Phy5645/Heisenberg Uncertainty Relation 3

Let us assume that a particle has the wavefunction,

$\psi (x)=\left({\frac {\pi }{a}}\right)^{-1/4}e^{-ax^{2}/2}.$

We now wish to verify the Heisenberg Uncertanity Principle for this case. To do so, 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 $\Delta x={\sqrt {\left\langle {x^{2}}\right\rangle -\left\langle {x}\right\rangle ^{2}}}.$

We will calculate the expectation values one by one.

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

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

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

Since the integral is of a Gaussian times a power of $x$, we are able to use the known results for such integrals.

Similarly to $\left\langle {x}\right\rangle ,$ $\left\langle {p}\right\rangle =0$ because the integrand will be an odd function as well.

$\left\langle {p^{2}}\right\rangle ={\sqrt {\frac {a}{\pi }}}\int {e^{-ax^{2}/2}}\left({\frac {\hbar }{i}}{\frac {\partial }{\partial x}}\right)^{2}e^{-ax^{2}/2}dx$

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

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

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

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

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

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

finally,

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

Back to Heisenberg Uncertainty Principle