Using Schwarz's Inequality to show an expectation value relationship of a particle

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?

• "\langle" and "\rangle" are what you are looking for Dec 4 '20 at 12:59
• This question is better suited on the Mathematics. Dec 4 '20 at 13:04

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.

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.

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 $$v=\sqrt P$$

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

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).

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.

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