The problems again and their solutions:
1) Solve for σ_x, σ_y and σ_z in terms of the orbital angular momenta, L(x), L(y) and L(z)
Solution: In terms of the angular momenta:
σ_x = L(x)/ ħ, σ_y = L(y)/ ħ and σ_z = L(z)/ ħ
2) Solve for the spin-momentum operators s(x), s(y) and s(z) in terms of σ_x, σ_y and σ_z. (State any assumptions made)
Basic assumption (borne out by experiment), is that the spin operators must satisfy the same commutation relations as for L(x), L(y) and L(z)
As we saw:
For L(x):
σ _x σ_y = - σ_y σ_x = i ħ σ_z
For L(y):
σ_y σ_z = - σ_z σ_y = i ħ σ_x
For L(z):
σ_z σ_x = -σ_x σ_z = i ħ σ_y
Then, similarly:
For s(x):
s _x s_y = - s_y s_x = i ħ s_z
For s(y):
s _y s_z = - s_z s_y = i ħ s_x
For s(z):
s_z s_x = -s_x s_z = i ħ s_y
And since the spin operator s(op)^2 = s(x)^2 + s(y)^2 + s(z)^2
Has eigenvalues = ½ (½ + 1) ħ^2 = 3 ħ^2/ 4 in both states, we have:
s(x) = ħ/2 (σ_x), s(y) = ħ/2 (σ_y) and s(z) = ħ/2 (σ_z)
3) Show that the TOTAL angular momentum vector, J(z) =
(ħ/i (@/@φ )+ ħ/2 ..... 0)
(0 ....... ħ/i (@/@ φ ) - ħ/2)
displays a Hermitian matrix, where φ is a spherical angle and @/@ φ denotes a partial derivative with respect to it.
Solution: Take the conjugate transpose which must equal to the original matrix if Hermitian.
Then M’ (conjugate) =
(ħ/i (@/@φ )- ħ/2 ..... 0)
(0 ....... ħ/i (@/@ φ ) + ħ/2)
And the transpose of this is ^M’ = M* =
(ħ/i (@/@φ )+ ħ/2 ..... 0)
(0 ....... ħ/i (@/@ φ ) - ħ/2)
So the matrix is Hermitian
4) Show by direct computation that: σ_x σ_y + σ_y σ_z =
(i……….i)
(i………-i)
Solution: This computation is shown worked out in the accompanying graphic (from Mathcad)
5) Compute the matrices for (L(x) + i L(y)) m.m' and (L(x) - i L(y)) m.m' if the angular momentum quantum number l = -1.
This is really a trick question! In fact, the lowest value the angular momentum quantum number l can have is l = 0 which is associated with the lowest energy subshell, s.. The angular momentum quantum number thereby increases with each subshell, s, p, d, f translating to l=0, l= 1, l=2 and l= 3 respectively.
1) Solve for σ_x, σ_y and σ_z in terms of the orbital angular momenta, L(x), L(y) and L(z)
Solution: In terms of the angular momenta:
σ_x = L(x)/ ħ, σ_y = L(y)/ ħ and σ_z = L(z)/ ħ
2) Solve for the spin-momentum operators s(x), s(y) and s(z) in terms of σ_x, σ_y and σ_z. (State any assumptions made)
Basic assumption (borne out by experiment), is that the spin operators must satisfy the same commutation relations as for L(x), L(y) and L(z)
As we saw:
For L(x):
σ _x σ_y = - σ_y σ_x = i ħ σ_z
For L(y):
σ_y σ_z = - σ_z σ_y = i ħ σ_x
For L(z):
σ_z σ_x = -σ_x σ_z = i ħ σ_y
Then, similarly:
For s(x):
s _x s_y = - s_y s_x = i ħ s_z
For s(y):
s _y s_z = - s_z s_y = i ħ s_x
For s(z):
s_z s_x = -s_x s_z = i ħ s_y
And since the spin operator s(op)^2 = s(x)^2 + s(y)^2 + s(z)^2
Has eigenvalues = ½ (½ + 1) ħ^2 = 3 ħ^2/ 4 in both states, we have:
s(x) = ħ/2 (σ_x), s(y) = ħ/2 (σ_y) and s(z) = ħ/2 (σ_z)
3) Show that the TOTAL angular momentum vector, J(z) =
(ħ/i (@/@φ )+ ħ/2 ..... 0)
(0 ....... ħ/i (@/@ φ ) - ħ/2)
displays a Hermitian matrix, where φ is a spherical angle and @/@ φ denotes a partial derivative with respect to it.
Solution: Take the conjugate transpose which must equal to the original matrix if Hermitian.
Then M’ (conjugate) =
(ħ/i (@/@φ )- ħ/2 ..... 0)
(0 ....... ħ/i (@/@ φ ) + ħ/2)
And the transpose of this is ^M’ = M* =
(ħ/i (@/@φ )+ ħ/2 ..... 0)
(0 ....... ħ/i (@/@ φ ) - ħ/2)
So the matrix is Hermitian
4) Show by direct computation that: σ_x σ_y + σ_y σ_z =
(i……….i)
(i………-i)
Solution: This computation is shown worked out in the accompanying graphic (from Mathcad)
5) Compute the matrices for (L(x) + i L(y)) m.m' and (L(x) - i L(y)) m.m' if the angular momentum quantum number l = -1.
This is really a trick question! In fact, the lowest value the angular momentum quantum number l can have is l = 0 which is associated with the lowest energy subshell, s.. The angular momentum quantum number thereby increases with each subshell, s, p, d, f translating to l=0, l= 1, l=2 and l= 3 respectively.
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