Go to the U of M home page
School of Physics & Astronomy
School of Physics and Astronomy Wiki

User Tools


classes:2009:fall:phys4101.001:q_a_1130

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision
Previous revision
classes:2009:fall:phys4101.001:q_a_1130 [2009/12/01 00:10] cwuclasses:2009:fall:phys4101.001:q_a_1130 [2009/12/16 00:44] (current) czhang
Line 24: Line 24:
 ====Dark Helmet 11/28 === ====Dark Helmet 11/28 ===
 Can we perhaps get another count of Pre-lecture Q/A points? Can we perhaps get another count of Pre-lecture Q/A points?
 +
 +=== The Doctor ===
 +Perhaps...
 +
 ====Schrodinger's Dog 11/29==== ====Schrodinger's Dog 11/29====
 On page 244, Griffiths mentions that he multiplies the degeneracy dk by 2 because of the spin. When we are talking about photons, I usually hear that this factor of 2 comes from the polarization of light, and the fact that it photons can only travel transversely. What is the story? Did we first think it was 2 when we working semi-classical(i.e. before modern QM) and then discovered spin, and accounted this factor of 2 because of spin in modern QM? On page 244, Griffiths mentions that he multiplies the degeneracy dk by 2 because of the spin. When we are talking about photons, I usually hear that this factor of 2 comes from the polarization of light, and the fact that it photons can only travel transversely. What is the story? Did we first think it was 2 when we working semi-classical(i.e. before modern QM) and then discovered spin, and accounted this factor of 2 because of spin in modern QM?
Line 44: Line 48:
 ==== Zeno 11/30 ==== ==== Zeno 11/30 ====
 Can someone explain where the reduced mass concept comes from? I remember using it in classical and astro, but I don't remember why it is useful or where and how it was derived. Can someone explain where the reduced mass concept comes from? I remember using it in classical and astro, but I don't remember why it is useful or where and how it was derived.
 +===chavez 12/01 2:50pm===
 +Read this: http://en.wikipedia.org/wiki/Reduced_mass.
 ===Schrodinger's Dog 11/30=== ===Schrodinger's Dog 11/30===
 It takes into account the nucleus, when looking at the atom, and taking into account the nucleus allows you to see the fine-structure splitting, which is discussed in chapter 6. They simply replace m with mu, and you get a split in the regular energy levels you see.  It takes into account the nucleus, when looking at the atom, and taking into account the nucleus allows you to see the fine-structure splitting, which is discussed in chapter 6. They simply replace m with mu, and you get a split in the regular energy levels you see. 
 +
 +=== Can 12/16 ===
 +It depends which particle you treated as stationary. Take the hydrogen atom as an example, for simplicity we treated the nucleus as stationary only the electron is rotating about the nucleus, which is not necessarily true.  remember the reduced mass equation is <math>m_r=\frac{m_e*m_p}{m_e+m_p}</math>, mass of electron is negligible compare to proton, thus normally the reduced mass of electron is just m_e, but if we treated the electron stationary and nucleus evolving about electron, things would be different.
 + 
  
 ==== Blackbox 11/30 10:50am ==== ==== Blackbox 11/30 10:50am ====
Line 56: Line 66:
  
 For a more non-mathematical description of subatomic particles I find wikipedia to be quite straightforward. But the particle classification system as it is now depends on why you're looking at them - are you concerned with energy distribution (fermions, bosons) or are you concerned with constituent parts of particles (ie. hadrons are made up of 3 quarks)? It's not like animal classification where one animal belongs to a single species; every particle is either a boson or fermion, so you can have a hadron that is a fermion or a collection of hadrons that make a boson. As physics typically goes, it is somewhat confusing until you can get an intuitive grasp for the odd way physicists do things. For a more non-mathematical description of subatomic particles I find wikipedia to be quite straightforward. But the particle classification system as it is now depends on why you're looking at them - are you concerned with energy distribution (fermions, bosons) or are you concerned with constituent parts of particles (ie. hadrons are made up of 3 quarks)? It's not like animal classification where one animal belongs to a single species; every particle is either a boson or fermion, so you can have a hadron that is a fermion or a collection of hadrons that make a boson. As physics typically goes, it is somewhat confusing until you can get an intuitive grasp for the odd way physicists do things.
- 
 ==== Esquire 11/30 13:57(Age of Letters)==== ==== Esquire 11/30 13:57(Age of Letters)====
 Dearest Wiki Readers, Dearest Wiki Readers,
Line 79: Line 88:
 I know the spin part of the wave function is sort of an "addition" to the x-dependent part of the wave function, it just wasn't included in Chapter 2 because we weren't considering spin at that point. I know the spin part of the wave function is sort of an "addition" to the x-dependent part of the wave function, it just wasn't included in Chapter 2 because we weren't considering spin at that point.
  
 +===Andromeda 12/01 8:05pm===
 +I think the time dependent wave function for two or more electrons is a combination of their space eigenfunction and spin eigenfunction. Both space eigenfunction and spin eigenfunction will have a time dependent part and can be one of the symmetric states or antisymmetric states. we just have to make the total wave function be antisymmetric for antisymmetric particles like electron, proton... and symmetric for symmetric particles like photons... 
 +==== Hardy 12/01 0:30 am====
  
 +Recently, We are dealing with the spin of the particle. It is strange that the result of our measurement has to turn out to be h/2. It let me think about the Postulate of QM. No matter how the wavefunction of the particle is, the results measurement have always to be the eigenvalues of the operators. Why the only possible values are its eigenvalues? It seems that the universe hates non-eigenvaules so much.
  
-==== Hardy 12/01 0:30 am===+=== nikif002 12/01 3:42 pm===
  
-RecentlyWe are dealing with the spin of the particleIt is strange that the result of our measurement has to turn out to be h/2It let me think about the Postulate of QM. No matter how the wavefunction of the particle is, the results measurement have always to be the eigenvalues of the operatorsWhy the only possible values are its eigenvaluesIt seems that the universe hates non-eigenvaules so much.+I thought I could answer thisbut I cannot. I scoured the book, and everything just comes down to the statements on page 106 that a measurement of an observable must yield one of the eigenvalues of that observableThis statement comes with no explanation, but Griffiths specifically states that this is unlike anything in classical physics. I guess it's just one of those things in QM that you have to accept. 
 + 
 +=== Yuichi === 
 +I guess you can say that this is a fundamental postulate for QM.  When a particle is trapped in an infinite square well, momentum (as well as the energy) can take only certain values, which is the eigenvalues of the momentum operator for the particle.  i.e. <math>p=\pm\frac{n\pi}{L}\hbar</math> The spin measurements that Hardy is talking about is no different from this case.  Or doesn't this analogy make any sense? 
 + 
 +====Andromeda 12/01 8:25 PM ==== 
 +How is <math/> p^2=1</math>??? i dont understand what he means by that. and how is its eigenvalues +/-1? 
 + 
 +=== The Doctor 12/02 1:45 AM === 
 +Where do you see this? 
 +===Andromeda 12/2 7:32 Am=== 
 +Oh sorrypage 205-the exchange operator 
 +===Jake22 12/2 7:42 pm=== 
 +Consider how operators can be treated as matrices and multiplied. When you multiply once by P, you interchange the two particles. If you multiply this product by P (obtaining <math/> p^2 f</math>), you interchange them again, which brings you to the same result that multiplying by 1 would haveFor the eigenvalues, consider the matrix that transforms a vector (a,b) to (b,a). It is easy to see that this matrix has eigenvalues of +/-1. 
 + 
 +====Pluto 4ever 12/01 4:19PM==== 
 +There is just one thing that concerns me slightly. When dealing with the sum of the states it says that <math>J_z = L_z + S_z</math> as implied by J = L + S. Does this also mean that <math>m_j = m_l + m_s</math> or do these z components only deal with the total m value? 
 +===nikif002 12/01 5:33PM=== 
 +<math>J_z = L_z + S_z</math> and <math>m_j = m_l + m_s</math> are equivalent statements, since <math>J_z = \hbar m_j</math>, <math>L_z = \hbar m_l</math> and <math>S_z = \hbar m_s</math>
  
 --------------------------------------- ---------------------------------------
classes/2009/fall/phys4101.001/q_a_1130.1259647830.txt.gz · Last modified: 2009/12/01 00:10 by cwu