Editing Uncertainty principle (section)

===Proof of the Kennard inequality using wave mechanics===

We are interested in the [[variance]]s of position and momentum, defined as
<math display="block">\sigma_x^2 = \int_{-\infty}^\infty x^2 \cdot |\psi(x)|^2 \, dx - \left( \int_{-\infty}^\infty x \cdot |\psi(x)|^2 \, dx \right)^2</math>
<math display="block">\sigma_p^2 = \int_{-\infty}^\infty p^2 \cdot |\varphi(p)|^2 \, dp - \left( \int_{-\infty}^\infty p \cdot |\varphi(p)|^2 \, dp \right)^2~.</math>

[[Without loss of generality]], we will assume that the [[expected value|means]] vanish, which just amounts to a shift of the origin of our coordinates. (A more general proof that does not make this assumption is given below.) This gives us the simpler form
<math display="block">\sigma_x^2 = \int_{-\infty}^\infty x^2 \cdot |\psi(x)|^2 \, dx</math>
<math display="block">\sigma_p^2 = \int_{-\infty}^\infty p^2 \cdot |\varphi(p)|^2 \, dp~.</math>

The function <math>f(x) = x \cdot \psi(x)</math> can be interpreted as a [[vector space|vector]] in a [[function space]]. We can define an [[inner product]] for a pair of functions ''u''(''x'') and ''v''(''x'') in this vector space:
<math display="block">\langle u \mid v \rangle = \int_{-\infty}^\infty u^*(x) \cdot v(x) \, dx,</math>
where the asterisk denotes the [[complex conjugate]].

With this inner product defined, we note that the variance for position can be written as
<math display="block">\sigma_x^2 = \int_{-\infty}^\infty |f(x)|^2 \, dx = \langle f \mid f \rangle ~.</math>

We can repeat this for momentum by interpreting the function <math>\tilde{g}(p)=p \cdot \varphi(p)</math> as a vector, but we can also take advantage of the fact that <math>\psi(x)</math> and <math>\varphi(p)</math> are Fourier transforms of each other. We evaluate the inverse Fourier transform through [[integration by parts]]:
<math display="block">\begin{align}
g(x) &= \frac{1}{\sqrt{2 \pi \hbar}} \cdot \int_{-\infty}^\infty \tilde{g}(p) \cdot e^{ipx/\hbar} \, dp \\
&= \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^\infty p \cdot \varphi(p) \cdot e^{ipx/\hbar} \, dp \\
&= \frac{1}{2 \pi \hbar} \int_{-\infty}^\infty \left[ p \cdot \int_{-\infty}^\infty \psi(\chi) e^{-ip\chi/\hbar} \, d\chi \right] \cdot e^{ipx/\hbar} \, dp \\
&= \frac{i}{2 \pi} \int_{-\infty}^\infty \left[ \cancel{ \left. \psi(\chi) e^{-ip\chi/\hbar} \right|_{-\infty}^\infty } - \int_{-\infty}^\infty \frac{d\psi(\chi)}{d\chi} e^{-ip\chi/\hbar} \, d\chi \right] \cdot e^{ipx/\hbar} \, dp \\
&= -i \int_{-\infty}^\infty \frac{d\psi(\chi)}{d\chi} \left[ \frac{1}{2 \pi}\int_{-\infty}^\infty \, e^{ip(x - \chi)/\hbar} \, dp \right]\, d\chi\\
&= -i \int_{-\infty}^\infty \frac{d\psi(\chi)}{d\chi} \left[ \delta\left(\frac{x - \chi }{\hbar}\right) \right]\, d\chi\\
&= -i \hbar \int_{-\infty}^\infty \frac{d\psi(\chi)}{d\chi} \left[ \delta\left(x - \chi \right) \right]\, d\chi\\
&= -i \hbar \frac{d\psi(x)}{dx} \\
&= \left( -i \hbar \frac{d}{dx} \right) \cdot \psi(x) ,
\end{align}</math>
where <math>v=\frac{\hbar}{-ip}e^{-ip\chi/\hbar}</math> in the integration by parts, the cancelled term vanishes because the wave function vanishes at both infinities and <math>|e^{-ip\chi/\hbar}|=1</math>, and then use the [[Dirac delta function#History|Dirac delta function]] which is valid because <math>\dfrac{d\psi(\chi)}{d\chi}</math> does not depend on ''p'' .

The term <math display="inline">-i \hbar \frac{d}{dx}</math> is called the [[momentum operator]] in position space. Applying [[Plancherel theorem|Plancherel's theorem]], we see that the variance for momentum can be written as
<math display="block">\sigma_p^2 = \int_{-\infty}^\infty |\tilde{g}(p)|^2 \, dp = \int_{-\infty}^\infty |g(x)|^2 \, dx = \langle g \mid g \rangle.</math>

The [[Cauchy–Schwarz inequality]] asserts that
<math display="block">\sigma_x^2 \sigma_p^2 = \langle f \mid f \rangle \cdot \langle g \mid g \rangle \ge |\langle f \mid g \rangle|^2 ~.</math>

The [[modulus squared]] of any complex number ''z'' can be expressed as
<math display="block">|z|^{2} = \Big(\text{Re}(z)\Big)^{2}+\Big(\text{Im}(z)\Big)^{2} \geq \Big(\text{Im}(z)\Big)^{2} = \left(\frac{z-z^{\ast}}{2i}\right)^{2}. </math>
we let <math>z=\langle f|g\rangle</math> and <math>z^{*}=\langle g\mid f\rangle</math> and substitute these into the equation above to get
<math display="block">|\langle f\mid g\rangle|^2 \geq \left(\frac{\langle f\mid g\rangle-\langle g \mid f \rangle}{2i}\right)^2 ~.</math>

All that remains is to evaluate these inner products.

<math display="block">\begin{align}
\langle f\mid g\rangle-\langle g\mid f\rangle  &= \int_{-\infty}^\infty \psi^*(x) \, x \cdot \left(-i \hbar \frac{d}{dx}\right) \, \psi(x) \, dx - \int_{-\infty}^\infty \psi^*(x) \, \left(-i \hbar \frac{d}{dx}\right) \cdot x \, \psi(x) \, dx \\
&= i \hbar \cdot \int_{-\infty}^\infty \psi^*(x) \left[ \left(-x \cdot \frac{d\psi(x)}{dx}\right) + \frac{d(x \psi(x))}{dx} \right] \, dx \\
&= i \hbar \cdot \int_{-\infty}^\infty \psi^*(x) \left[ \left(-x \cdot \frac{d\psi(x)}{dx}\right) + \psi(x) + \left(x \cdot \frac{d\psi(x)}{dx}\right)\right] \, dx \\
&= i \hbar \cdot \int_{-\infty}^\infty \psi^*(x) \psi(x) \, dx \\
&= i \hbar \cdot \int_{-\infty}^\infty |\psi(x)|^2 \, dx \\
&= i \hbar
\end{align}</math>

Plugging this into the above inequalities, we get
<math display="block">\sigma_x^2 \sigma_p^2 \ge |\langle f \mid g \rangle|^2 \ge \left(\frac{\langle f\mid g\rangle-\langle g\mid f\rangle}{2i}\right)^2 = \left(\frac{i \hbar}{2 i}\right)^2 = \frac{\hbar^2}{4}</math>
and taking the square root
<math display="block">\sigma_x \sigma_p \ge \frac{\hbar}{2}~.</math>

with equality if and only if ''p'' and ''x'' are linearly dependent. Note that the only ''physics'' involved in this proof was that <math>\psi(x)</math> and <math>\varphi(p)</math> are wave functions for position and momentum, which are Fourier transforms of each other. A similar result would hold for ''any'' pair of conjugate variables.