# PlanetPhysics/Something Related to Heisenberg Uncertainty Principle

We will find the Fourier transform
   ${\begin{matrix}F(\omega ):={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }f(t)e^{-i\omega t}\,dt\end{matrix}}$
of the Gaussian bell-shaped function
   ${\begin{matrix}f(t)\;=\;Ce^{-at^{2}}\end{matrix}}$
where $C$ and $a$ are positive constants.\\
We get first $F(\omega )={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }Ce^{-at^{2}}e^{-i\omega t}\,dt\;=\;{\frac {C}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{-at^{2}-i\omega t}\,dt.$ Completing the square in $-at^{2}-i\omega t\,=\,-a\left(t^{2}+{\frac {i\omega t}{a}}\right)\,=\,-a\left(t+{\frac {i\omega }{2a}}\right)^{2}-{\frac {\omega ^{2}}{4a}}$ and substituting\, ${\sqrt {a}}\left(t+{\frac {i\omega }{2a}}\right)\,:=\,z$,\, we may write
   ${\begin{matrix}F(\omega )\,=\,{\frac {C}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }e^{-a\left(t+{\frac {i\omega }{2a}}\right)^{2}}e^{-{\frac {\omega ^{2}}{4a}}}\,dt\,=\,{\frac {C}{\sqrt {2\pi a}}}e^{-{\frac {\omega ^{2}}{4a}}}\int _{l}e^{-z^{2}}\,dz,\end{matrix}}$
where $l$ is a line of the complex plane parallel to the real axis and passing through the point \,$z={\frac {i\omega }{2{\sqrt {a}}}}$.\, Now we can show that the integral $I_{y}\,:=\,\int _{l}e^{-z^{2}}\,dz=\int _{-\infty }^{\infty }e^{-(x+iy)^{2}}\,dx$ does not depend on $y$ at all.\, In fact, we have Failed to parse (unknown function "\sijoitus"): {\displaystyle \frac{\partial I_y}{\partial y} \,=\, \int_{-\infty}^\infty\frac{\partial}{\partial y} e^{-(x+iy)^2}dx \,=\, -2i\int_{-\infty}^\infty e^{-(x+iy)^2}(x+iy)\,dx \,=\, i\!\sijoitus{x\,=-\infty}{\quad \infty}\!e^{-(x+iy)^2} \,=\, i\!\sijoitus{x\,=-\infty}{\quad \infty}\!e^{-x^2+y^2}e^{-2ixy} \,=\,0.}
Hence we may evaluate $I_{y}$ as $I_{y}\,=\,I_{0}=\int _{-\infty }^{\infty }e^{-x^{2}}\,dx\;=\;{\sqrt {\pi }}$ (see the "area under Gaussian curve").\, Putting this value to (3) yields the result
   ${\begin{matrix}F(\omega )\;=\;{\frac {C}{\sqrt {2a}}}e^{-{\frac {\omega ^{2}}{4a}}}.\end{matrix}}$
Thus, we have gotten another Gaussian bell-shaped function (4) corresponding to the given Gaussian bell-shaped function (2).\\
Interpretation. \, One can take for the breadth of the bell the portion of the abscissa axis, outside which the ordinate drops under the maximum value divided by $e$, for example.\, Then, for the bell (2) one writes $Ce^{-at^{2}}=Ce^{-1},$ whence\, $t={\frac {1}{\sqrt {a}}}$\, giving, by evenness of the function, the breadth\, $\Delta t={\frac {2}{\sqrt {a}}}$.\, Similarly, the breadth of the bell (4) is\, $\Delta \omega =4{\sqrt {a}}$.\, We see that the product
   ${\begin{matrix}\Delta t\cdot \Delta \omega =8\end{matrix}}$
has a constant value.\, One can show that any other shape of the graphs of $f$ and $F$ produces a relation similar to (5).\, The breadths are thus inversely proportional.\\
If $t$ is the time and $f$ is the action of a force on a system of oscillators with their natural frequencies, then in the formula $f(t)\;=\;{\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }F(\omega )e^{i\omega t}\,d\omega$ of the inverse Fourier transform, $F(\omega )$ represents the amplitude of the oscillator with angular frequency $\omega$.\, One can infer from (5) that the more localised is (smaller $\Delta t$ ) the external force in time, the more spread out (greater $\Delta \omega$ ) is its spectrum of frequencies, i.e. the greater is the amount of the oscillators the force has excited with roughly the same amplitude.\, If, conversely, one wants to achieve better selectivity, i.e. to compress the spectrum to a narrower range of frequencies, then one has to spread out the external action in time.\, The impossibility to simultaneously localise the action in time and also enhance the selectivity of the action is one of the manifestations of the quantum-mechanical uncertainty principle (or quantum `Principle of Indetermination'), which plays a fundamental role in modern physics.