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= Past news of persistent value = == Past news of persistent value ==
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  * I guess I will see many of you tomorrow during my office hours ('''noon-2PM'''). In addition, please feel free to ask your questions on the forum or by email. '''Good (= sincere) discussions will really protect you in my course''', since (1) you are a learner&mdash;I/you/anybody should never expect you be a knower already (although you could be close or already there)&mdash;and (2) your willingness to learn will make people (esp., me) respect you very much.&mdash;~-''<<DateTime(2013-10-20T20:10:29-0700)>>''-~

  * Office hours: Monday <<color(12-2 PM)>>, Thursday, Friday, 1-2 PM, or OBA. (Syllabus updated.)&mdash;~-''<<DateTime(2013-10-13T19:34:38-0700)>>''-~
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= Past news of purely archival value = == Past news of purely archival value ==

Past news of persistent value

  • I guess I will see many of you tomorrow during my office hours (noon-2PM). In addition, please feel free to ask your questions on the forum or by email. Good (= sincere) discussions will really protect you in my course, since (1) you are a learner—I/you/anybody should never expect you be a knower already (although you could be close or already there)—and (2) your willingness to learn will make people (esp., me) respect you very much.—8:10PM, Oct 20, 2013

  • Office hours: Monday 12-2 PM, Thursday, Friday, 1-2 PM, or OBA. (Syllabus updated.)—7:34PM, Oct 13, 2013

  • There are some discussions of homework problems in the forum. Check them out!—10:46AM, Oct 05, 2013

  • Welcome back, students!

Past news of purely archival value

  • For problem 4 of homework 2, it would be sufficient to prove that $\langle n^{(0)} | {\underline n}^{(j+1)} \rangle = 0$ for any $j = 0, 1, 2, ...$. The ket $| {\underline n}^{(j+1)} \rangle$ is defined in this image. In the same image, you can find other information. For example, four completely equivalent statements, important for this problem, are noted in the green rectangle. The equivalence of these four statements can be proven based on Eqs. 3.19 and 3.20, alone. This is kind of demonstrated in this image, and you are not required to demonstrate it yourself. Just prove that $\langle n^{(0)} | {\underline n}^{(j+1)} \rangle = 0$ for any $j = 0, 1, 2, ...$ (using the proof by induction).3:35PM, Oct 14, 2013