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<item xml:lang="es">
		<title>Funci&#243;n de onda de una part&#237;cula cu&#225;ntica como combinaci&#243;n lineal de los estados estacionarios (8466)</title>
		<link>http://ejercicios-fyq.com/Funcion-de-onda-de-una-particula-cuantica-como-combinacion-lineal-de-los</link>
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		<dc:date>2025-05-29T11:34:42Z</dc:date>
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		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Ecuaci&#243;n Schr&#246;dinger</dc:subject>
		<dc:subject>Funci&#243;n de onda</dc:subject>

		<description>
&lt;p&gt;Considera una part&#237;cula cu&#225;ntica de masa &#171;m&#187; confinada en un pozo de potencial unidimensional infinito en el intervalo . En el instante t = 0, la funci&#243;n de onda de la part&#237;cula viene dada por: &lt;br class='autobr' /&gt; &lt;br class='autobr' /&gt;
a) Normaliza la funci&#243;n de onda inicial y verifica que ya est&#225; normalizada. &lt;br class='autobr' /&gt;
b) Expresa como una combinaci&#243;n lineal de los estados estacionarios del pozo infinito. &lt;br class='autobr' /&gt;
c) Determina la funci&#243;n de onda en un tiempo t &gt; 0. &lt;br class='autobr' /&gt;
d) Calcula la probabilidad de que, al medir la energ&#237;a, se obtenga el (&#8230;)&lt;/p&gt;


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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Considera una part&#237;cula cu&#225;ntica de masa &#171;m&#187; confinada en un pozo de potencial unidimensional infinito en el intervalo &lt;img src='http://ejercicios-fyq.com/local/cache-vignettes/L93xH18/db921407c58886ead57ef43e7f1adfac-b6685.png?1787430695' style='vertical-align:middle;' width='93' height='18' alt=&#034;0 \leq x \leq L&#034; title=&#034;0 \leq x \leq L&#034; /&gt;. En el instante t = 0, la funci&#243;n de onda de la part&#237;cula viene dada por:&lt;/p&gt;
&lt;p&gt;
&lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-vignettes/L224xH52/3f49bbefa9b2e07427d05c16ef3800da-6a053.png?1787430695' style='vertical-align:middle;' width='224' height='52' alt=&#034;\Psi(x, 0) = \sqrt{\frac{30}{L^5}} \, x (L - x)&#034; title=&#034;\Psi(x, 0) = \sqrt{\frac{30}{L^5}} \, x (L - x)&#034; /&gt;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;a) Normaliza la funci&#243;n de onda inicial y verifica que ya est&#225; normalizada.&lt;/p&gt;
&lt;p&gt;b) Expresa &lt;img src='http://ejercicios-fyq.com/local/cache-vignettes/L61xH23/57b582839fc8384f7195f9066219374a-da523.png?1787430695' style='vertical-align:middle;' width='61' height='23' alt=&#034;\Psi(x, 0)&#034; title=&#034;\Psi(x, 0)&#034; /&gt; como una combinaci&#243;n lineal de los estados estacionarios &lt;img src='http://ejercicios-fyq.com/local/cache-vignettes/L50xH23/0384894ad2cad3e86d932dbdb7202511-8059e.png?1787430695' style='vertical-align:middle;' width='50' height='23' alt=&#034;\psi_n(x)&#034; title=&#034;\psi_n(x)&#034; /&gt; del pozo infinito.&lt;/p&gt;
&lt;p&gt;c) Determina la funci&#243;n de onda &lt;img src='http://ejercicios-fyq.com/local/cache-vignettes/L59xH23/7958d280ff667bd2eb74e389b9e8bb1d-16dec.png?1787430695' style='vertical-align:middle;' width='59' height='23' alt=&#034;\Psi(x, t)&#034; title=&#034;\Psi(x, t)&#034; /&gt; en un tiempo t &gt; 0.&lt;/p&gt;
&lt;p&gt;d) Calcula la probabilidad de que, al medir la energ&#237;a, se obtenga el valor correspondiente al primer estado excitado (n = 2).&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) La funci&#243;n de onda inicial estar&#225; normalizada cuando cumpla la condici&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/74d83981dcbb5ac46342991fae33c0e3.png' style=&#034;vertical-align:middle;&#034; width=&#034;212&#034; height=&#034;54&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\int_{0}^{L} |\Psi(x, 0)|^2\ dx = 1}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\int_{0}^{L} |\Psi(x, 0)|^2\ dx = 1}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Escribes la ecuaci&#243;n sustiyendo la funci&#243;n de onda: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/d380354b0642b7832863c214b2e394ba.png' style=&#034;vertical-align:middle;&#034; width=&#034;484&#034; height=&#034;69&#034; alt=&#034;\int_{0}^{L} \left( \sqrt{\frac{30}{L^5}} \, x (L - x) \right)^2 dx = \color[RGB]{2,112,20}{\bm{\frac{30}{L^5} \int_{0}^{L} x^2 (L - x)^2\ dx}}&#034; title=&#034;\int_{0}^{L} \left( \sqrt{\frac{30}{L^5}} \, x (L - x) \right)^2 dx = \color[RGB]{2,112,20}{\bm{\frac{30}{L^5} \int_{0}^{L} x^2 (L - x)^2\ dx}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si desarrollas el cuadrado y divides en tres integrales: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/052bace1481ebdfd7b966aa53d779630.png' style=&#034;vertical-align:middle;&#034; width=&#034;447&#034; height=&#034;54&#034; alt=&#034;\frac{30}{L^5} \left[ L^2 \int_{0}^{L} x^2 \, dx - 2L \int_{0}^{L} x^3 \, dx + \int_{0}^{L} x^4 \, dx \right] = 1&#034; title=&#034;\frac{30}{L^5} \left[ L^2 \int_{0}^{L} x^2 \, dx - 2L \int_{0}^{L} x^3 \, dx + \int_{0}^{L} x^4 \, dx \right] = 1&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Las integrales son inmediatas y las calculas entre los l&#237;mites de integraci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/9fbd1e95e7c488b711dee201f9aa0601.png' style=&#034;vertical-align:middle;&#034; width=&#034;512&#034; height=&#034;55&#034; alt=&#034;\frac{30}{L^5} \left( L^2\cdot \frac{L^3}{3} - 2L\cdot \frac{L^4}{4} + \frac{L^5}{5} \right) = \color[RGB]{0,112,192}{\bm{\frac{30}{L^5} \left( \frac{L^5}{3} - \frac{L^5}{2} + \frac{L^5}{5} \right)}}&#034; title=&#034;\frac{30}{L^5} \left( L^2\cdot \frac{L^3}{3} - 2L\cdot \frac{L^4}{4} + \frac{L^5}{5} \right) = \color[RGB]{0,112,192}{\bm{\frac{30}{L^5} \left( \frac{L^5}{3} - \frac{L^5}{2} + \frac{L^5}{5} \right)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si sacas factor com&#250;n &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/bee16925f81c32af7649b0798ec0e7ae.png' style=&#034;vertical-align:middle;&#034; width=&#034;20&#034; height=&#034;19&#034; alt=&#034;L^5&#034; title=&#034;L^5&#034; /&gt; y simplificas: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/8237f800da8c3f9fd5b1f393f9101993.png' style=&#034;vertical-align:middle;&#034; width=&#034;562&#034; height=&#034;54&#034; alt=&#034;\frac{30\cdot \cancel{L^5}}{\cancel{L^5}} \left( \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right) = 30 \left( \frac{10}{30} - \frac{15}{30} + \frac{6}{30} \right) = \fbox{\color[RGB]{192,0,0}{\bm{30 \left( \frac{1}{30} \right) = 1}}}&#034; title=&#034;\frac{30\cdot \cancel{L^5}}{\cancel{L^5}} \left( \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right) = 30 \left( \frac{10}{30} - \frac{15}{30} + \frac{6}{30} \right) = \fbox{\color[RGB]{192,0,0}{\bm{30 \left( \frac{1}{30} \right) = 1}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Como puedes ver, la funci&#243;n de onda est&#225; normalizada. &lt;br/&gt; &lt;br/&gt; b) La ecuaci&#243;n de los estados estacionarios de un pozo infinito es: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/6603ae6c6e1bbe7601218ca4e625ebff.png' style=&#034;vertical-align:middle;&#034; width=&#034;271&#034; height=&#034;56&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\psi_n(x) = \sqrt{\frac{2}{L}}\cdot sen \left( \frac{n \pi x}{L} \right) }}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\psi_n(x) = \sqrt{\frac{2}{L}}\cdot sen \left( \frac{n \pi x}{L} \right) }}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Tienes que expresar la funci&#243;n de onda como combinaci&#243;n lineal de los estados estacionarios: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/62bfdf01cdba5772ee24b568fd94c83a.png' style=&#034;vertical-align:middle;&#034; width=&#034;241&#034; height=&#034;60&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{\Psi(x, 0) = \sum_{n=1}^{\infty} c_n\cdot \psi_n(x)}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{\Psi(x, 0) = \sum_{n=1}^{\infty} c_n\cdot \psi_n(x)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Los coeficientes &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/6f58730f154756d9dc7efb13fc938933.png' style=&#034;vertical-align:middle;&#034; width=&#034;19&#034; height=&#034;15&#034; alt=&#034;c_n&#034; title=&#034;c_n&#034; /&gt; los calculas de esta manera: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/7019abec8dc6385fb18ff9ca80096a5b.png' style=&#034;vertical-align:middle;&#034; width=&#034;676&#034; height=&#034;56&#034; alt=&#034;c_n = \int_{0}^{L} \psi_n^*(x)\cdot \Psi(x, 0)\ dx = \sqrt{\frac{2}{L}}\cdot \sqrt{\frac{30}{L^5}} \int_{0}^{L} x\cdot (L - x)\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx&#034; title=&#034;c_n = \int_{0}^{L} \psi_n^*(x)\cdot \Psi(x, 0)\ dx = \sqrt{\frac{2}{L}}\cdot \sqrt{\frac{30}{L^5}} \int_{0}^{L} x\cdot (L - x)\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si operas con las constantes que est&#225; fuera del integrando tienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/7473043a08b243cad69cab32cfbd79df.png' style=&#034;vertical-align:middle;&#034; width=&#034;387&#034; height=&#034;54&#034; alt=&#034;c_n = \frac{2 \sqrt{15}}{L^3} \int_{0}^{L} (Lx - x^2)\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx&#034; title=&#034;c_n = \frac{2 \sqrt{15}}{L^3} \int_{0}^{L} (Lx - x^2)\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La resoluci&#243;n de la integral la haces en dos partes: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/0bbb93f42f526b017b79dd841a02231f.png' style=&#034;vertical-align:middle;&#034; width=&#034;554&#034; height=&#034;105&#034; alt=&#034;\left L\ \int_{0}^{L} x\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx = \dfrac{L^2 (-1)^{n+1}}{n \pi} \atop \int_{0}^{L} x^2\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx = \dfrac{2L^3 (-1)^{n+1}}{n \pi} - \dfrac{L^3 (2 - n^2 \pi^2 (-1)^n)}{n^3 \pi^3} \right \}&#034; title=&#034;\left L\ \int_{0}^{L} x\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx = \dfrac{L^2 (-1)^{n+1}}{n \pi} \atop \int_{0}^{L} x^2\cdot sen\ \left( \frac{n \pi x}{L} \right)\ dx = \dfrac{2L^3 (-1)^{n+1}}{n \pi} - \dfrac{L^3 (2 - n^2 \pi^2 (-1)^n)}{n^3 \pi^3} \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Combinas los t&#233;rminos anteriores y tienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/a0e3c5da74602fd6a2a34982021711ad.png' style=&#034;vertical-align:middle;&#034; width=&#034;560&#034; height=&#034;54&#034; alt=&#034;c_n = \frac{2 \sqrt{15}}{\cancel{L^3}}\cdot \cancel{L^3} \left( \frac{(-1)^{n+1}}{n \pi} - \frac{2\cdot (-1)^{n+1}}{n \pi} + \frac{(2 - n^2 \pi^2 (-1)^n)}{n^3 \pi^3} \right)&#034; title=&#034;c_n = \frac{2 \sqrt{15}}{\cancel{L^3}}\cdot \cancel{L^3} \left( \frac{(-1)^{n+1}}{n \pi} - \frac{2\cdot (-1)^{n+1}}{n \pi} + \frac{(2 - n^2 \pi^2 (-1)^n)}{n^3 \pi^3} \right)&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si tienes en cuenta que para valores pares de &#171;n&#187; los t&#233;rminos se cancelan y simplificas, obtienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/e6d8d177177f3de76378443068737276.png' style=&#034;vertical-align:middle;&#034; width=&#034;237&#034; height=&#034;53&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{c_n = \frac{4 \sqrt{15}}{n^3 \pi^3} \left( 1 - (-1)^n \right)}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{c_n = \frac{4 \sqrt{15}}{n^3 \pi^3} \left( 1 - (-1)^n \right)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Por lo tanto, para valores pares de &#171;n&#187; los coeficientes son nulos y para valores impares de &#171;n&#187; obtienes: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/04c2a1b221d1d61b754b6118114834ff.png' style=&#034;vertical-align:middle;&#034; width=&#034;112&#034; height=&#034;44&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{c_n = \frac{8 \sqrt{15}}{n^3 \pi^3}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{c_n = \frac{8 \sqrt{15}}{n^3 \pi^3}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; c) La funci&#243;n de onda en funci&#243;n del tiempo, escrita como combinaci&#243;n lineal de los estados estacionarios, es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/03b030ad0729d2fb05ce484362f06f3e.png' style=&#034;vertical-align:middle;&#034; width=&#034;492&#034; height=&#034;64&#034; alt=&#034;\color[RGB]{192,0,0}{\bm{\Psi(x, t) = \sum_{n=1, 3, 5, \dots} \frac{8\sqrt{15}}{n^3\pi^3} \sqrt{\frac{2}{L}}\cdot sen \left( \frac{n \pi x}{L} \right) e^{\frac{-i E_n t}{\hbar}}}}&#034; title=&#034;\color[RGB]{192,0,0}{\bm{\Psi(x, t) = \sum_{n=1, 3, 5, \dots} \frac{8\sqrt{15}}{n^3\pi^3} \sqrt{\frac{2}{L}}\cdot sen \left( \frac{n \pi x}{L} \right) e^{\frac{-i E_n t}{\hbar}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; donde el t&#233;rmino &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/f68207972fe0c39be7798431a8afcc29.png' style=&#034;vertical-align:middle;&#034; width=&#034;24&#034; height=&#034;20&#034; alt=&#034;E_n&#034; title=&#034;E_n&#034; /&gt; es: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/f2f52bd80229fffb3461b7c22699af1c.png' style=&#034;vertical-align:middle;&#034; width=&#034;133&#034; height=&#034;51&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; d) Como &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/d42957487f34bea1601b6333cada1826.png' style=&#034;vertical-align:middle;&#034; width=&#034;58&#034; height=&#034;20&#034; alt=&#034;c_n = 0&#034; title=&#034;c_n = 0&#034; /&gt; para cualquier valor par de &#171;n&#187;, la probabilidad de encontrar &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/8e3b512c2f053602a180ee612fd581a6.png' style=&#034;vertical-align:middle;&#034; width=&#034;18&#034; height=&#034;15&#034; alt=&#034;E_2&#034; title=&#034;E_2&#034; /&gt; es nula, es decir: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='http://ejercicios-fyq.com/local/cache-TeX/647b31c714d943188eaf53f7bde46f47.png' style=&#034;vertical-align:middle;&#034; width=&#034;196&#034; height=&#034;35&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P(E_2) = |c_2|^2 = 0}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P(E_2) = |c_2|^2 = 0}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Esto ocurre porque la funci&#243;n de onda es una combinaci&#243;n de estados estacionarios impares, como has calculado en el segundo apartado del problema.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

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<item xml:lang="es">
		<title>Probabilidad de encontrar una part&#237;cula en una zona de una caja unidimensional (8103)</title>
		<link>http://ejercicios-fyq.com/Probabilidad-de-encontrar-una-particula-en-una-zona-de-una-caja-unidimensional</link>
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		<dc:date>2023-12-01T05:33:26Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Ecuaci&#243;n Schr&#246;dinger</dc:subject>

		<description>
&lt;p&gt;Una part&#237;cula se mueve en el interior de una caja unidimensional de longitud a y potencial infinito en sus extremos calcular la probabilidad de encontrar la part&#237;cula a del lado izquierdo de la caja y para que valor del estado cu&#225;ntico n es m&#225;xima esta probabilidad.&lt;/p&gt;


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&lt;a href="http://ejercicios-fyq.com/Ecuacion-Schrodinger" rel="tag"&gt;Ecuaci&#243;n Schr&#246;dinger&lt;/a&gt;

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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Una part&#237;cula se mueve en el interior de una caja unidimensional de longitud &lt;i&gt;a&lt;/i&gt; y potencial infinito en sus extremos calcular la probabilidad de encontrar la part&#237;cula a &lt;img src='http://ejercicios-fyq.com/local/cache-vignettes/L6xH21/8b72600e177e2cb1546d4cbc3074c364-7fc41.png?1787523926' style='vertical-align:middle;' width='6' height='21' alt=&#034;\textstyle{1\over 8}&#034; title=&#034;\textstyle{1\over 8}&#034; /&gt; del lado izquierdo de la caja y para que valor del estado cu&#225;ntico &lt;i&gt;n&lt;/i&gt; es m&#225;xima esta probabilidad.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Como nuestro problema es unidimensional, vamos a considerar solo la componente &#171;x&#187; del sistema y la posici&#243;n de la part&#237;cula tendr&#225; que variar entre los puntos 0 y &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/df9b21c88db63bfacd059d96a51d2904.png' style=&#034;vertical-align:middle;&#034; width=&#034;7&#034; height=&#034;18&#034; alt=&#034;\textstyle{a\over 8}&#034; title=&#034;\textstyle{a\over 8}&#034; /&gt;. La funci&#243;n de onda que vamos a considerar, por lo tanto, ser&#225;: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/bf0bf03978f26a55ce5b9a7e7d0adfb5.png' style=&#034;vertical-align:middle;&#034; width=&#034;122&#034; height=&#034;43&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{\sqrt{\frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{\sqrt{\frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La probabilidad que debes calcular la integral del producto de la conjugada de la funci&#243;n de onda por la propia funci&#243;n de onda. En este caso, al ser unidireccional, coinciden ambas ecuaciones: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/8c59d2bc3ebe335d38f51586c2edacb0.png' style=&#034;vertical-align:middle;&#034; width=&#034;417&#034; height=&#034;43&#034; alt=&#034;P(0, \frac{a}{8}) = \int_0^{\frac{a}{8}}\psi^*\cdot \psi\cdot dx = \int_0^{\frac{a}{8}} \frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)\cdot \frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)&#034; title=&#034;P(0, \frac{a}{8}) = \int_0^{\frac{a}{8}}\psi^*\cdot \psi\cdot dx = \int_0^{\frac{a}{8}} \frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)\cdot \frac{2}{a}}sen\left(\frac{n\pi x}{a}\right)&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si multiplicas ambas funciones y sacas del integrando las constante, tienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/011e4194657d4d3b42d5a4dbb55ae97a.png' style=&#034;vertical-align:middle;&#034; width=&#034;267&#034; height=&#034;44&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{P(0, \frac{a}{8}) = \frac{2}{a}\int_0^{\frac{a}{8}} sen^2\left(\frac{n\pi x}{a}\right)\cdot dx}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{P(0, \frac{a}{8}) = \frac{2}{a}\int_0^{\frac{a}{8}} sen^2\left(\frac{n\pi x}{a}\right)\cdot dx}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Si aplicas la siguiente igualdad trigonom&#233;trica puedes hacer la integral m&#225;s f&#225;cil: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/208974ff322c92446c792719f834abe0.png' style=&#034;vertical-align:middle;&#034; width=&#034;140&#034; height=&#034;34&#034; alt=&#034;sen^2 \alpha = \frac{1-cos\ 2\alpha}{2}&#034; title=&#034;sen^2 \alpha = \frac{1-cos\ 2\alpha}{2}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La integral anterior, al sacar la constante &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/26da16dcd83476d49b0255dcbae8635b.png' style=&#034;vertical-align:middle;&#034; width=&#034;12&#034; height=&#034;45&#034; alt=&#034;\textstyle{1\over 2}&#034; title=&#034;\textstyle{1\over 2}&#034; /&gt; y operar con la otra que estaba fuera del integrando, queda como: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/246647ba8bb3a6d8a0299325376edd35.png' style=&#034;vertical-align:middle;&#034; width=&#034;321&#034; height=&#034;44&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{P(0, \frac{a}{8}) = \frac{1}{a} \int_0^{\frac{a}{8}} \left[1 - cos\ \left(\frac{2n\pi x}{a}\right) \right]\cdot dx}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{P(0, \frac{a}{8}) = \frac{1}{a} \int_0^{\frac{a}{8}} \left[1 - cos\ \left(\frac{2n\pi x}{a}\right) \right]\cdot dx}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Puedes dividir la integral en dos integrales; una de ellas es inmediata y la otra debe ser resuelta por sustituci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/1187ad4d7640fdd21ef0df024ea35722.png' style=&#034;vertical-align:middle;&#034; width=&#034;335&#034; height=&#034;50&#034; alt=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{a}\left[ \int_0^{\frac{a}{8}} dx - \int_0^{\frac{a}{8}} cos\ \left(\dfrac{2n\pi x}{a}\right)\cdot dx\right]&#034; title=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{a}\left[ \int_0^{\frac{a}{8}} dx - \int_0^{\frac{a}{8}} cos\ \left(\dfrac{2n\pi x}{a}\right)\cdot dx\right]&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El resultado de las integrales que obtienes es: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/af8a21137cf9622d64bf70d9fe52b9ff.png' style=&#034;vertical-align:middle;&#034; width=&#034;314&#034; height=&#034;40&#034; alt=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{a} \left[x |_0^{\frac{a}{8}} - \frac{a}{2n\pi}\cdot sen\ \left(\frac{2n\pi x}{a}\right) |_0^{\frac{a}{8}\right]&#034; title=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{a} \left[x |_0^{\frac{a}{8}} - \frac{a}{2n\pi}\cdot sen\ \left(\frac{2n\pi x}{a}\right) |_0^{\frac{a}{8}\right]&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Aplicas los l&#237;mites superior e inferior en cada caso y obtienes: &lt;br/&gt; &lt;br/&gt; &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/642d3f91d6f6bebcac110f5ee95163da.png' style=&#034;vertical-align:middle;&#034; width=&#034;444&#034; height=&#034;38&#034; alt=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{\cancel{a}}\cdot \frac{\cancel{a}}{8} - \frac{1}{\cancel{a}}\cdot \frac{\cancel{a}}{2n\pi}\cdot sen\ \frac{2n\pi \cancel{a}}{8 \cancel{a}} = \color[RGB]{0,112,192}{\bm{\frac{1}{8} - \frac{1}{2n\pi}\cdot sen\ \frac{n\pi}{4}}}&#034; title=&#034;P\left(0, \frac{a}{8}\right) = \frac{1}{\cancel{a}}\cdot \frac{\cancel{a}}{8} - \frac{1}{\cancel{a}}\cdot \frac{\cancel{a}}{2n\pi}\cdot sen\ \frac{2n\pi \cancel{a}}{8 \cancel{a}} = \color[RGB]{0,112,192}{\bm{\frac{1}{8} - \frac{1}{2n\pi}\cdot sen\ \frac{n\pi}{4}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La soluci&#243;n depende de la funci&#243;n seno, con lo que el resultado no es &#250;nico y es necesario hacer el an&#225;lisis de los posibles valores de &#171;n&#187; para obtener las probabilidades. &lt;br/&gt; &lt;br/&gt; &lt;u&gt;Para valores impares de &#171;n&#187;&lt;/u&gt;. &lt;br/&gt; &lt;br/&gt; Haces dos divisiones de valores impares de &#171;n&#187;: &lt;br/&gt; &lt;br/&gt; n = 1, 3, 9, 11, 17, 19... &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/9d8a6a2dc1ee39fdb2237f7d85862654.png' style=&#034;vertical-align:middle;&#034; width=&#034;154&#034; height=&#034;34&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} - \frac{\sqrt{2}}{4n\pi}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} - \frac{\sqrt{2}}{4n\pi}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; n = 5, 7, 13, 15, 21, 23... &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/a1e73dcd8a23f10b3b0c0e815ea4e59f.png' style=&#034;vertical-align:middle;&#034; width=&#034;154&#034; height=&#034;34&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} + \frac{\sqrt{2}}{4n\pi}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} + \frac{\sqrt{2}}{4n\pi}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;u&gt;Para valores pares de &#171;n&#187;&lt;/u&gt;. &lt;br/&gt; &lt;br/&gt; En este caso la divisi&#243;n la haces en tres grupos de valores: &lt;br/&gt; &lt;br/&gt; n = 2, 10, 18, 26... &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/548257a4a0fae783d70d4e7d28528b89.png' style=&#034;vertical-align:middle;&#034; width=&#034;154&#034; height=&#034;30&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} - \frac{1}{2n\pi}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} - \frac{1}{2n\pi}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; n = 4, 8, 12, 16, 20, 22... &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/e6fa40cfbd9af3730b1886eb0ecd7935.png' style=&#034;vertical-align:middle;&#034; width=&#034;105&#034; height=&#034;30&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; n = 6, 14, 22, 30... &lt;img src='http://ejercicios-fyq.com/local/cache-TeX/5eb96ffd8c03b7bbe6d495bdebb3e56b.png' style=&#034;vertical-align:middle;&#034; width=&#034;154&#034; height=&#034;30&#034; alt=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} + \frac{1}{2n\pi}}}}&#034; title=&#034;\fbox{\color[RGB]{192,0,0}{\bm{P\left(0, \frac{a}{8}\right) = \frac{1}{8} + \frac{1}{2n\pi}}}}&#034; /&gt;&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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