Angular Momentum Operators

In classical mechanics, the vector angular momentum, L, of a particle of position vector ${\bf r}$ and linear momentum ${\bf p}$ is defined as

$${\bf L} = {\bf r}\times {\bf p}.$$ (526)

It follows that

$$L_x = y p_z - z p_y,$$ (527)

$$L_y = z p_x - x p_z,$$ (528)

$$L_z = x p_y - y p_x.$$ (529)

Let us, first of all, consider whether it is possible to use the above expressions as the definitions of the operators corresponding to the components of angular momentum in quantum mechanics, assuming that the $x_i$ and $p_i$ (where $x_1\equiv x$, $p_1\equiv p_x$, $x_2\equiv y$, etc.) correspond to the appropriate quantum mechanical position and momentum operators. The first point to note is that expressions (527)-(529) are unambiguous with respect to the order of the terms in multiplicative factors, since the various position and momentum operators appearing in them all commute with one another [see Eqs. (483)]. Moreover, given that the $x_i$ and the $p_i$ are Hermitian operators, it is easily seen that the $L_i$ are also Hermitian. This is important, since only Hermitian operators can represent physical variables in quantum mechanics (see Sect. 4.6). We, thus, conclude that Eqs. (527)-(529) are plausible definitions for the quantum mechanical operators which represent the components of angular momentum.

Let us now derive the commutation relations for the $L_i$. For instance,

$$[L_x, L_y] = [y p_z-z p_y, z p_x-x p_z] = y p_x [p_z,z] + x p_y [z,p_z]$$

$$= {\rm i} \hbar (x p_y-y p_x) = {\rm i} \hbar L_z,$$ (530)

where use has been made of the definitions of the $L_i$ [see Eqs. (527)-(529)], and commutation relations (481)-(483) for the $x_i$ and $p_i$. There are two similar commutation relations: one for $L_y$ and $L_z$, and one for $L_z$ and $L_x$. Collecting all of these commutation relations together, we obtain

$$[L_x, L_y] = {\rm i} \hbar L_z,$$ (531)

$$[L_y, L_z] = {\rm i} \hbar L_x,$$ (532)

$$[L_z,L_x] = {\rm i} \hbar L_y.$$ (533)

By analogy with classical mechanics, the operator $L^2$, which represents the magnitude squared of the angular momentum vector, is defined

$$L^2 = L_x^{ 2} + L_y^{ 2} + L_z^{ 2}.$$ (534)

Now, it is easily demonstrated that if $A$ and $B$ are two general operators then

$$[A^2,B]= A [A, B] +[A,B] A.$$ (535)

Hence,

$$[L^2,L_x] = [L_y^{ 2},L_x] + [L_z^{ 2},L_x]$$

$$= L_y [L_y,L_x] + [L_y,L_x] L_y + L_z [L_z,L_x] + [L_z, L_x] L_z$$

$$= {\rm i} \hbar\left(-L_y L_z - L_z L_y + L_z L_y + L_y L_z\right) = 0,$$ (536)

where use has been made of Eqs. (531)-(533). In other words, $L^2$ commutes with $L_x$. Likewise, it is easily demonstrated that $L^2$ also commutes with $L_y$, and with $L_z$. Thus,

$$[L^2, L_x]= [L^2, L_y] = [L^2,L_z] = 0.$$ (537)

Recall, from Sect. 4.10, that in order for two physical quantities to be (exactly) measured simultaneously, the operators which represent them in quantum mechanics must commute with one another. Hence, the commutation relations (531)-(533) and (537) imply that we can only simultaneously measure the magnitude squared of the angular momentum vector, $L^2$, together with, at most, one of its Cartesian components. By convention, we shall always choose to measure the $z$-component, $L_z$.

Finally, it is helpful to define the operators

$$L_\pm = L_x\pm {\rm i} L_y.$$ (538)

Note that $L_+$ and $L_-$ are not Hermitian operators, but are the Hermitian conjugates of one another (see Sect. 4.6): i.e.,

$$(L_\pm)^\dag = L_\mp,\ [0.5ex]$$ (539)

Moreover, it is easily seen that

$$L_+ L_- = (L_x+{\rm i} L_y) (L_x-{\rm i} L_y) = L_x^{ 2} + L_y^{ 2} - {\rm i} [L_x, L_y] = L_x^{ 2} + L_y^{ 2} + \hbar L_z$$

$$= L^2 - L_z^{ 2} + \hbar L_z.$$ (540)

Likewise,

$$L_- L_+ = L^2 - L_z^{ 2} -\hbar L_z,$$ (541)

giving

$$[L_+, L_-]= 2 \hbar L_z.$$ (542)

We also have

$$[L_+,L_z] = [L_x,L_z] + {\rm i} [L_y,L_z] = -{\rm i} \hbar L_y - \hbar L_x= - \hbar L_+,$$ (543)

and, similarly,

$$[L_-,L_z]= \hbar L_-.$$ (544)