# Question #4655a

Sep 1, 2017

Before we proceed into a formal derivation, we must try to understand what $\Delta x$ and $\Delta p$ actually mean.

They are infact the standard deviations in measurements of $x$ and $p$ respectively and may be defined as in probability theory as,

$\Delta x = \sqrt{< {\left(x - < x >\right)}^{2} >}$

$\Delta p = \sqrt{< {\left(p - < p >\right)}^{2} >}$

Now, for two complex valued functions $f$ and $g$, there holds the following inequality,

$\int | f \left(x\right) {|}^{2} \mathrm{dx} \int | g \left(x\right) {|}^{2} \mathrm{dx} \ge$ $\frac{1}{4} {\left[\int \left(\dot{f} g + \dot{g} f\right) \mathrm{dx}\right]}^{2}$

where $\dot{f}$ and $\dot{g}$ are complex conjugates.

For simplicity, assume that $< x > = 0$ and $< p > = 0$

then for, $f = \hat{p} \psi = - i \overline{h} \frac{\partial \psi}{\partial x}$ and $g = i \overline{h} \psi$

we have, $\int \dot{f} f \mathrm{dx} = {\overline{h}}^{2} \int \frac{\partial \dot{\psi}}{\partial x} \frac{\partial \psi}{\partial x} \mathrm{dx}$
$\implies \int \dot{f} f \mathrm{dx} = {\overline{h}}^{2} \left[- \int \dot{\psi} \frac{{\partial}^{2} \psi}{\partial \psi} ^ 2 \mathrm{dx}\right]$

In the last step, integration by parts has been done and the first term is put equal to zero since wave functions go to zero when $x$ does to infinity.

$\int \dot{f} f \mathrm{dx} = \int \dot{\psi} {\left(- i \overline{h} \frac{\partial}{\partial x}\right)}^{2} \psi \mathrm{dx}$
$\implies \int \dot{f} f \mathrm{dx} = < {p}^{2} >$ where $\hat{p} = - i \overline{h} \frac{\partial}{\partial x}$ is the one dimensional momentum operator.

Now, $\int \dot{g} g \mathrm{dx} = \int \dot{\psi} {x}^{2} \psi \mathrm{dx}$
$\implies \int \dot{g} g \mathrm{dx} = < {x}^{2} >$

and $\int \left(\dot{f} g + \dot{g} f\right) \mathrm{dx} = - \overline{h} \int \frac{\partial \dot{\psi}}{\partial x} x \psi \mathrm{dx} - \overline{h} \int \frac{\partial \psi}{\partial x} x \dot{\psi} \mathrm{dx}$
$\implies \int \left(\dot{f} g + \dot{g} f\right) \mathrm{dx} = - \overline{h} \int \frac{\partial}{\partial x} \left(\dot{\psi} x \psi\right) \mathrm{dx} + \overline{h} \int \dot{\psi} \psi \mathrm{dx}$

Now, the first term when integrated shall go to zero because wavefunctions go to zero as x goes to infinity.

$\implies \int \left(\dot{f} g + \dot{g} f\right) \mathrm{dx} = \overline{h}$

Therefore, using the inequality stated previously,

$\int | f \left(x\right) {|}^{2} \mathrm{dx} \int | g \left(x\right) {|}^{2} \mathrm{dx} \ge$ $\frac{1}{4} {\left[\int \left(\dot{f} g + \dot{g} f\right) \mathrm{dx}\right]}^{2}$

$\implies {\left(\Delta x\right)}^{2} \cdot {\left(\Delta p\right)}^{2} \ge \frac{1}{4} \left({\overline{h}}^{2}\right)$

Thus, $\Delta x \cdot \Delta p \ge \frac{\overline{h}}{2}$

Which is the uncertainty principle.

This may also be derived for the more general case that $< x >$ and $< p >$ not zero by taking,

$f = \left(p - < p >\right) \psi$ and $g = \left(x - < x >\right) \psi$.
The evaluation of the result would be somewhat laborious so I skipped doing it myself.

References -
1) Quantum Mechanics : Theory and Applications by A Ghatak and S Lokanathan
2) Introduction to Quantum Mechanics by DJ Griffiths