Problem 5.9 In example 5.1 and problem 5.5(b) we ignored spin (or, if you prefer we assumed the particles are in the same spin state). (a) Do it now for particles of spin 1/2. Construct the four lowest-energy configurations
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In Example 5.1 and Problem $5.5(\mathrm{b})$ we ignored spin (or, if you prefer, we assumed the particles are in the same spin state). (a) Do it now for particles of spin $1 / 2$. Construct the four lowest-energy configurations, and specify their energies and degeneracies. Suggestion: Use the notation $\psi_{n_{1} n_{2}}|s m\rangle,$ where $\psi_{n_{1} n_{2}}$ is defined in Example 5.1 and $| s m)$ in Section $4.4 .3 . ?$ (b) Do the same for spin 1. Hint. First work out the spin-1 analogs to the spin-1/2 singlet and triplet configurations, using the Clebsch-Gordan coefficients; note which of them are symmetric and which antisymmetric.
Problem 5: 8 points. Two spin-half fermions are in some external potential with single-particle energy levels En and normalized wavefunctions ̄̄n (̄̄0 is the ground state etc.). Their only interaction is through their spins, in the form Hint = ̄̄S₁ · S₂ where S₁,₂ are the spins of the fermions and ̄̄ a real constant. a) The fermions are in the normalized state ̄̄ = ̄̄₀(̄̄₁)̄̄₀(̄̄₂)̄̄ where ̄̄₁,₂ are the coordinates of the fermions and ̄̄ is a state of their spins. Determine ̄̄ by expressing in terms of the basis states |uu⟩, |ud⟩, |du⟩, |dd⟩ where, as usual |uu⟩ is the state with both spins having z-component +ħ/2 etc. b) Find the energy of the state of part (a) (it is an energy eigenstate). c) The fermions are now in the normalized state ̄̄' = (̈̄₀(̄̄₁)̈̄₁(̄̄₂) − ̈̄₀(̄̄₂)̈̄₁(̄̄₁))̄̄' with ̄̄' a new state of their spins which is an eigenstate of the z-component of the total spin Sz = S1z + S2z. Determine the possible states ̄̄' and give the value of Sz for each. d) Find the energy of the states of part (c) and their degeneracy.
Adi S.
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: α|1,1,2> + β|1,2,1> + γ|2,1,1> + δ|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 Phi(1,2,3) = (1/√3)[Phi(1,2)Phi(3) - Phi(1,3)Phi(2) + Phi(2,3)Phi(1)] where Phi(i,j) is the wave function of two particles in the n=1 state and the singlet spin configuration, Phi(i,j) = (1/√2)(|up_i down_j> - |down_i up_j>), and Phi(i) is the wave function of the ith particle in the n=2 spin-up state: Phi(i) = psi_2(x_i)|up_i>.
Sri K.
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