Phy5645/Heisenberg Uncertainty Relation 3: Difference between revisions

Revision as of 16:23, 10 April 2013

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 $x$ , we are able to use the known results for such integrals.

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

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

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

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

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

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

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

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

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

finally,

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

Back to Heisenberg Uncertainty Principle

@@ Line 1: / Line 1: @@
-Let's say that a particle has wavefunction : <math>\Psi (x)=(\frac{\pi }{a})^{-1/4}e^{-ax^{2}/2}</math>
+Let's say that a particle has wavefunction : <math>\Psi (x)=\left (\frac{\pi }{a}\right )^{-1/4}e^{-ax^{2}/2}</math>
 and we are trying to verify Heisenberg Uncertanity relation.
-In order to verify the uncertanity relation, we need to find these elements,
+In order to verify the uncertanity relation, 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>
+and
+<math>\Delta x=\sqrt{\left \langle {x^{2}} \right \rangle -\left \langle {x} \right \rangle ^{2}}.</math>
-<math>\vartriangle p=\left \langle {p^{2}} \right \rangle -\left \langle {p} \right \rangle ^{2}</math>
+Lets start by calculating the expectation values one by one.
-and
-<math>\vartriangle x=\left \langle {x^{2}} \right \rangle -\left \langle {x} \right \rangle ^{2}</math>.
-Lets start by calculating one by one.
+<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>\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.
+since the integrand is odd and thus the integral over all space is zero.
 <math>\left \langle {x^{2}} \right \rangle =\left \langle {\Psi \left |{x^{2}} \right |\Psi } \right \rangle </math>
-<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>
+<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>
-Since the integral is Gaussian integral, we used Gaussian integral results.
-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.
-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 =\left \langle {\Psi \left |{p} \right |\Psi } \right \rangle =0</math> because the integrand will be an odd function as well.
 <math>\left \langle {p^{2}} \right \rangle =\left \langle {\Psi \left |{p^{2}} \right |\Psi } \right \rangle </math>
-<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 }} \int {e^{-ax^{2}/2}} \left (\frac{\hbar }{i}\frac{\partial}{\partial x}\right )^{2}e^{-ax^{2}/2}dx</math>
-<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}} \frac{\partial}{\partial x}\left (-\frac{2ax}{2}e^{-ax^{2}/2}\right )\,dx</math>
-<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})\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>\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)\int {e^{-ax^{2}}} dx\text{ +}\sqrt {\frac{a}{\pi }} (-\hbar ^{2}a^{2})\int {x^{2}e^{-ax^{2}}} dx</math>
-<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>
+<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>
 <math>\text{=}(\hbar ^{2}a)-\frac{\hbar ^{2}a}{2}=\frac{\hbar ^{2}a}{2}</math>
-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>
 and
-<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>
 finally,
-<math>\sqrt {\vartriangle p\vartriangle x} =\sqrt {\frac{\hbar ^{2}a}{2}\frac{1}{2a}} =\sqrt {\frac{\hbar ^{2}}{4}} =\frac{\hbar }{2}</math>.
+<math>\Delta p\,\Delta x =\hbar\sqrt{\frac{a}{2}}\frac{1}{\sqrt{2a}} =\sqrt {\frac{\hbar ^{2}}{4}} =\frac{\hbar }{2}</math>.
-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.
 Back to [[Heisenberg Uncertainty Principle]]

Phy5645/Heisenberg Uncertainty Relation 3: Difference between revisions

Revision as of 16:23, 10 April 2013

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