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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