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--- classes:2009:fall:phys4101.001:q_a_1113 [2009/11/12 23:51] – fitch
+++ classes:2009:fall:phys4101.001:q_a_1113 [2009/11/30 09:10] (current) – x500_bast0052
@@ Line 1: / Line 1: @@
-===== Nov 13 (Fri)  =====
+===== Nov 13 (Fri)  Quiz 3 up to Hydrogen  =====
 **Return to Q&A main page: [[Q_A]]**\\
 **Q&A for the previous lecture: [[Q_A_1111]]**\\
@@ Line 11: / Line 11: @@
 ====Andromeda 11/11/09 8:41====
 Is <math> C_0 </math> in equation 4.28 equal to 2?
-====Ralph 11/12/09 11:06 pm====
+===Ralph 11/12/09 11:06 pm===
-I think that's the normalization constant (assuming you mean equation 4.82 rather than 4.28?) so in order to find it, you would integrate <math>\int_0^{\infty}|R_{20}(r)|^2 r^2 dr = 1</math> and solve.  <math> c_0 = -\frac{2 a^4 e^{\frac{a r}{2}}}{-2+2 a^2-a r}</math>
+I think that's the normalization constant (assuming you mean equation 4.82 rather than 4.28?) so in order to find it, you would integrate <math>\int_0^{\infty}|R_{20}(r)|^2 r^2 dr = 1</math> and solve.  I did this with a computer algebra system and got <math> \text{c0}\to \frac{a^4}{-1+a^2}</math>, but I'm not sure if this is right because it seems like <math>a</math> should be able to be 1...
 ====Yuichi ====
@@ Line 43: / Line 43: @@
 ===Esquire 11/12 8:38pm===
 Oy! I also found the Bohr radius. Mate.
+===Devlin===
+I got the Bohr radius as well.
 ===Super Hot Guy 11/12/09 12:16===
@@ Line 49: / Line 52: @@
 ===Dark Helmet 11/03===
 Half an angstrom here as well
+===liux0756 11/13===
+Do we need to calculate the value of the quantity or we can just express it using <math>\hbar, e, m_e, \eps_0</math>?
+=== Spherical Chicken ===
+It seems fair to me that we'd be able to leave it in constants form.  The rest of the test is pretty calculator independent... so I wouldn't imagine that if we had the constants correct we'd be penalized for not multiplying the actual values.
 ====Andromeda 11/12 2:10====
@@ Line 75: / Line 85: @@
 === joh04684 11/12/09 ===
 Yeah, that's what I ended up getting for my energies too
+=== Yuichi ===
+Those equations actually lead to the same relation between "x" and "y."  That's why you can take one or the other equation to get the right relation between them.  If they are NOT redundant, then the only answer will be x=y=0.  To avoid this result, we required that the determinant <math>det(H-\lambda I)=0</math>.  Since the equation to obtain the eigenvector, <math>v=\begin{bmatrix}x\\ y \end{bmatrix}</math> is given by <math>(H-\lambda I)v=0</math>, if the determinant is NOT zero, the matrix, <math>det(H-\lambda I)=0</math> has an inverse matrix, M.  By applying M from the left, <math>M(H-\lambda I)v=M*0</math>.  Since the two matrices on the LHS are inverse of each other, their product is an identity matrix, I.  The RHS is zero.  Then you will get <math>v=0</math>.
 ===Hydra 9pm===
 I got the same for my energies too, but confused as to how to determine the eigenstates.
 Has anybody worked on #4 yet?
- ===Andromeda 11/12 9:54===
+===Andromeda 11/12 9:54===
 I worked it out twice and got different answers each time but i am more confident on my second answer which is 33a^2
@@ Line 87: / Line 100: @@
 I believe you are right, Daniel.  It wouldn't make sense to do an integral over only one of the variables.
+===liux0756 11/13===
+I got my answer which is 12a^2, but I am not confident~~
 ====Esquire 11/12/09 AD (Information Age) 8:15pm====
@@ Line 94: / Line 110: @@
 ===Spherical Harmonic Star date 63241.5 ===
-er.. chicken…. I believe that in fact the operator has to be square – the rules of matrix multiplication say that a 1×4 * 4×4 give 1×4. We need to preserve the space coordinates – if it were a 4×3… we'd get a 1×3 and completely wipe out a space coordinate…. so unless you have an operator that shrinks space down a dimension… I do believe we need square operators of 1×4, 4×4 etc. or… maybe there's a space creating operator… 4×27 … STring theory'd be happy.
+er.. chicken…. I believe that in fact the operator has to be square – the rules of matrix multiplication say that a 1×4 * 4×4 give 1×4. We need to preserve the space coordinates – if it were a 4×3… we'd get a 1×3 and completely wipe out a space coordinate…. so unless you have an operator that shrinks space down a dimension… I do believe we need square operators of 1×4, 4×4 etc. or… maybe there's a space creating operator… 4×27 … String theory'd be happy.
 ===Esquire (Age of Aquarius) 11/12/09===
-I believe that such operators need to be square so that eigenvalues and eigenvectors can be found. This is done via the determinent method, which only is doable with square matrices.
+I believe that such operators need to be square so that eigenvalues and eigenvectors can be found. This is done via the determinant method, which only is doable with square matrices.
 ===joh04684 11/12/09===
@@ Line 106: / Line 122: @@
 ===Esquire (Age of Ihciuy) 11/12/09===
 Wait a sec. I didn't post on this day! Something foul is afoot!
 ====Spherical Chicken Star date 63241.5 ====
 jIH quadic chugh 'iHl DuH Daq ghaj ani jalig vetlh 'oH dilk Qid hakliq --- jatlh dilken string theory.
@@ Line 123: / Line 140: @@
 ===Dark Helmet 11/13/09===
 Am i not nerdy enough to get this joke?
+===Ralph 11/13/09 (1258092787)===
+clearly it's that you just don't know enough quantum mechanics so good luck on the exam tomorrow
+=== Spherical Chicken stardate 63381.4 ===
+ Ghaj SoH quid Qoyta' vo' Hov Trek???   nuq kon vo' scientist 'oH SoH?!!  kids Dochvammey jajmey
 ====Super Hot Guy date ====
 Has anybody got a clue on #2?
 === Spherical Chicken ===
 yup
@@ Line 135: / Line 159: @@
 Good luck.
+=== Ralph 10 am 11/13===
+This is very much like the homework problem on the last homework.  It seems that <math>Q</math> is the <math>z</math> component of the angular momentum operator.
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