By Hakim V., Ambegaokar V.

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29b) Indeed, this is the weighted average of energy. For H given in a different representation, with a different choice of basis states, one finds the same expectation value, as will be checked at the end of Sect. 1. The only energy measurement that we have met so far is that of the Zeeman effect; therefore, let us find the expectation value of energy for this case. 3 Expectation Values 23 B-field turned on because the Zeeman splitting leaves the mean energy of the mstates unaffected. Does this mean that the different energies measured in the Zeeman effect are not expectation values?

We first assume that the atoms are in zero magnetic field, and we go directly to Step 3. Step 3, polarization analysis: Light from unpolarized atoms is emitted isotropically, due to rotational symmetry, while light from polarized atoms usually has an anisotropic angular distribution. In our example of a ji = 1 to jf = 0 decay shown in Fig. 2, the reemitted π light in a mi = 0 to mf = 0 transition has the angular distribution of a classical dipole antenna 1 I (θ) = I0 sin2 θ, π light. 7) 2 This distribution is rotationally symmetric about the axis z of polarization of the incoming light, as shown in Fig.

46b) This means that for B along z, the energy eigenfunctions simultaneously are spin eigenfunctions. 47a) 0 −Bz a . 7). If initially the system is in a state ψ(0) = a(0)α + b(0)β, after time t it is in the state ψ(t) = a(t)α + b(t)β, with a(t) = a(0)e−iωL t/2 , b(t) = b(0)eiωL t/2 . 48) The probability of being spin up or spin down does not change in the process, 2 2 2 pup (t) = a(t) = a(0) , 2 pdown (t) = b(t) = b(0) . 48), merely acquires an additional dynamical phase δ± (t) = ∓ 12 ωL t , or, for a time dependent field Bz (t), δ± (t) = ∓ 1 2 t ωL t dt .