Phy5645/Heisenberg Uncertainty Relation 3

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

In order to verify the uncertanity relation, we need to find the uncertainties in position and momentum, $\Delta p={\sqrt {\left\langle {p^{2}}\right\rangle -\left\langle {p}\right\rangle ^{2}}}$ and $\Delta x={\sqrt {\left\langle {x^{2}}\right\rangle -\left\langle {x}\right\rangle ^{2}}}.$

Lets start by calculating the expectation values one by one.

$\langle {x}\rangle =\left\langle {\Psi \left|{x}\right|\Psi }\right\rangle =\int \limits _{-\infty }^{\infty }{x\left|{\Psi (x)}\right|^{2}\,dx}={\sqrt {\frac {a}{\pi }}}\int _{-\infty }^{\infty }xe^{-ax^{2}}\,dx=0$

since the integrand is odd and thus the integral over all space is zero.

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

Since the integral is of a Gaussian times a power of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} , we are able to use the known results for such integrals.

Similarly to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \langle {x} \right \rangle, } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \langle {p} \right \rangle =\left \langle {\Psi \left |{p} \right |\Psi } \right \rangle =0} because the integrand will be an odd function as well.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left \langle {p^{2}} \right \rangle =\left \langle {\Psi \left |{p^{2}} \right |\Psi } \right \rangle }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt {\frac{a}{\pi }} \int {e^{-ax^{2}/2}} \left (\frac{\hbar }{i}\frac{\partial}{\partial x}\right )^{2}e^{-ax^{2}/2}dx}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt {\frac{a}{\pi }} (-\hbar ^{2})\int {e^{-ax^{2}/2}} \frac{\partial}{\partial x}\left (-\frac{2ax}{2}e^{-ax^{2}/2}\right )\,dx}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\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}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\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}

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

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

Combining these results, we obtain $\Delta p=\hbar {\sqrt {\frac {a}{2}}}$ and $\Delta x={\frac {1}{\sqrt {2a}}}$

finally,

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

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