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Schwarz's Inequality

\vert\left\langle{\psi_1\vert\psi_2}\right\rangle{}\vert^2 \...
\end{displaymath} (1)

Written out explicitly
\left[{\int_a^b \psi_1(x)\psi_2(x)\,dx}\right]^2\leq \int_a^b [\psi_1(x)]^2\,dx \int_a^b [\psi_2(x)]^2\,dx,
\end{displaymath} (2)

with equality Iff $g(x)=\alpha f(x)$ with $\alpha$ a constant. To derive, let $\psi(x)$ be a Complex function and $\lambda$ a Complex constant such that $\psi(x)\equiv f(x)+\lambda g(x)$ for some $f$ and $g$. Since $\int\psi^*\psi\,dx\geq 0$,

\int \psi^*\psi\,dx = \int f^*f\,dx+\lambda\int f^*g\,dx+\lambda^*\int g^*f\,dx+\lambda\lambda^* \int g^*g\,dx \geq 0,
\end{displaymath} (3)

with equality when $\psi(x)=0$. Set
\lambda=-\int g^*f\,dx
\end{displaymath} (4)

so that
\lambda^*=-\int f^*g\,dx.
\end{displaymath} (5)

Plugging (5) and (4) into (3) then gives
$\int f^*f\,dx\int g^*g\,dx-\int g^*f\,dx\int f^*g\,dx $
$-\int f^*g\,dx\int g^*f\,dx + \int g^*f\,dx\int f^*g\,dx\geq 0\quad$ (6)

\int g^* f\,dx\int f^*g\,dx\leq \int f^*f\,dx\int g^*g\,dx
\end{displaymath} (7)

\left\vert{\int g^*f\,dx}\right\vert = \left\vert{\int f^*g\,dx}\right\vert \leq \int f^*f\,dx\int g^*g\,dx
\end{displaymath} (8)

\vert\left\langle{f\vert g}\right\rangle{}\vert^2\leq \left\...
...f\vert f}\right\rangle{}\left\langle{g\vert g}\right\rangle{}.
\end{displaymath} (9)

Bessel's Inequality can be derived from this.


Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 11, 1972.

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 527-529, 1985.

© 1996-9 Eric W. Weisstein