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Chapter 9 Angular Momentum
Quantum Mechanical Angular Momentum Operators
Classical angular momentum is a vector quantity denoted ~L = ~r X p~. A common mnemonic to calculate the components is
~L =
ˆi ˆj ˆk x y z px py pz
ypz − zpy
i +
zpx − xpz
j +
xpy − ypx
j
= Lxˆi + Lyˆj + Lzˆj.
Let’s focus on one component of angular momentum, say Lx = ypz − zpy. On the right side of the equation are two components of position and two components of linear momentum. Quantum mechanically, all four quantities are operators. Since the product of two operators is an operator, and the difference of operators is another operator, we expect the components of angular momentum to be operators. In other words, quantum mechanically
Lx = YPz − ZPy , Ly = ZPx − X Pz , Lz = X Py − YPx.
These are the components. Angular momentum is the vector sum of the components. The sum of operators is another operator, so angular momentum is an operator. We have not encountered an operator like this one, however, this operator is comparable to a vector sum of operators; it is essentially a ket with operator components. We might write
∣ L > =
Lx Ly Lz
YPz − ZPy ZPx − X Pz X Py − YPx
A word of caution concerning common notation—this is usually written just L, and the ket/vector nature of quantum mechanical angular momentum is not explicitly written but implied.
Equation (9-1) is in abstract Hilbert space and is completely devoid of a representation. We will want to pick a basis to perform a calculation. In position space, for instance
X → x, Y → y, and Z → z,
and
Px → −i¯h
∂x
, Py → −i¯h
∂y
, and Pz → −i¯h
∂z
Equation (9–1) in position space would then be written
∣ L > =
−i¯hy (^) ∂z∂ + i¯hz (^) ∂y∂ −i¯hz (^) ∂x∂ + i¯hx (^) ∂z∂ −i¯hx (^) ∂y∂ + i¯hy (^) ∂x∂
The operator nature of the components promise difficulty, because unlike their classical analogs which are scalars, the angular momentum operators do not commute.
Example 9–1: Show the components of angular momentum in position space do not commute.
Let the commutator of any two components, say
[
Lx, Ly
]
, act on the function x. This means
[ Lx, Ly
]
x =
Lx Ly − Ly Lx
x
−i¯hy
∂z
∂y
−i¯hz
∂x
∂z
x −
−i¯hz
∂x
∂z
−i¯hy
∂z
∂y
x
−i¯hy
∂z
∂y
− i¯hz
−i¯hz
∂x
∂z
− i¯h
y
= −¯h^2 y 6 = 0,
therefore Lx and Ly do not commute. Using functions which are simply appropriate posi- tion space components, other components of angular momentum can be shown not to commute similarly.
Example 9–2: What is equation (9–1) in the momentum basis?
In momentum space, the operators are
X → i¯h
∂px
, Y → i¯h
∂py
, and Z → i¯h
∂pz
and Px → px, Py → py , and Pz → pz.
Equation (9–1) in momentum space would be written
∣ L > =
i¯h (^) ∂p∂y pz − i¯h (^) ∂p∂z py i¯h (^) ∂p∂z px − i¯h (^) ∂p∂x pz i¯h (^) ∂p∂x py − i¯h (^) ∂p∂y px
Canonical Commutation Relations in Three Dimensions
We indicated in equation (9–3) the fundamental canonical commutator is [ X , P
]
= i¯h.
This is fine when working in one dimension, however, descriptions of angular momentum are generally three dimensional. The generalization to three dimensions^2 ,^3 is
[ Xi, Xj
]
(^2) Cohen-Tannoudji, Quantum Mechanics (John Wiley & Sons, New York, 1977), pp 149 – 151. (^3) Sakurai, Modern Quantum Mechanics (Addison–Wesley Publishing Company, Reading, Mas-
sachusetts; 1994), pp 44 – 51.
Example 9–4: Develop a relation for
[
A B, C
]
in terms of commutators of individual operators. [ A B, C
]
= A B C − C A B
= A B C − A C B + A C B − C A B
= A
B C − C B
A C − C A
B
= A
[
B, C
]
[
A, C
]
B.
Example 9–5: Develop a relation for
[
A B, C D
]
in terms of commutators of individual operators.
Using the result of example 9–3, [ A B, C D
]
[
A B, C
]
D + C
[
A B, D
]
and using the result of example 9–4 on both of the commutators on the right,
[ A B, C D
]
A
[
B, C
]
[
A, C
]
B
D + C
A
[
B, D
]
[
A, D
]
B
= A
[
B, C
]
D +
[
A, C
]
B D + C A
[
B, D
]
+ C
[
A, D
]
B,
which is the desired result.
Angular Momentum Commutation Relations
Given the relations of equations (9–3) through (9–5), it follows that [ Lx, Ly
]
= i¯h Lz ,
[
Ly , Lz
]
= i¯h Lx, and
[
Lz , Lx
]
= i¯h Ly. (9 − 7)
Example 9–6: Show
[
Lx, Ly
]
= i¯h Lz. [ Lx, Ly
]
[
Y Pz − Z Py , Z Px − X Pz
]
Y Pz − Z Py
Z Px − X Pz
Z Px − X Pz
Y Pz − Z Py
= Y Pz Z Px − Y Pz X Pz − Z Py Z Px + Z Py X Pz − Z PxY Pz + Z PxZ Py + X Pz Y Pz − X Pz Z Py
=
Y Pz Z Px −Z PxY Pz
Z Py X Pz −X Pz Z Py
Z PxZ Py − Z Py Z Px
X Pz Y Pz − Y Pz X Pz
[
Y Pz , Z Px
]
[
Z Py , X Pz
]
[
Z Px, Z Py
]
[
X Pz , Y Pz
]
Using the result of example 9–5, the plan is to express these commutators in terms of individual operators, and then evaluate those using the commutation relations of equations (9–3) through (9– 5). In example 9–5, one commutator of the products of two operators turns into four commutators. Since we start with four commutators of the products of two operators, we are going to get 16
commutators in terms of individual operators. The good news is 14 of them are zero from equations (9–3), (9–4), and (9–5), so will be struck.
[ Lx, Ly
]
= Y
[
Pz , Z
]
Px +
[
Y, Z
]
Pz Px
+ Z Y
[
Pz , Px
/ ]
+ Z
[
Y, Px
]
Pz
+ Z
[
Py , X
]
Pz
[
Z, X
]
Py Pz
+ X Z
[
Py , Pz
/ ]
+ X
[
Z, Pz
]
Py
+ Z
[
Px, Z
]
Py
[
Z, Z
]
Px Py
+ Z Z
[
Px, Py
]
+ Z
[
Z, Py
]
Px
+ X
[
Pz , Y
]
Pz
[
X , Y
]
Pz Pz
+ Y X
[
Pz , Pz
/ ]
+ Y
[
X , Pz
]
Pz
= Y
[
Pz , Z
]
Px + X
[
Z, Pz
]
Py = Y
− i¯h
Px + X
i¯h
Py = i¯h
X Py − Y Px
= i¯hLz.
The other two relations,
[
Ly , Lz
]
= i¯h Lx and
[
Lz , Lx
]
= i¯h Ly can be calculated using similar procedures.
A Representation of Angular Momentum Operators
We would like to have matrix operators for the angular momentum operators Lx, Ly , and Lz. In the form Lx, Ly , and Lz , these are abstract operators in an infinite dimensional Hilbert space. Remember from chapter 2 that a subspace is a specific subset of a general complex linear vector space. In this case, we are going to find relations in a subspace C^3 of an infinite dimensional Hilbert space. The idea is to find three 3 X 3 matrix operators that satisfy relations (9–7), which are [ Lx, Ly
]
= i¯h Lz ,
[
Ly , Lz
]
= i¯h Lx, and
[
Lz , Lx
]
= i¯h Ly.
One such group of objects is
Lx =
(^) ¯h, Ly = √^1 2
0 −i 0 i 0 −i 0 i 0
(^) ¯h, Lz =
(^) ¯h. (9 − 8)
You have seen these matrices in chapters 2 and 3. In addition to illustrating some of the math- ematical operations of those chapters, they were used when appropriate there, so you may have a degree of familiarity with them here. There are other ways to express these matrices in C^3. Relations (9–8) are dominantly the most popular. Since the three operators do not commute, we arbitrarily have selected a basis for one of them, and then expressed the other two in that basis. Notice Lz is diagonal. That means the basis selected is natural for Lz. The terminology usually used is the operators in equations (9–8) are in the Lz basis.
We could have selected a basis which makes Lx or Ly , and expressed the other two in terms of the natural basis for Lx or Ly. If we had done that, the operators are different than
Example 9–8: Show L^2 = 2¯h^2 I.
L^2 = < L
∣ L >
(^) ¯h, √^1 2
0 −i 0 i 0 −i 0 i 0
(^) ¯h,
(^) ¯h
√^1 2
(^) ¯h
√^1 2
0 −i 0 i 0 −i 0 i 0
(^) ¯h
(^) ¯h
(^) ¯h √^1 2
(^) ¯h + √^1 2
0 −i 0 i 0 −i 0 i 0
(^) ¯h √^1 2
0 −i 0 i 0 −i 0 i 0
(^) ¯h
(^) ¯h
(^) ¯h
(^) ¯h^2 +^1 2
(^) ¯h^2 +
(^) ¯h^2
(^) ¯h^2 +
(^) ¯h^2 +
(^) ¯h^2
(^) ¯h^2 +
(^) ¯h^2 =
(^) ¯h^2
= 2¯h^2 I.
Complete Set of Commuting Observables ...A Discussion about
Operators which do not Commute....
The intent of this section is to appreciate non–commutivity from a new perspective, and explain “what can be done about it” if the non–commuting operators represent physical quanti- ties we want to measure. The following toy example is adapted from Quantum Mechanics and Experience^4.
We want two operators which do not commute. We are deliberately using simple operators in an effort to focus on principles. In a two dimensional linear vector space, the property of “hardness” is modelled
Hard =
(^4) Albert, Quantum Mechanics and Experience (Harvard University Press, Cambridge, Mas-
sachusetts, 1992), pp 30–33.
and has eigenvalues of ± 1 and eigenvectors
| 1 >hard =
and | − 1 >hard =
Let’s also consider the “color” operator,
Color =
with eigenvalues of ± 1 and eigenvectors
| 1 >color =
and | − 1 >color =
Note that a “hardness” or “color” eigenvector is a superposition of the eigenvectors of the other property, i.e.,
| 1 >hard =
| 1 >color +
| − 1 >color
| − 1 >hard =
| 1 >color −
| − 1 >color
| 1 >color =
| 1 >hard +
| − 1 >hard
| − 1 >color =
| 1 >hard −
| − 1 >hard
Hardness is a superposition of color states and color is a superposition of hardness states. That is the foundation of incompatibility, or non–commutivity. Each measurable state is a linear combi- nation or superposition of the measurable states of the other property. To disturb one property is to disturb both properties.
Also in chapter 3, we indicated if two Hermitian operators commute, there exists a basis of common eigenvectors. Conversely, if they do not commute, there is no basis of common eigenvec- tors. We conclude there is no common eigenbasis for the “hardness” and “color” operators.
This is exactly the status of the three angular momentum component operators, except there are three vice two operators which do not commute with one another. None of the component operators commutes with any other. There is no common basis of eigenvectors between any two, so can be no common eigenbasis between all three.
Back to the hardness and color operators. If we can find an operator with which both commute, say the two dimensional identity operator I, we can ascertain the eigenstate of the system. If we
measure an eigenvalue of 1 for color, the eigenstate is proportional to
, were we to operate
on this with the identity operator, the eigenstate of system is either
or
. If we then
measure with the hardness operator, the eigenvalue will be 1 if the state was
, or − 1 if
the state was
. We have effectively removed the indeterminacy of the system by including
comes from application of Lz. A significant portion of the reason to address angular momentum and explain the concept of a complete set of commuting observables now is for use in the next chapter on the hydrogen atom.
Ladder Operators for Angular Momentum
We are going to address angular momentum, like the SHO, from both a linear algebra and a differential equation perspective. We are going to assume rotational invariance, or spherical symmetry, so we have H, L^2 , and Lz as a complete set of commuting observables. We will address linear algebra arguments first. And we will work only with the components and L^2 , saving the Hamiltonian for the next chapter.
The four angular momentum operators are related as
L^2 = L^2 x + L^2 y + L^2 z ⇒ L^2 − L^2 z = L^2 x + L^2 y.
The sum of the two components L^2 x + L^2 y would appear to factor
( Lx + iLy
Lx − iLy
and they would if the factors were scalars, but they are operators which do not commute, so this is not factoring. Just like the SHO, it is a good mnemonic, nevertheless.
Example 9–9: Show L^2 x + L^2 y 6 =
Lx + iLy
Lx − iLy
Lx + iLy
Lx − iLy
= L^2 x − iLxLy + iLy Lx + L^2 y = L^2 x + L^2 y − i
LxLy − Ly Lx
= L^2 x + L^2 y − i
[
Lx, Ly
]
= L^2 x + L^2 y − i
i¯hLz
= L^2 x + L^2 y + ¯hLz 6 = L^2 x + L^2 y ,
where the expression in the next to last line is a significant intermediate result, and we will have reason to refer to it.
Like the SHO, the idea is to take advantage of the commutation relations of equations (9–7). We will use the notation
L+ = Lx + iLy , and L− = Lx − iLy , (9 − 12)
which together are often denoted L±. We need commutators for L±, which are
[ L^2 , L±
]
[
Lz , L±
]
= ±¯h L±. (9 − 14)
Example 9–10: Show
[
L^2 , L+
]
[
L^2 , L+
]
[
L^2 , Lx + iLy
]
[
L^2 , Lx
]
[
L^2 , Ly
]
= 0 + i(0) = 0.
Example 9–11: Show
[
Lz , L+
]
= ¯h L+. [ Lz , L+
]
[
Lz , Lx + iLy
]
[
Lz , Lx
]
[
Lz , Ly
]
= i¯hLy + i
− i¯hLx
= ¯h
Lx + iLy
= ¯hL+.
We will proceed essentially as we did the the raising and lowering operators of the SHO. Since L^2 and Lz commute, they share a common eigenbasis.
Example 9–12: Show L^2 and Lz commute.
[ L^2 , Lz
]
[
L^2 x + L^2 y + L^2 z , Lz
]
[
L^2 x, Lz
]
[
L^2 y , Lz
]
[
L^2 z , Lz
/ ]
[
Lx Lx, Lz
]
[
Ly Ly , Lz
]
= Lx
[
Lx, Lz
]
[
Lx, Lz
]
Lx + Ly
[
Ly , Lz
]
[
Ly , Lz
]
Ly = Lx
− i¯hLy
− i¯hLy
Lx + Ly
i¯hLx
i¯hLx
Ly
− i¯hLx Ly + i¯hLx Ly
− i¯hLy Lx + i¯hLy Lx
where we have used the results of example 9–4 and two of equations (9–7) in the reduction.
We assume L^2 and Lz will have different eigenvalues when they operate on the same basis vector, so we need two indices for each basis vector. The first index is the eigenvalue for L^2 , we will use α for the eigenvalue, and the second index is the eigenvalue for Lz , denoted by β. If we had a third commuting operator, for instance H which we will add in the next chapter, we would need three eigenvalues to uniquely identify each ket. Here we are considering two commuting operators, so we need two indices representing the eigenvalues of the two commuting operators.
Considering just L^2 and Lz here, the form of the eigenvalue equations must be
L^2
α, β> = α
α, β>, (9 − 15)
Lz
∣α, β> = β
∣α, β>, (9 − 16)
where
∣α, β> is the eigenstate, α is the eigenvalue of L^2 , and β is the eigenvalue of Lz.
Equation (9–14)/example 9–11 give us
[ Lz , L+
]
= Lz L+ − L+ Lz = ¯h L+
⇒ Lz L+ = L+ Lz + ¯h L+.
Using this in equation (9–16),
Lz L+
α, β> =
L+ Lz + ¯h L+
α, β> = L+ Lz
∣α, β> +¯h L+
∣α, β>
= L+ β
α, β> +¯h L+
α, β> (9 − 17)
β + ¯h
L+
∣α, β>.
where we have assumed orthonormality of eigenstates in equation (9–19). The condition that the difference in equation (9–20) is non–negative is from the fact the braket is expressible in terms of a sum of L^2 x and L^2 y , as seen in equation (9–18). Both Lx and Ly are Hermitian, so their eigenvalues are real. The sum of the squares of the eigenvalues, corresponding to operations by L^2 x and L^2 y in equation (9–18), must be non–negative. In mathematical vernacular, L^2 x and L^2 y are positive definite.
Equation (9–20) is equivalent to α ≥ β^2 , which means β is bounded for a given value of α. Therefore there is an eigenstate |α, βmax> which cannot be raised, and another eigenstate |α, βmin> which cannot be lowered. In other words, we have a ladder which has a bottom, like the SHO, and a top, unlike the SHO. In a calculation similar to example 9–9,
L−L+ = L^2 x + L^2 y − ¯h = L^2 − L^2 z − ¯hLz ,
so
L−L+|α, βmax> = ~ 0 ⇒
L^2 − L^2 z − ¯hLz
|α, βmax> = 0 (9 − 21) ⇒ L^2 |α, βmax> −L^2 z |α, βmax> −¯hLz |α, βmax> = 0 ⇒ α|α, βmax> −βmax^2 |α, βmax> −¯h βmax|α, βmax> = 0 ⇒
α − βmax^2 − ¯h βmax
|α, βmax> = 0 ⇒ α − βmax^2 − ¯h βmax = 0 ⇒ α = β^2 max + ¯h βmax. (9 − 22)
Similarly, L+L−|α, βmin> = ~ 0
⇒ α = βmax^2 − ¯h βmax. (9 − 23)
Equating equations (9–22) and (9–23), we get
β^2 max + ¯hβmax − β^2 min + ¯hβmin = 0.
This is quadratic in both βmax and βmin, and to solve the equation, we will use the quadratic formula to solve for βmax, or
βmax = −
¯h ±
¯h^2 − 4(−βmin^2 + ¯h βmin)
¯h ±
4 β^2 min − 4¯h βmin + ¯h^2
= −
¯h ±
(2βmin − ¯h)^2
¯h ±
(2βmin − ¯h)
⇒ βmax = −βmin, βmin − ¯h. (9 − 24)
The case βmax = −βmin is the maximum sep- aration case. It gives us the top and bottom of the ladder. We assume the rungs of the ladder are
If we had started out with J± = Jx ± iJy , or S± = Sx ± iSy , we would have come out with exactly the same result. In fact, this is the problem we solved, except using the symbol L vice J or S.
We will reinforce in chapter 13 that spin can have half integral values, or values of multiples of ¯h/2. Since spin can be half integral, values of total angular momentum can be half integral. When we use symbols such as L^2 and Li, we get the information contained in the commutation relations, independent of whatever symbols we choose. Had we used explicit representations, such as equations (9–8) and (9–9), we would get the same information, however, limited by the representation. In that case, only integral values would be possible, though the form of the result analogous to equation (9–25) would remain the same. Using l as the quantum number for orbital angular momentum, the eigenvalue for orbital angular momentum squared is
α = ¯h^2 l(l + 1). (9 − 27)
A comment about the picture and notation is appropriate. The first impression is that this is similar to classical mechanics. The earth orbits the sun and has orbital angular momentum in that regard, and also spins on its axis so has spin angular momentum as well. It is tempting to apply this picture to a quantum mechanical system, say an electron in a hydrogen atom. It simply does not apply. The electron is not a small ball spinning on its axis as it orbits the proton. Per the first chapter, an electron is not a particle, it is not a wave, it is an electron. There is no classical analogy for an electron, and many of the manifestations of quantum mechanical angular momentum are similarly not classical analogs.
Equation (9–26) says the total is the sum of the parts, but it is an operator equation which in Dirac notation is |J > = |L> + |S>. Since earlier development was in this form, it may be useful to assist you to realize that each of these three operators has three components which are also each operators. Equation (9-26) is standard notation nevertheless.
Eigenvalue Solution for the Z Component of Orbital
Angular Momentum
We have calculated the eigenvalue of L^2 , but still need to find the eigenvalue of Lz. We know one of the possible eigenvalues of Lz is zero from the last of equations (9–8), the explicit representations, regardless of the eigenstate. We have also calculated
Lz
L+
∣α, β> )^ = (β + ¯h)(L+
∣α, β> ).
If we start with an eigenstate that has the z component of angular momentum equal to zero,
Lz
L+
α, 0 >
0 + ¯h
L+
α, 0 >
= ¯h
L+
α, 0 >
so ¯h is the next eigenvalue. Using ¯h as the eigenvalue,
Lz
L+
∣α, ¯h>
¯h + ¯h
L+
∣α, ¯h>
= 2¯h
L+
∣α, ¯h>
so 2¯h is the next eigenvalue. If we use this as an eigenvalue,
Lz
L+
∣α, 2¯h> )^ = (¯h + 2¯h)(L+
∣α, 2¯h> )^ = 3¯h(L+
∣α, 2¯h> ),
and 3¯h is the next eigenvalue up the ladder. We can continue, and will attain integral values of ¯h. But we cannot continue forever, because we determined β is bounded by the eigenvalue of L^2. What is the maximum value? We go back to figure 9–1 and the result from this figure is
βmax =
n¯h 2
where we want only integral values for the orbital angular momentum, so this becomes
βmax = l¯h.
Were we to do the same calculation with the lowering operator, that is
Lz
L−
∣α, 0 > )^ = −¯h(L−
∣α, 0 > ),
we step down the ladder in increments of −¯h until we get to βmin. Remember βmin also has a minimum, which is of the same magnitude but negative or
βmin = −l¯h.
So we have eigenvalues which climb to l¯h and drop to −l¯h in integral increments of ¯h. The eigenvalue of the z component of angular momentum is just an integer times ¯h, from minimum to maximum values. The symbol conventionally used to denote this integer is m, so
Lz
∣α, β> = m¯h
∣α, β>, −l < m < l
is the eigenvalue/eigenvector equation for the z component of angular momentum. The quantum number m, occasionally denoted ml, is known as the magnetic quantum number.
Eigenvalue/Eigenvector Equations for Orbital Angular
Momentum
If we use l vice α to denote the state of total angular momentum, realizing l itself is not an eigenvalue of L^2 , and m to denote the state of the z component of angular momentum, realizing the eigenvalue of Lz is actually m¯h, the eigenvalue/eigenvector equations for L^2 and Lz are L^2
l, m> = ¯h^2 l(l + 1)
l, m>, (9 − 28)
Lz
∣l, m> = m¯h
∣l, m>, −l < m < l (9 − 29)
which is the conventional form of the two eigenvalue/eigenvector equations for L^2 and Lz.
Angular Momentum Eigenvalue Picture for Eigenstates
What is |l, m>? It is an eigenstate of the commuting operators L^2 and Lz. The quantum numbers l and m are not eigenvalues. The corresponding eigenvalues are ¯h^2 l(l + 1) and m¯h. Were we to use eigenvalues in the ket, the eigenstate would look like |¯h^2 l(l + 1), m¯h>. But just l and m uniquely identify the state, and that is more economical, so only the quantum numbers are conventionally used. This is essentially the same sort of convenient shorthand used to denote an eigenstate of a SHO |n>, vice using the eigenvalue
∣(n + 12 )¯hω>.
Only one quantum number is needed to uniquely identify an eigenstate of a SHO, but two are needed to uniquely identify an eigenstate of angular momentum. Because the angular momentum component operators do not commute, a complete set of commuting observables are needed. Each of the component operators commutes with L^2 , so we use it and one other, which is Lz chosen by convention. One quantum number is needed for each operator in the complete set. Multiple quantum numbers used to identify a ket denote a complete set of commuting observables is needed.
Remember that a system is assumed to exist in a linear combination of all possible eigenstates until we measure. If we measure, what are the possible outcomes? Possible outcomes are the eigenvalues. For a given value of the orbital angular momentum quantum number, the magnetic quantum number can assume integer values ranging from −l to l. The simplest case is l = 0 ⇒ m = 0 is the only possible value of the magnetic quantum number. The possible outcomes of a measurement of such a system are eigenvalues of ¯h^2
= 0¯h^2 or just 0 for L^2 , and m¯h =
¯h or just 0 for Lz as well, corresponding to figure 9–2.a.
Figure 9 − 2 .a. l = 0. Figure 9 − 2 .b. l = 1. Figure 9 − 2 .c. l = 2.
If we somehow knew l = 1, which could be ascertained by a measurement of ¯h^2
= 2¯h^2 for L^2 , the possible values of the magnetic quantum number are m = − 1 , 0, or 1, so the eigenvalues which could be measured are −¯h, 0, or ¯h for Lz , per figure 9–2.b. If we measured ¯h^2
= 6¯h^2 for L^2 , we would know we had l = 2, and the possible values of the magnetic quantum number are m = − 2 , − 1 , 0 , 1, or 2, so the eigenvalues which could be measured are −2¯h, −¯h, 0 , ¯h, or 2¯h for Lz , per figure 9–2.c. Though the magnetic quantum number is bounded by the orbital angular momentum quantum number, the orbital angular momentum quantum number is not bounded, so we can continue indefinitely. Notice there are 2l + 1 possible values of m for every value of l.
A semi–classical diagram is often used. A simple interpretation of |l, m> is that it is a vector quantized in length of (^) ∣ ∣L
∣L
L^2 = ¯h
l(l + 1).
