropriately chosen functions ϕ k, such that ψ = D c k ϕ k, k=1 where D is the dimension of the space. If our chosen functions are orthonormal, then ϕ k ϕ l = δ k,l, where δ k,l is the Kroenecker delta, i.e. is equal to 1 if k = l, and 0 otherwise. NOTE: This is like the equation for probability we defined above. Operating on these new functions yields: Âϕ k = D a lk ϕ l. (4.31) k=1

20 Consequently, equation 4.21 can be rewritten as Â{ψ} = Â c k ϕ k = a lk c k ϕ l = d l ϕ l k k l l (4.32) where d l = k a lk c k d = (A) c (4.33) where the latter equation is in matrix form. Examples of (A) are (A) = a 11 a 12 a 21 a 22 (4.34) for D= 2 and for D= 3. (A) = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 (4.35) In general, the elements of a matrix are obtained

21 using the definition < ϕ j  ϕ k >= l a lk < ϕ j ϕ l >= l a lk δ jl = a jk (4.36) where we use the Dirac notation (see box 1). NOTE: This is just another way of writing the expectation value of the operator the functions ϕ k. Â, operating on Box 1. Dirac Notation Dirac notation is a short form to save having to write out the integral: < m n >= dxψm(x)ψ n (x). Since changing the representation of an operator into matrix form should not change its properties, the commutation relations given in equation 4.28 should still be valid, i.e. [(I x ), (I y )] = i(i z ) [(I y ), (I z )] = i(i x ) [(I z ), (I x )] = i(i y )

22 (4.37) where the () are there to emphasize that we are writing the equations in terms of matrices. We can use these relations and the convention that the (I z ) matrix is diagonal to derive the three operators in matrix representation. The dimension D of our matrices is determined by the spin quantum number, which in chapter 1 we defined as m J = J, J + 1,...J, (4.38) such that D = 2J + 1. This corresponds to the number of energy levels we drew for a given value of J. NOTE: In many textbooks, J is replaced by I in order to emphasize that we are talking about nuclear spins. Recall, that J is the property of a nucleus. Let us assume, that (I z ) and (I x ) are given by (I z ) = a 0 0 b (4.39)

23 and (I x ) = c d d e. (4.40) We can get (I y ) from the relation i[(i z ), (I x )] = i ac ad ca db bd be d a eb = i(a b) 0 d d 0 = (I y ) (4.41) Using this definition of (I y ), we can write 0 (a b)d i[(i y ), (I z )] = (a b) (a b)d 0 = = (I x ) (4.42) c d d e Using this last relation, we see therefore that c = e = 0

24 (a b) 2 = 1 (4.43) Using the final commutation relation, i[(i x ), (I y )] = 2dd 0 0 2dd = a 0 = (I z ), (4.44) 0 b we can solve for a, b, and d: yields a = 1 2 Finally, given that and b = 1 2. a = b a b = 1 (4.45) 2dd = 1 2, (4.46) we obtain d = 1 2 as well. Therefore, we can write the matrix form of (I z ), (I x ),

25 and (I y ), as (I z ) = (I x ) = (I y ) = i 2 i 2 0 (4.47) These three matrices represent the Pauli matrices for spin 1/2. In general, we should be able to find functions ϕ k = φ m such that Î z φ mj = m J φ mj. (4.48) Given the equation above, we say that φ mj are eigenfunctions and m J are eigenvalues, because the result of operating on φ mj is a constant times the same function φ mj.

26 Therefore in matrix representation, (I z ) = < φ J Îz φ J > < φ J Îz φ J 1 >... < φ J Îz φ J > < φ J 1 Îz φ J > < φ J 1 Îz φ J 1 >... < φ J 1 Îz φ J > < φ J Îz φ J > < φ J Îz φ J 1 >... < φ J Îz φ J > (4.49) or after evaluating the expectation values, J J (I z ) =. (4.50) J Putting in J = 1 2, we see that we get the same (I z) defined in At this point, we can introduce two new operators Î + and Î, which are defined as Î + = Îx + iîy Î = Îx iîy (4.51) These operators have the special property that Î + φ mj = J(J + 1) m J (m J + 1)φ mj +1

27 Î φ mj = J(J + 1) m J (m J 1)φ mj 1, (4.52) i.e. the Î+ operator raises the magnetic quantum number by 1 and Î lowers the magnetic quantum number by one. Let s see what this means by way of the spin 1/2 example: Eigenfunctions: φ 1/2, φ 1/2. (see box 2) Therefore, Î + φ 1/2 = Î + φ 1/2 = Î φ 1/2 = Î φ 1/2 = J(J + 1) 1 2 ( )φ 1/2+1 = 0 J(J + 1) 1 2 ( )φ 1/2+1 = φ 1/2 = φ 1/2 J(J + 1) 1 2 (1 2 1)φ 1/2 1 J(J + 1) 1 2 ( 1 2 1)φ 1/2 1 = 0 (4.53) where J = 1/2.

28 Box 2. Special Case: Spin 1/2 For spin 1/2, the eigenfunctions φ 1/2 and φ 1/2 are often written in terms of two new symbols: φ 1/2 = α φ 1/2 = β. In matrix form, for spin 1/2, the raising operator, Î +, and lowering operator, Î, are (I + ) = (4.54) and (I ) = (4.55) We can write Îx and Îy in terms of the raising and lowering operators: Î x = 1 2 [Î+ + Î ] Î y = 1 2i [Î+ Î ]

29 < φ i Î φ j >= J(J + 1) j(j 1)δ j 1,i (4.57) (4.56) We can also see from the spin 1/2 case, that in general, the elements of (I + ) and (I ) are given by < φ i Î+ φ j >= J(J + 1) j(j + 1)δ j+1,i So if we recall that for a general operator in matrix form as (A) = Â, written < φ J Â φ J > < φ J Â φ J 1 >... < φ J Â φ J > < φ J 1 Â φ J > < φ J 1 Â φ J 1 >... < φ J 1 Â φ J > < φ J Â φ J > < φ J Â φ J 1 >... < φ J Â φ J > then the equations in 4.57 tell us that (I + ) = 0 < φ J Î+ φ J 1 > < φ J 1 Î+ φ J 2 > with the non-zero terms in blue and similarly (4.58) (4.59)

30 (I ) = < φ J 1 Î φ J > < φ J 2 Î φ J 1 > with the non-zero terms in again blue. (4.60) Therefore equations 4.50, 4.56, 4.59 and 4.60 can be used to derive all five operators in matrix form for any m J. As practice for your midterm, try this for spin 1!!! 4.5 Spin Hamiltonian In the previous sections, we saw that the Hamilton operator can be used to describe the total energy of a system (recall: the particle in a box section). In order to describe a molecule on which we want to perform an NMR experiment, i.e. a molecule in a magnetic field B 0 (and possibly an electric field E), we therefore need to describe the energy of the system by writing the Hamiltonian: Ĥ = Ên kin +Êe kin +Ên pot +Êe pot +Êen pot +ĤQ +ĤLL +... (4.61)

31 where the terms can be visualized by decomposing the molecule in terms of its properties, i.e. momentum, charge, position, angular momentum, electron spin angular momentum,... or graphically: T^ V^ e e LL SS H^ H^ p^e e βl ^ e gβ^s e Electrons E H^ Q H^ Q B 0 ze Q p^n n γ hi ^ T^ V^ n n II n H^ n Nuclei Although this Hamiltonian is very descriptive, it is also too cumbersome. Therefore, we must try to simplify it: 1 apply the Born-Oppenheimer approximation, which assumes that nuclear motion can be

32 neglected. 2 for the magnetic properties of the molecule, we need only to consider unpaired electron and nuclear spins, such that the spin Hamiltonian ĤS is Ĥ S = < ψ( q, s e, s n ) Ĥ ψ( q, s e, s n ) > Ĥ S = ĤS ( Ŝi, Î l ) A i Ŝi + l B l Îl Ĥ S = i + i + l j k Ŝi (C ij ) Ŝ j Îl (D lk ) Î k Ŝi (E il ) Î l. (4.62) + i l We will see in the NMR Interactions section what some of these coefficients (vectors and matrices) are. For example, (D lk ) includes terms such as scalar spin-spin coupling, dipolar coupling, and quadrupolar coupling.

33 In any case, this equation clearly shows that it is possible, with a spin Hamiltonian, to describe both interactions of a spin with its environment (terms in Î l ) and interactions between spins (terms in Îl X Î k ). Recall, that the whole reason for switching to a quantum mechanical picture is that Bloch equations cannot be used to describe spins interacting with other spins (i.e. coupled spins). SUMMARY Up this point, in terms of NMR experiments, we have described: 1 the spin angular momentum, in both operator and matrix form; 2 the spin Hamiltonian. What we want is a quantum mechanical equivalent to d M dt. So we still need to describe the total magnetization in our sample quantum mechanically, as well as an equation of motion. 4.6 Density Operator/Matrix As we saw in Chapter 1, if you place a single spin

34 (let s take J = 1/2) in a magnetic field B 0, then there are 2J + 1 states (corresponding to two energy levels for spin 1/2), which are described by the eigenfunctions φ 1/2 and φ 1/2 (or α and β). In a NMR sample, we typically have on the order of spins (i.e. an ensemble), some being close to state α >, some being close to state β >, while the rest are in intermediate states. Therefore, we say that we are not dealing with pure states, but with a superposition of states. In this case, the wavefunction for our ensemble is given by ψ = k c k ϕ k. and the expectation value for a general operator  is <  > = < ψ  ψ > < ψ ψ > = < ψ  ψ > (4.63) since the probability < ψ ψ > is 1. NOTE: This is different from equation 4.36, which gives us the elements of the matrix (A). Here we are dealing with the total wavefunction.

35 Putting in the definition of ψ into the equation for <  >, we get <  > = ψ Â{ψ} ( ) ( = c l ϕ l Â{ c k ϕ k )} l = c l c k ϕ l Â{ϕ k } l k = c l c k < ϕ l  ϕ k > l k = c l c k a lk l k (4.64) We can define a new parameter ρ kl as the product ρ kl = c l c k (4.65) k such that <  > = k = k ρ kl a lk l ((ρ)(a)) kk

36 = T r{(ρ)(a)} = T r{ˆρâ} (4.66) with the trace (Tr) being the sum of the diagonal elements, and (ρ) is the density matrix and ˆρ is the density operator. NOTE: This ρ is different from the ρ n used in the definition of the probability. The implication of the last line in the above equation (or equivalently the line above that) in NMR is that any macroscopic observable (e.g. magnetization) can be deduced from two operators: 1 one which represents the observable being measured; 2 another representing the state of the entire spin ensemble, independent of the number of spins in it. In other words, rather than having to define microscopic states for each of our spins, we can used a single operator ˆρ. For spin 1/2, we can have a look at a few specific

37 expectation values: < φ 1/2 ˆρ φ 1/2 > = c α c α < φ 1/2 ˆρ φ 1/2 > = c β c β (4.67) which represent the population of the lower and upper energy levels for a two spin system, respectively. The line over the terms on the right hand side of the equation represents an average over all spins in the ensemble. The other two possible terms, < φ 1/2 ˆρ φ 1/2 > = c α c β < φ 1/2 ˆρ φ 1/2 > = c β c α (4.68) represent coherences, i.e. that a net spin polarization exists in the direction perpendicular to the external magnetic field. This requires that in our sample, we have some spins which are in superposition states. In other words, these spins do not point along z but in the xy-plane, as shown below.

38 M.H. Levitt, Spin Dynamics, p We will see in the solution NMR section how coherences are used. 4.7 Equations of motion in MR The equation of motion in quantum systems is given by the time-dependent Schrödinger equation, ψ t = i Ĥψ. (4.69) h

39 Special case: time-dependent Schrödinger equation When we can separate the wave function into a timeindependent component and a time-dependent term, i.e. ψ k ( q, Q, s e, s n, t) = ψ k ( q, Q, s)e ie kt/ h. (4.70) then the time-dependent Schrödinger equation simplifies to Ĥ ψ k = E k ψk, (4.71) which is the time-independent Schrödinger equation. Given the definition for the density matrix in equation 4.65, we can write the differential with respect to time as: ρ kl t = c k c t c l l + c k t (4.72) Now, given the time-dependent Schrödinger equation above and the definition of the NMR wavefunction ψ, we can determine that c k t = i h < ϕ k Ĥ ϕ j > c j j

40 c k t = i h = ī h H kj c j j j < ϕ j Ĥ ϕ k > c j = ī h H jk c j j (4.73) Therefore equation 4.72 can be rewritten as ρ kl t = c k c t c l l + c k t = i H kj c j c l + ī h h j = i h [ j H kj ρ jl j c k c j H jl j ρ kj H jl ] (4.74) Rewritting in operator form, we can see that the term in [ ] is a commutation relation, therefore dˆρ(t) = i ] [Ĥ, ˆρ(t). (4.75) dt h This equation is equivalent to the time-dependent Schrödinger equation and is known as the

41 Liouville-von Neumann equation. The solution of this equation is ˆρ(t) = e iĥt/ h ˆρ(0)e iĥt/ h (4.76) With this, we can finally describe a time-dependent expectation value for a general operator  as <  > (t) = T r{e iĥt/ h ˆρ(0)e iĥt/ h Â} (4.77) With these two last equations, we can describe how our spin system evolves over time, given the initial state of our spin system ˆρ(0), under a Hamiltonian Ĥ. This is exactly what NMR simulation packages like GAMMA ( and SIMPSON ( bionmr/software/index.php) use. Here is an example of a GAMMA program to simulate the outcome of a one-pulse experiment in solution NMR:

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