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| <math>\Delta x=\sqrt{\left \langle {x^{2}} \right \rangle -\left \langle {x} \right \rangle ^{2}}.</math> | | <math>\Delta x=\sqrt{\left \langle {x^{2}} \right \rangle -\left \langle {x} \right \rangle ^{2}}.</math> |
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| Lets start by calculating the expectation values one by one.
| | We will calculate the expectation values one by one. |
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| <math>\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 </math> | | <math>\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 </math> |
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| since the integrand is odd and thus the integral over all space is zero. | | since the integrand is odd and thus the integral over all space is zero. |
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| <math>\left \langle {x^{2}} \right \rangle =\left \langle {\Psi \left |{x^{2}} \right |\Psi } \right \rangle </math> | | <math>\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}</math> |
| <math>=\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}</math>
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| Since the integral is of a Gaussian times a power of <math>x</math>, we are able to use the known results for such integrals. | | Since the integral is of a Gaussian times a power of <math>x</math>, we are able to use the known results for such integrals. |
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| Similarly to <math>\left \langle {x} \right \rangle, </math> <math>\left \langle {p} \right \rangle =\left \langle {\Psi \left |{p} \right |\Psi } \right \rangle =0</math> because the integrand will be an odd function as well. | | Similarly to <math>\left \langle {x} \right \rangle, </math> <math>\left \langle {p} \right \rangle=0</math> because the integrand will be an odd function as well. |
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| <math>\left \langle {p^{2}} \right \rangle =\left \langle {\Psi \left |{p^{2}} \right |\Psi } \right \rangle </math> | | <math>\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</math> |
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| <math>=\sqrt {\frac{a}{\pi }} \int {e^{-ax^{2}/2}} \left (\frac{\hbar }{i}\frac{\partial}{\partial x}\right )^{2}e^{-ax^{2}/2}dx</math>
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| <math>=\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</math> | | <math>=\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</math> |
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| Combining these results, we obtain <math>\Delta p=\hbar\sqrt{\frac{a}{2}}</math> | | Combining these results, we obtain <math>\Delta p=\hbar\sqrt{\frac{a}{2}}</math> |
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| <math>\Delta x=\frac{1}{\sqrt{2a}}</math> | | <math>\Delta x=\frac{1}{\sqrt{2a}}.</math> |
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| finally,
| | Finally, |
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| <math>\Delta p\,\Delta x =\hbar\sqrt{\frac{a}{2}}\frac{1}{\sqrt{2a}} =\sqrt {\frac{\hbar ^{2}}{4}} =\frac{\hbar }{2}</math>. | | <math>\Delta p\,\Delta x =\hbar\sqrt{\frac{a}{2}}\frac{1}{\sqrt{2a}}=\frac{\hbar}{2}.</math> |
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| Back to [[Heisenberg Uncertainty Principle]] | | Back to [[Heisenberg Uncertainty Principle#Problems|Heisenberg Uncertainty Principle]] |
Latest revision as of 13:24, 18 January 2014
Let us assume that a particle has the wavefunction,
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,
and
We will calculate the expectation values one by one.
since the integrand is odd and thus the integral over all space is zero.
Since the integral is of a Gaussian times a power of 
, we are able to use the known results for such integrals.
Similarly to 
because the integrand will be an odd function as well.
Combining these results, we obtain 
and
Finally,
Back to Heisenberg Uncertainty Principle