griffiths d.j. introduction to quantum mechanics 2ed(solutions)

griffiths d.j. introduction to quantum mechanics 2ed(solutions)

ID:31462500

大小:3.73 MB

页数:303页

时间:2019-01-10

griffiths d.j. introduction to quantum mechanics 2ed(solutions)_第1页
griffiths d.j. introduction to quantum mechanics 2ed(solutions)_第2页
griffiths d.j. introduction to quantum mechanics 2ed(solutions)_第3页
griffiths d.j. introduction to quantum mechanics 2ed(solutions)_第4页
griffiths d.j. introduction to quantum mechanics 2ed(solutions)_第5页
资源描述:

《griffiths d.j. introduction to quantum mechanics 2ed(solutions)》由会员上传分享,免费在线阅读,更多相关内容在学术论文-天天文库

1、ContentsPreface21TheWaveFunction32Time-IndependentSchrödingerEquation143Formalism624QuantumMechanicsinThreeDimensions875IdenticalParticles1326Time-IndependentPerturbationTheory1547TheVariationalPrinciple1968TheWKBApproximation2199Time-DependentPerturbation

2、Theory23610TheAdiabaticApproximation25411Scattering26812Afterword282AppendixLinearAlgebra283ndst2Edition–1EditionProblemCorrelationGrid2992PrefaceThesearemyownsolutionstotheproblemsinIntroductiontoQuantumMechanics,2nded.Ihavemadeeveryefforttoinsurethattheya

3、reclearandcorrect,buterrorsareboundtooccur,andforthisIapologizeinadvance.Iwouldliketothankthemanypeoplewhopointedoutmistakesinthesolutionmanualforthefirstedition,andencourageanyonewhofindsdefectsinthisonetoalertme(griffith@reed.edu).I’llmaintainalistoferrataon

4、mywebpage(http://academic.reed.edu/physics/faculty/griffiths.html),andincorporatecorrectionsinthemanualitselffromtimetotime.IalsothankmystudentsatReedandatSmithformanyusefulsuggestions,andaboveallNeelakshSadhoo,whodidmostofthetypesetting.Attheendofthemanualt

5、hereisagridthatcorrelatestheproblemnumbersinthesecondeditionwiththoseinthefirstedition.DavidGriffithsc2005PearsonEducation,Inc.,UpperSaddleRiver,NJ.Allrightsreserved.Thismaterialisprotectedunderallcopyrightlawsastheycurrentlyexist.Noportionofthismaterialmayb

6、ereproduced,inanyformorbyanymeans,withoutpermissioninwritingfromthepublisher.CHAPTER1.THEWAVEFUNCTION3Chapter1TheWaveFunctionProblem1.1(a)j2=212=441.11j2=j2N(j)=(142)+(152)+3(162)+2(222)+2(242)+5(252)N1416434=(196+225+768+968+1152+3125)==459.571.141

7、4j∆j=j−j1414−21=−71515−21=−6(b)1616−21=−52222−21=12424−21=32525−21=411σ2=(∆j)2N(j)=(−7)2+(−6)2+(−5)2·3+(1)2·2+(3)2·2+(4)2·5N141260=(49+36+75+2+18+80)==18.571.1414√σ=18.571=4.309.(c)j2−j2=459.571−441=18.571.[Agreeswith(b).]c2005PearsonEducation,In

8、c.,UpperSaddleRiver,NJ.Allrightsreserved.Thismaterialisprotectedunderallcopyrightlawsastheycurrentlyexist.Noportionofthismaterialmaybereproduced,inanyformorbyanymeans,withoutpermissioninwritingfromthepublishe

当前文档最多预览五页,下载文档查看全文

此文档下载收益归作者所有

当前文档最多预览五页,下载文档查看全文
温馨提示:
1. 部分包含数学公式或PPT动画的文件,查看预览时可能会显示错乱或异常,文件下载后无此问题,请放心下载。
2. 本文档由用户上传,版权归属用户,天天文库负责整理代发布。如果您对本文档版权有争议请及时联系客服。
3. 下载前请仔细阅读文档内容,确认文档内容符合您的需求后进行下载,若出现内容与标题不符可向本站投诉处理。
4. 下载文档时可能由于网络波动等原因无法下载或下载错误,付费完成后未能成功下载的用户请联系客服处理。