Angular Momentum set II

Transcription

1 Angular Momentum set II PH - QM II Sem, 7-8 Problem : Using the commutation relations for the angular momentum operators, prove the Jacobi identity Problem : [ˆL x, [ˆL y, ˆL z ]] + [ˆL y, [ˆL z, ˆL x ]] + [ˆL z, [ˆL x, ˆL y ]] Consider a system with total angular momentum j. The y component of the angular momentum operator is given by the matrix Ĵ y i i i i a What are the possible values that we will obtain when measuring Ĵy? b Calculate Ĵz, Ĵ z and J z if the system is in the state j y. c Repeat b for Ĵx, Ĵ x and J x for the same state. Problem : Consider a particle with angular momentum j /. a Find the matrices representing the operators Ĵ, Ĵ x, Ĵ y and Ĵz in the {, m } basis. b Using these matrices, show that Ĵx, Ĵy and Ĵz satisfy the commutation relation [Ĵx, Ĵy] i Ĵz. c Calculate the mean values of Ĵx and Ĵ x in the state ψ d Calculate Ĵx Ĵy in the state ψ and verify that this product satisfies Heisenberg s uncertainty principle..

2 Problem : Consider the Pauli matrices σ x i, σ y i and σ z a Verify that σ x σ y σ z I where I is the identity matrix. b Calculate the commutators [σ x, σ y ], [σ y, σ z ] and [σ z, σ x ]. c Calculate the anticommutator {σ x, σ y } σ x σ y + σ y σ x. d Show that e iθσy I cos θ + iσ y sin θ. Problem 5: Consider a spin- particle whose Hamiltonian is given by: Ĥ ε Ŝ x Ŝ y ε Ŝ z, where varepsilon is a real constant having the dimensions of energy. a Find the matrix form for the Hamiltonian and diagonalise it to find the energy eigenvalues. b Find the energy eigenstates and show that the energy levels are doubly degenerate.

3 Problem Same as Question in Problem Set. Problem Set Solution Problem Angular Momentum for j Bosons, assuming j s Given the system has total angular momentum j m,,. The joint eigenstates of Ĵ and Ĵz are j, m {,,,,, }. The matrix representations of the various angular momentum operators with respect to this eigenbasis is given by Ĵ x h Ĵ h Ĵ y h Ĵ + h i i i i Ĵ z h Ĵ h Notice that Ĵz and Ĵ are diagonal matrices as they are represented by their own eigenkets with the diagonal entries as their eigenvalues respectively. Recall that Ĵ z j, m m h j, m Eigenvalues m h { h,, h}diagonal entries of matrix Ĵzwith respective eigenvectors {,, },, Ĵ j, m jj + h j, m Eigenvalues jj + h h diagonal entries of matrix Ĵz a The possible values of J y are the eigenvalues of its operator Ĵy. Let Ĵy M i i i i h M where The characteristic equation detλi M gives λ i i λ i i λ λ i i λ i λ λ λ, ± where the above is evaluated using the Sarrus rule. Hence the possible values of J y are j y h,, h and the respective normalised eigenvectors are i b Given the system is in the state that gives j y h. This imply that the system is in state. Hence J z Ĵz Ĵz h Ĵ z Ĵ z h Ĵ z Ĵz h i i i i i h

4 c Similarly Note Hence J x Ĵx Ĵx h Ĵ x Ĵ x h 8 Ĵ x Ĵx h i i As a result J x J z h, it satisfies Heisenberg uncertainty inequality. i i h i Since the system is in the eigenstate of Ĵy and Ĵx, Ĵz do not commute with Ĵy, this imply Ĵx Ĵz ii Ĵ Ĵ x + Ĵ y + Ĵ z commutes with Ĵx, Ĵy, Ĵz. Problem Angular Momentum for j Fermions, assuming j s Given an operator Â, its matrix representation in basis { φ i,... φ n } is given by the n n matrix φ  φ φ  φ n.  ij φ i  φ. j φ n  φ φ n  φ n Consider a particle with angular momentum j m,,,. The joint eigenstates of Ĵ and Ĵz are j, m, m {,,,,,,, }. The matrix representations of the various angular momentum operators with respect to this eigenbasis are given by Ĵ x h Ĵ x h 7 7 Ĵy i h Ĵ y h Ĵz h 7 7 Ĵ z h Ĵ 5 h Ĵ+ h Ĵ h { } Ĵ z j, m m h j, m Eigenvalues m h diagonal entries of matrix h, h, h, h Ĵz with respective eigenvectors { },,,,,, Ĵ j, m jj + h j, m Eigenvalues jj + h 5 h diagonal entries of matrix Ĵ a As shown above. 9 9

5 b [Ĵx, Ĵy] i h i h c Ĵx ψ Ĵx ψ h Ĵ x ψ Ĵ x ψ h 7 7 Note Since ψ Hence Ĵx d Similarly Ĵy ψ Ĵy ψ i h Ĵ y ψ Ĵ y ψ h Note Since ψ Hence Ĵy h is an eigenvector of Ĵz and Ĵx, Ĵz do not commute Ĵx. Ĵ x Ĵx h 7. As a result Ĵx Ĵy Ĵ y Ĵy h h is an eigenvector of Ĵz and Ĵy, Ĵz do not commute Ĵy. h 7 7 h h h. Recall in Problem Set Question 6, we show that Ĵx Ĵy m h. Here the state ψ has m. i hĵz Problem Pauli Matrices a Identity i σx σy σz I i i Note iσ x σ y σ z i I i

6 Hence σ x σ y σ z iσ x σ y σ z I b Commutators i i [σ x, σ y ] σ x σ y σ y σ x i iσ i i z i i [σ y, σ z ] σ y σ z σ z σ y i iσ i i x i [σ z, σ x ] σ z σ x σ x σ z i iσ i y Notice the cyclic relationship in the commutators of Pauli matrices. In general [σ i, σ j ] iε ijk σ k where ε ijk is the Levi-Civita symbol if i, j, k is an even permutation of x, y, z ε ijk if i, j, k is an odd permutation of x, y, z where i, j, k {x, y, z} if i j or j k or k i c Anticommutators {σ x, σ y } σ x σ y + σ y σ x i i i + i matrix In general {σ i, σ j } Iδ ij where i, j, k {x, y, z}. d Euler s equation Using Euler s equation e iθ cos θ + i sin θ we have e iθσj cosθσ j + i sinθσ j [ θσ j! + θσ j! ] [ I [ θ! + θ! iσ j θ θ σj! ] i [θσ j θσ j + θσ j 5 ] +...! 5! + θ5 σ j 5! ] ] I [ θ! + θ! iσ j [θ θ! + θ5 5! ] zero I cos θ + iσ j sin θ where we have use the series expansion for sine, cosine and the result that σ j I for j,,. Problem 5 Spin Particle Given the Hamiltonian Ĥ ε ] [Ŝ h x Ŝy Ŝz ε [ h Ŝ x Ŝ] The angular momentum matrix for s is solved in Problem above. Substituting into the Hamiltonian we have: 9 Ĥ ε 9 where the matrix representation is written with respect to the common eigenbasis vectors for Ŝz and Ŝ. Notice that the Hamiltonian is Hermitian, i.e. Ĥ Ĥ. To diagonalizing Ĥ we first find its eigenvalues and eigenvectors. The energy eigenvalues are E E ε and E E ε with energy eigenstates as

7 E E E E respectively. Thus the energy levels are doubly degenerated. The Hamiltonian Ĥ is diagonalize via ˆD ˆP Ĥ ˆP where ˆP ˆD ε h ˆP Observe that the column vectors of ˆP are formed by eigenvectors of Ĥ with the diagonal entries of the diagonal matrix ˆD being the respective eigenvalues. Notice also that matrix ˆP is not unitary, i.e. ˆP ˆP. Suppose we now normalised the eigenvectors E E and diagonalize Ĥ via E E where ˆD Û ĤÛ Û ˆD ε h Û Here, the column vectors of Û are formed by the normalised eigenvectors of Ĥ. Notice that the matrix Û is unitary, i.e. Û Û. Alternative Shorter Solution Given Ĥ ε ] [Ŝ h x Ŝy Ŝz ε [ h Ŝ x Ŝ] { Let j, m : j, m,..., } be the common eigenbasis vectors for both Ŝx and Ŝ. Then their matrix representation with respect to this basis will be diagonal as follow Ŝ x h Ŝ z h 9 9 The Hamiltonian matrix representation will then also be diagonal Ĥ ε and Ŝ 5 h 5

8 The diagonal entries are the energy eigenvalues E E ε and E E ε with normalised energy eigenvectors E E E E 6