Bloch equations

For the wavefunction of a particle in a periodic potential, seeBloch's theorem.

In physics and chemistry, specifically innuclear magnetic resonance(NMR),magnetic resonance imaging(MRI), andelectron spin resonance(ESR), theBloch equationsare a set of macroscopic equations that are used to calculate the nuclear magnetizationM= (M_x,M_y,M_z) as a function of time whenrelaxation timesT₁andT₂are present. These arephenomenologicalequations that were introduced byFelix Blochin 1946.Sometimes they are called theequations of motionof nuclear magnetization. They are analogous to theMaxwell–Bloch equations.

In the laboratory (stationary) frame of reference

Under the effect of the external fieldB,the magnetization vectorMrelaxes to its equilibrium configuration while precessing around the magnetic field.

LetM(t) = (M_x(t),M_y(t),M_z(t)) be the nuclear magnetization. Then the Bloch equations read:

${\frac {dM_{x}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{x}-{\frac {M_{x}(t)}{T_{2}}}$

${\frac {dM_{y}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{y}-{\frac {M_{y}(t)}{T_{2}}}$

${\frac {dM_{z}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{z}-{\frac {M_{z}(t)-M_{0}}{T_{1}}}$

where γ is thegyromagnetic ratioandB(t) = (B_x(t),B_y(t),B₀+ ΔB_z(t)) is themagnetic fieldexperienced by the nuclei. Thezcomponent of the magnetic fieldBis sometimes composed of two terms:

• one,B₀,is constant in time,
• the other one, ΔB_z(t), may be time dependent. It is present inmagnetic resonance imagingand helps with the spatial decoding of the NMR signal.

M(t) ×B(t) is thecross productof these two vectors. M₀is the steady state nuclear magnetization (that is, for example, when t → ∞); it is in thezdirection.

Physical background

With no relaxation (that is bothT₁andT₂→ ∞) the above equations simplify to:

${\frac {dM_{x}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{x}$

${\frac {dM_{y}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{y}$

${\frac {dM_{z}(t)}{dt}}=\gamma (\mathbf {M} (t)\times \mathbf {B} (t))_{z}$

or, in vector notation:

${\frac {d\mathbf {M} (t)}{dt}}=\gamma \mathbf {M} (t)\times \mathbf {B} (t)$

This is the equation forLarmor precessionof the nuclear magnetizationMin an external magnetic fieldB.

The relaxation terms,

$\left(-{\frac {M_{x}}{T_{2}}},-{\frac {M_{y}}{T_{2}}},-{\frac {M_{z}-M_{0}}{T_{1}}}\right)$

represent an established physical process of transverse and longitudinal relaxation of nuclear magnetizationM.

As macroscopic equations

Main article:Magnetic resonance (quantum mechanics)

These equations are notmicroscopic:they do not describe the equation of motion of individual nuclear magnetic moments. Those are governed and described by laws ofquantum mechanics.

Bloch equations aremacroscopic:they describe the equations of motion of macroscopic nuclear magnetization that can be obtained by summing up all nuclear magnetic moment in the sample.

Alternative forms

Opening the vector product brackets in the Bloch equations leads to:

${\frac {dM_{x}(t)}{dt}}=\gamma \left(M_{y}(t)B_{z}(t)-M_{z}(t)B_{y}(t)\right)-{\frac {M_{x}(t)}{T_{2}}}$

${\frac {dM_{y}(t)}{dt}}=\gamma \left(M_{z}(t)B_{x}(t)-M_{x}(t)B_{z}(t)\right)-{\frac {M_{y}(t)}{T_{2}}}$

${\frac {dM_{z}(t)}{dt}}=\gamma \left(M_{x}(t)B_{y}(t)-M_{y}(t)B_{x}(t)\right)-{\frac {M_{z}(t)-M_{0}}{T_{1}}}$

The above form is further simplified assuming

$M_{xy}=M_{x}+iM_{y}{\text{ and }}B_{xy}=B_{x}+iB_{y}\,$

wherei=√−1.After some algebra one obtains:

${\frac {dM_{xy}(t)}{dt}}=-i\gamma \left(M_{xy}(t)B_{z}(t)-M_{z}(t)B_{xy}(t)\right)-{\frac {M_{xy}(t)}{T_{2}}}$.

${\frac {dM_{z}(t)}{dt}}=i{\frac {\gamma }{2}}\left(M_{xy}(t){\overline {B_{xy}(t)}}-{\overline {M_{xy}}}(t)B_{xy}(t)\right)-{\frac {M_{z}(t)-M_{0}}{T_{1}}}$

where

${\overline {M_{xy}}}=M_{x}-iM_{y}$.

is the complex conjugate ofM_xy.The real and imaginary parts ofM_xycorrespond toM_xandM_yrespectively. M_xyis sometimes calledtransverse nuclear magnetization.

Matrix form

The Bloch equations can be recast in matrix-vector notation:

${\frac {d}{dt}}\left({\begin{array}{c}M_{x}\\M_{y}\\M_{z}\end{array}}\right)=\left({\begin{array}{ccc}-{\frac {1}{T_{2}}}&\gamma B_{z}&-\gamma B_{y}\\-\gamma B_{z}&-{\frac {1}{T_{2}}}&\gamma B_{x}\\\gamma B_{y}&-\gamma B_{x}&-{\frac {1}{T_{1}}}\end{array}}\right)\left({\begin{array}{c}M_{x}\\M_{y}\\M_{z}\end{array}}\right)+\left({\begin{array}{c}0\\0\\{\frac {M_{0}}{T_{1}}}\end{array}}\right)$

In a rotating frame of reference

In a rotating frame of reference, it is easier to understand the behaviour of the nuclear magnetizationM.This is the motivation:

Solution of Bloch equations withT₁,T₂→ ∞

Assume that:

• att= 0 the transverse nuclear magnetizationM_xy(0) experiences a constant magnetic fieldB(t) = (0, 0,B₀);
• B₀is positive;
• there are no longitudinal and transverse relaxations (that isT₁andT₂→ ∞).

Then the Bloch equations are simplified to:

${\frac {dM_{xy}(t)}{dt}}=-i\gamma M_{xy}(t)B_{0}$,

${\frac {dM_{z}(t)}{dt}}=0$.

These are two (not coupled)linear differential equations.Their solution is:

$M_{xy}(t)=M_{xy}(0)e^{-i\gamma B_{0}t}$,

$M_{z}(t)=M_{0}={\text{const}}\,$.

Thus the transverse magnetization,M_xy,rotates around thezaxis withangular frequencyω₀= γB₀in clockwise direction (this is due to the negative sign in the exponent). The longitudinal magnetization,M_zremains constant in time. This is also how the transverse magnetization appears to an observer in thelaboratory frame of reference(that is to astationary observer).

M_xy(t) is translated in the following way into observable quantities ofM_x(t) andM_y(t): Since

$M_{xy}(t)=M_{xy}(0)e^{-i\gamma B_{z0}t}=M_{xy}(0)\left[\cos(\omega _{0}t)-i\sin(\omega _{0}t)\right]$

then

$M_{x}(t)={\text{Re}}\left(M_{xy}(t)\right)=M_{xy}(0)\cos(\omega _{0}t)$,

$M_{y}(t)={\text{Im}}\left(M_{xy}(t)\right)=-M_{xy}(0)\sin(\omega _{0}t)$,

where Re(z) and Im(z) are functions that return the real and imaginary part of complex numberz.In this calculation it was assumed thatM_xy(0) is a real number.

Transformation to rotating frame of reference

This is the conclusion of the previous section: in a constant magnetic fieldB₀alongzaxis the transverse magnetizationM_xyrotates around this axis in clockwise direction with angular frequency ω₀.If the observer were rotating around the same axis in clockwise direction with angular frequency Ω,M_xyit would appear to her or him rotating with angular frequency ω₀- Ω. Specifically, if the observer were rotating around the same axis in clockwise direction with angular frequency ω₀,the transverse magnetizationM_xywould appear to her or him stationary.

This can be expressed mathematically in the following way:

• Let (x,y,z) the Cartesian coordinate system of thelaboratory(orstationary)frame of reference,and
• (x′,y′,z′) = (x′,y′,z) be a Cartesian coordinate system that is rotating around thezaxis of the laboratory frame of reference with angular frequency Ω. This is called therotating frame of reference.Physical variables in this frame of reference will be denoted by a prime.

Obviously:

$M_{z}'(t)=M_{z}(t)\,$.

What isM_xy′(t)? Expressing the argument at the beginning of this section in a mathematical way:

$M_{xy}'(t)=e^{+i\Omega t}M_{xy}(t)\,$.

Equation of motion of transverse magnetization in rotating frame of reference

What is the equation of motion ofM_xy′(t)?

${\frac {dM_{xy}'(t)}{dt}}={\frac {d\left(M_{xy}(t)e^{+i\Omega t}\right)}{dt}}=e^{+i\Omega t}{\frac {dM_{xy}(t)}{dt}}+i\Omega e^{+i\Omega t}M_{xy}(t)=e^{+i\Omega t}{\frac {dM_{xy}(t)}{dt}}+i\Omega M_{xy}'(t)$

Substitute from the Bloch equation in laboratory frame of reference:

${\begin{aligned}{\frac {dM_{xy}'(t)}{dt}}&=e^{+i\Omega t}\left[-i\gamma \left(M_{xy}(t)B_{z}(t)-M_{z}(t)B_{xy}(t)\right)-{\frac {M_{xy}(t)}{T_{2}}}\right]+i\Omega M_{xy}'(t)\\&=\left[-i\gamma \left(M_{xy}(t)e^{+i\Omega t}B_{z}(t)-M_{z}(t)B_{xy}(t)e^{+i\Omega t}\right)-{\frac {M_{xy}(t)e^{+i\Omega t}}{T_{2}}}\right]+i\Omega M_{xy}'(t)\\&=-i\gamma \left(M_{xy}'(t)B_{z}(t)-M_{z}(t)B_{xy}'(t)\right)+i\Omega M_{xy}'(t)-{\frac {M_{xy}'(t)}{T_{2}}}\\\end{aligned}}$

But by assumption in the previous section:B_z′(t) =B_z(t) =B₀+ ΔB_z(t) andM_z(t) =M_z′(t). Substituting into the equation above:

${\begin{aligned}{\frac {dM_{xy}'(t)}{dt}}&=-i\gamma \left(M_{xy}'(t)(B_{0}+\Delta B_{z}(t))-M_{z}'(t)B_{xy}'(t)\right)+i\Omega M_{xy}'(t)-{\frac {M_{xy}'(t)}{T_{2}}}\\&=-i\gamma B_{0}M_{xy}'(t)-i\gamma \Delta B_{z}(t)M_{xy}'(t)+i\gamma B_{xy}'(t)M_{z}'(t)+i\Omega M_{xy}'(t)-{\frac {M_{xy}'(t)}{T_{2}}}\\&=i(\Omega -\omega _{0})M_{xy}'(t)-i\gamma \Delta B_{z}(t)M_{xy}'(t)+i\gamma B_{xy}'(t)M_{z}'(t)-{\frac {M_{xy}'(t)}{T_{2}}}\\\end{aligned}}$

This is the meaning of terms on the right hand side of this equation:

• i(Ω - ω₀)M_xy′(t) is the Larmor term in the frame of reference rotating with angular frequency Ω. Note that it becomes zero when Ω = ω₀.
• The -iγ ΔB_z(t)M_xy′(t) term describes the effect of magnetic field inhomogeneity (as expressed by ΔB_z(t)) on the transverse nuclear magnetization; it is used to explainT₂^*.It is also the term that is behindMRI:it is generated by the gradient coil system.
• TheiγB_xy′(t)M_z(t) describes the effect of RF field (theB_xy′(t) factor) on nuclear magnetization. For an example see below.
• -M_xy′(t) /T₂describes the loss of coherency of transverse magnetization.

Similarly, the equation of motion ofM_zin the rotating frame of reference is:

${\frac {dM_{z}'(t)}{dt}}=i{\frac {\gamma }{2}}\left(M'_{xy}(t){\overline {B'_{xy}(t)}}-{\overline {M'_{xy}}}(t)B'_{xy}(t)\right)-{\frac {M_{z}-M_{0}}{T_{1}}}$

Time independent form of the equations in the rotating frame of reference

When the external field has the form:

$B_{x}(t)=B_{1}\cos \omega t$

$B_{y}(t)=-B_{1}\sin \omega t$

$B_{z}(t)=B_{0}$,

We define:

$\epsilon:=\gamma B_{1}$and$\Delta:=\gamma B_{0}-\omega$,

and get (in the matrix-vector notation):

${\frac {d}{dt}}\left({\begin{array}{c}M'_{x}\\M'_{y}\\M'_{z}\end{array}}\right)=\left({\begin{array}{ccc}-{\frac {1}{T_{2}}}&\Delta &-\epsilon \\-\Delta &-{\frac {1}{T_{2}}}&\epsilon \\-\epsilon &-\epsilon &-{\frac {1}{T_{1}}}\end{array}}\right)\left({\begin{array}{c}M'_{x}\\M'_{y}\\M'_{z}\end{array}}\right)+\left({\begin{array}{c}0\\0\\{\frac {M_{0}}{T_{1}}}\end{array}}\right)$

Simple solutions

Relaxation of transverse nuclear magnetizationM_xy

Assume that:

• The nuclear magnetization is exposed to constant external magnetic field in thezdirectionB_z′(t) =B_z(t) =B₀.Thus ω₀= γB₀and ΔB_z(t) = 0.
• There is no RF, that isB_xy' = 0.
• The rotating frame of reference rotates with an angular frequency Ω = ω₀.

Then in the rotating frame of reference, the equation of motion for the transverse nuclear magnetization,M_xy'(t) simplifies to:

${\frac {dM_{xy}'(t)}{dt}}=-{\frac {M_{xy}'}{T_{2}}}$

This is a linear ordinary differential equation and its solution is

$M_{xy}'(t)=M_{xy}'(0)e^{-t/T_{2}}$.

whereM_xy'(0) is the transverse nuclear magnetization in the rotating frame at timet= 0. This is the initial condition for the differential equation.

Note that when the rotating frame of reference rotatesexactlyat the Larmor frequency (this is the physical meaning of the above assumption Ω = ω₀), the vector of transverse nuclear magnetization,M_xy(t) appears to be stationary.

Relaxation of longitudinal nuclear magnetizationM_z

Assume that:

• The nuclear magnetization is exposed to constant external magnetic field in thezdirectionB_z′(t) =B_z(t) =B₀.Thus ω₀= γB₀and ΔB_z(t) = 0.
• There is no RF, that isB_xy' = 0.
• The rotating frame of reference rotates with an angular frequency Ω = ω₀.

Then in the rotating frame of reference, the equation of motion for the longitudinal nuclear magnetization,M_z(t) simplifies to:

${\frac {dM_{z}(t)}{dt}}=-{\frac {M_{z}(t)-M_{z,\mathrm {eq} }}{T_{1}}}$

This is a linear ordinary differential equation and its solution is

$M_{z}(t)=M_{z,\mathrm {eq} }-[M_{z,\mathrm {eq} }-M_{z}(0)]e^{-t/T_{1}}$

whereM_z(0) is the longitudinal nuclear magnetization in the rotating frame at timet= 0. This is the initial condition for the differential equation.

90 and 180° RF pulses

Assume that:

• Nuclear magnetization is exposed to constant external magnetic field inzdirectionB_z′(t) =B_z(t) =B₀.Thus ω₀= γB₀and ΔB_z(t) = 0.
• Att= 0 an RF pulse of constant amplitude and frequency ω₀is applied. That isB'_xy(t) =B'_xyis constant. Duration of this pulse is τ.
• The rotating frame of reference rotates with an angular frequency Ω = ω₀.
• T₁andT₂→ ∞. Practically this means that τ ≪T₁andT₂.

Then for 0 ≤t≤ τ:

${\begin{aligned}{\frac {dM_{xy}'(t)}{dt}}=i\gamma B_{xy}'M_{z}(t)\end{aligned}}$

${\frac {dM_{z}(t)}{dt}}=i{\frac {\gamma }{2}}\left(M'_{xy}(t){\overline {B'_{xy}}}-{\overline {M'_{xy}}}(t)B'_{xy}\right)$

See also

• TheBloch–Torrey equationis a generalization of the Bloch equations, which includes added terms due to the transfer of magnetization by diffusion.

Further reading

• Charles Kittel,Introduction to Solid State Physics,John Wiley & Sons, 8th edition (2004),ISBN978-0-471-41526-8.Chapter 13 is on Magnetic Resonance.