Griffith's Example 4.3: Larmor Precession We Will Be Considering the Spin of an Electron

Griffith's Example 4.3: Larmor Precession We will be considering the spin of an electron. The operator for the component of spin in the rˆ = (sin cosijˆˆ sin sin cosˆ ) k cos sin ei cos(½ ) leading to Sˆ  which has an eigen-spinor  rˆ  rˆ i   i sin e cos sin(½ )e i rˆ sin(½ )e with eigenvalue ½  and    with eigenvalue -½ . We can imagine a cos(½ ) state in which the spin is initially in the x-z plane and makes an angle  relative to the cos(½ ) z axis so (0)  . sin(½ )

First we consider an applied magnetic field in the z direction providing interaction ˆ  ˆ energy: H  Bkozo  SB   Bmos. Using the standard eigenstates of Sz, ½0 B ˆ  o  H     0½  Bo 

Note that the difference between the energies of the spin up and spin down states is Bo. We expect things to oscillate at frequencies corresponding to the energy differences so the frequency for this problem is Bo, the Larmor frequency. The frequencies associated with the two states have a magnitude of one-half of the Larmor frequency; the difference between the frequencies is the relevant frequency for oscillation of probability density or, in this case, precessing.

  The hamiltonian is diagonal so the spin z up and down states are the eigenfunctions. a ½it ½0 Bo a  cos(½ )e  Hiˆ    it () t t   b  ½it  0½  Bo b t sin(½ )e  where  = Bo . The expectation values can be computed with the results:    Sz(t) = cos() ( /2); Sx(t) = sin() ( /2) sin(t);Sy(t) = -sin() ( /2) cos(t)

Exercise: Compute Sz(t) and Sx(t) for(t) given above.  Extending Example 4.3: (NOT YET CHECKED; look at overall sign. RECALL Griffiths  is NEGATIVE. I also use  for the Larmor frequency. FOR THIS XTENDED EXAMPLE ONLY:  is ½ . ) A small magnetic field oscillating at  ˆ ˆ frequency  is added in the y direction.  B Bkoo Bcos(  t ) j. The full hamiltonian is now  ˆ  ˆˆ H  BSBSB zo  y ocos( t ).

½½cB iBos() t ˆ oo H   ½ciB ooos()½ t B

This problem is getting difficult so let us think about it a little. Suppose  = 0. That would just be a static magnetic field directed at and angle of about  relative to the z direction. The spin would precess about this new field direction. As  is small, the z component would be about cos() times its value before the small perturbation was added so not much would change in the optical pumping1 experiment. The high m states would remain highly populated so little light would be absorbed. The time dependence must be important.

dSz The plan of attach is to compute dSz  dt dt to find conditions under which the expectation value of Sz would change indicating a relative change in the population of the spin up and spin down states. it a ½½cBiBtooos()cos(½ )e   Hiˆ    i t   t ½cos()½iB  t B it b oosin(½ )e  t   Using: = ½ B . d  t dt   o  iecos(½ )it  cos( te )sin(½  ) it dd t  itit iesin(½ )  cos( te )cos(½  ) More pain! ††ˆˆ†ˆ SSdSdSSdzz so z ( ) z z iecos(½ )it  cos( teie ) sin(½  )it ,  sin(½ )it  cos( t ) cos(½  ) eit † ˆ  ()dSz 

10cos(½ ) e  it  ½ dt =  it 01 sin(½ ) e iecos(½ )it  cos( teie ) sin(½  )it ,  sin(½ )it  cos( te ) cos(½  ) it  cos(½  ) eit ½ dt  it  sin(½ ) e

1 This example is related to the Teachspin Optical Pumping experiment. Just ignore the references if you are not familiar with the experiment. 22it22it ½ [ite cos (½ ) cos( ) sin(½ ) cos(½ ) i sin (½ ) cos( te ) sin(½ ) cos(½ ) ]dt

22it it i½dt ½ cos(tee ) sin(½ ) cos(½ ) dt

1 it  i t 22 it  it i½dt4 sin(½ ) cos(½ ) ee  e  e dt The first factor is imaginary, and it would cancel the corresponding piece from † ˆ  Sdz  if I were dedicated enough to compute it. Rather than compute it, we † ˆ † ˆ † ˆ recognize that  Sdz  is the Hermitian conjugate of ()dS z  . As  Sdz  is just an inner product and hence a complex scalar value, its conjugate is its complex conjugate. † it  i t 22 it  it ˆ ½1 sin(½ ) cos(½ ) ee e e Sdz i dt  4    dt It follows that: ˆ dSz  cos(tt )sin(½ )cos(½  ) cos[( 2  ) ] cos[( 2 )t ]dt

For electrons: sz = ½  {Prob() – Prob()}

Changing sz redistributes probability among the m levels. J More generally: Jmz    JJProbabilty(m )  mJ The expression has a factor of the small parameter . It will not accomplish much unless it can be integrated for a long time coherently. In the case that  = 2  =  Bo, the term grows linearly in time meaning that Sz is changing so the populations of the states of the various magnetic quantum numbers are changing. We conclude that applying a second field transverse to the main magnetic field that is oscillating at  =  Bo, the Larmor precession frequency, causes a steady tipping of the magnetic moments thus redistributing them among the various m levels. (This redistribution meets the requirement to see addition absorption in the optical pumping experiment.) In the optical pumping experiment, the expectation value of the magnetic moment lies parallel to the field prior to the magnetic field sweeping to a value such that the condition  Bo =  is met. The net spin then is rotated away from the field direction and for a time T2 called the de-coherence time, the transverse part of the dipole has a finite expectation value that rotates at the Larmor frequency  Bo. This transverse rotating moment can be detected by coils mounted in the same manner as the RF tipping coils by monitoring the Faraday potential induced in the coils. The de- coherence time T2 is the time that the contributions to the transverse moment decay away or de-phase due to inhomogeneities in the field and collisions. After the transverse components have decayed, the spin expectation value exhibits only the decreased Sz. If the optical pumping light is turned off, the spin expectation value Sz decays to zero with a characteristic time T1, the longitudinal relaxation time. The transverse relaxation (or de-phasing) time T2 cannot exceed T1 as magnitude decay and de-phasing contribute to the decay of the transverse components.

Exercise: The expression for dSz vanishes for  = 0 (which is correct). As this is the case for the optical pumping experiment, we should consider the classical analog for a moment. Suppose a magnetic moment lies along the z direction, and there is a magnetic field in the y direction. What is the direction of the torque? Which component of the moment has an initial non-zero change? Prepare a sketch of a magnetic moment precessing around a magnetic field in the z direction and making an angle  with respect to it. Assume that a small field in the y direction is applied. Identify the directions of the additional torque when the moment lies in the x-z half- plane and when it lies in the –(x-z) half-plane. Does it appear that a static field in the y direction would cause a steady tipping of the moment? What about a rotating field that instantaneously points in the direction of dS ?

Exercise: Argue that a field Bo cos(t) ˆj can be represented as the sum of two counter rotating fields in the x-y plane.

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