BU, PHYSICS 451

Excerpt: ... PY 451: Quantum Physics I Problem Set 12 Due date: Thursday, April 30 2009 by 5pm in the graders' box 1. (Griffiths 4.35) Quarks carry spin 1/2. Three quarks bind together to make a baryon (such as the proton or neutron); two quarks (actually a quark and anti-quark) bind to make a meson (such as the pion or kaon). Assume that the quarks are in their ground states so that the orbital angular momentum is zero. (a) What spin quantum numbers are possible for baryons? (b) What spin quantum numbers are possible for mesons? 2. By following the steps outlined in lecture, determine by detailed calculation the spin matrices Sx , Sy , and Sz for a spin-1 particle. For the ambitious students, note that this same calculation can be extended to a particle with arbitrary spin quantum number s. You are invited to try this latter calculation; the answer is given in problem 4.53 of Griffiths. 3. (part of Griffiths 4.36) Two particles with spin quantum numbers 1 and 2 are at rest in a configuration such that the total spin qu ...

Delaware, PHYSICS 424

Excerpt: ... Lecture 25 Addition of angular momenta Ground state of hydrogen: it has one proton with spin and one electron with spin ( orbital angular momentum is zero). What is the total angular momentum of the hydrogen atom? Total spin Electron's spin, acts only on electron's spin states Proton's spin, acts only on proton's spin states The z components just add together and quantum number m for the composite system is simply Combination of two spin particles can carry a total spin of s =1 or s = 0, depending on whether they occupy the triplet or singlet configuration. Three states with spin s = 1, m = 1, 0, -1 This is called a triplet configuration. and one state with spin s = 0, m = 0: This is called a singlet configuration. How do we prove that it is true? Lecture 25 Page 1 L25. P2 To prove the above, we need to show that for triplet states since and for singlet states since This can be shown by direct calculation (see page 186 of the textbook for the proof). In general, it you combine any angular momentum va ...

East Los Angeles College, PHYS 3002

Excerpt: ... 19 19.1 Parity, Charge Conjugation and CP Intrinsic Parity In the same way that nuclear states have parity, so hadrons, which are bound states of quarks (and antiquarks) have parity. This is called intrinsic parity, and under a parity inversion the wavefunction for a hadron acquires a factor P {P } (r) = = {P } (-r) = {P } {P } (r). can take the values 1 noting that applying the parity operator twice must bring us back to the original state. The lighter baryons (for which there is zero orbital angular momentum ) have positive intrinsic parity. On the other hand, antiquarks have the opposite parity from quarks. This means that the light antibaryons have negative parity. It also means that the light mesons, such as pions and kaons, which are bound states of a quark and an antiquark have negative intrinsic parity. The lightest spin-one mesons, such as the -meson, also have zero orbital angular momentum and thus they too have negative intrinsic parity - they have spin-one because of the alignment of the spin ...

University of Florida , PHY 3101

Excerpt: ... other words, the energy degeneracy with respect to magnetic quantum number ml of the H atom is broken with the application of an external magnetic field. It is possible to see the effect of the quantization of angular momentum, because the energy levels will be split into 2l + 1 different levels, and the line spectra, therefore, will be split also. This is known as the normal Zeeman effect. It gives fine structure to spectral lines. The energy splitting between two levels is given by: E eh E = B Bml = B with m l = 1 2m E For a 1 Tesla field, this splitting is 5.79 10 5 eV, which is very small. n=2 ml = 1 ml = 0 ml = 1 l =1 n=1 D. Acosta Page 3 ml = 0 1/3/2001 PHY3101 Modern Physics Lecture Notes Spin 1 Spin and the Stern-Gerlach Experiment The normal Zeeman effect is not always seen. It turns out that orbital angular momentum is not the complete story, and that there exists intrinsic angular momentum. Lets consider the Stern-Gerlach experiment of 1922. In that experimen ...

Washington, CHEM 152

Excerpt: ... tum number, and ranges from 1 to infinity. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). ml, the "z component" of orbital angular momentum . Ranges in value from -l to 0 to l. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). m, the "z component" of orbital angular momentum . Ranges in value from -l to 0 to l. We can then characterize the wavefunctions based on the quantum numbers (n, l, m). Orbital Shapes Let's take a look at the lowest energy orbital, the "1s" orbital (n = 1, l = 0, m = 0) 1 1s = p Z e a o 3 2 - Z r a0 3 1 = p -s Z 2 e a o a0 is referred to as the Bohr radius, and = 0.529 1 2 Z - 18 E n = - 2.178x10 J 2 - 2.178x10- 18 J = n 1 Orbital Shapes (cont.) Note that the "1s" wavefunction ...

Caltech, PH 0304

Excerpt: ... in the expansion of the eikx expansion). Consider transitions between states of specified initial and final orbital angular momentum , i and f , and initial and final projections of the orbital angular momentum on the z-axis, mi and mf . In the electric dipole approximation, what transitions are permitted (e.g., what are the permitted values of f and mf if i and mi are given)? 19 ...

University of Iowa, C 004132

Excerpt: ... Coupling quantum mechanics yield these energy levels 1 E l,s,j = hcA{ j(j + 1) - l(l + 1) - s(s + 1)} 2 Aisthespin - orbitcouplingconstant Term Symbols Term symbols - method for labeling electronic states in a multi-electron system Contains three pieces of information Central capital letter (S, P, D, .) indicates total orbital angular momentum , quantum # (Q.N.) = L. Left superscript gives the multiplicity of the term. Right superscript is the value of the total angular momentum, quantum # = J. 1 Rules for Term Symbols- part.1 Total orbital angular momentum Q.N., L, gives the magniture of the orbital angular momentum from {L(L+1)}1/2h. It has 2L+1 orientations distinguished by the mL Q.N. where mL= -L, -L+1, . , +L. L is obtained by coupling individual orbital angular momenta using a Clebsch-Gordon Series from total orbital angular momentum eigenfunctions L = l1+l2, l1+l2-1, ., |l1-l2|. L = 0, 1, 2, 3, 4, . (written S, P, D, F, G, .) Total Orbital Angular Momentum of a p and d e ...

East Los Angeles College, MAB 503

Excerpt: ... Nuclear Physics II PROBLEM CLASS 1 1. Show that, in addition to the static electric multipole moments discussed so far, the next highest order multipole moment expected in nuclei would be described by an expression of the form 1 2 2 2 4 e (r , ' ) 5z 7z 6r + 3r dV 8 (you may assume that P3 ( x ) = ( ( ) ) 1 1 (5 x 3 3 x ) and P4 ( x ) = (35 x 4 30 x 2 + 3) ) 2 8 2. Show that the size of the spin-orbit splitting in nuclei is proportional to (2l+1) where l is the orbital angular momentum quantum number 3. Show, using the method of counting m-states, that it is not possible to generate an odd-spin state with two identical particles in the same shell-model level. Do this by following this procedure separately for two different shell model levels, and analysing the result. 4. Electric quadrupole moments: Comparison of equation I.10 with the previous equation in the lecture notes shows that the effective quadrupole moment measured for a single proton in a level with spin j is differen ...

Washington, CHEM 152

Excerpt: ... E n = - 2 2 2 = -2.178x10-18 J 2 n 8e0 h n n is the principle quantum number, and ranges from 1 to infinity. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). ml, the z component of orbital angular momentum . Ranges in value from -l to 0 to l. H-atom wavefunctions (cont.) In solving the Schrodinger Equation, two other quantum numbers become evident: l, the orbital angular momentum quantum number. Ranges in value from 0 to (n-1). m, the z component of orbital angular momentum . Ranges in value from -l to 0 to l. We can then characterize the wavefunctions based on the quantum numbers (n, l, m). Orbital Shapes Lets take a look at the lowest energy orbital, the 1s orbital (n = 1, l = 0, m = 0) 1 y1s = p Z e ao 3 2 -Z r a0 1 = p Z 2 -s e ao 3 a0 is referred to as ...

UNC Charlotte, CHEM 2141

Excerpt: ... Answer: 4d, since 0=s, 1=p, 2=d and the values of m range from -l to +l =-2,-1,0,1,2 S -orbitals S orbitas (n = 1, l = 0, ml =0) have forrm: -r a0 = e a 0 3 There is no angular dependence thus function is spherically symmetric. 2 Volume element between r and r+ r spheres is: 4r r=V 2 2 2 The probability of finding the electron between r and r+dr is V =4r r 4r2 2 = radial distribution fn. Shell Probability: What's probability of finding electron between a0 and (a0+ 1 pm) Use this for illustration 13.3 in which r = 1 pm, a0 = 52.9 pm, and is given by the above expression , thus the probability of an S-orbital electron between a0 and a0+1pm: Note that max value of is 0 when r = 0 a0 =P(r) the radial distribution function occurs at r = a0 for 1S orbital 3 = 1 however maximum value of 4r22 P & d orbitals P orbitals have l = 1 Existence of nodal plane for all orbitals with l>0 (p,d,f.) is due to nonzero orbital angular momentum being interpreted as the electron circulating about the nucleus, ...