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Show that unless l=s, L and S cannot be exactly opposite: that is, show that at its minimum possible value. Forwhich j=l-s. The magnitude of the total angular momentum is strictly greater than the difference L-Sbetween the magnitudes of the orbital and intrinsic angular momentum vectors.

Short Answer

Expert verified

The magnitude J of the total angular momentum is strictly greater than the differenceL-S between the magnitudes of the orbital and intrinsic angular momentum vectors.

Step by step solution

01

Identification of the given data 

The given data can be listed below as:

  • The total angular momentum鈥檚 magnitude is J .
  • The orbital angular momentum is L .
  • The spin angular momentum is S.
  • The minimum value of the total angular momentum鈥檚 magnitude is j .
  • The minimum value of the orbital angular momentum is l.
  • The minimum value of the intrinsic angular momentum is s .
02

Significance of the angular momentum

The angular momentum is described as the quantity of the rotation of a particular system. The quantity of the rotation is described as the product of the angular velocity and the moment of inertia of a system.

03

Determination of the proof of the problem statement

The equation of the total angular momentum is expressed as:

J=jj+1

Here, J is the total angular momentum鈥檚 magnitude, j is the minimum value of the total angular momentum鈥檚 magnitude and is the Planck鈥檚 constant.

The equation of the orbital angular momentum is expressed as:

L=ll+1

Here, L is the orbital angular momentum, l is the minimum value of orbital angular momentum andis the Planck鈥檚 constant.

The equation of the spin angular momentum is expressed as:

S=ss+1

Here, is the spin angular momentum, is the minimum value of spin angular momentum and is the Planck鈥檚 constant.

The equation of the minimum value of the square of the total angular momentum is expressed as:

jm2=l-sl-s+12=ll+1+ss+1-2s-2sl2=ll+1+ss+1-2sl+12鈥(颈)

The equation of the difference amongst the spin and orbital angular momentum is expressed as:

L-S=ll+1-ss+1L-S2=2ll+1+ss+1-2sll+1s+1鈥(颈颈)

Comparing the equation (i) and (ii), it can be identified that

sl+1<lsl+1s+1

Squaring the above equation, the equation can be expressed as:

sl+1<ls+1sl+s<sl+l

Hence, l>sas then .jm>L-S

Thus, magnitude J of the total angular momentum is strictly greater than the difference L-Sbetween the magnitudes of the orbital and intrinsic angular momentum vectors.

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Most popular questions from this chapter

Exercise 44 gives an antisymmetric multiparticle state for two particles in a box with opposite spins. Another antisymmetric state with spins opposite and the same quantum numbers is n(x1)n2(x2)nn(x1)n(x2)

Refer to these states as 1 and 11. We have tended to characterize exchange symmetry as to whether the state's sign changes when we swap particle labels. but we could achieve the same result by instead swapping the particles' stares, specifically theandin equation (8-22). In this exercise. we look at swapping only parts of the state-spatial or spin.

(a) What is the exchange symmetric-symmetric (unchanged). antisymmetric (switching sign). or neither-of multiparticle states 1 and Itwith respect to swapping spatial states alone?

(b) Answer the same question. but with respect to swapping spin states/arrows alone.

(c) Show that the algebraic sum of states I and II may be written(n(x1)n'(x2)n'(x1)n(x2))(+)

Where the left arrow in any couple represents the spin of particle 1 and the right arrow that of particle?

(d) Answer the same questions as in parts (a) and (b), but for this algebraic sum.

(e) ls the sum of states I and 11 still antisymmetric if we swap the particles' total-spatial plus spin-states?


(f) if the two particles repel each other, would any of the three multiparticle states-l. II. and the sum-be preferred?

Explain.

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Question: In the Stern-Gerlach experiment how much would a hydrogen atom emanating from a 500 K oven(KE=32kBT)be deflected in traveling 1 m through a magnetic field whose rate of change is 10 T/m?

Here we consider adding two electrons to two "atoms," represented as finite wells. and investigate when the exclusion principle must be taken into account. In the accompanying figure, diagram (a) shows the four lowest-energy wave functions for a double finite well that represents atoms close together. To yield the lowest energy. the first electron added to this system must have wave function Aand is shared equally between the atoms. The second would al so have function Aand be equally shared. but it would have to be opposite spin. A third would have function B. Now consider atoms far a part diagram(b) shows, the bumps do not extend much beyond the atoms - they don't overlap-and functions Aand Bapproach equal energy, as do functions Cand D. Wave functionsAandBin diagram (b) describe essentially identical shapes in the right well. while being opposite in the left well. Because they are of equal energy. sums or differences ofandare now a valid alternative. An electron in a sum or difference would have the same energy as in either alone, so it would be just as "happy" inrole="math" localid="1659956864834" A,B,A+B, orA- B. Argue that in this spread-out situation, electrons can be put in one atom without violating the exclusion principle. no matter what states electrons occupy in the other atom.

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