{"id":180933,"date":"2025-01-09T08:26:17","date_gmt":"2025-01-09T08:26:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=180933"},"modified":"2025-01-09T08:26:20","modified_gmt":"2025-01-09T08:26:20","slug":"consider-the-hermiticity-of-the-following-operators","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/09\/consider-the-hermiticity-of-the-following-operators\/","title":{"rendered":"Consider the Hermiticity of the following operators"},"content":{"rendered":"\n

Consider the Hermiticity of the following operators: (a) Is the operator<\/p>\n\n\n\n

dx<\/p>\n\n\n\n

d<\/p>\n\n\n\n

?<\/p>\n\n\n\n

a Hermitian operator? Prove your answer (b) Is the operator<\/p>\n\n\n\n

dx<\/p>\n\n\n\n

2<\/p>\n\n\n\n

d<\/p>\n\n\n\n

2<\/p>\n\n\n\n

?<\/p>\n\n\n\n

a Hermitian operator? Prove your answer. (c) Is the Hamiltonian operator<\/p>\n\n\n\n

H<\/p>\n\n\n\n

=-<\/p>\n\n\n\n

2m<\/p>\n\n\n\n

?<\/p>\n\n\n\n

2<\/p>\n\n\n\n

?<\/p>\n\n\n\n

dx<\/p>\n\n\n\n

2<\/p>\n\n\n\n

d<\/p>\n\n\n\n

2<\/p>\n\n\n\n

?<\/p>\n\n\n\n

+V(x) a Hermitian operator if V(x) is a real-valued function? Prove your answer . Hints: For (a) and (b), you may want to use integration by parts, and you may presume that any well-behaved function ?(x) and its derivative<\/p>\n\n\n\n

dx<\/p>\n\n\n\n

d?(x)<\/p>\n\n\n\n

?<\/p>\n\n\n\n

vanish at \u00b18.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

(a) Is the operator ddx\\frac{d}{dx} a Hermitian operator?<\/h3>\n\n\n\n

To determine if ddx\\frac{d}{dx} is Hermitian, we must check if the following condition holds for all well-behaved wave functions \u03c8(x)\\psi(x) and \u03d5(x)\\phi(x): \u222b\u2212\u221e\u221e\u03c8\u2217(x)d\u03d5(x)dx\u2009dx=\u222b\u2212\u221e\u221e(d\u03c8(x)dx)\u2217\u03d5(x)\u2009dx\\int_{-\\infty}^{\\infty} \\psi^*(x) \\frac{d\\phi(x)}{dx} \\, dx = \\int_{-\\infty}^{\\infty} \\left( \\frac{d\\psi(x)}{dx} \\right)^* \\phi(x) \\, dx<\/p>\n\n\n\n

We perform an integration by parts: \u222b\u2212\u221e\u221e\u03c8\u2217(x)d\u03d5(x)dx\u2009dx=[\u03c8\u2217(x)\u03d5(x)]\u2212\u221e\u221e\u2212\u222b\u2212\u221e\u221ed\u03c8\u2217(x)dx\u03d5(x)\u2009dx\\int_{-\\infty}^{\\infty} \\psi^*(x) \\frac{d\\phi(x)}{dx} \\, dx = \\left[ \\psi^*(x) \\phi(x) \\right]_{-\\infty}^{\\infty} – \\int_{-\\infty}^{\\infty} \\frac{d\\psi^*(x)}{dx} \\phi(x) \\, dx<\/p>\n\n\n\n

Given that \u03c8(x)\\psi(x) and \u03d5(x)\\phi(x) vanish at infinity, the boundary term [\u03c8\u2217(x)\u03d5(x)]\u2212\u221e\u221e=0\\left[ \\psi^*(x) \\phi(x) \\right]_{-\\infty}^{\\infty} = 0. Thus, we are left with: \u222b\u2212\u221e\u221e\u03c8\u2217(x)d\u03d5(x)dx\u2009dx=\u2212\u222b\u2212\u221e\u221ed\u03c8\u2217(x)dx\u03d5(x)\u2009dx\\int_{-\\infty}^{\\infty} \\psi^*(x) \\frac{d\\phi(x)}{dx} \\, dx = – \\int_{-\\infty}^{\\infty} \\frac{d\\psi^*(x)}{dx} \\phi(x) \\, dx<\/p>\n\n\n\n

Now, compare this result with the right-hand side of the Hermiticity condition: \u222b\u2212\u221e\u221e(d\u03c8(x)dx)\u2217\u03d5(x)\u2009dx\\int_{-\\infty}^{\\infty} \\left( \\frac{d\\psi(x)}{dx} \\right)^* \\phi(x) \\, dx<\/p>\n\n\n\n

The two expressions are not identical because of the negative sign. Thus, ddx\\frac{d}{dx} is not<\/strong> a Hermitian operator.<\/p>\n\n\n\n

(b) Is the operator d2dx2\\frac{d^2}{dx^2} a Hermitian operator?<\/h3>\n\n\n\n

Now, let’s check if d2dx2\\frac{d^2}{dx^2} is Hermitian. We need to verify the following: \u222b\u2212\u221e\u221e\u03c8\u2217(x)d2\u03d5(x)dx2\u2009dx=\u222b\u2212\u221e\u221e(d2\u03c8(x)dx2)\u2217\u03d5(x)\u2009dx\\int_{-\\infty}^{\\infty} \\psi^*(x) \\frac{d^2\\phi(x)}{dx^2} \\, dx = \\int_{-\\infty}^{\\infty} \\left( \\frac{d^2\\psi(x)}{dx^2} \\right)^* \\phi(x) \\, dx<\/p>\n\n\n\n

Performing integration by parts twice:<\/p>\n\n\n\n

    \n
  1. First integration by parts: \u222b\u2212\u221e\u221e\u03c8\u2217(x)d2\u03d5(x)dx2\u2009dx=[\u03c8\u2217(x)d\u03d5(x)dx]\u2212\u221e\u221e\u2212\u222b\u2212\u221e\u221ed\u03c8\u2217(x)dxd\u03d5(x)dx\u2009dx\\int_{-\\infty}^{\\infty} \\psi^*(x) \\frac{d^2\\phi(x)}{dx^2} \\, dx = \\left[ \\psi^*(x) \\frac{d\\phi(x)}{dx} \\right]_{-\\infty}^{\\infty} – \\int_{-\\infty}^{\\infty} \\frac{d\\psi^*(x)}{dx} \\frac{d\\phi(x)}{dx} \\, dx<\/li>\n\n\n\n
  2. The boundary term vanishes, and we are left with: \u2212\u222b\u2212\u221e\u221ed\u03c8\u2217(x)dxd\u03d5(x)dx\u2009dx- \\int_{-\\infty}^{\\infty} \\frac{d\\psi^*(x)}{dx} \\frac{d\\phi(x)}{dx} \\, dx<\/li>\n\n\n\n
  3. Apply integration by parts again to the remaining integral: \u2212\u222b\u2212\u221e\u221ed\u03c8\u2217(x)dxd\u03d5(x)dx\u2009dx=[d\u03c8\u2217(x)dx\u03d5(x)]\u2212\u221e\u221e\u2212\u222b\u2212\u221e\u221ed2\u03c8\u2217(x)dx2\u03d5(x)\u2009dx- \\int_{-\\infty}^{\\infty} \\frac{d\\psi^*(x)}{dx} \\frac{d\\phi(x)}{dx} \\, dx = \\left[ \\frac{d\\psi^*(x)}{dx} \\phi(x) \\right]_{-\\infty}^{\\infty} – \\int_{-\\infty}^{\\infty} \\frac{d^2\\psi^*(x)}{dx^2} \\phi(x) \\, dx<\/li>\n<\/ol>\n\n\n\n

    Again, the boundary term vanishes, leaving: \u222b\u2212\u221e\u221e\u03c8\u2217(x)d2\u03d5(x)dx2\u2009dx=\u222b\u2212\u221e\u221e(d2\u03c8(x)dx2)\u2217\u03d5(x)\u2009dx\\int_{-\\infty}^{\\infty} \\psi^*(x) \\frac{d^2\\phi(x)}{dx^2} \\, dx = \\int_{-\\infty}^{\\infty} \\left( \\frac{d^2\\psi(x)}{dx^2} \\right)^* \\phi(x) \\, dx<\/p>\n\n\n\n

    Thus, d2dx2\\frac{d^2}{dx^2} is a Hermitian<\/strong> operator.<\/p>\n\n\n\n

    (c) Is the Hamiltonian operator H=\u2212\u210f22md2dx2+V(x)H = -\\frac{\\hbar^2}{2m} \\frac{d^2}{dx^2} + V(x) a Hermitian operator if V(x)V(x) is a real-valued function?<\/h3>\n\n\n\n

    The Hamiltonian operator consists of two parts: the kinetic term \u2212\u210f22md2dx2-\\frac{\\hbar^2}{2m} \\frac{d^2}{dx^2} and the potential term V(x)V(x).<\/p>\n\n\n\n

      \n
    • We already established that the kinetic term \u2212\u210f22md2dx2-\\frac{\\hbar^2}{2m} \\frac{d^2}{dx^2} is Hermitian.<\/li>\n\n\n\n
    • The potential term V(x)V(x) is a real-valued function. To check if it is Hermitian, we need to verify if:<\/li>\n<\/ul>\n\n\n\n

      \u222b\u2212\u221e\u221e\u03c8\u2217(x)V(x)\u03d5(x)\u2009dx=\u222b\u2212\u221e\u221e(\u03c8(x))\u2217V(x)\u03d5(x)\u2009dx\\int_{-\\infty}^{\\infty} \\psi^*(x) V(x) \\phi(x) \\, dx = \\int_{-\\infty}^{\\infty} \\left( \\psi(x) \\right)^* V(x) \\phi(x) \\, dx<\/p>\n\n\n\n

      Since V(x)V(x) is real, this condition holds trivially, as V(x)V(x) does not involve derivatives, and the integrals will be symmetric.<\/p>\n\n\n\n

      Thus, the Hamiltonian operator HH is a Hermitian<\/strong> operator, given that V(x)V(x) is real-valued.<\/p>\n\n\n\n

      Conclusion:<\/h3>\n\n\n\n
        \n
      • ddx\\frac{d}{dx} is not a Hermitian operator.<\/li>\n\n\n\n
      • d2dx2\\frac{d^2}{dx^2} is a Hermitian operator.<\/li>\n\n\n\n
      • The Hamiltonian operator H=\u2212\u210f22md2dx2+V(x)H = -\\frac{\\hbar^2}{2m} \\frac{d^2}{dx^2} + V(x) is a Hermitian operator if V(x)V(x) is a real-valued function.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

        Consider the Hermiticity of the following operators: (a) Is the operator dx d ? a Hermitian operator? Prove your answer (b) Is the operator dx 2 d 2 ? a Hermitian operator? Prove your answer. (c) Is the Hamiltonian operator H =- 2m ? 2 ? dx 2 d 2 ? +V(x) a Hermitian operator […]<\/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-180933","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180933","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=180933"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180933\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=180933"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=180933"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=180933"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}