User:Suman Chatterjee DHEP/sandbox

As a magnetic dipole, take a spin ${\tfrac {1}{2}}$ system, say proton; according to formulation of Quantum mechanics state of the system, denoted by :$|\Psi (t)\rangle$ , evolve by the action of an unitary operator $e^{-i{{\hat {H}}\hbar }/t}$  ; they obey Schrödinger equation -

                        $i\hbar {\frac {\partial }{\partial t}}\Psi ={\hat {H}}\Psi$

States with definite energy evolve in time with phase $e^{-iEt/\hbar }$ ,( $|\Psi (t)\rangle =|\Psi (0)\rangle e^{-iEt/\hbar }$ ) where E is the energy of the state, since the probability of finding the system in that state in this state $|\langle x|\Psi (t)\rangle |^{2}$= $|\langle x|\Psi (0)\rangle |^{2}$ -is independent of time, such states are termed as stationary states , so if a system is prepared in a staionary state, (i.e. one of the eigenstates of the Hamiltonian operator) ,then P(t)=1,i.e. it remains in that state indefinitely. But this is the case for isolated systems only. When a system, in one of it's stationary states, is perturbed, it is forced to change its state from the previous one, so it is no longer an eigenstate of the complete Hamiltonian of the system. This same phenomena happens in Magnetic resonance for a spin ${\tfrac {1}{2}}$ system in magnetic field.

Hamiltonian for a magnetic dipole $\mathbf {m}$ (associated with a spin ${\tfrac {1}{2}}$ particle) in a magnetic field $\mathbf {B_{0}} {\hat {z}}$ is-

                        ${\hat {H}}=-\mathbf {m} .\mathbf {B_{0}}$  =$-{\tfrac {\hbar }{2}}\gamma .\sigma _{z}.\mathbf {B_{0}}$ =$-{\tfrac {\hbar }{2}}.\omega _{0}.{\begin{bmatrix}1&0\\0&-1\end{bmatrix}}$

Here $\omega _{0}$(=$\gamma B_{0}$) is the larmor precession frequency of the dipole for $\mathbf {B_{0}}$ magnetic field and $\sigma _{z}$ is Pauli z matrix. So, eigenvalues of ${\hat {H}}$ are $-{\tfrac {\hbar }{2}}\omega _{0}$ and ${\tfrac {\hbar }{2}}\omega _{0}$. If now, the system is perturbed by a weak magnetic field $\mathbf {B_{1}}$, rotating counterclockwise in x-y plane (normal to $\mathbf {B_{0}}$) with angular frequency $\omega .$, so that $\mathbf {B_{1}} ={\hat {i}}B_{1}cos{\omega t}-{\hat {j}}B_{1}sin{\omega t}$ , then ${\begin{bmatrix}1\\0\end{bmatrix}}$ and ${\begin{bmatrix}0\\1\end{bmatrix}}$ are not eigenstates of the hamiltonian, which is modified into

      ${\hat {H}}={\begin{pmatrix}\mathbf {B_{0}} &\mathbf {B_{1}} \ e^{\omega it}\\\mathbf {B_{1}} e^{-\omega it}&\mathbf {-B_{0}} \end{pmatrix}}$

But it is inconvenient to deal with time-dependent hamiltonian, to make ${\hat {H}}$ time-independent ,we choose new reference frame rotating with $\mathbf {B_{1}}$ ,i.e. use rotational operator ${\hat {R}}(t)$ on $|\Psi (t)\rangle$, which amounts to basis change in hilbert space. Using this on Schrödinger equation , we have Hamiltonian in mew basis as-

               ${\hat {H^{\prime }}}=R(t){\hat {H}}R(t)^{\dagger }+{\tfrac {\hbar }{2}}\omega \sigma _{z}$

Writting ${\hat {R(t)}}$ in the basis of $\sigma _{z}$ as-

               ${\hat {R}}(t)={\begin{pmatrix}e^{-i{\omega }t/2}&0\\0&e^{i{\omega }t/2}\end{pmatrix}}$

Now, using this form of the hamiltonian in new basis is found

           ${\hat {H^{\prime }}}={\tfrac {\hbar }{2}}{\begin{pmatrix}\Delta \omega &-\omega _{1}\\-\omega _{1}&-\Delta \omega \end{pmatrix}}$     where $\Delta \omega =\omega -\omega _{0}$ and $\omega _{1}=\gamma B_{1}$

Now this hamiltonian looks exactly similar to that of a two state system with unperturbed energies ${\tfrac {\hbar }{2}}\Delta \omega$ & $-{\tfrac {\hbar }{2}}\Delta \omega$ with a perturbation expressed by ${\tfrac {\hbar }{2}}{\begin{pmatrix}0&-\omega _{1}\\-\omega _{1}&0\end{pmatrix}}$; Then, according to Rabi oscillation, if we start with ${\begin{bmatrix}1\\0\end{bmatrix}}$ state, i.e. dipole in parallel to $\mathbf {B_{0}}$ with energy $-{\tfrac {\hbar }{2}}\omega _{0}$, the probability that it will transit to ${\begin{pmatrix}0\\1\end{pmatrix}}.$ state (i.e. it will flip) is

     $P_{12}={\frac {|\omega _{1}^{2}|}{|\Delta \omega ^{2}+\omega _{1}^{2}|}}sin^{2}[{\sqrt {\omega ^{2}+\Delta \omega ^{2}}}t/2]$

Now if $\omega$=$\omega _{0}$, i.e.$\mathbf {B_{1}}$ rotates with larmor frequency ofn the dipole in $\mathbf {B_{0}}$ magnetic field, then at some point of time, say t= ${\frac {|(2n+1)\Pi )|}{|{\sqrt {\omega ^{2}+\Delta \omega ^{2}}}|}}$, it will completely flip into another (previous) energy eigenstate ${\begin{bmatrix}0\\1\end{bmatrix}}$ ;i.e. diplole can be completely flipped. When $\omega \not =\omega _{0}$, then there is very small probability of change of energy state, so in the previous case we had Resonance.

Plot of Probability Amplitude of Spin Flip at Resonance

Plot of Probability Amplitude of Spin Flip in absense of Resonance

This resonance condition is used for measurement of magnetic moment of a dipole, magnetic field at a point in space, etc.

A Special Case to Show Some of The Applications

A special case where two levels ,between which oscillation takes place, are unstable is considered, suppose they have same life time $\tau$ . Now if atoms are excited at a constant ,say n/time, to 1st state, then some of them decay and rest have a probability $P_{1}2$to make transition to 2nd state, so in time interval between t and (t+dt) no. of atoms which jump to the 2nd state from 1st one is $n(1-e^{-t/\tau })P_{1}2dt$, so at time t no. of atoms in the 2nd state is-

                        $dN=n.e^{-t/\tau }.(1-e^{-t/\tau })P_{12}dt$
                                    =$n.e^{-t/\tau }.P_{12}dt$ 

Now,rate of decay from state 2 depends on no. of atoms, got collected in that state from all the previous intervals, so no. of atoms at state 2 is $\int _{-\infty }^{0}ne^{-t/\tau }P_{12}\ dt$; Now rate of decay of atoms from state 2 is proportional to no. of atoms present in that state, const of proportionality is decay const $\lambda$ , Now performing the integration rate of decay of atoms from state 2 is obtained as

                                 $(n/2)\omega ^{2}/(\delta \omega ^{2}+\omega _{1}^{2}+1/\tau ^{2})$

From this expression many interesting points can be exploited, such as-

1. If uniform magnetic field $B_{0}$ is varied (so that $\omega _{0}$ in $\delta \omega$) is varied a Lorentz curve is obtained, detecting the peak of that curve, abscissa of it gives $\omega _{0}$, so now $\omega$(angular frequency of rotation of $\mathbf {B} _{1}$ = $\gamma$ $(B_{0})_{max}$ ,

Plot of Rate of Decay with Variation of Uniform Magnetic Field $B_{0}$

so from known value of $\omega$ and $(B_{0})_{max}$, gyromagnetic ratio $\gamma$ of the dipole can be measured; by this method we can measure nuclear spin where all electronic spins are balanced. Correct measurement of nuclear magnetic moment helps to understand the character of nuclear force.

1. If $\gamma$ is known, varying $\omega$, value of $B_{0}$ can be obtained. This measurement technique is extremely precise and is used in sensitive magnetometers. Using this technique ,value of magnetic field acting at a particular lattice site by its environment inside a crystal can be obtained.

Plot of Half-Width of Lorentz Curve with Changing $B_{1}$

1. By measuring half-width of the curve ,d=${\sqrt {\omega _{1}^{2}+1/\tau ^{2}}}$, for several values of $\omega _{1}$ (i.e. of $B_{1}$), we can plot d vs $\omega _{1}$, and by extrapolating this line for $\omega _{1}$, life-time of unstable states can be obtained from the intercept.

Rabi's Method to Measure Magnetic Resonance

File:Rabi method 2.png

Apparatus Suggested and Used by Rabi

Existence of spin angular momentum of electron were discovered experimentally in Stern-Gerlach experiment, in which a beam of neutral atoms with one electron in valence shell,carrying no orbital momentum (from the view-point of quantum mechanics) was passed through a inhomogeneous magnetic field and observing the splitting of the beam, but this process was not very accurate due to small deflection of atoms, resulting a considerable uncertainty in measured value. Rabi's method was an improvement over Stern-Gerlach apparatus. As shown in the figure, source emits a beam of neutral atoms, having spin angular momentum $\hbar /2$. The beam passes through a series of three magnets aligned in a line. Magnet1 produces a inhomogeneous magnetic field with a high gradient${\frac {\partial B}{\partial z}}$ (as in Stern-Gerlach aparatus), so the atoms having 'upward' spin (with $S_{z}=\hbar /2$) will deviate downward (path 1),i.e. to the region of less magnetic field B, to minimize energy , and atoms with 'downward' spin with $S_{z}=-\hbar /2$) will deviate upward similarly(path 2).Beams are passed through slit 1, to reduce effect of source beyond. Now, magnet 2 produces only uniform magnetic field in vertical direction giving no force on atomic beam, and magnet 3 is actually inverted magnet 1. In region between poles of magnet 3 ,atoms having 'upward' spin gets upward push and atoms having 'downward' spin feels downward push, so their path remains 1 & 2 respectively. These beams pass through a second slit S2, and arrive at detector and get detected.

Now, if a horizontal rotating field $B_{1}$,angular frequency of rotation $\omega _{1}$ is applied in the region between poles of magnet 2, produced by oscillating current in circular coils then there is a probability for the atoms passing through there from one spin state to another ($S_{z}=+\hbar /2->-\hbar /2$ and vice-versa), when $\omega _{1}$=$\omega _{p}$, Larmor frequency of precession of magnetic moment in B. The atoms, transited from 'upward' to 'downward' spin will experience a downward force while passing through magnet 3, and will follow path 1'. Similarly,atoms, transited from 'downward' to 'upward' spin will follow path 2', and these atoms will not reach detector, causing a minimum in detector count. So, if angular frequency $\omega _{1}$ of $B_{1}$ is varied continuously, then a minimum in detector current will be obtained (when,$\omega _{1}$=$\omega _{p}$). From this known value of $\omega _{1}$($=geB/{2\hbar }$, where g is 'Lande g factor'),'Lande g factor' is obtained which will enable one to have correct value of magnetic moment $\mu (=gq\hbar /{4m})$. This experiment, suggested and performed by Rabi is more sensitive and accurate compared to that of Stern-Gerlach.

Correspondence Between Classical and Quantum Mechanical Explanation

Though the notion of spin angular momentum arises only in quantum mechanics and has no classical analogue , magnetic resonance phenomena can be explained in the light of Classical physics satisfactorily to some extent. When viewed from the reference frame attached to the rotating field ,it seems that magnetic dipole precess around a net magnetic field $(\Delta \omega {\hat {z}}-\omega _{1}{\hat {X}})/\gamma$, where ${\hat {z}}$ is the unit vector along uniform magnetic field $B_{0}$ and ${\hat {X}}$ is same in the direction of rotating field $B_{1}$ and $\delta \omega =\omega -\omega _{0}$.

Proof of Classical Expression

Classical picture of precession As Classical Electrodynamics tells that torque on a magnetic dipole of moment $\mathbf {m}$ is $\mathbf {m}$ × $\mathbf {B}$ ,so its equation of motion is - ${\frac {d\mathbf {L} }{dt}}=$$\mathbf {m}$ × $\mathbf {B}$ ,(where $\mathbf {L}$ is the angular momentum associated with dipole) so - ${\frac {d\mathbf {m} }{dt}}=\gamma$$\mathbf {m}$ × $\mathbf {B}$ For considered case dipole is under the action of magnetic field $\mathbf {B}$ and $\mathbf {B_{1}}$, hence ${\frac {d\mathbf {m} }{dt}}=\gamma$$\mathbf {m}$ × $\mathbf {B} +\mathbf {B_{1}}$ It is easier to solve it by transforming co-ordinate system to OXYZ in which $\mathbf {B_{1}}$ becomes OX axis, in that frame - ${\frac {d\mathbf {m^{'}} }{dt}}={\frac {d\mathbf {m} }{dt}}+$$\mathbf {m}$ × $\mathbf {\omega }$ here $\mathbf {\omega } =\omega {\hat {z}}$ Using $\omega _{0}=\gamma B$ and $\omega _{1}=\gamma B_{1}$, one can see that- ${\frac {d\mathbf {m^{'}} }{dt}}=$$\mathbf {m}$×${(\mathbf {\omega _{0}} +\mathbf {\omega _{1}} -\mathbf {\omega } )}$ $=$$\mathbf {m}$× ($\mathbf {\Delta \omega }$ $+$ $\mathbf {\omega _{1}}$) so, here effective field becomes :$B_{effective}=\Delta \mathbf {\omega } -\mathbf {\omega _{1}} ={\Delta \omega }\mathbf {\hat {z}} -\omega _{1}\mathbf {\hat {X}}$

So when $\omega =\omega _{0}$, there is a high amplitude of precession and so magnetic moment can be completely flipped. So there is a nice correspondence between classical and quantum mechanical predictions, which can be viewed as an example of Bohr Correspondence principle , which states that quantum mechanical phenomena when predicted in classical regime should match with classical result. Now, origin of this correspondence is that evolution of expectation value of magnetic moment is identical to that obtained by classical reasoning. As we know, expectation value of magnetic moment $<\mathbf {m} >=\gamma <\mathbf {S} >$ ; Now time evolution of $<\mathbf {m} >$ is given by

       $i\hbar {\frac {d}{dt}}<\mathbf {m} >=<[\mathbf {m} ,{\hat {H}}]>$;
               ${\hat {H}}=-\mathbf {m} .\mathbf {B(t)}$

so, $[m_{i},{\hat {H}}]=[m_{i},-m_{j}B_{j}]=[\gamma \mathbf {S} _{i},-\gamma \mathbf {S} _{j}\mathbf {B} _{j}]=-\gamma ^{2}[\mathbf {S} _{i},\mathbf {S} _{j}\mathbf {B} _{j}]=-\gamma ^{2}i\hbar [{\mathbf {S} _{k}\mathbf {B} _{j}-\mathbf {S} _{k}\mathbf {B} _{j}}],(i\neq j,k)$

So, $[m_{i},{\hat {H}}]=i\hbar \gamma [\mathbf {B} _{j}\mathbf {m} _{k}-\mathbf {B} _{j}\mathbf {m} _{k}],{\frac {d}{dt}}<\mathbf {m(t)} >=\gamma$ $<\mathbf {m} (t)>$ × $<\mathbf {B} (t)>$

which looks exactly similar to equation of motion of magnetic moment $\mathbf {m}$ in classical mechanics -

                  ${\frac {d}{dt}}\mathbf {m(t)} =\gamma$ $\mathbf {m(t)}$ × $\mathbf {B} (t)$ 

This analogy in mathematical equation for the evolution of magnetic moment and it's expectation value facilitates to understand the phenomena without a background of quantum mechanics.

MRI -A Use of Magnetic Resonance

A beautiful and handy application of magnetic resonance is MRI, here spin angular momentum of proton is used. Most available source for proton is water, with hydrogen nucleus comprised only of proton. So, if a strong magnetic field $B$ is applied on water, there will be a splitting of energy associated with spin angular momentum ,-($+\gamma \hbar B/2$) and -($-\gamma \hbar B/2$) , using $E=-\mathbf {\mu } .\mathbf {B}$ .

Now, according to Boltzmann distribution (no. of systems having energy $E$ out of $N_{0}$ at temperature $T$ is $N_{0}e^{-E/kT}$ where k is Boltzmann constant) lower energy level, associated with spin $\hbar /2$ is more populated than the other. So, in presence of rotating magnetic field more no. of protons flip from state with $S_{z}=+\hbar /2$ to $S_{z}=-\hbar /2$ than the other way, causing sharp absorption of microwave or radio-wave radiation (from rotating field) . When rotating magnetic field is withdrawn, protons tend to reequilibrate with Boltzmann distribution , so there are a transition of some protons from higher energy level to lower ones, emitting microwave or radio-wave radiation of sharp frequency.

Instead of nuclear spin, spin angular momentum of unpaired electron is used in EPR (Electron paramagnetic resonance) to detect free radicals, etc.

Is Magnetic Resonance a Quantum Phenomena ?

Phenomenon of magnetic resonance is rooted on the existence of spin angular momentum of a quantum system and it's specific orientation w.r.t. applied magnetic field, both the cases have no ground in Classical approach and can be understood by only using Quantum mechanics. Some people claim that purely quantum phenomena are those which can not be explained by Classical approach upto any extent , but phenomena,in microscopic domain, which are to some extent be described by Classical analogy are not really quantum phenomena. But for magnetic resonance, as basic elements have no classical origin, hence though analogy can be made with Classical Larmor precession , it should be treated as a quantum phenomena .