Modern 2:Exam 1 Overview

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Latest revision as of 21:44, 13 March 2006

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I will drop your lowest exam grade. Second hourly exam is next Monday. Recall, the hourlies make up 40% of the grade. Course requirements

concerned $1.9 = 2$ and $π = 3$ so $1.9π = 6$

lots of people messed up exponents when doing division. $109 / 10 - 9 = 10 18$
break up complicated numerical factors involving multiplications and divisions into small pieces. First do the numerator, then do the denominator and then get your final estimate.
lots of people messed up: $| e I k x | 2$ . This is 1!
some people used $e^{I k_0 x}$ in the wavefunction and some used $e^{-Ik_0 x}$ .

I didn't take off for this. But it does change the sign of the momentum.

of the distribution is $x 0$ . You can show this by integration, but it should be sufficiently well known that you merely state it as a fact.

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+{{Start Hierarchy|link=Course Wikis|title=Course Wikis}}
+{{Hierarchy Item|link=Physics Course Wikis|title=Physics Course Wikis}}
+{{Hierarchy Item|link=Modern 2|title=Modern 2}}
+{{End Hierarchy}}
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+I will drop your lowest exam grade.  Second hourly exam is next Monday.  Recall,
+the hourlies make up 40% of the grade.
+[[Course requirements]]
+=== tips and tricks ===
+* I don't expect exactitude in  numerical calculations.  So as far as I'm
+concerned <math>1.9 = 2 </math> and <math>\pi = 3 </math>  so  <math>1.9  \pi = 6</math>
+* lots of people messed up exponents when doing division.  <math>10^9 / 10^{-9} = 10^{18} </math>
+* break up complicated numerical factors involving multiplications and divisions into small pieces.  First do the numerator, then do the denominator and then get your final estimate.
+* lots of people messed up: <math> |e^{I k x}|^2 </math>.  This is 1!
+* some people used <math> e^{I k_0 x} </math> in the wavefunction and some used <math> e^{-Ik_0 x} </math>.
+I didn't take off for this.  But it does change the sign of the momentum.
+* A Gaussian <math>e^{(x-x_0)^2/2\sigma^2}</math> is centered at <math>x_0</math>.  I.e., the mean
+of the distribution is <math>x_0</math>.  You '''can'''
+show this by integration, but it should be sufficiently well known that you merely state it as
+a fact.
+* About 95% of you plotted <math>|\psi|^2</math> and not <math> Re(\psi)</math>.