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--- classes:2009:fall:phys4101.001:lec_notes_0930 [2009/09/30 16:47] – prestegard
+++ classes:2009:fall:phys4101.001:lec_notes_0930 [2009/10/02 00:07] (current) – yk
@@ Line 5: / Line 5: @@
 **previous lecture note: [[lec_notes_0928]]**\\
 **next lecture note: [[lec_notes_1005]]**\\
+**Important concepts for quiz 1: [[quiz_1_1002]]**\\
 **Main class wiki page: [[home]]**
-Please try to include the following
+//This note complemented my actual lecture with proper factors so that "~" becomes "=", and some parts that perhaps I was not clear (from your faces) were more clearly and completely explained.  Thank you for you the wonderful job.  Yuichi//
-  * main points understood, and expand them - what is your understanding of what the points were.
-    * expand these points by including many of the details the class discussed.
-  * main points which are not clear.  - describe what you have understood and what the remain questions surrounding the point(s).
-    * Other classmates can step in and clarify the points, and expand them.
-  * How the main points fit with the big picture of QM.  Or what is not clear about how today's points fit in in a big picture.
-  * wonderful tricks which were used in the lecture.
 === Main Points ===
 == Quiz Questions ==
-The only question asked about the quiz was regarding how to determine the expectation values of the momentum and position for the simple harmonic oscillator by using the ladder operators.  It is possible to write the position and momentum operators in terms of the ladder operators:
+It was asked whether knowledge of the analytical solution to the harmonic oscillator potential would be needed for the test.  It is possible that it will, but not likely due to time constraints.  It would probably still be good to have some idea of how it works, even if you can't reproduce it entirely.
+The only other question asked about the quiz was regarding how to determine the expectation values of the momentum and position for the simple harmonic oscillator by using the ladder operators.  It is possible to write the position and momentum operators in terms of the ladder operators:
 <p> = <math>i \sqrt{\frac{mw\hbar}{2}}(a_+ - a_-)</math>
@@ Line 42: / Line 38: @@
 Yuichi suggested that we try Problem 2.20 in Griffiths to gain a better understanding.
-== to be finished later tonight ==
+==Group Velocity vs. Phase Velocity==
+The book defines the group velocity as <math>v_g=\frac{d\omega}{dk}</math>, but what does that mean?  What is the group velocity?  What makes it different from the phase velocity?
+The real part of the wavefunction of a wave packet looks like a sinusoidal function bound within an envelope.  The wavefunction can move faster or slower than the envelope.  The speed of the envelope is the group velocity, and the speed of the waves inside the envelope is the phase velocity.
+The group velocity represents the velocity that we would measure, <math>\frac{d<x>}{dt}</math>.  One may wonder how this can be if <x> is at a place where the wavefunction plotted is 0; the answer is that the plotted part of the wavefunction does not include the imaginary component. The envelope represents the magnitude of the wavefunction (the real and imaginary parts added in quadrature, <math>\|\Psi\|=\sqrt{\Psi^{*}\Psi}=\sqrt{\(Re\(\Psi\)\)^2 + \(Im\(\Psi\)\)^2}</math>).
+So why does <math>v_g = \frac{d\omega}{dt}</math>?  Why would it be different in different systems?  For a quantum mechanical particle, <math>\omega \propto E\propto \frac{p^2}{2m}\propto k^2</math>.  For light, <math>\omega \propto k</math>.  For water waves, <math>\omega \propto \sqrt{k}</math>.
+Suppose that the wave packet has waves corresponding to <math>e^{ikx}</math> and <math>e^{i(k+\Delta k)x}</math>.
+<math>\Psi (x, 0)=e^{ikx}+e^{i(k+\Delta k)x} = e^{i(\frac{k+k+\Delta k}{2})x}\(e^{-i\frac{\Delta k}{2} x}+e^{i\frac{\Delta k}{2} x}\) \\ = \[cos( \frac{(2k+\Delta k)}{2}x ) +i sin(\frac{(2k+\Delta k)}{2}x)\]\(2cos(\frac{\Delta k}{2} x)\)</math>
+The second cosine in the last expression is the envelope, while the first term represents the phase inside the envelope.
+We can also see that <math>\Psi (x, 0) = e^{ikx}+e^{i(k+\Delta k)x}=\(cos(kx)+cos((k+\Delta k)x)\)+i\(sin(kx) + sin((k+\Delta k)x)\)</math>.  <math>\|\Psi\|</math> will then be at a maximum when the two waves are in phase, <math>kx_g=(k+\Delta k)x_g</math>, where <math>x_g</math> is the x corresponding to a maximum in the envelope.  Taking time into consideration gives
+<math>kx_g - \omega t = (k+\Delta k)x_g - (\omega + \Delta \omega)t</math>
+At <math>t=0</math>, this gives <math>x_g = 0</math>, and at <math>t=\Delta t</math>, it gives <math>x_g=\frac{\Delta \omega}{\Delta k}\Delta t</math>.  Then, we can see that
+<math>v_g\approx \frac{\Delta x_g}{\Delta t}\approx \frac{d\omega}{dk}</math>
+with the approximations becoming exact in the limit of small <math>\Delta</math>.
@@ Line 49: / Line 68: @@
 **To go back to the lecture note list, click [[lec_notes]]**\\
 **previous lecture note: [[lec_notes_0928]]**\\
-**next lecture note: [[lec_notes_1002]]**\\
+**next lecture note: [[lec_notes_1005]]**\\
+**Important concepts for quiz 1: [[quiz_1_1002]]**\\

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