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--- classes:2009:fall:phys4101.001:q_a_1209 [2009/12/09 10:54] – x500_sohnx020
+++ classes:2009:fall:phys4101.001:q_a_1209 [2009/12/19 17:23] (current) – x500_sohnx020
@@ Line 20: / Line 20: @@
 Another question about [4.151].
 What if I switch alpha and beta, we would get <math>\chi_{-}^{x}=\begin{pmatrix} -1/sqrt2\\1/sqrt2 \end{pmatrix} </math>.
-Is this the same as 4.151? I mean Kai will be different finally.
+Is this the same as 4.151? I mean Kai will be different finally?
+===Yuichi===
+Your <math>\chi_{-}^{x}</math> times -1 is 4.151.  So the difference is in the normalization factor whose absolute value is the same, so they are can considered the same function.
 ==== Zeno 12/8 11:45am ====
@@ Line 31: / Line 34: @@
 ==== Zeno 12/9 10:15AM ====
 So in discussion we saw that the first order correction term for the harmonic oscillator was the same as the first term when writing the new Hamiltonian and doing a Taylor expansion. The TA noted that this is a particularly special case. My question is: how do we know how accurate the correction terms are and is there a way to predict when a simple Taylor expansion will produce the same results as the more tedious correction term formula?
+===Yuichi===
+If the magnitude of E's is much smaller than unperturbed energies, E's (or more accurately, the differences, E_0-E_1, E_1-E_2, etc.) the chances are the E" are even smaller.  So the series E+E'+E"+... will converge.  Being physicists, we don't usually focus too much on if the series can be proven to converge, but do the calculations and see if the corrections make the difference between the calculation (before the perturbation) and experimental results get smaller.  If it does, we are usually happy.  If the convergence is marginal, probably it is too slow to be producing good results with only the first and second order perturbation corrections.
+====Green Suit 12/9 3:24====
+On page 260 Griffiths writes, "If you're faced with degenerate states, **look around** for some hermitian operator <math>A</math> that commutes with <math>H^o</math> and <math>H'</math>"
+How does one "look around" for such an operator? i.e. problem 6.7.d
+===prest121 12/9 6:15pm===
+I don't know if this is very helpful, but I think the operator is hopefully more obvious in a specific problem.  I haven't worked through 6.7, though.
+===Pluto 4ever 12/9 9:13PM===
+I think it just means that you should try implementing hermitian conjugates that you know or some form of them that would would with the given Hamiltonians. At least that's my take on it.
+===Yuichi===
+This is an elegant way to solve a problems for sure, but I don't think it is an efficient way to find the solution necessarily.   If you come across with such an operator, your life is easy, but if I were you, if this discovery does not happen, I would use more dumb but sure way, which may require finding eigenvalues and eigenvectors.  Once that's done, with the wisdom of hindsight, you may be able to find such an operator.  The right operator for 6.7 is not something you are very familiar with, and many of you may not think of it.
+====Jake22 12/15 4:30====
+What are some suggested alternatives to using 4.178 for describing a two level system (like one of states spin up and spin down) that may allow the two particle state to be expressed without entanglement, as in a composition of one-particle states?
+=== Blackbox 5:30 ===
+.178 shows only one state  with m=0 carries s=0 which is singlet case. I don't think there may exist other good alternatives for this espression.
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