Lecture #6 NMR in Hilbert Space

Transcription

1 Lecture #6 NMR in Hilbert Space Topics Review of spin operators Single spin in a magnetic field: longitudinal and transverse magnetiation Ensemble of spins in a magnetic field RF excitation Handouts and Reading assignments van de Ven, section 1.10, Appendices B.1 and C. F. Bloch, W. Hansen, and M. Packard, Nuclear Induction. Phys. Rev., 69: 127, 1946 E. M. Purcell, H. C. Torrey and and R. V. Pound. Resonance absorption by nuclear magnetic moments in a solid. Phys. Rev., 69: 37, Miller, Chapter 12: pp (optional). 1

2 Isolated Spin in a Magnetic Field Goal: Find the appropriate wavefunction ψ(t) that describes a system consisting of a nucleus (spin = 1/2) in a uniform magnetic field. Procedure: Given: Find ˆ H (t). Schrödinger s Equation: t ψ(t) = i ˆ H (t) ψ(t) Solve for ψ. Compute quantities of interest: e.g. components of magnetic moment ˆ µ x, ˆ µ y, and ˆ µ. Later we ll show these correspond to the familiar quantities M x, M y, and M. 2

3 Review: Spin, Angular Momentum, and Magnetic Moment Spin, angular momentum, and magnetic moment operators are linearly related. ˆ µ p = γl ˆ p = γ!ˆ I p, p = { x, y,} magnetic moment angular momentum spin eigenkets of ˆ I Spin 1/2 particle: I ˆ α = α and I ˆ β = 1 2 β Matrix representation in { α, β } basis: I, = 1 # 1 0 & I x, = 1 " 0 1% $ ' I y, = 1 # 0 i& % ( % ( 2 # 1 0& 2$ i 0 ' 2$ 0 1' 3

4 from which it follows that Spin Operators General case (based on properties of angular momentum): " ˆ $ # [ I ˆ, I ˆ ] x y = iˆ I p n % ' I ˆ & q = I [ I ˆ, I ˆ ] y = iˆ I x I ˆ x, I ˆ y, and I ˆ (* [ I ˆ p, I ˆ q ], n odd ) I ˆ + * q, n even superoperation notation [ I ˆ, I ˆ ] x = iˆ I y commute cyclically: p,q = x,y,; p q Let s also define a new operator I ˆ 2 corresponding to the total angular momentum (magnitude). I ˆ 2 = I ˆ ˆ x I x + I ˆ ˆ y I y + I ˆ ˆ I which is easily shown to satisfy [ I ˆ 2, I ˆ ] x = [ I ˆ 2, I ˆ ] y = [ I ˆ 2, I ˆ ] = 0 4

5 Spin Operators From the commutator relations, one can derive the corresponding eigenkets and eignevalues of I ˆ 2 and I ˆ (note, since operators commute, they have a common set of eigenkets). Spectrum of I ˆ 2 = I(I +1) for I integer multiple of 1/2 Spectrum of I ˆ = m for m = I, I +1,,I 1,I I is known as the spin quantum number. m is known as the magnetic quantum number. Hence, spin/angular momentum/magnetic moment of elementary particles (e.g. electrons, protons, etc) are quantied in magnitude and along a projection onto any one axis. Formally, eigenkets of Î are written as: I, m where for I = ½: α = 1 2, and β = 1 2, 1 2. shorthand notation 5

6 Space quantiation! In a magnetic field B = B 0ˆ, magnetic moment is quantied in (remember Stern-Gerlach experiment). Pictorial drawings for spin 1/2 nuclei (e.g. 1 H, 31 P, 13 C) Picture 1: cones (used by de Graaf) spin up, parallel! B = B 0ˆ spin down, anti-parallel! B = B 0ˆ angle between spin and magnetic field $ ' m θ = cos 1 & % I( I +1) ) ( discrete! Picture 2: polariation (used by Levitt) Single spin µ θ r µ θ = 54.7!, I = 1, m = µ = 1 γ!, " µ = 3 γ! 2 2 = α µ θ r µ θ =125.3!, I = 1, m = µ = 1 γ!, " µ = 3 γ! 2 2 What s missing from these drawings? Arrow depicts the direction for which magnetic moment is well defined, i.e. = 1 γ! with 2 probability 1.0 = β Note: β α If you were to measure I x, I y, and I for an individual spin, what would you get? 6

7 The Hamiltonian Total energy of the system is given by the expected value of H ˆ ( t). E =! H ˆ =! ψ H ˆ ψ (remember, we defined H ˆ as energy/!) Classically, the potential energy of a dipole in a magnetic field is: E =! µ B! = µ B 0 (assumes field is in direction) Substituting the operator corresponding to µ, yields the quantum mechanical Hamiltonian operator. H ˆ = γb oˆ I = ω 0ˆ I Thus, the spectrum of H ˆ is discrete with eigenvalues 1 ω and 2 0 eigenkets α and β. E ΔE =!ω 0 Zeeman splitting B 0 7

8 Solving Schrödinger s Equation H ˆ = γb 0ˆ, hence Solution: I ( ) = e it ˆ H ˆ t ( ) is time independent. Schrödinger s Equation: ( ) H ψ t ψ 0 $ ψ( t) = E ˆ + ( ith ˆ ( ) + it H ˆ ) 2 ( + it H ˆ ) 3 ' & ) +! & 2! 3! ψ( 0) ) % ( Above equation implies expanding ψ in terms of eigenkets of would be helpful. Most general solution then given by: t ψ(t) = i ˆ H ψ(t) ˆ H ψ( t) = c α e i( φ α +γb o t / 2) α + c β e i ( φ β γb ot / 2) β c α, c β, φ α, and φ β real constants where c α 2 + c β 2 =1. 8

9 Longitudinal Magnetiation Wavefunction is ψ( t) = c α e i ( φ α +γb o t / 2 ) α + c β e i ( φ β γb o t / 2 ) β. Longitudinal magnetiation ˆ µ =!γ ψ I ˆ ψ =!γ ( 2 c 2 2 α c ) β ( ) where probability finding the system in state α % ( % =!γ 2 P α P β $ & P α P β How do we find P α and P β? ' ) $ & β ' ( ). 9

10 Boltman Distribution Probability P n of finding a system in a specific state n is dependent on the energy E n as given by the Boltmann distribution Boltmann constant P n = 1 Z e E n / kt Z = e E i where / kt. absolute temperature e E n / kt 1 E n / kt partition function (normaliation factor) i sum over all possible energies NMR Energies ( E n = "ω 0 /2) much smaller than kt. Thus high temperature approximation Hence ˆ µ =!γ 2 P α P β ( ) =!2 γ 2 B & ) 0 =!γ 2 $!ω 0 ' % 2kT ( 4kT factor of two from Z term (compare Lecture 2, slide 14) 10

11 Transverse Magnetiation Some useful equations: I ˆ x α = 1 β 2 Letting Δφ=φ β -φ α, yields (after some algebra) ˆ µ x =!γ ψ I ˆ x ψ =!γ 2 c αc β e i ( ω 0t +Δφ ) + c β c α e +i ω 0t +Δφ Similarly... ˆ µ y =!γ ψ I ˆ y ψ ˆ I x β = 1 2 α Aren t φ α and φ β arbitrary? ˆ I y α = i 2 β ˆ I y β = i 2 α ( ( )) ( ) =!γc α c β cos ω 0 t + Δφ =!γc α c β sin( ω 0 t + Δφ) Larmor precession! y Δφ! µ xy ω 0 t x free precession 11

12 Ensemble of Identical Spins Consider an ensemble of N independent spins with φ α and φ β (and by extension Δφ) randomly distributed. average over ensemble ˆ µ x = ˆ µ y = 0 Physical pictures for a collection of spins in states α and β : + Cones B 0 ˆ µ = N!2 γ 2 B 0 4kT Polariation B 0 M = N V! 2 γ 2 B 0 4kT = ρ!2 γ 2 B 0 4kT spins/volume 12

13 RF Excitation In order to get transverse magnetiation, we need to establish some phase relationship (coherence) among spins. In the presence of a rotating magnetic field, the Hamiltonian is: H ˆ (t) = ω 0ˆ I ω ˆ 1 I x cosωt I ˆ y sinωt where ψ " = e iωt I ˆ ψ and H ˆ " = e iωt ˆ ( ) ω 1 = γb 1. I H ˆ e iωt I ˆ = e iωt ˆ I Using Schrödinger s equation and the chain rule for differentiation: ψ # = ih ˆ t eff ψ # Time independent RF excitation H ˆ ( t) is periodic change to rotating frame of reference. where ˆ H eff ˆ H = (ω 0 ω)ˆ I ω 1ˆ I x Effective field in the rotating frame just like the classical case (Change of basis) 13

14 RF Excitation Assuming RF pulse is on resonance (i.e. ω=ω 0 ), of a constant pulse of length τ is: ψ "( τ) = e iτ ˆ H eff " Case 1: ω 1 τ =180 o Cones ψ 0 ( ) = c α e iφ α cos 1 2 ω 1τ ( ) = i c α e iφ α β + c β e iφ β α ψ " τ 180 ψ "( τ) at the end [ ( ) α + isin( 1 ω 2 1τ ) β ] ( ) β + isin( 1 ω 2 1τ ) α [ ] + c β e iφ β cos 1 2 ω 1τ ( ) α ( ) ˆ µ =!γ 2 c 2 2 β c α α B 0 Polariation β 180 o β M inverted 14

15 General case: ψ "( τ) = e iτ ˆ H eff " ψ 0 RF Excitation ( ) = c α e iφ α cos 1 2 ω 1τ Case 2: γ 1 B 1 τ = 90 o (about x axis) ( ) = 2 ψ " τ 90 ) c 2 α e iφ α + ic β e iφ β + * [ ( ) α + isin( 1 ω 2 1τ ) β ] ( ) β + isin( 1 ω 2 1τ ) α [ ] + c β e iφ β cos 1 2 ω 1τ ( ) α + ( c β e iφ β + ic α e iφ ) α β,. - Single Spin ˆ µ x =!γc α c β cosδφ ( ) ˆ µ y = 1!γ c α c β ˆ µ =!γc α c β sinδφ Ensemble average ˆ µ x = 0 ˆ µ y = 1!γ c α c β ˆ µ = 0 ( ) as expected 15

16 90 x o Cone Diagram B 0 90 x! RF Excitation In summary RF pulse causes: 1) equaliation of probabilities of α, β states y { } M = 0. { } 2) a phase coherence between α, β states generating M y. 1 2!ω 0 E 1 2!ω 0 Energy Diagram Transitions occur via emission of absorption of a photon of energy!ω 0. M Bloch M y Purcell There is something subtle, yet fundamentally wrong, about the above diagrams. What is it? 16

17 Linear Superposition of States Consider the following two examples: System 1: N α and N β spins with ψ α = α and ψ β = β respectively such that N = N α + N β, N α N = c 2 α, and N β N = c 2 β. Implies that a given spin has probabilities c 2 2 α and c β being in state α and β respectively. Polariation α Cones B 0 B 0 of β However, System 1 virtually never occurs in practice! It is wrong to claim that all spins are either spin up of spin down. 17

18 Linear Superposition of States System 2: N spins each with wavefunction ψ = c α e iφ α α + c β e iφ β β. Does NOT imply that a given spin has probabilities c 2 2 α and c β of being in state α and β respectively. Cones : picture doesn t work Polariation : works better, but still not very realistic Here, spins are almost fully polaried in 18

19 Linear Superposition of States System 2 spins are described by a linear superposition of states as opposed to the statistical mixture of states in System 1. Example: If we insist that each spin is always either spin up or spin down (System 1), then for all spins: {c α, c β }={1, 0} or {c α, c β }={0, 1}. Hence this system could never generate any transverse magnetiation. ( ) ˆ µ x =!γc α c β cos ω 0 t + Δφ product always = 0 ( ) ˆ µ y =!γc α c β sin ω 0 t + Δφ ˆ µ x = ˆ µ x = 0 independent of any phase coherences For System 2, all spins have perfect phase coherence. 19

20 Actual MR Experiments In a real NMR experiment, we actually deal with a statistical mixture of spins each of which is described by a linear superposition of states (topic for next lecture). System 3: N spins with wavefunctions ψ i = c αi e iφ α i α + c βi e iφ β i β where i=1,..n, c c 2 αi + c 2 αi, c βi,φ αi, and φ βi real constants for which βi =1. Polariation diagram Energy diagram E + ˆ I = population deficit = population excess ˆ I At typical magnetic fields and temperatures, spins are polaried almost isotropically in space, with the term almost referring to a slight preference for the + component (~10ppm for 1 H, B 0 = 3 Tesla, T = 37 C) We ll make use of this alternative energy diagram when studying relaxation. 20

21 Summary Quantum mechanical derivations show that ˆ µ x, ˆ µ y, and ˆ faithfully reproduce the classically-derived behavior of M x, M y, and M (e.g. Larmor precession, RF excitation, etc). Rigorous but with limited intuition. Subsequent lectures will show that Liouville Space description of NMR and, in particular, the Product Operator Formalism is. Mathematically easier. Retains intuition associated with classical vector formulation. Readily extended to the case of interacting spins (coupling). µ 21

22 Next Lecture: NMR in Liouville Space 22