L x op = -i h [y (¶/¶x ) – x (¶/¶y)]
L y op = -i h [z (¶/¶x ) – x (¶/¶z)]
L z op = -i h [x (¶/¶y ) – y (¶/¶x)]
In spherical coordinates.
Ans.
L x op =
iħ [sin f (¶ / ¶ q ) + cot q cos
f (¶ / ¶ f)]
L y op =
iħ [- cos f (¶ / ¶ q ) + cot q sin f (¶ / ¶ f)]
L z op =
- iħ (¶ / ¶ f)
2) Consider the form: [H, L x ]
Show that [H, L x ] = yZ - zY and give the condition for H and L x to commute.
Soln.
Hamiltonian H = p 2/ 2m + V(r)
[H, L x
] = [ p 2/ 2m + V(r) , L x ] = 1/ 2m [ p 2 , L x ] + [V(r) , L x ]
[p 2 , L x ] = [ p x 2 , L x ] + [ p y 2 , L x ] + [ p z 2 , L x ]
[ p x 2 , L x ] = p x [ p x , L x ] + [ p x , L x ] p x
[ p x , L x ] = [ p x , y p z - z p y ] = 0
And, for constant angular momentum:
[ p x , L x ] = 0 = [H, L x ]
Then: [H, L x ] = yZ - zY
REM:
Z = p z = -i h (¶/¶z ) and Y = p y = -i h (¶/¶y )
H and L
x do commute ([H, L x ] = 0) since a key property of central potential problems is that the angular momentum operators commute with the Hamiltonian .
Then: [L x , H] = 0, And [H, L x ] = 0
The above commutator implies that the ^L i operators are conserved in central potentials.
3) (a)
(L’ x - i L’ y)
=
h (0...0)
(1....0)
(1....0)
And: (L’ x
+ i L’ y)
=
h (0...1)
(0...0)
(0...0)
Hence the matrices are found to be Hermitian,
which means the matrix is equal to its conjugate transpose.
b) If ℓ = 1 calculate (L’ x + i L’ y) m m’
Soln.
(L’ x + i L’ y) m m’ =
ħ [ (ℓ - m’ ) (ℓ + m’ + 1 ) ]1/2 exp (i m ℓ j) d m m’+1
= ħ [ (1 - 0 ) (1 + 0 + 1 ) ]1/2 exp (i 0 j) [1]
= ħ [Ö2 ]
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