Time period of precession of Sx about B

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SUMMARY

The discussion focuses on the time period of precession of the spin operator Sx about a uniform magnetic field B for an electron in the state 1/√2{1,1}. The Hamiltonian matrix is defined as {μBB,0,0,-μBB}, leading to time evolution expressed as 1/√2{1,0}exp(-iEt/ħ) + 1/√2{0,1}exp(iEt/ħ), where E = μBB. The minimum time for a spin flip of Sx is established as t = πħ/2E, resulting in the conclusion that the time period of precession T is 2t = πħ/μBB. The expectation value of Sx is confirmed to oscillate as ħ/2(cos(2μBBt/ħ)).

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Apashanka
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Consider an electron for which l=0 is kept in a uniform magnetic field B.
For which the hamiltonian matrix is {μBB,0,0,-μBB}
now if the electron is in the state 1/√2{1,1}(e.g in the eigenstates of Sx eigenvalue ħ/2}
If this state is time evolved
1/√2{1,0}exp(-iEt/ħ)+1/√2{0,1}exp(iEt/ħ)
where E=μBB
The minimum time for the state becomes
1/√2{1,-1}(e.g eigenvalue -ħ/2 of Sx) spin flip of Sx is t=πħ/2E or t=πħ/2μBB
Is 2t =T(the time period of precession of Sx about B)??
 
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Examine the time dependence of the expectation value ##\langle \Psi_+ | S_x|\Psi_+ \rangle## where ##|\Psi_+ \rangle## is the time-evolved eigenstate corresponding to eigenvalue ##+\hbar/2.##
 
kuruman said:
Examine the time dependence of the expectation value ##\langle \Psi_+ | S_x|\Psi_+ \rangle## where ##|\Psi_+ \rangle## is the time-evolved eigenstate corresponding to eigenvalue ##+\hbar/2.##
Yes it's ħ/2(cos(2μBBt/ħ)
 
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