Question

Problem 5.10 For two spin-1/2 particles you can construct symmetric and antisymmetric spin states (the triplet and singlet combinations, respectively). For three spin-1/2 particles you can construct symmetric combinations (the quadruplet, in Problem 4.65), but no completely antisymmetric configuration is possible. (a) Prove it. Hint: The "bulldozer" method is to write down the most general linear combination: ?|1,1,2> + ?|1,2,1> + ?|2,1,1> + ?|1,2,2>. What does antisymmetry under 1 ? 2 tell you about the coefficients? (Note that the eight terms are mutually orthogonal.) Now invoke antisymmetry under 2 ? 3. (b) Suppose you put three identical noninteracting spin-1/2 particles in the infinite square well. What is the ground state for this system, what is its energy, and what is its degeneracy? Note: You can't put all three in the position state |1> (why not?); you'll need two in |1> and the other in |2>. But the symmetric configuration |1>|1>|2> is no good (because there's no antisymmetric spin combination to go with it), and you can't make a completely antisymmetric combination of those three terms. In this case you simply cannot construct an antisymmetric product of a spatial state and a spin state. But you can do it with an appropriate linear combination of such products. Hint: Form the Slater determinant (Problem 5.8) whose top row is |1>|2>|3>. (c) Show that your answer to part (b), properly normalized, can be written in the form ?(1,2,3) = 1/?3 [?(1,2)?(3) - ?(1,3)?(2) + ?(2,3)?(1)] where ?(i,j) is the wave function of two particles in the n=1 state and the singlet spin configuration, ?(i,j) = (1/?2)(|up_i down_j> - |down_i up_j>), and ?(i) is the wave function of the ith particle in the n=2 spin-up state: ?(i) = ?_2(x_i)|up_i>.

          Problem 5.10 For two spin-1/2 particles you can construct symmetric and antisymmetric spin states (the triplet and singlet combinations, respectively). For three spin-1/2 particles you can construct symmetric combinations (the quadruplet, in Problem 4.65), but no completely antisymmetric configuration is possible.

(a) Prove it. Hint: The "bulldozer" method is to write down the most general linear combination:
?|1,1,2> + ?|1,2,1> + ?|2,1,1> + ?|1,2,2>.
What does antisymmetry under 1 ? 2 tell you about the coefficients? (Note that the eight terms are mutually orthogonal.) Now invoke antisymmetry under 2 ? 3.

(b) Suppose you put three identical noninteracting spin-1/2 particles in the infinite square well. What is the ground state for this system, what is its energy, and what is its degeneracy? Note: You can't put all three in the position state |1> (why not?); you'll need two in |1> and the other in |2>. But the symmetric configuration |1>|1>|2> is no good (because there's no antisymmetric spin combination to go with it), and you can't make a completely antisymmetric combination of those three terms. In this case you simply cannot construct an antisymmetric product of a spatial state and a spin state. But you can do it with an appropriate linear combination of such products. Hint: Form the Slater determinant (Problem 5.8) whose top row is |1>|2>|3>.

(c) Show that your answer to part (b), properly normalized, can be written in the form

?(1,2,3) = 1/?3 [?(1,2)?(3) - ?(1,3)?(2) + ?(2,3)?(1)]

where ?(i,j) is the wave function of two particles in the n=1 state and the singlet spin configuration,

?(i,j) = (1/?2)(|up_i down_j> - |down_i up_j>),

and ?(i) is the wave function of the ith particle in the n=2 spin-up state: ?(i) = ?_2(x_i)|up_i>.

Problem 5.10 For two spin-1/2 particles you can construct symmetric and antisymmetric spin states (the triplet and singlet combinations, respectively). For three spin-1/2 particles you can construct symmetric combinations (the quadruplet, in Problem 4.65), but no completely antisymmetric configuration is possible.

(a) Prove it. Hint: The "bulldozer" method is to write down the most general linear combination:
?|1,1,2> + ?|1,2,1> + ?|2,1,1> + ?|1,2,2>.
What does antisymmetry under 1 ? 2 tell you about the coefficients? (Note that the eight terms are mutually orthogonal.) Now invoke antisymmetry under 2 ? 3.

(b) Suppose you put three identical noninteracting spin-1/2 particles in the infinite square well. What is the ground state for this system, what is its energy, and what is its degeneracy? Note: You can't put all three in the position state |1> (why not?); you'll need two in |1> and the other in |2>. But the symmetric configuration |1>|1>|2> is no good (because there's no antisymmetric spin combination to go with it), and you can't make a completely antisymmetric combination of those three terms. In this case you simply cannot construct an antisymmetric product of a spatial state and a spin state. But you can do it with an appropriate linear combination of such products. Hint: Form the Slater determinant (Problem 5.8) whose top row is |1>|2>|3>.

(c) Show that your answer to part (b), properly normalized, can be written in the form

?(1,2,3) = 1/?3 [?(1,2)?(3) - ?(1,3)?(2) + ?(2,3)?(1)]

where ?(i,j) is the wave function of two particles in the n=1 state and the singlet spin configuration,

?(i,j) = (1/?2)(|upi downj> - |downi upj>),

and ?(i) is the wave function of the ith particle in the n=2 spin-up state: ?(i) = ?2(xi)|upi>.

Added by Chad W.

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