{"id":258710,"date":"2025-07-18T09:20:30","date_gmt":"2025-07-18T09:20:30","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=258710"},"modified":"2025-07-18T09:20:32","modified_gmt":"2025-07-18T09:20:32","slug":"consider-a-particle-whose-wave-function-is-given-by-i%cb%86x-ae-ax","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/18\/consider-a-particle-whose-wave-function-is-given-by-i%cb%86x-ae-ax\/","title":{"rendered":"Consider a particle whose wave function is given by \u00cf\u02c6(x) = Ae^(-ax)"},"content":{"rendered":"\n

Consider a particle whose wave function is given by \u00cf\u02c6(x) = Ae^(-ax) (A) What is the value of A if this wave function is normalized? (B) What is the expectation value of x for this particle? Let’s apply the normalization condition: \u00e2\u02c6\u00ab|\u00cf\u02c6(x)|^2 dx = \u00e2\u02c6\u00ab|Ae^(-ax)|^2 dx = \u00e2\u02c6\u00abA^2e^(-2ax) dx \u00e2\u02c6\u00ab|\u00cf\u02c6(x)|^2 dx = A^2 \u00e2\u02c6\u00abe^(-2ax) dx \u00e2\u02c6\u00ab|\u00cf\u02c6(x)|^2 dx = A^2 * (-1\/2a) * e^(-2ax) + C To normalize the wave function, the integral of |\u00cf\u02c6(x)|^2 over all space must equal 1. Therefore: A^2 * (-1\/2a) * e^(-2ax) + C = 1 Solving for A: A^2 * (-1\/2a) * e^(-2ax) = 1 – C A^2 = (1 – C) * (-2a) * e^(2ax) A = \u00e2\u02c6\u0161(1 – C) * (-2a) * e^(2ax)<\/a> To find the expectation value of x, we use the formula: \u00e2\u0178\u00a8x\u00e2\u0178\u00a9 = \u00e2\u02c6\u00abx|\u00cf\u02c6(x)|^2 dx \u00e2\u0178\u00a8x\u00e2\u0178\u00a9 = \u00e2\u02c6\u00abx * A^2e^(-2ax) dx \u00e2\u0178\u00a8x\u00e2\u0178\u00a9 = A^2 * \u00e2\u02c6\u00abxe^(-2ax) dx \u00e2\u0178\u00a8x\u00e2\u0178\u00a9 = A^2 * (-1\/2a) * xe^(-2ax) + A^2 * (1\/4a^2) * e^(-2ax) + D \u00e2\u0178\u00a8x\u00e2\u0178\u00a9 = A^2 * (-1\/2a) * xe^(-2ax) + A^2 * (1\/4a^2) * e^(-2ax) + D<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s go through the problem step by step.<\/p>\n\n\n\n

Part A: Normalizing the Wave Function<\/h3>\n\n\n\n

The wave function is given by:\u03c8(x)=Ae\u2212ax\\psi(x) = A e^{-ax}\u03c8(x)=Ae\u2212ax<\/p>\n\n\n\n

To normalize the wave function, we need to apply the normalization condition, which requires that the integral of \u2223\u03c8(x)\u22232|\\psi(x)|^2\u2223\u03c8(x)\u22232 over all space equals 1. This ensures that the total probability is 1.<\/p>\n\n\n\n

The integral of \u2223\u03c8(x)\u22232|\\psi(x)|^2\u2223\u03c8(x)\u22232 is:\u222b\u2212\u221e\u221e\u2223\u03c8(x)\u22232dx=1\\int_{-\\infty}^{\\infty} |\\psi(x)|^2 dx = 1\u222b\u2212\u221e\u221e\u200b\u2223\u03c8(x)\u22232dx=1<\/p>\n\n\n\n

First, let’s compute \u2223\u03c8(x)\u22232|\\psi(x)|^2\u2223\u03c8(x)\u22232:\u2223\u03c8(x)\u22232=\u2223Ae\u2212ax\u22232=A2e\u22122ax|\\psi(x)|^2 = |A e^{-ax}|^2 = A^2 e^{-2ax}\u2223\u03c8(x)\u22232=\u2223Ae\u2212ax\u22232=A2e\u22122ax<\/p>\n\n\n\n

Now, we integrate:\u222b0\u221eA2e\u22122axdx=1\\int_{0}^{\\infty} A^2 e^{-2ax} dx = 1\u222b0\u221e\u200bA2e\u22122axdx=1<\/p>\n\n\n\n

The integral of e\u22122axe^{-2ax}e\u22122ax is:\u222be\u22122axdx=\u221212ae\u22122ax\\int e^{-2ax} dx = \\frac{-1}{2a} e^{-2ax}\u222be\u22122axdx=2a\u22121\u200be\u22122ax<\/p>\n\n\n\n

Now substitute this into the integral:A2\u222b0\u221ee\u22122axdx=A2[\u221212ae\u22122ax]0\u221eA^2 \\int_0^\\infty e^{-2ax} dx = A^2 \\left[\\frac{-1}{2a} e^{-2ax}\\right]_{0}^{\\infty}A2\u222b0\u221e\u200be\u22122axdx=A2[2a\u22121\u200be\u22122ax]0\u221e\u200b<\/p>\n\n\n\n

At x=\u221ex = \\inftyx=\u221e, e\u22122axe^{-2ax}e\u22122ax goes to zero, and at x=0x = 0x=0, e\u22122ax=1e^{-2ax} = 1e\u22122ax=1. So, the integral becomes:A2(0\u2212(\u221212a))=A212aA^2 \\left( 0 – \\left( \\frac{-1}{2a} \\right) \\right) = A^2 \\frac{1}{2a}A2(0\u2212(2a\u22121\u200b))=A22a1\u200b<\/p>\n\n\n\n

For normalization, this must equal 1:A212a=1A^2 \\frac{1}{2a} = 1A22a1\u200b=1<\/p>\n\n\n\n

Solving for AAA:A2=2aA^2 = 2aA2=2aA=2aA = \\sqrt{2a}A=2a\u200b<\/p>\n\n\n\n

So, the normalization constant is:A=2aA = \\sqrt{2a}A=2a\u200b<\/p>\n\n\n\n

Part B: Expectation Value of xxx<\/h3>\n\n\n\n

The expectation value of xxx, denoted \u27e8x\u27e9\\langle x \\rangle\u27e8x\u27e9, is given by the formula:\u27e8x\u27e9=\u222b0\u221ex\u2223\u03c8(x)\u22232dx\\langle x \\rangle = \\int_{0}^{\\infty} x |\\psi(x)|^2 dx\u27e8x\u27e9=\u222b0\u221e\u200bx\u2223\u03c8(x)\u22232dx<\/p>\n\n\n\n

Substitute \u2223\u03c8(x)\u22232=A2e\u22122ax|\\psi(x)|^2 = A^2 e^{-2ax}\u2223\u03c8(x)\u22232=A2e\u22122ax:\u27e8x\u27e9=\u222b0\u221exA2e\u22122axdx\\langle x \\rangle = \\int_{0}^{\\infty} x A^2 e^{-2ax} dx\u27e8x\u27e9=\u222b0\u221e\u200bxA2e\u22122axdx<\/p>\n\n\n\n

Now, we need to integrate xe\u22122axx e^{-2ax}xe\u22122ax. We can solve this using integration by parts. Let:<\/p>\n\n\n\n

    \n
  • u=xu = xu=x, so du=dxdu = dxdu=dx<\/li>\n\n\n\n
  • dv=e\u22122axdxdv = e^{-2ax} dxdv=e\u22122axdx, so v=\u221212ae\u22122axv = \\frac{-1}{2a} e^{-2ax}v=2a\u22121\u200be\u22122ax<\/li>\n<\/ul>\n\n\n\n

    Using integration by parts:\u222bxe\u22122axdx=\u2212x2ae\u22122ax+\u222b12ae\u22122axdx\\int x e^{-2ax} dx = \\frac{-x}{2a} e^{-2ax} + \\int \\frac{1}{2a} e^{-2ax} dx\u222bxe\u22122axdx=2a\u2212x\u200be\u22122ax+\u222b2a1\u200be\u22122axdx<\/p>\n\n\n\n

    The second integral is:\u222be\u22122axdx=\u221212ae\u22122ax\\int e^{-2ax} dx = \\frac{-1}{2a} e^{-2ax}\u222be\u22122axdx=2a\u22121\u200be\u22122ax<\/p>\n\n\n\n

    So, putting everything together:\u222bxe\u22122axdx=\u2212x2ae\u22122ax+14a2e\u22122ax\\int x e^{-2ax} dx = \\frac{-x}{2a} e^{-2ax} + \\frac{1}{4a^2} e^{-2ax}\u222bxe\u22122axdx=2a\u2212x\u200be\u22122ax+4a21\u200be\u22122ax<\/p>\n\n\n\n

    Evaluating this at x=0x = 0x=0 and x=\u221ex = \\inftyx=\u221e, we find:[\u2212x2ae\u22122ax+14a2e\u22122ax]0\u221e\\left[\\frac{-x}{2a} e^{-2ax} + \\frac{1}{4a^2} e^{-2ax}\\right]_0^\\infty[2a\u2212x\u200be\u22122ax+4a21\u200be\u22122ax]0\u221e\u200b<\/p>\n\n\n\n

    At x=\u221ex = \\inftyx=\u221e, the terms go to 0, and at x=0x = 0x=0, they also vanish, except for the second term:14a2\\frac{1}{4a^2}4a21\u200b<\/p>\n\n\n\n

    Thus, the expectation value becomes:\u27e8x\u27e9=A2\u22c514a2\\langle x \\rangle = A^2 \\cdot \\frac{1}{4a^2}\u27e8x\u27e9=A2\u22c54a21\u200b<\/p>\n\n\n\n

    Now, substitute A2=2aA^2 = 2aA2=2a into this:\u27e8x\u27e9=(2a)\u22c514a2\\langle x \\rangle = (2a) \\cdot \\frac{1}{4a^2}\u27e8x\u27e9=(2a)\u22c54a21\u200b<\/p>\n\n\n\n

    Simplifying:\u27e8x\u27e9=12a\\langle x \\rangle = \\frac{1}{2a}\u27e8x\u27e9=2a1\u200b<\/p>\n\n\n\n

    So, the expectation value of xxx is:\u27e8x\u27e9=12a\\langle x \\rangle = \\frac{1}{2a}\u27e8x\u27e9=2a1\u200b<\/p>\n\n\n\n

    Final Answers:<\/h3>\n\n\n\n
      \n
    • The normalization constant A=2aA = \\sqrt{2a}A=2a\u200b.<\/li>\n\n\n\n
    • The expectation value of xxx is \u27e8x\u27e9=12a\\langle x \\rangle = \\frac{1}{2a}\u27e8x\u27e9=2a1\u200b.<\/li>\n<\/ul>\n\n\n\n
      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Consider a particle whose wave function is given by \u00cf\u02c6(x) = Ae^(-ax) (A) What is the value of A if this wave function is normalized? (B) What is the expectation value of x for this particle? Let’s apply the normalization condition: \u00e2\u02c6\u00ab|\u00cf\u02c6(x)|^2 dx = \u00e2\u02c6\u00ab|Ae^(-ax)|^2 dx = \u00e2\u02c6\u00abA^2e^(-2ax) dx \u00e2\u02c6\u00ab|\u00cf\u02c6(x)|^2 dx = A^2 \u00e2\u02c6\u00abe^(-2ax) dx […]<\/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-258710","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258710","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=258710"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258710\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258710"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258710"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258710"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}