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		<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>https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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='https://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>


		

	</item>
<item xml:lang="es">
		<title>Constante de normalizaci&#243;n y probabilidad de estar en el estado fundamental sabiendo la funci&#243;n de onda (8396)</title>
		<link>https://ejercicios-fyq.com/Constante-de-normalizacion-y-probabilidad-de-estar-en-el-estado-fundamental</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Constante-de-normalizacion-y-probabilidad-de-estar-en-el-estado-fundamental</guid>
		<dc:date>2025-02-12T04:31:53Z</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>Funci&#243;n de onda</dc:subject>
		<dc:subject>Constante normalizaci&#243;n</dc:subject>

		<description>
&lt;p&gt;Una part&#237;cula de masa &#171;m&#187; est&#225; confinada en una caja unidimensional de longitud &#171;L&#187;, con paredes infinitamente altas, es decir, con potencial infinito fuera de la caja. La funci&#243;n de onda inicial de la part&#237;cula es: &lt;br class='autobr' /&gt; &lt;br class='autobr' /&gt;
a) Determina la constante de normalizaci&#243;n &#171;A&#187;. &lt;br class='autobr' /&gt;
b) Encuentra la probabilidad de que la part&#237;cula se encuentre en el estado fundamental (n=1).&lt;/p&gt;


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&lt;a href="https://ejercicios-fyq.com/Funcion-de-onda" rel="tag"&gt;Funci&#243;n de onda&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Constante-normalizacion" rel="tag"&gt;Constante normalizaci&#243;n&lt;/a&gt;

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 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Una part&#237;cula de masa &#171;m&#187; est&#225; confinada en una caja unidimensional de longitud &#171;L&#187;, con paredes infinitamente altas, es decir, con potencial infinito fuera de la caja. La funci&#243;n de onda inicial de la part&#237;cula es:&lt;/p&gt;
&lt;p&gt;
&lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L335xH53/5198c5a1d37d7b6573bf695c9eac185c-63e92.png?1787436711' style='vertical-align:middle;' width='335' height='53' alt=&#034;\Psi(x,0) = \left\{ {A\cdot sen(\frac{\pi\cdot x}{L}),\ \ 0\leq x \leq L \atop 0,\ \ \ \ \ \text{en~otro~caso}}&#034; title=&#034;\Psi(x,0) = \left\{ {A\cdot sen(\frac{\pi\cdot x}{L}),\ \ 0\leq x \leq L \atop 0,\ \ \ \ \ \text{en~otro~caso}}&#034; /&gt;&lt;/p&gt;
&lt;/p&gt;
&lt;p&gt;a) Determina la constante de normalizaci&#243;n &#171;A&#187;.&lt;/p&gt;
&lt;p&gt;b) Encuentra la probabilidad de que la part&#237;cula se encuentre en el estado fundamental (n=1).&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 normalizaci&#243;n de la funci&#243;n de onda debe cumplir: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/4b24279652ae337bcf461da64ec482c9.png' style=&#034;vertical-align:middle;&#034; width=&#034;181&#034; height=&#034;54&#034; alt=&#034;\int_0^L |\Psi(x,0)|^2 dx = 1&#034; title=&#034;\int_0^L |\Psi(x,0)|^2 dx = 1&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El cuadrado de la funci&#243;n de onda, en el intervalo de la integral, y la integral que tienes que resolver son: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6d4b186521e259f39ae53f5bb1133f82.png' style=&#034;vertical-align:middle;&#034; width=&#034;577&#034; height=&#034;54&#034; alt=&#034;{\color[RGB]{0,112,192}{\bm{|\Psi(x,0)|^2 = A^2 \sen^2 \left(\frac{\pi x}{L}\right)}}}\ \to\ {\color[RGB]{2,112,20}{\bm{\int_0^L A^2 \sen^2 \left(\frac{\pi x}{L}\right) dx = 1}}}&#034; title=&#034;{\color[RGB]{0,112,192}{\bm{|\Psi(x,0)|^2 = A^2 \sen^2 \left(\frac{\pi x}{L}\right)}}}\ \to\ {\color[RGB]{2,112,20}{\bm{\int_0^L A^2 \sen^2 \left(\frac{\pi x}{L}\right) dx = 1}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Se trata de una integral con integrando trignom&#233;trico y puedes resolverla si tienes en cuenta la ecuaci&#243;n del cuadrado del seno: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/951d7112f71ab61883ebe8719b7d454c.png' style=&#034;vertical-align:middle;&#034; width=&#034;220&#034; height=&#034;49&#034; alt=&#034;\color[RGB]{0,112,192}{\bm{\sen^2 \alpha = \frac{1 - \cos(2\alpha)}{2}}}&#034; title=&#034;\color[RGB]{0,112,192}{\bm{\sen^2 \alpha = \frac{1 - \cos(2\alpha)}{2}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Sustituyes en la funci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/b6f4c03fcec18892a8fa2a678067b3c6.png' style=&#034;vertical-align:middle;&#034; width=&#034;472&#034; height=&#034;54&#034; alt=&#034;\int_0^L A^2 \sen^2 \left(\frac{\pi x}{L}\right) dx = \frac{A^2}{2} \int_0^L \left[1 - \cos\left(\frac{2\pi x}{L}\right)}\right] dx&#034; title=&#034;\int_0^L A^2 \sen^2 \left(\frac{\pi x}{L}\right) dx = \frac{A^2}{2} \int_0^L \left[1 - \cos\left(\frac{2\pi x}{L}\right)}\right] dx&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Para hacer la integral la separas en dos integrales: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6e3e9843f3331f06d8158b125b8ce79b.png' style=&#034;vertical-align:middle;&#034; width=&#034;340&#034; height=&#034;54&#034; alt=&#034;A^2 \int_0^L \frac{1}{2} dx - A^2 \int_0^L \frac{\cos\left(\frac{2\pi x}{L}\right)}{2} dx = 1&#034; title=&#034;A^2 \int_0^L \frac{1}{2} dx - A^2 \int_0^L \frac{\cos\left(\frac{2\pi x}{L}\right)}{2} dx = 1&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La primera integral es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e39052681d884fd49e9cc8e5b4c7183b.png' style=&#034;vertical-align:middle;&#034; width=&#034;179&#034; height=&#034;54&#034; alt=&#034;A^2 \int_0^L \frac{1}{2} dx = \color[RGB]{0,112,192}{\bm{A^2 \frac{L}{2}}}&#034; title=&#034;A^2 \int_0^L \frac{1}{2} dx = \color[RGB]{0,112,192}{\bm{A^2 \frac{L}{2}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La integral trigonom&#233;trica es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/ae064947559cf17effada58003bbcbfc.png' style=&#034;vertical-align:middle;&#034; width=&#034;494&#034; height=&#034;58&#034; alt=&#034;\frac{A^2}{2} \int_0^L \cos\left(\frac{2\pi x}{L}\right) dx = \frac{A^2\cdot L}{4\pi}\cdot \left[\sen\left(\frac{2\pi\cdot x}{L}\right)\right]_0^L = \color[RGB]{0,112,192}{\bf 0}&#034; title=&#034;\frac{A^2}{2} \int_0^L \cos\left(\frac{2\pi x}{L}\right) dx = \frac{A^2\cdot L}{4\pi}\cdot \left[\sen\left(\frac{2\pi\cdot x}{L}\right)\right]_0^L = \color[RGB]{0,112,192}{\bf 0}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La constante de normalizaci&#243;n ser&#225;: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0cc35dd1edc5fadd9743fdf2435edea0.png' style=&#034;vertical-align:middle;&#034; width=&#034;243&#034; height=&#034;53&#034; alt=&#034;\frac{A^2\cdot L}{2} = 1\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{A = \sqrt{\frac{2}{L}}}}}&#034; title=&#034;\frac{A^2\cdot L}{2} = 1\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{A = \sqrt{\frac{2}{L}}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) La funci&#243;n de onda del estado fundamental de una part&#237;cula en una dimensi&#243;n es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/90328f377f7ba4d9b3a2dd855c1463b7.png' style=&#034;vertical-align:middle;&#034; width=&#034;206&#034; height=&#034;52&#034; alt=&#034;\psi_1 = \sqrt{\frac{2}{L}}\cdot \sen\left(\frac{\pi\cdot x}{L}\right)&#034; title=&#034;\psi_1 = \sqrt{\frac{2}{L}}\cdot \sen\left(\frac{\pi\cdot x}{L}\right)&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Observa que es de la misma forma que la funci&#243;n de onda y eso va a ser muy relevante. La probabilidad de que est&#233; en el estado fundamental ser&#225;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c91d1d64fe852f06e524328d4855d9fa.png' style=&#034;vertical-align:middle;&#034; width=&#034;742&#034; height=&#034;70&#034; alt=&#034;{\color[RGB]{2,112,20}{\bm{P_1 = \left| \int_0^L \psi_1^*(x) \Psi(x,0) dx \right|^2}}}\ \to\ P_1 = \left| \int_0^L \sqrt{\frac{2}{L}} \sen\left(\frac{\pi x}{L}\right) \cdot \sqrt{\frac{2}{L}} \sen\left(\frac{\pi x}{L}\right) dx \right|^2&#034; title=&#034;{\color[RGB]{2,112,20}{\bm{P_1 = \left| \int_0^L \psi_1^*(x) \Psi(x,0) dx \right|^2}}}\ \to\ P_1 = \left| \int_0^L \sqrt{\frac{2}{L}} \sen\left(\frac{\pi x}{L}\right) \cdot \sqrt{\frac{2}{L}} \sen\left(\frac{\pi x}{L}\right) dx \right|^2&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Simplificas la expresi&#243;n anterior y resuelves: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/33bdabdebd5c1ada9ace8c21efb00719.png' style=&#034;vertical-align:middle;&#034; width=&#034;463&#034; height=&#034;59&#034; alt=&#034;P_1 = \left| \int_0^L \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right) dx \right|^2 = \frac{2}{L}\cdot {\color[RGB]{0,112,192}{\bm{\frac{L}{2}}}}\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{P_1 = 1}}}&#034; title=&#034;P_1 = \left| \int_0^L \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right) dx \right|^2 = \frac{2}{L}\cdot {\color[RGB]{0,112,192}{\bm{\frac{L}{2}}}}\ \to\ \fbox{\color[RGB]{192,0,0}{\bm{P_1 = 1}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
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