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This is a question from Cambridge Tripos for 1st years Natural Science students which I just can not solve. I have spent hours on it, and I am going around in circles.

A particle is located between $x$ and $x + dx$ with probability $P(x)dx$ . If:

$$\langle |x|^n\rangle \equiv \int_{-\infty}^{\infty} |x|^n P(x) dx $$

Show that, using Schwarz's Inequality,

$$\langle |x|\rangle^2 \le \langle |x|^2\rangle $$

where the pointed brackets denote expectation value.

Using Schwarz's Inequality, I get stuck at:

$$\left( \int_{-\infty}^{\infty} |x|P(x) dx \right)^2 \le \left( \int_{-\infty}^{\infty} (|x |)^2 dx \right)\left( \int_{-\infty}^{\infty} (P(x))^2 dx \right)$$

How can I manipulate this further to the end result?

I can (albeit shakily) prove it if I say as the particle does not have a negative variance that:

$ \langle |x|^2\rangle - \langle |x|\rangle^2 \ge 0 $ but Schwarz's Inequality is nowhere to be seen.

What is my next step?

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    $\begingroup$ "\langle" and "\rangle" are what you are looking for $\endgroup$
    – ohneVal
    Dec 4 '20 at 12:59
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    $\begingroup$ This question is better suited on the Mathematics. $\endgroup$ Dec 4 '20 at 13:04
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The Cauchy-Schwartz inequality, $|u\cdot v|^2 \leq (u\cdot u)(v\cdot v)$, is explicitly phrased for inner products. Luckily, your left-hand side, $$ ⟨|x|⟩ = \left<\psi\middle||x|\middle|\psi\right>, $$ admits a bunch of different interpretations for the two vectors, i.e., with $v=|x|\left|\psi\right>$, $v=\sqrt{|x|}\left|\psi\right>$, $v=\left|\psi\right>$, and probably a bunch of others. Try formulating those explicit inner products and exploring the inequalities that result from the Cauchy-Schwartz standard when you apply it to them. Since your target relationship is of Cauchy-Schwartz form, it has to come from an argument of that structure.

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Let me remind ourselves of the fact that $$\langle f, g \rangle = \int f(x)g(x)P(x)dx$$ will be an inner product if $P(x)$ is a probability distribution and that norms are computed as $$\| f \| = \sqrt{\langle f,f\rangle} = \sqrt{\int f^2(x)P(x)dx}$$

Let us prove first that

$$\langle |x|\rangle \leq \sqrt{\langle |x|^2\rangle }\tag{1}\label{step},$$ once we have that you can multiply it with itself to get your inequality. We can prove the expression above, using three ingredients, the Cauchy-Schwartz inequality already mentioned, but also the triangle inequality for integrals and the fact that $P$ is a probability distribution:

$$\begin{align} \langle |x|\rangle &= |\langle \sqrt{x}, \sqrt{x}\,\rangle | \\ &\leq \| \sqrt{x} \,\|^2 \qquad\qquad \text{Cauchy-Schwartz}\\ &= \left[\left( \int |x|P(x)dx \right)^2 \right]^{1/2} \qquad\qquad \text{Norm def and swapped exponents}\\ &\leq \left[ \int |x|^2P^2(x)dx \right]^{1/2} \qquad\qquad \text{Triangle inequality for integrals}\\ &\leq \left[ \int |x|^2P(x)dx \right]^{1/2} \qquad\qquad \text{since } P^2(x) \leq P(x) \\ &=\sqrt{\langle |x|^2\rangle} & \end{align}$$

This proves Eq.~\eqref{step} and now by multiplying it with itself you get your result.

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It's very simple. Schwarz says (as you presumably know) $(\int uv \, dx )^2\leq \int u^2 \, dx \int v^2 \, dx$

Then the trick is to use $u=|x|\sqrt P$ and $v=\sqrt P$

$(\int |x| P \, dx )^2\leq \int |x|^2 P \, dx \int P \, dx$

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For the Cauchy-Schwartz inequality you need to have some inner product that you can use. The inner product you used $$\left\langle f,g\right\rangle=\int f(x)g(x)dx$$ is valid, but the definition $$\left\langle f,g\right\rangle=\int f(x)g(x)P(x)dx$$ also gives an inner product, as long as $P(x)$ is strictly positive. You then get $$\left(\int f(x)g(x)P(x)dx\right)^2\le\left(\int f^2(x)dx\right) \left(\int g^2(x)dx\right)$$ which you can use (with appropriate choice of $f,g$ to solve your question).

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The Schwarz's inequality for intergation is: $$(\int f(x)g(x) \text{d}x)^2\le\int f^2(x)\text{d}x\int g^2(x)\text{d}x........(a)$$ We can write $\langle |x|^2 \rangle$ in a new from: $$ \langle |x|^2 \rangle = \int_{-\infty}^{+\infty}|x^2|P(x)\text{d}x = \int_{-\infty}^{+\infty}|x^2|P(x)\text{d}x\int_{-\infty}^{+\infty}1\cdot P(x)\text{d}x $$ This is just the right hand side of (a) with $f(x) = |x|\sqrt{P(x)}$ and $g(x) = \sqrt{P(x)}$. So the right hand side of (a) will be: $$ (\int|x|P(x))^2 = (\langle|x|\rangle)^2 $$ This is just the result we want.

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  • $\begingroup$ Sorry, it should be the left hand side of (a) that is $(\int|x|P(x)\text{d}x)^2$, not right hand side. $\endgroup$
    – Xzy
    Dec 5 '20 at 0:19

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