Download Introductory Quantum Optics by Christopher Gerry, Peter Knight PDF

By Christopher Gerry, Peter Knight

This ordinary creation to the topic of quantum optics, the learn of the quantum mechanical nature of sunshine and its interplay with topic, is sort of solely focused on the quantized electromagnetic box. The textual content is designed for upper-level undergraduates taking classes in quantum optics who've already taken a direction in quantum mechanics, and for first- and moment- yr graduate scholars.

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Wave functions of the form 1 ψm (φ) = √ eimφ , 2π m = 0, ±1, ±2, . . 201) where Lˆ z ψm (φ) = hmψm (φ) . ) But φ itself is not a periodic function, having the range −∞ < φ < ∞. In fact the fluctuations φ can be greater than 2π and for a number state, Eq. 199) would seem to imply that φ → ∞, a nonsensical behavior. A possible solution to this problem is to introduce a periodic coordinate (φ) [10] behaving in a discontinuous fashion according to Fig. 4. 35 36 Field quantization Φ Fig. 4. The periodic function , arbitrarily chosen to be discontinuous.

214) 37 38 Field quantization the first of which is similar to Eq. 48). 216) and where the means the root-mean-square deviation. 217) n≥1 . 219) which would seem to correspond to a phase angle equally likely to have any value in the range 0 to 2π. ) In any case, the right-hand sides of Eqs. 215) are zero for a number state as required in order that the uncertainty relations be satisfied. 220) are given by |φ = ∞ einφ |n . 221) n=0 These states are not normalizable nor are they orthogonal as the scalar product of |φ and |φ is not the delta function δ(φ − φ ).

The Dirac approach fails because of the underlying assumption that a Hermitian phase operator φˆ actually exists. There are in fact two reasons for the failure of ˆ should be a the Dirac approach. If φˆ exists as a Hermitian operator, then exp(i φ) unitary operator. From Eqs. 191) i φˆ † aˆ † = (e ) . 193) 1 ˆ ˆ (ei φ )(ei φ )† = aˆ aˆ † = 1. 194) ˆ ˆ but ˆ is not a unitary operator from which it follows that φˆ is not So, in fact, exp(i φ) Hermitian. The root of the problem is that the operator nˆ has a spectrum bounded from below; it does not include the negative integers.

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