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| Let's say that a particle has wavefunction : <math>\Psi (x)=(\frac{\pi }{a})^{-1/4}e^{-ax^{2}/2}</math> | | Let us assume that a particle has the wavefunction, |
| and we are trying to verify Heisenberg Uncertanity relation.
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| In order to verify the uncertanity relation, we need to find these elements,
| | <math>\psi (x)=\left (\frac{\pi }{a}\right )^{-1/4}e^{-ax^{2}/2}.</math> |
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| <math>\vartriangle p=\left \langle {p^{2}} \right \rangle -\left \langle {p} \right \rangle ^{2}</math> | | 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, |
| | <math>\Delta p=\sqrt{\left \langle {p^{2}} \right \rangle -\left \langle {p} \right \rangle ^{2}}</math> |
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| <math>\vartriangle x=\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 one by one.
| | We will calculate the expectation values one by one. |
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| <math>\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} </math> since it is an odd function and its integral over all the space is zero. | | <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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| <math>\left \langle {x^{2}} \right \rangle =\left \langle {\Psi \left |{x^{2}} \right |\Psi } \right \rangle </math>
| | since the integrand is odd and thus the integral over all space is zero. |
| <math>=\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}</math>
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| Since the integral is Gaussian integral, we used Gaussian integral results.
| | <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> |
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| Just as <math>\left \langle {x} \right \rangle </math>, also <math>\left \langle {p} \right \rangle =\left \langle {\Psi \left |{p} \right |\Psi } \right \rangle =0</math> because it is an odd function as well.
| | 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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| If we look at <math>\left \langle {p^{2}} \right \rangle </math>,
| | 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}} (\frac{\hbar }{i}\frac{\Game }{\Game x})^{2}e^{-ax^{2}/2}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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| <math>=\sqrt {\frac{a}{\pi }} (-\hbar ^{2})\int {e^{-ax^{2}/2}} \frac{\Game }{\Game x}(\frac{-2ax}{2}e^{-ax^{2}/2})dx</math> | | <math>=\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</math> |
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| <math>\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</math> | | <math>=\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</math> |
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| <math>\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</math> | | <math>=\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}}} </math> |
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| <math>\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}}} </math>
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| <math>\text{=}(\hbar ^{2}a)-\frac{\hbar ^{2}a}{2}=\frac{\hbar ^{2}a}{2}</math> | | <math>\text{=}(\hbar ^{2}a)-\frac{\hbar ^{2}a}{2}=\frac{\hbar ^{2}a}{2}</math> |
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| So, results are;
| | Combining these results, we obtain <math>\Delta p=\hbar\sqrt{\frac{a}{2}}</math> |
| <math>\vartriangle p=\left \langle {p^{2}} \right \rangle -\left \langle {p} \right \rangle ^{2}=\frac{\hbar ^{2}a}{2}</math> | |
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| <math>\vartriangle x=\left \langle {x^{2}} \right \rangle -\left \langle {x} \right \rangle ^{2}=\frac{1}{2a}</math> | | <math>\Delta x=\frac{1}{\sqrt{2a}}.</math> |
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| finally,
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| <math>\sqrt {\vartriangle p\vartriangle x} =\sqrt {\frac{\hbar ^{2}a}{2}\frac{1}{2a}} =\sqrt {\frac{\hbar ^{2}}{4}} =\frac{\hbar }{2}</math>.
| | Finally, |
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| 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.
| | <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]] |
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