Difference between revisions of "- Saturation"

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(Created page with "===Saturation=== Previously, we considered the interaction between the spins and lattice and found the characteristic relaxation time <math>T_1</math>. That is, when the spin...")
 
(Saturation)
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We now investigate the ''steady-state behavior'' of the spin-lattice system with an applied transverse RF field.
 
We now investigate the ''steady-state behavior'' of the spin-lattice system with an applied transverse RF field.
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In the absence of the transverse RF field, the differential equation describing the time variation of <math>n</math>, the excess number of nuclei in the lower state is <math>\tfrac{dn}{dt}=\tfrac{n_{_0}-n}{T_{_1}}</math>. This makes sense because if <math>T_S>T</math> then <math>\tfrac{dn}{dt}</math> is greater than zero (<math>n<n_{_0}</math>) and there's a net flow of nuclei into the lower state with a characteristic time constant <math>T_{_1}</math>.
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When the transverse RF field is present, another term must be added to account for the induced upward transitions corresponding to a net absorption of energy so that we get <math>\tfrac{dn}{dt}=\tfrac{n_{_0}-n}{T_{_1}}-2nP</math>, where <math>P</math> is the probability per unit time that a nuclei makes an upward transition. A steady state is reached when <math>\tfrac{dn}{dt}=0</math> so that a steady-state value of <math>n_{_{ss}}</math> is given by <math>\tfrac{n_{_{ss}}}{n_{_0}}=\tfrac{1}{1+2PT_{_1}}</math>.
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we must now evaluate the transition probability <math>P</math>. If a transverse RF field is applied with the correct sense of rotation at the resonant frequency the probability of a transition in unit time between the states is given (with the help of Fermi's golden rule) by
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<math>P_{_{m\rightarrow m'}}=\tfrac{1}{2}\gamma^2H_{_1}^1|<m|I|m'>|^2g(\nu)</math>, where <math><m|I|m'></math> is the appropriate matrix element of the nuclear spin operator.
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For <math>m'=m-1</math> it can be shown that <math>|<m|I|m'>|^2=\tfrac{1}{2}(I+m)(I-m+1)</math>. For the case of <math>I=\tfrac{1}{2}</math> this reduces to
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<math>P=\tfrac{1}{4}\gamma^2H_{_1}^2g(\nu)</math> so that we have
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<math>\tfrac{n_{_{ss}}}{n_{_0}}=[1+\tfrac{1}{2}\gamma^2H_{_1}^2T_{_1}g(\nu)]^{-1}\equiv Z</math>.
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If an RF field is applied, whose amplitude <math>H_{_1}</math> is large, <math>\tfrac{n_{_{ss}}}{n_{_0}}</math> becomes quite small; the spin temperature <math>T_{_S}</math> becomes very high and the spin system is said to be ''saturated''. Remembering that <math>T_{_2}=\tfrac{1}{2}g(\nu)_{_{max}}</math> we can write
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<math>\tfrac{n_{_{ss}}}{n_{_0}}=[1+\gamma^2H_{_1}^2T_{_1}T_{_2}]^{-1}=Z_{_0}</math>, where <math>Z_{_0}</math> is the value of the saturation factor at the maximum of the lineshape function <math>g(\nu)</math>.
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Recall that, in thermal equilibrium, the lattice temperature <math>T</math> is defined in terms of <math>n_{_0}</math> as <math>T=\tfrac{N\mu H_{_0}}{n_{_0}k}</math>. Similarly, when the system is not in thermal equilibrium, the spin temperature <math>T_{_S}</math> is related to <math>n</math> as <math>T_{_S}=\tfrac{N\mu H_{_0}}{kn}</math>. These two results yield
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<math>T_{_S}=T\tfrac{n_{_0}}{n_{_{ss}}}=\frac{T}{Z}</math>.

Revision as of 12:19, 15 February 2019

Saturation

Previously, we considered the interaction between the spins and lattice and found the characteristic relaxation time . That is, when the spin temperature is greater than the lattice temperature how long does it take for the spin temperature to return to the lattice temperature? Answer: .

We then discussed spin-spin interactions which led to the notion of inhomogeneous broadening by a spatially-varying local magnetic field. This led to the concept of a phase-memory time constant . That is, when two nearly-resonant spins are excited then allowed to freely decay (RF field off) how long before the two spins are out of phase? Answer: .

We now investigate the steady-state behavior of the spin-lattice system with an applied transverse RF field.

In the absence of the transverse RF field, the differential equation describing the time variation of , the excess number of nuclei in the lower state is . This makes sense because if then is greater than zero () and there's a net flow of nuclei into the lower state with a characteristic time constant .

When the transverse RF field is present, another term must be added to account for the induced upward transitions corresponding to a net absorption of energy so that we get , where is the probability per unit time that a nuclei makes an upward transition. A steady state is reached when so that a steady-state value of is given by .

we must now evaluate the transition probability . If a transverse RF field is applied with the correct sense of rotation at the resonant frequency the probability of a transition in unit time between the states is given (with the help of Fermi's golden rule) by

, where is the appropriate matrix element of the nuclear spin operator.

For it can be shown that . For the case of this reduces to

so that we have

.

If an RF field is applied, whose amplitude is large, becomes quite small; the spin temperature becomes very high and the spin system is said to be saturated. Remembering that we can write

, where is the value of the saturation factor at the maximum of the lineshape function .

Recall that, in thermal equilibrium, the lattice temperature is defined in terms of as . Similarly, when the system is not in thermal equilibrium, the spin temperature is related to as . These two results yield

.