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Hi I was reading Introductory Quantum Optics by Gerry and Knight, particularly the part about higher degrees of coherence (Section 5.4), and in there they write:

Now according to Cauchy’s inequality applied to a pair of measurements at times $t_1$ and $t_2$, we have: $$2I(t_{1}) I(t_{2}) ≤ I(t_{1})^{2} I(t_{2})^{2} .$$

It is not clear to me how the Cauchy-Schwarz inequality is used here and where exactly this equation comes from, could someone please explain this for me?

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  • $\begingroup$ I believe that Gerry and Knight actually made a mistake on this equation and missed a plus sign on the RHS between the two intensities and however I am still struggling to understand this conceptually $\endgroup$ Commented Mar 13, 2021 at 20:47

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Cauchy-Schwarz inequality state that in any inner product space: ${\displaystyle \left|\left\langle \mathbf {u} ,\mathbf {v} \right\rangle \right|^{2}\leq \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle}$ for any u,v.

In the special case of L2 space (the inner product space of function that we use in quantom mechanics) the inner product between two functions $f(x)$ and $g(x)$ is the integral :${\displaystyle \int _{\mathbb {R} ^{n}}f(x){\overline {g(x)}}\,dx}$ and it follows that Cauchy-Schwarz inequality in this case become: ${\displaystyle \left|\int _{\mathbb {R} ^{n}}f(x){\overline {g(x)}}\,dx\right|^{2}\leq \int _{\mathbb {R} ^{n}}|f(x)|^{2}\,dx\int _{\mathbb {R} ^{n}}|g(x)|^{2}\,dx.}$

In quantum mechanics, the expectation values of any measurable quantity (correspond to $I(t_1),I(t_2)$) is given by the integral $I_i=∫_{R^n}|I`(x,t_i)|dx$ when $I'$ is the operator of the measurable value $I$. Similarly, The measurable value $I(t_1)I(t_2)$ So when applying Cauchy-Schwarz inequality we get: $I(t_1)I(t_2)≤I(t_1)^2I(t_2)^2$.

The 2 factor though look suspicious. Can you provide more details about I operator?

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  • $\begingroup$ Sorry for the late response, but there was no quantum mechanics involved in this section in the Gerry and Knight textbook. Also there should be a plus sign between the squares of your result as there is in Quantum Theory of Light by Loudon $\endgroup$ Commented Mar 20, 2021 at 14:43
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I believe you are correct - Gerry and Knight does have a typo. I believe it should be the sum of the two squared intensities, which comes from the Cauchy-Schwartz inequality for real numbers. We can show this by considering intensity to be a real value, such that

$$ (I(t_1) - I(t_2))^2 \geq 0 $$ $$ I^2(t_1) + I^2(t_2) - 2 I(t_1) I(t_2) \geq 0 $$ $$ I^2(t_1) + I^2(t_2) \geq I(t_1) I(t_2) $$

I would recommend checking out Fox's Quantum Optics, p143 (problem 6.3/6.4), for a step-by-step derivation of these limits, as well as Loudon's Quantum Theory of Light, p108 (eq 3.7.3) for verification.

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  • $\begingroup$ From line 2 to line 3 of your equations how did you get rid of the factor of 2? Also what about dn? $\endgroup$ Commented Mar 4, 2022 at 19:43

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