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Time-dependent Hamiltonian 31 Now, by applying the time-evolution operator on the state given by Eq. 22) The integral over k in the above equation can be done using the usual trick of completing the square in the sum of the exponents. We obtain 1 Ψ(x, t) = C 2 π(σ 2 + i t 2m ) exp (2σ 2 k0 + ix)2 − σ 2 k02 . 27) x2 xc 8σ 4 k0 σ ˜2 σ2 . 28) Notice that xc (t) is the classical trajectory of a free particle moving with momentum p0 = k 0 . In Fig. 1 the real part of Ψ(x, t) is shown as a function of x for xc (t) = 0, 10, and 20 and for k0 = 20, and in Fig.

As we will see later on in chapter 23, this Hamiltonian corresponds to the Zeeman term for spin-1/2, where the magnetic field is along the x-axis and we used the basis labeled by the spin eigenstates along the z-axis. In such a case, g represents the product of the external magnetic field times the magnetic moment due to spin. Problem: Now, we wish to solve the following problem: Given that we prepare the system to be in state |1 at time t = 0, what is the probability to be found in state |1 if a measurement is carried out at a later time t?

2) In order to prove this we will take advantage of the so-called Cauchy–Schwartz inequality: | χ|ψ |2 ≤ χ|χ φ|φ . 3) Let us apply this inequality using as |χ and |ψ the following, |χ = (Aˆ − ˆ− |φ = (B Aˆ )|ψ ≡ α ˆ |ψ , ˆ . 5) This leads to the following inequality: ˆ |2 , (∆A)2 (∆B)2 ≥ | ψ|ˆ αβ|ψ 1 1 αβˆ − βˆα ˆ )|ψ + ψ|(ˆ αβˆ + βˆα ˆ )|ψ |2 = | ψ|(ˆ 2 2 1 1 ˆ ˆ = | ψ|[ˆ α, β]|ψ |2 + | ψ|{ˆ α, β}|ψ |2 , 4 4 where Practical Quantum Mechanics. First Edition. Efstratios Manousakis. c Efstratios Manousakis 2016.