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Ladder Operators
----------------
The following calculations equally apply for
angular momentum, L; spin angular momentum, S;
and the total angular momentum, J where J = L + S
(vector addition). Therefore, we can substitute L
with either S or J in the following equations.
Consider L.
Classically:
L = R x P - L points along spin axis.
In cartesian coords
L_{x} = yp_{z} - zp_{y}
L_{y} = zp_{x} - xp_{z}
L_{z} = xp_{y} - yp_{x}
QM Commutators:
[x_{i},x_{j}] = 0
[p_{i},p_{j}] = 0
[x,p_{x}] = ih
or in general
[x_{i},p_{j}] = ihδ_{ij}
[L_{x},L_{y}] = [yp_{z} - zp_{y}, zp_{x} - xp_{z}]
= -ihyp_{x} + ihxp_{y}
= ih(xp_{y} - yp_{x})
= ihL_{z}
[L_{y},L_{z}] = ihL_{x}
[L_{z},L_{x}] = ihL_{y}
What this means is that you cannot simultaneously
measure angular momentum in more than one direction.
Introduce raising and lowering (ladder) operators
similar to annihilation and creation operators.
L_{+} = L_{x} + iL_{y} and L_{-} = L_{x} - iL_{y}
[L_{+},L_{z}] = [L_{x} + iL_{y},L_{z}]
_{ } = [L_{x},L_{y}] + i[L_{y},L_{z}]
_{ } = ihL_{y} + iihL_{x}
_{ } = ihL_{y} - hL_{x}
_{ } = -h(L_{x} + iL_{y})
_{ } = -hL_{+}
Similarly, we can show:
[L_{-},L_{z}] = +hL_{-}
What do these operators do? Consider a state,
m, such that L_{z}|m> = m|m>. The eigenvalue, m,
is the value of the angular momentum along the
z axis (i.e. the magnetic quantum number, m_{l}).
Lets now form a new state by applying L_{+} to |m>
and again measuring the value of the angular
momentum along the z axis. Therefore, we can
write:
L_{z}(L_{+}|m>) = ?
Now consider the commutator [L_{+},L_{z}]. From
before,
[L_{+},L_{z}] = L_{+}L_{z} - L_{z}L_{+} = -hL_{+}
∴ L_{z}L_{+} = L_{+}L_{z} + hL_{+}
Therefore,
L_{z}(L_{+}|m>) = L_{+}L_{z}|m> + hL_{+}|m>
L_{z}(L_{+}|m>) = L_{+}mh|m> + hL_{+}|m> since L_{z}|m> = mh|m>
= (m+1)hL_{+}|m>
Therefore, L_{+}|m> = (m+1)hL_{+}|m> which means that
the operator L_{+} raises m by h. Similarly, we
can show that L_{-}|m> = (m-1)hL_{-}|m> lowers m by h.
Therefore he spectrum of L_{z} is gapped by integers.
Obviously, m cannot go on forever so we need to
define what is the maximum and min value for a
specific particle.
z z
+ + <-- m_{max}
| |
+3/2 + 2
| |
+1/2 + 1
| |
+0 + 0
| |
+-1/2 +-1
| |
+-3/2 +-2
| |
+ + <-- m_{min}
Fermion Boson
The L^{2} Operator
---------------
The square of the magnitude of the angular momentum
operator, L, is given by:
L^{2} = L_{x}^{2} + L_{y}^{2} + L_{z}^{2}
Now L_{-}L_{+} = (L_{x} - iL_{y})(L_{x} + iL_{y})
_{ } = L_{x}^{2} + L_{y}^{2} + i(L_{x}L_{y} - L_{y}L_{x})
_{ } = L_{x}^{2} + L_{y}^{2} + iihL_{z}
_{ } = L_{x}^{2} + L_{y}^{2} - hL_{z}
Therefore,
L^{2} = L_{-}L_{+} + L_{z}^{2} + hL_{z}
Now operate on m_{max}
L^{2}|m_{max}> = L_{-}L_{+}|m_{max}> + L_{z}^{2}|m_{max}> + hL_{z}|m_{max}>
L_{+}|m_{max} is not allowed, therefore,
L^{2}|m_{max}> = 0 + m_{max}^{2}|m_{max}> + hm_{max}|m_{max}>
= (m_{max}^{2} + m_{max})|m_{max}>
Classically, we would expect that L_{z}^{2}|m_{max}> = m_{max}^{2}|m_{max}>.
This corresponds to the situation where the angular
momentum is aligned along the positive z axis.
Likewise, the minimum occurs when the momentum is
aligned along the negative z axis. The total number
of states is therefore 2m_{max} + 1 where m_{max}
defines the type of particle.
The fact that we have an extra m_{max} term in the
above reflects the commutation relationships that
state we cannot measure more than one component
of momentum at a time. In essence, the term
reflects the uncertainty associated with the
angular momentum in the x and y directions that
cannot be measured.
Can you measure the z component of angular
momentum and the magnitude of the angular
momentum simultaneously? Consider,
[L^{2},L_{z}]|m_{max}>
This is equivalent to,
[L_{-}L_{+} + L_{z}^{2} + L_{z},L_{z}]|m_{max}>
[(0 + m_{max}^{2} + m_{max}),m_{max}]
Since all real numbers commute, the commutator
is equal to 0. Therefore, it is possible to
measure L^{2} and L_{z} at the same time.
The L Matrices
--------------
From the note on Clebsch-Gordan coefficients, for
l = 1, the operator matrices are:
- -
_{ } | 0 1 0 |
L_{x} = h/√2| 1 0 1 |
_{ } | 0 1 0 |
- -
- -
_{ } | 0 1 0 |
L_{y} = h/√2| -1 0 1 |
_{ } | 0 1 0 |
- -
- -
_{ } | 1 0 0 |
L_{z} = h| 0 0 0 |
_{ } | 0 0 -1 |
- -
Using the notation |l,m> we can define the 3
eigenstates as follows:
- - - - - -
| 1 | | 0 | | 0 |
|1,1> = | 0 | |1,0> = | 1 | |1,-1> = | 0 |
| 0 | | 0 | | 1 |
- - - - - -
For L_{z}, m = -1, 0, 1 we get:
- - - - - -
| 1 0 0 || 1 | | 1 |
h| 0 0 0 || 0 | = +1| 0 |h
| 0 0 -1 || 0 | | 0 |
- - - - - -
- - - - - -
| 1 0 0 || 0 | | 0 |
h| 0 0 0 || 1 | = 0| 1 |h
| 0 0 -1 || 0 | | 0 |
- - - - - -
- - - - - -
| 1 0 0 || 0 | | 0 |
h| 0 0 0 || 0 | = -1| 0 |h
| 0 0 -1 || 1 | | 1 |
- - - - - -
L^{2} = L_{x}^{2} + L_{y}^{2} + L_{z}^{2} = 2h^{2}I
In the process of solving the Schrodinger equation
for the hydrogen atom, it is found that the orbital
angular momentum is quantized according to the
relationship:
L^{2} = l(l + 1)h^{2} = 2h^{2}I
Where L now represents the orbital angular momentum
in units of h.
The operators for spin 1/2 are the Pauli matrices.
Similar calculations to the above yield:
S^{2} = S_{x}^{2} + S_{y}^{2} + S_{z}^{2} = (3h^{2}/4)I
Where S_{i} = (h/2)σ_{i}
also,
S^{2} = s(s + 1)h^{2} = 3h^{2}/4