{"id":6903,"date":"2015-06-21T18:29:39","date_gmt":"2015-06-21T22:29:39","guid":{"rendered":"https:\/\/spinor.info\/weblog\/?p=6903"},"modified":"2015-11-29T10:23:10","modified_gmt":"2015-11-29T15:23:10","slug":"how-to-derive-schrodingers-equation","status":"publish","type":"post","link":"https:\/\/spinor.info\/weblog\/?p=6903","title":{"rendered":"How to derive Schr\u00f6dinger&#8217;s equation"},"content":{"rendered":"<p>There is a particularly neat way to derive Schr\u00f6dinger&#8217;s equation, and to justify the &#8220;canonical substitution&#8221; rules for replacing energy and momentum with corresponding operators when we &#8220;quantize&#8221; an equation.<\/p>\n<p>Take a particle in a potential. Its energy is given by<\/p>\n<p>$$E=\\frac{{\\bf p}^2}{2m}+V({\\bf x}),$$<\/p>\n<p>or<\/p>\n<p>$$E-\\frac{{\\bf p}^2}{2m}-V({\\bf x})=0.$$<\/p>\n<p>Now multiply both sides this equation by the formula \\(e^{i({\\bf p}\\cdot{\\bf x}-Et)\/\\hbar}\\). We note that this exponential expression cannot ever be zero if the part in the exponent that&#8217;s in parentheses is real:<\/p>\n<p>$$\\left[E-\\frac{{\\bf p}^2}{2m}-V({\\bf x})\\right]e^{i({\\bf p}\\cdot{\\bf x}-Et)\/\\hbar}=0.$$<\/p>\n<p>So far so good. But now note that<\/p>\n<p>$$Ee^{i({\\bf p}\\cdot{\\bf x}-Et)\/\\hbar}=i\\hbar\\frac{\\partial}{\\partial t}e^{i({\\bf p}\\cdot{\\bf x}-Et)\/\\hbar},$$<\/p>\n<p>and similarly,<\/p>\n<p>$${\\bf p}^2e^{i({\\bf p}\\cdot{\\bf x}-Et)\/\\hbar}=-\\hbar^2{\\boldsymbol\\nabla}e^{i({\\bf p}\\cdot{\\bf x}-Et)\/\\hbar}.$$<\/p>\n<p>This allows us to rewrite the previous equation as<\/p>\n<p>$$\\left[i\\hbar\\frac{\\partial}{\\partial t}+\\hbar^2\\frac{{\\boldsymbol\\nabla}^2}{2m}-V({\\bf x})\\right]e^{i({\\bf p}\\cdot{\\bf x}-Et)\/\\hbar}=0.$$<\/p>\n<p>Or, writing \\(\\Psi=e^{i({\\bf p}\\cdot{\\bf x}-Et)\/\\hbar}\\) and rearranging:<\/p>\n<p>$$i\\hbar\\frac{\\partial}{\\partial t}\\Psi=-\\hbar^2\\frac{{\\boldsymbol\\nabla}^2}{2m}\\Psi+V({\\bf x})\\Psi,$$<\/p>\n<p>which is the good old Schr\u00f6dinger equation.<\/p>\n<p>The method works for an arbitrary, generic Hamiltonian, too. Given<\/p>\n<p>$$H({\\bf p})=E,$$<\/p>\n<p>we can write<\/p>\n<p>$$\\left[E-H({\\bf p})\\right]e^{i({\\bf p}\\cdot{\\bf x}-Et)\/\\hbar}=0,$$<\/p>\n<p>which is equivalent to<\/p>\n<p>$$\\left[i\\hbar\\frac{\\partial}{\\partial t}-H(-i\\hbar{\\boldsymbol\\nabla})\\right]\\Psi=0.$$<\/p>\n<p>So if this equation is identically satisfied for a classical system with Hamiltonian \\(H\\), what&#8217;s the big deal about quantum mechanics? Well&#8230; a classical system satisfies \\(E-H({\\bf p})=0\\), where \\(E\\) and \\({\\bf p}\\) are eigenvalues of the differential operators \\(i\\hbar\\partial\/\\partial t\\) and \\(-i\\hbar{\\boldsymbol\\nabla}\\), respectively. Schr\u00f6dinger&#8217;s equation, on the other hand, remains valid\u00a0in the general case, not just for the eigenvalues.<\/p>\n<fb:like href='https:\/\/spinor.info\/weblog\/?p=6903' send='true' layout='standard' show_faces='true' width='450' height='65' action='like' colorscheme='light' font='lucida grande'><\/fb:like>","protected":false},"excerpt":{"rendered":"<p>There is a particularly neat way to derive Schr\u00f6dinger&#8217;s equation, and to justify the &#8220;canonical substitution&#8221; rules for replacing energy and momentum with corresponding operators when we &#8220;quantize&#8221; an equation. Take a particle in a potential. Its energy is given by $$E=\\frac{{\\bf p}^2}{2m}+V({\\bf x}),$$ or $$E-\\frac{{\\bf p}^2}{2m}-V({\\bf x})=0.$$ Now multiply both sides this equation by <a href='https:\/\/spinor.info\/weblog\/?p=6903' class='excerpt-more'>[&#8230;]<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"class_list":["post-6903","post","type-post","status-publish","format-standard","hentry","category-physics","category-3-id","post-seq-1","post-parity-odd","meta-position-corners","fix"],"_links":{"self":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts\/6903","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=6903"}],"version-history":[{"count":17,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts\/6903\/revisions"}],"predecessor-version":[{"id":6920,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=\/wp\/v2\/posts\/6903\/revisions\/6920"}],"wp:attachment":[{"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6903"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6903"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/spinor.info\/weblog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6903"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}