Phy5645/Heisenberg Uncertainty Relation 3

Let's say that a particle has wavefunction : $\Psi (x)=({\frac {\pi }{a}})^{-1/4}e^{-ax^{2}/2}$ and we are trying to verify Heisenberg Uncertanity relation.

In order to verify the uncertanity relation, we need to find these elements,

$\vartriangle p=\left\langle {p^{2}}\right\rangle -\left\langle {p}\right\rangle ^{2}$ and $\vartriangle x=\left\langle {x^{2}}\right\rangle -\left\langle {x}\right\rangle ^{2}$.

Lets start by calculating one by one.

$\left\langle {x}\right\rangle =\left\langle {\Psi \left|{x}\right|\Psi }\right\rangle =\int \limits _{-\alpha }^{\infty }{x\left|{\Psi (x)}\right|^{2}dx}={\sqrt {\frac {a}{\pi }}}\int {xe^{-ax^{2}}dx=0}$ since it is an odd function and its integral over all the space is zero.

$\left\langle {x^{2}}\right\rangle =\left\langle {\Psi \left|{x^{2}}\right|\Psi }\right\rangle$ $=\int \limits _{-\infty }^{\infty }{x^{2}\left|{\Psi (x)}\right|^{2}dx}={\sqrt {\frac {a}{\pi }}}\int {x^{2}e^{-ax^{2}}dx=}{\sqrt {\frac {a}{\pi }}}{\frac {1}{2}}{\sqrt {\frac {\pi }{a^{3}}}}={\frac {1}{2a}}$

Since the integral is Gaussian integral, we used Gaussian integral results.

Just as $\left\langle {x}\right\rangle$, also $\left\langle {p}\right\rangle =\left\langle {\Psi \left|{p}\right|\Psi }\right\rangle =0$ because it is an odd function as well.

If we look at $\left\langle {p^{2}}\right\rangle$,

$\left\langle {p^{2}}\right\rangle =\left\langle {\Psi \left|{p^{2}}\right|\Psi }\right\rangle$

$={\sqrt {\frac {a}{\pi }}}\int {e^{-ax^{2}/2}}({\frac {\hbar }{i}}{\frac {\Game }{\Game x}})^{2}e^{-ax^{2}/2}dx$

$={\sqrt {\frac {a}{\pi }}}(-\hbar ^{2})\int {e^{-ax^{2}/2}}{\frac {\Game }{\Game x}}({\frac {-2ax}{2}}e^{-ax^{2}/2})dx$

${\text{=}}{\sqrt {\frac {a}{\pi }}}(-\hbar ^{2})\int {e^{-ax^{2}/2}}\left\lbrace {-ae^{-ax^{2}/2}+\left({\frac {-2ax}{2}}\right)\left({\frac {-2ax}{2}}\right)e^{-ax^{2}/2}}\right\rbrace dx$

${\text{=}}{\sqrt {\frac {a}{\pi }}}(\hbar ^{2}a)\int {e^{-ax^{2}}}dx{\text{ +}}{\sqrt {\frac {a}{\pi }}}(-\hbar ^{2}a^{2})\int {x^{2}e^{-ax^{2}}}dx$

${\text{=}}{\sqrt {\frac {a}{\pi }}}(\hbar ^{2}a){\sqrt {\frac {\pi }{a}}}+{\sqrt {\frac {a}{\pi }}}(-\hbar ^{2}a^{2}){\frac {1}{2}}{\sqrt {\frac {\pi }{a^{3}}}}$

${\text{=}}(\hbar ^{2}a)-{\frac {\hbar ^{2}a}{2}}={\frac {\hbar ^{2}a}{2}}$

So, results are; $\vartriangle p=\left\langle {p^{2}}\right\rangle -\left\langle {p}\right\rangle ^{2}={\frac {\hbar ^{2}a}{2}}$ and $\vartriangle x=\left\langle {x^{2}}\right\rangle -\left\langle {x}\right\rangle ^{2}={\frac {1}{2a}}$

finally,

${\sqrt {\vartriangle p\vartriangle x}}={\sqrt {{\frac {\hbar ^{2}a}{2}}{\frac {1}{2a}}}}={\sqrt {\frac {\hbar ^{2}}{4}}}={\frac {\hbar }{2}}$.

This is basic problem about an Uncertainty realtion. It basically provides that the more distribution we get around x, the smaller distribution we get around momentum and vice versa.