# 6.3: The Three Components of Angular Momentum Cannot be Measured Simultaneously with Arbitrary Precision

Consider a particle described by the Cartesian coordinates $$(x, y, z)\equiv {\bf x}$$ and their conjugate momenta $$(p_x, p_y, p_z)\equiv {\bf p}$$. The classical definition of the orbital angular momentum of such a particle about the origin is $${\bf L} = {\bf x}\times{\bf p}$$, giving (i.e., via the cross product):

$\hat{L}_x = y\, p_z - z\, p_y, \label{6.3.1a}$

$\hat{L}_y = z\, p_x - x\, p_z \label{6.3.1b}$

$\hat{L}_z = x\,p_y - y \,p_x \label{6.3.1c}$

Let us assume that the operators $$(\hat{L}_x, \hat{L}_y, \hat{L}_z)\equiv {\bf L}$$ which represent the components of orbital angular momentum in quantum mechanics can be defined in an analogous manner to the corresponding components of classical angular momentum. In other words, we are going to assume that the above equations specify the angular momentum operators in terms of the position and linear momentum operators.

In Cartesian coordinates, the three components of orbital angular momentum can be written

$\hat{L}_x = -{\rm i}\,\hbar\left(y\,\dfrac{\partial}{\partial z} - z\,\dfrac{\partial} {\partial y}\right) \label{6.3.2a}$

$\hat{L}_y = -{\rm i}\,\hbar\left(z\,\dfrac{\partial}{\partial x} - x\,\dfrac{\partial} {\partial z}\right) \label{6.3.2b}$

$\hat{L}_z = -{\rm i}\,\hbar\left(x\,\dfrac{\partial}{\partial y} - y\,\dfrac{\partial} {\partial x}\right) \label{6.3.2c}$

using the Schrödinger representation. Transforming to standard spherical polar coordinates,

\begin{align} x &= r \,\sin\theta\, \cos\varphi \label{6.3.3a} \\[5pt] y &= r\, \sin\theta\, \sin\varphi \label{6.3.3b} \\[5pt] z &=r \cos \theta \label{ 6.3.3c} \end{align}

we obtain

\begin{align} \hat{L}_x &= {\rm i}\,\hbar\,\left(\sin\varphi\, \dfrac{\partial}{\partial \theta} + \cot\theta \cos\varphi\,\dfrac{\partial}{\partial \varphi}\right) \label{6.3.4a} \\[5pt] \hat{L}_y &= -{\rm i} \,\hbar\,\left(\cos\varphi\, \dfrac{\partial}{\partial\theta} -\cot\theta \sin\varphi \,\dfrac{\partial}{\partial \varphi}\right) \label{6.3.4b} \\[5pt] \hat{L}_z &= -{\rm i}\,\hbar\,\dfrac{\partial}{\partial\varphi} \label{6.3.4c} \end{align}

We can introduce a new operator $$\hat{L}^2$$:

\begin{align} \hat{L}^2 &= L_x^{\,2}+L_y^{\,2}+L_z^{\,2} \label{6.3.5} \\[5pt] &= - \hbar^2\left( \dfrac{1}{\sin\theta}\dfrac{\partial}{\partial \theta} \sin \theta \dfrac{\partial}{\partial \theta} + \dfrac{1}{\sin^2\theta}\dfrac{\partial^2} {\partial\varphi^2}\right) \label{6.3.6} \end{align}

The eigenvalue problem for $$\hat{L}^2$$ takes the form

$\hat{L}^2 | \psi \rangle = \lambda \,\hbar^2 | \psi \rangle \label{6.3.6a}$

where $$\psi(r, \theta, \varphi)$$ is the wavefunction, and $$\lambda$$ is a number. Let us write

$\psi(r, \theta, \varphi) = R(r) \,Y(\theta, \varphi) \label{6.3.6b}$

By definition,

$\color{red}L^2 \,Y_{l\,m_l} = l\,(l+1)\,\hbar^2\,Y_{l\,m_l} \label{6.3.9}$

where $$l$$ is an integer. This is an important conclusion that argues the angular momentum is quantized with the square of the magnitude of the angular momentum only capable of assume one of the discrete set of values (Equation $$\ref{6.3.9}$$). From this, the amplitude of angular momentum can be expressed

$\color{red} |L| =\sqrt{L^2} = \sqrt{l(l+1)} \hbar \label{6.3.10}$

Warning

We often refer to a particle in a state with angular momentum quantum number $$l$$ as having angular momentum $$l$$, rather than saying that it has angular momentum of $$\sqrt{l(l+1)} \hbar$$ magnitude, primarily since it is awkward to say quickly.

The properties of spherical harmonics that the z-component of the angular momentum ($$L_z$$) is also quantized and can only assume a one of a discrete set of values

$\color{red}L_z \,Y_{l\,m} = m\,\hbar\,Y_{l\,m} \label{6.3.11}$

where $$m_l$$ is an integer lying in the range $$-l\leq m \leq l$$.

Thus, the wavefunction $$\psi(r, \theta, \varphi) = R(r) \,Y_{l,m}(\theta, \phi)$$, where $$R$$ is a general function, has all of the expected features of the wavefunction of a simultaneous eigenstate of $$L^2$$ and $$L_z$$ belonging to the quantum numbers $$l$$ and $$m$$.

• $$l$$ is sometimes called "azimuthal quantum number" or "orbital quantum number"
• $$m_l$$ is sometimes called "magnetic quantum number"

### Simultaneous Measurements

Note that $$L_x$$, $$L_y$$, and $$L_z$$ can, in principle, be measured. However, to determine if they can be measured simultaneously with infinite precision, these operators must commute. Remember that the fundamental commutation relations satisfied by the position and linear momentum operators are:

\begin{align} [x_i, x_j] &=0 \label{6.3.12} \\[5pt] [p_i, p_j] &=0 \label{6.3.13} \\[5pt] [x_i, p_j] &= {\rm i}\,\hbar \,\delta_{ij} \label{6.3.14} \end{align}

where $$i$$ and $$j$$ stand for either $$x$$, $$y$$, or $$z$$. Consider the commutator of the operators $$L_x$$ and $$L_z$$ :

\begin{align} [L_x, L_y] & = [(y\,p_z-z\,p_y), (z\,p_x-x \,p_z)] \\[5pt] &= y\,[p_z, z]\,p_x + x\,p_y\,[z, p_z] \label{6.3.15} \\[5pt] &= {\rm i}\,\hbar\,(-y \,p_x+ x\,p_y) \\[5pt] &= {\rm i}\,\hbar\, L_z \label{6.3.16} \end{align}

The cyclic permutations of the above result yield the fundamental commutation relations satisfied by the components of an orbital angular momentum:

$[L_x, L_y] = {\rm i}\,\hbar\, L_z \label{6.3.17a}$

$[L_y, L_z] = {\rm i}\,\hbar\, L_x \label{6.3.17b}$

$[L_z, L_x] = {\rm i}\,\hbar\, L_y \label{6.3.17c}$

These can be summed up more succinctly by writing

${\bf L}\times {\bf L} = {\rm i}\,\hbar \,{\bf L} \label{6.3.18}$

Example $$\PageIndex{1}$$: Commutators

Show that the $$\hat{L}^2$$ and $$\hat{L}_x$$ operators commute.

Solution

We want to confirm that $$[\hat{L}^2, \hat{L}_x] = 0$$ that from Equation $$\ref{6.3.5}$$ this can be expanded

$[\hat{L}^2, \hat{L}_x] = [\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2, \hat{L}_x] \nonumber$

from the properties of commutators, this can be expanded

$[\hat{L}^2, \hat{L}_x] = [\hat{L}_x^2, \hat{L}_x] + [\hat{L}_y^2, \hat{L}_x] + [\hat{L}_z^2 , \hat{L}_x] \nonumber$

However,

$[\hat{L}_x^2, \hat{L}_x] = \hat{L}_x^2 \hat{L}_x - \hat{L}_x \hat{L}_x^2 = \hat{L}_x \hat{L}_x \hat{L}_x - \hat{L}_x \hat{L}_x \hat{L}_x = 0\nonumber$

So

\begin{align*} [\hat{L}^2, \hat{L}_x] &= [\hat{L}_y^2, \hat{L}_x] + [\hat{L}_z^2, \hat{L}_x]\nonumber \\ &= \hat{L}_y^2 \hat{L}_x - \hat{L}_x \hat{L}_y^2 + \hat{L}_z^2 \hat{L}_x - \hat{L}_x \hat{L}_z^2 \nonumber \\ &= \hat{L}_y \hat{L}_y \hat{L}_x - \hat{L}_x \hat{L}_y \hat{L}_y + \hat{L}_z \hat{L}_z \hat{L}_x - \hat{L}_x \hat{L}_z \hat{L}_z \nonumber \end{align*}

Lets look at some related forms which can be used to simplify the above expression. The first two terms can and final two terms can be rewritten as different commutators

\begin{align*} [\hat{L}_y , \hat{L}_x] &= \hat{L}_y + \hat{L}_y [\hat{L}_y , \hat{L}_x] \nonumber \\ &= (\hat{L}_y \hat{L}_x - \hat{L}_x \hat{L}_y) \hat{L}_y + \hat{L}_y (\hat{L}_y \hat{L}_x - \hat{L}_x \hat{L}_y) \nonumber \\ &= \cancel{\hat{L}_y \hat{L}_x \hat{L}_y} - \hat{L}_x \hat{L}_y \hat{L}_y + \hat{L}_y \hat{L}_y \hat{L}_x - \cancel{\hat{L}_y \hat{L}_x \hat{L}_y} \nonumber \end{align*}

The first & fourth terms cancel, giving

$[\hat{L}_y , \hat{L}_x] \hat{L}_y + \hat{L}_y [\hat{L}_y , \hat{L}_x] = \hat{L}_y \hat{L}_y \hat{L}_x - \hat{L}_x \hat{L}_y \hat{L}_y \nonumber$

Similarly,

$[Lz , \hat{L}_x] \hat{L}_z + \hat{L}_z [\hat{L}_z, \hat{L}_x] = \hat{L}_z \hat{L}_z \hat{L}_x - \hat{L}_x \hat{L}_z \hat{L}_z \nonumber$

So,

\begin{align*} [\hat{L}^2 , \hat{L}_x] &= [\hat{L}_y , \hat{L}_x] \hat{L}_y + \hat{L}_y [\hat{L}_y , \hat{L}_x] + [\hat{L}_z , \hat{L}_x] \hat{L}_z + \hat{L}_z [\hat{L}_z , \hat{L}_x]\nonumber \\[5pt] &= - i\hbar \hat{L}_z \hat{L}_y - i\hbar \hat{L}_y \hat{L}_z + i\hbar \hat{L}_y \hat{L}_z + ih \hat{L}_z \hat{L}_y = 0 \nonumber \end{align*}

One can also show similarly that

$[\hat{L}^2, \hat{L}_y] = [\hat{L}^2, \hat{L}_z] = 0 \nonumber$

The three commutation relations (Equations $$\ref{6.3.17a}$$ - $$\ref{6.3.17c}$$) are the foundation for the whole theory of angular momentum in quantum mechanics. Whenever we encounter three operators having these commutation relations, we know that the dynamical variables that they represent have identical properties to those of the components of an angular momentum (which we are about to derive). In fact, we shall assume that any three operators that satisfy the commutation relations (Equations $$\ref{6.3.17a}$$ - $$\ref{6.3.17c}$$) represent the components of some sort of angular momentum.

In fact, we shall assume that any three operators that satisfy the commutation relations (Equations $$\ref{6.3.17a}$$ - $$\ref{6.3.17c}$$) represent the components of some sort of angular momentum.

The fact that $$L_x$$ is known with certainty, but $$L_x$$ and $$L_y$$ are unknown; therefore every classical vector with the appropriate length and z-component can drawn, forming a cone (Figure $$\PageIndex{1}$$). The expected value of the angular momentum for a given ensemble of systems in the quantum state characterized by $$l$$ and $$m_l$$ could be somewhere on this cone while it cannot be defined for a single system (since the components of $$L$$ do not commute with each other.

Figure $$\PageIndex{1}$$: Illustration of the vector model of orbital angular momentum. A set of states with quantum numbers $$l=2$$, and $$m_l = -2, -1, 0, +1, +2$$. Image used with permission from Wikipedia.

### Summary

In the quantum world, angular momentum is quantized. The square of the magnitude of the angular momentum (determined by the eigenvalues of the $$\hat{L}^2$$ operator) can only assume one of the discrete set of values

$l(l + 1)\hbar^2 \nonumber$

with $$l = 0, 1, 2, ... \nonumber$$

The z-component of the angular momentum (i.e., projection of $$L$$ onto the $$z$$-axis) is also quantized with

$L_z= m \hbar \nonumber$

with

$m_l = -l, 0-1, ..., 0, ... +l +1, l \nonumber$

for a given value of $$l$$. Hence, $$l$$ and $$m_l$$ are the angular momentum quantum number and the magnetic quantum number, respectively.

If two operators commute then share a common set of eigenfunctions!

Assume that $$a$$ is an eigenfunction of $$\hat{A}$$ with eigenvalue $$|a \rangle$$. Then since $$\hat{A}$$ and $$\hat{B}$$ commute, then

$\hat{A}\hat{B} |a \rangle = \hat{B}\hat{A} |a \rangle = a \hat{B} |a \rangle$

Therefore, $$\hat{B} |a \rangle$$ must be an eigenfunction of $$\hat{A}$$ with eigenvalue $$a$$ just like $$|a \rangle$$ itself is. That is essentially saying that $$|a \rangle$$ is an eigenfunction of $$\hat{B}$$.

A key example of this is since $$\hat{L}^2$$ and $$\hat{L}_x$$ commute (Example $$\PageIndex{1}$$) then both operators share the same eigenstates. Hence, we do no need to solve two eignevalue problems:

$\hat{L}^2 | \psi \rangle = \lambda | \psi \rangle$

and

$\hat{L}_x | \psi \rangle = \beta| \psi \rangle$

If we solve one, we then know the eigenvalues ($$| \psi \rangle$$) for the other!