Normalized Wave Function for a Particle in an Infinite Potential Well<\/h3>\n\n\n\n
For a particle trapped in a one-dimensional infinite potential well (or box) of width ( L ), the potential ( V(x) ) is defined as:<\/p>\n\n\n\n
- \n
- ( V(x) = 0 ) for ( 0 \\leq x \\leq L )<\/li>\n\n\n\n
- ( V(x) = \\infty ) for ( x < 0 ) and ( x > L )<\/li>\n<\/ul>\n\n\n\n
Derivation of the Wave Function:<\/h4>\n\n\n\n
The Schr\u00f6dinger equation for this system is:
[<\/p>\n\n\n\n- \n
- \\frac{\\hbar^2}{2m} \\frac{d^2 \\psi}{dx^2} = E \\psi
]
Inside the well, the potential energy ( V(x) = 0 ), so the equation simplifies to:
[
\\frac{d^2 \\psi}{dx^2} = -\\frac{2mE}{\\hbar^2} \\psi
]
Let ( k^2 = \\frac{2mE}{\\hbar^2} ). Thus, the equation becomes:
[
\\frac{d^2 \\psi}{dx^2} = -k^2 \\psi
]
The general solution to this differential equation is:
[
\\psi(x) = A \\sin(kx) + B \\cos(kx)
]
Boundary conditions dictate that ( \\psi(0) = 0 ) and ( \\psi(L) = 0 ). Applying these:<\/li>\n\n\n\n - ( \\psi(0) = 0 ) implies ( B = 0 ) (since ( \\cos(0) = 1 ))<\/li>\n\n\n\n
- ( \\psi(L) = 0 ) leads to ( A \\sin(kL) = 0 )<\/li>\n<\/ul>\n\n\n\n
Since ( A \\neq 0 ) for a non-trivial solution, ( \\sin(kL) = 0 ) which means ( kL = n\\pi ) where ( n ) is a positive integer. Hence, ( k = \\frac{n\\pi}{L} ).<\/p>\n\n\n\n
The solutions are therefore:
[
\\psi_n(x) = A_n \\sin\\left(\\frac{n\\pi x}{L}\\right)
]
where ( n = 1, 2, 3, \\ldots ).<\/p>\n\n\n\nNormalization:<\/h4>\n\n\n\n
The wave function must be normalized:
[
\\int_0^L |\\psi_n(x)|^2 \\, dx = 1
]
Substitute ( \\psi_n(x) = A_n \\sin\\left(\\frac{n\\pi x}{L}\\right) ):
[
\\int_0^L |A_n|^2 \\sin^2\\left(\\frac{n\\pi x}{L}\\right) \\, dx = 1
]
Using ( \\int_0^L \\sin^2\\left(\\frac{n\\pi x}{L}\\right) \\, dx = \\frac{L}{2} ), we find ( |A_n|^2 \\frac{L}{2} = 1 ), so ( |A_n|^2 = \\frac{2}{L} ).<\/p>\n\n\n\nThus, ( A_n = \\sqrt{\\frac{2}{L}} ) and the normalized wave function is:
[
\\psi_n(x) = \\sqrt{\\frac{2}{L}} \\sin\\left(\\frac{n\\pi x}{L}\\right)
]<\/p>\n\n\n\nThis normalized wave function satisfies the boundary conditions, the differential equation, and the normalization condition, fully describing the state of a quantum particle in an infinite potential well.<\/p>\n\n\n\n
I’ll now generate an image to visually represent the wave function inside an infinite potential well.<\/p>\n\n\n\n
Here’s the visual representation of the quantum wave functions inside a one-dimensional infinite potential well, showing different energy states as sine waves confined within the well. Each wave represents a different quantum state, differentiated by color and wavelength.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1124.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Derive an expression for the normalized wave function of a particle trapped in an infinite deep potential well. The Correct Answer and Explanation is : Normalized Wave Function for a Particle in an Infinite Potential Well For a particle trapped in a one-dimensional infinite potential well (or box) of width ( L ), the potential […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-193775","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=193775"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193775\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193775"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193775"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- \\frac{\\hbar^2}{2m} \\frac{d^2 \\psi}{dx^2} = E \\psi