Where I noted:
"The brilliance of the early quantum mechanicians lay in substituting the operators ( e.g Eop, p^ or p op ) for the corresponding quantities of the original classical Hamiltonian, then multiplying through by the wave function y :
H op y = Eop y
H op y = Eop y
In this way, a drastically simplified basic quantum mechanical equation could be obtained, which could then be expanded once one substituted the operators, i.e.:
p op = -i h (¶/ ¶ x)
Eop = i h (¶ /¶t)
Other (3D system) operators can be seen in the quantum mechanical equation:
T op y = Eop y
Where T op is the kinetic energy operator which is written in 3D as:
T op = p^ p^ / 2m = [p^ x 2 + p^ y 2 + p^ z 2 ] / 2m
= - ħ2/ 2m [ ¶2 / ¶ x2 + ¶2 / ¶ y2 + ¶2 / ¶ z2 ) = - ħ2/ 2m Ñ 2
Where Ñ is the Laplacian operator.
Angular momentum operators are also important in quantum mechanics. The physical significance of the angular momentum quantum number (ℓ) is to convey the shape of the probability density cloud or orbital for a given atom. The numbering rule for ℓ is directly contingent on the value for n. Thus, for any given n, then ℓ must be such that it has integral values from 0 to (n -1). This means if n = 2, then ℓ can have (n- 1) = (2 -1) = 1. But if n =1, then: ℓ = (n – 1) = 0.
And as before we use L to distinguish the total angular momentum from the angular momentum quantum number ℓ since the magnitude of L is: L = (ħ) Ö [ℓ (ℓ + 1)]
It is also useful here to note the role of the Pauli spin matrices (s x etc.), e.g.
Angular momentum operators are also important in quantum mechanics. The physical significance of the angular momentum quantum number (ℓ) is to convey the shape of the probability density cloud or orbital for a given atom. The numbering rule for ℓ is directly contingent on the value for n. Thus, for any given n, then ℓ must be such that it has integral values from 0 to (n -1). This means if n = 2, then ℓ can have (n- 1) = (2 -1) = 1. But if n =1, then: ℓ = (n – 1) = 0.
We now look at the total angular momentum,
which would be:
L = [ℓ (ℓ +
1)] 1/2 (ħ)
Consider now
the commutator relations for the respective components L
x and L y :
[L
x , L y ] = [y p x - zp y , zp x - x p z ]
= ħ / i
= [y p x - xp y ,] = i ħ L z
Thus:
[L x , L
y ] ≠ 0
From
classical mechanics we saw for the angular momentum :
ℓ
= r x
p = r
p q = r p sin q
Where r and
p are the magnitudes of the position of the particle relative to the origin and
its momentum, respectively. But in
quantum mechanics, say for rectilinear coordinates ,we have:
L =
-i
h [i j k ]
[x y z]
[¶/¶x ¶/¶y ¶/¶ z]
Where: L z = -i
h [x ¶/¶y - y ¶/¶x ]
And as before we use L to distinguish the total angular momentum from the angular momentum quantum number ℓ since the magnitude of L is: L = (ħ) Ö [ℓ (ℓ + 1)]
It is
straightforward to show the classical rectangular components of L are:
L x = y p x - z p y
L y = z p x - xp z
L z = x p y - y p x
Where x, y,
z are the components of r, and p x , p y and p z
are the components of p.
In quantum
mechanics the angular momentum operators are obtained by replacing the momentum
components from the classical case by their quantum mechanical equivalents,
i.e.
p x
= -i h (¶/¶x )
p y
= -i h (¶/¶y )
p z
= -i h (¶/¶z )
The quantum
mechanical angular momentum operators are then written:
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)]
It is also useful here to note the role of the Pauli spin matrices (s x etc.), e.g.
L’ x = h /2 ( s x )
L’ y = h /2 ( s y )
L’ z = h /2 ( s z )
Where:
s x
=
(0.......1)
(1.......0);
(1.......0);
s y
=
(0.......-i)
(i.........0)
(i.........0)
s z
=
(1.......0)
(0......-1)
(0......-1)
And we note
here each of these matrices is Hermitian.
(See e.g. Hermitian Matrices and Orbital Angular Momentum )
(See e.g. Hermitian Matrices and Orbital Angular Momentum )
Further,
since L’ x and L’ y are Hermitian
then (L’ x + i L’ y) and (L’ x - i L’ y) are Hermitian
conjugates.
It can also
be shown:
L x L y =
- h 2 (y
¶/¶z ) – (z ¶/¶y )( z
¶/¶x – x ¶/¶z)
And:
[L, L y ] = i h 2 L
z
[ L y , L z ] =
i h 2 L
x
[ L z , L x ] =
i h 2 L
y
i.e. none of
the components of angular momentum commute with each other.
Further:
[ L x , L y ] L y
= i h L y L z
And:
L y [ L x , L y ] =
i h L y L z
Problems for the Physics enthusiast:
1) Write
each of the angular momentum operators:’
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.
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.
3) (a) Show that (L’ x + i L’ y) and (L’ x - i L’ y) are Hermitian conjugates.
b) A matrix form for one of the angular momentum Hermitan conjugates may be written:
b) A matrix form for one of the angular momentum Hermitan conjugates may be written:
(L’ x + i L’ y) m m’
=
ħ [ (ℓ
- m’ ) (ℓ + m’ + 1 ) ]1/2 exp (i m ℓ j) d m m’+1
Where
d
m m’
+1 is the Kronecker delta applied to the
quantum numbers m and m’ + 1. The table
below may be used to compute the appropriate value for the case given.
If
ℓ = 1 calculate (L’ x + i L’ y) m m’
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