Chapter 3: Angular momentum

Note

It’s not merely that you don’t know all three components of the angular momentum L; there simply aren’t three components – a particle just cannot have a determinate angular momentum vector, any more than it can simultaneously have a determinate position and momentum. – David Griffith, Reed College

Summary

Attention

In Chapter 1: Stern-Gerlach Experiments, we introduced the ket vector ( $\ket{\psi}$ ) and found that the Stern-Gerlach experiments yield observations that cannot be explained using classical arguments. In Chapter 2: Rotation of basis states and matrix mechanics, we learned how a ket vector can be modified using operators. In particular, while building a (unitary) operator that rotates a spin-state around an axis, we introduced generators of rotation that turned out to be Hermitian operators corresponding to the spin angular momentum (the requirement for hermiticity was due solely to the fact the rotations are unitary operations). By correspond, we mean that the eigenvectors and eigenvalues of those operators are what we actually measure in the SG experiments. In this chapter, we take a deeper dive into the properties of those operators and, in particular, why SG experiments showed we can’t have a knowledge of two projections of the spin angular momentum along two axes at a given time.

First, we use a geometrical argument that states that when applying two rotations consecutively, the order of the operations matters (unless the rotations are performed along the same axis). It is useful to stress that the importance of the order of the operations is already present in usual geometry. The importance of the order is formalized by the so-called commutators. We use the expression for rotation of spin states to account for the fact that the rotations do not commute. This translates into key relationships between the different spin angular momenta, e.g.,

$\comm{\hat{J}_x}{\hat{J}_y}=i\hbar\hat{J}_z, ~ \comm{\hat{J}_y}{\hat{J}_z}=i\hbar\hat{J}_x,~ \comm{\hat{J}_z}{\hat{J}_x}=i\hbar\hat{J}_y$

We also have (with $\mathbf{\hat{J}}^2=\hat{J}_x^2+\hat{J}_y^2+\hat{J}_z^2$ ):

$\comm{\mathbf{\hat{J}}^2}{\hat{J}_x}=\comm{\mathbf{\hat{J}}^2}{\hat{J}_y}=\comm{\mathbf{\hat{J}}^2}{\hat{J}_z}=0.$

Second, we remember, from elementary linear algebra, that two operators that commute share the same eigenvectors (in the general case of degenerate eigenvalues, we can find linear combinations of the eigenvectors corresponding to a given eigenvalue that are eigenvectors for both operators, see box below for more information on this central point needed to understand quantum mechanics). The interesting result has the corollary that states that if two operators do not commute, they do not share the same eigenvectors. As a result, we realize that this explains one of the most curious results of SG experiments: there is no common eigenvectors shared by $\hat{S}_x$ and $\hat{S}_z$ , or, using the methaphoric language used in class: one can’t know the color of the monster’s shirt and the color of its socks at the same time.

Third, we examine what the actual eigenvalues of the angular momentum operators are. To achieve this, we introduce the raising and lowering operators (very handy operators that will be very useful as we progress in the course) to find that the eigenvalues are components of the angular momentum operators can only assume integer or half-integer values. For a spin j-system, the eigenvalues of $\hat{S}_x$ takes all values from $-j\hbar$ and $+j\hbar$ . The corresponding eigenvalue of $\mathbf{\hat{J}}^2$ is $j(j+1)\hbar^2$ with degeneracy $2j+1$ . The fact the largest eigenvalue for the projection of the angular momentum is smaller than the length of the total angular momentum is a direct consequence of the fact we can’t know the value of two or more projections at any given time.

There is an elegant way to further realize the importance of the commutation relation regarding the interpretation of quantum mechanics, and in particular the concept of uncertainty. Indeed, using Schwartz inequality applied for the Hilbert space (i.e., the space where quantum state vectors live), we find that the product of the uncertainty of two Hermitian operators is related to the expectation value of the commutator between those two operators.

Note

Two operators that commute share the same eigenvectors

If two operators commute, then there exists a basis that is simultaneously an eigenbasis for both operators.

In the case when there is degeneracy however, any linear combination of two eigenvectors is also an eigenvector of that operator, but that linear combination might not be an eigenvector of the second operator (this is explained with an example in the box below). Another way to say this is that we have a fredom of choice *within the subspace of eigenvectors corresponding to a given eigenvalue).

Note

Degeneracy

In quantum mechanics, an eigenvalue is degenerate if it corresponds to two or more different measurable eigenstates. An example of degeneracy is the concept of multiplicity, which applies specifically to degenerate (or, in general quasi-degenerate) spin states.

For example, the following matrix

$\hat{A}=\left(\begin{array}{lll}1 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 2\end{array}\right)$

has eigenvalues $\lambda_1=1$ , $\lambda_2=2$ , and $\lambda_3=2$ . Thus, the eigenvalue 2 has a twofold degeneracy.

How can we chose the eigenvectors in case of degeneracy? In the example above, the eigenstate for $\lambda_1$ is unambiguously given by

$\mathbf{e}_1=\left(\begin{array}{l}1 \\0 \\0\end{array}\right)$

However, we have a choice regarding $\mathbf{e}_2$ and $\mathbf{e}_3$ as any linear combination of them will yield the correct eigenvalue.

$\mathbf{e}_2=c_1\left(\begin{array}{l}0 \\1 \\0\end{array}\right)+c_2\left(\begin{array}{l}0 \\0 \\1\end{array}\right)$

and

$\mathbf{e}_3=d_1\left(\begin{array}{l}0 \\1 \\0\end{array}\right)+d_2\left(\begin{array}{l}0 \\0 \\1\end{array}\right),$

where the coefficients must be chosen so that the two vectors are normalized (2 constraints) and orthogonal (1 constraint) . We see that even with those three conditions, we still have one degree of freedom to determine the eigenstates (note: this degree of freedom corresponds to the rotation of the y and z axes in the yz plane, in this example).

So, when we say that there exists a common set of eigenvectors for two operators that commute, it does not mean that all eigenvectors will do the job. Instead, it means we can always find a linear combination that will yield the desired result of a common set of eigenvectors. This is a common difficulty for students. Make sure you understand this well.

In the case of our angular momentum operators, it is always possible to find a set of eigenstates that diagonalize (note: expressing an operator in the eigenvector basis yields a diagonal matrix; thus “diagonalizing” and “finding eigenstates” are synonymous) $\hat{\mathbf{J}}^{2}$ and $\hat{J}_z$ or $\hat{\mathbf{J}}^{2}$ and $\hat{J}_y$ or $\hat{\mathbf{J}}^{2}$ and $\hat{J}_x$ . However, the linear combination used in one case does not work with the other cases since the operators $\hat{J}_x$ , $\hat{J}_y$ , $\hat{J}_z$ do not commute.

Learning Material

Copy of Slides

The slides for Chapter 3 are available in pdf format here: 📂.

Screencast

The last three sections of this chapter are available in a separate video you can access here.

Key Learning Points

Much of quantum mechanics is formalized using Linear Algebra, and in particular properties of commutators:
- Definition: $\comm{\hat{A}}{\hat{B}}=\hat{A}\hat{B}-\hat{B}\hat{A}$
- Permutation: $\comm{\hat{B}}{\hat{A}}=-\comm{\hat{A}}{\hat{B}}$
- Commutator of products: $\comm{\hat{A}}{\hat{B}\hat{C}}=\hat{B}\comm{\hat{A}}{\hat{C}}+\comm{\hat{A}}{\hat{B}}\hat{C}$
Angular Momentum vector: $\hat{\mathbf{J}}=\hat{J}_{x} \mathbf{i}+\hat{J}_{y} \mathbf{j}+\hat{J}_{z} \mathbf{k}$
Square of total momentum: $\hat{\mathbf{J}}^{2}=\hat{\mathbf{J}} \cdot \hat{\mathbf{J}}=\hat{J}_{x}^{2}+\hat{J}_{y}^{2}+\hat{J}_{z}^{2}$
Commutation relations:
- $\comm{\hat{J}_x}{\hat{J}_y}=i\hbar\hat{J}_z, ~ \comm{\hat{J}_y}{\hat{J}_z}=i\hbar\hat{J}_x,~ \comm{\hat{J}_z}{\hat{J}_x}=i\hbar\hat{J}_y$
- $\left[\hat{\mathbf{J}}^{2}, \hat{J}_{x}\right]=0,~\left[\hat{\mathbf{J}}^{2}, \hat{J}_{y}\right]=0,~\left[\hat{\mathbf{J}}^{2}, \hat{J}_{z}\right]=0$
The relations in #4 show that while we cannot find common eigenvectors between different components of the angular momentum, we can find eigenstates common to, e.g., $\hat{\mathbf{J}}^{2}$ and $\hat{J}_z$ :

$\begin{array}{l}\hat{\mathbf{J}}^{2}|\lambda, m\rangle=\lambda \hbar^{2}|\lambda, m\rangle \\\hat{J}_{z}|\lambda, m\rangle=m \hbar|\lambda, m\rangle\end{array}$

As described in Chapter 1: Stern-Gerlach Experiments, we used labels to write the ket states, on the basis of the knowledge we have, that is: the corresponding eigenvalues of $\hat{\mathbf{J}}^{2}$ and $\hat{J}_z$ .
Eigenvalues of $\hat{J}_z$ and $\hat{\mathbf{J}}^{2}$ :

$\begin{array}{l} \hat{\mathbf{J}}^{2}|j, m\rangle=j(j+1) \hbar^{2}|j, m\rangle \\ \hat{J}_{z}|j, m\rangle=m \hbar|j, m\rangle \quad m=j, j-1, j-2, \ldots,-j+1,-j \end{array}$
Raising and Lowering operators:
- Definition: $\hat{J}_{\pm}=\hat{J}_{x} \pm i \hat{J}_{y}$
- Properties: these operators are neither Hermitian nor unitary.
- Effect on eigenstates of $\hat{J}_{z}$ :
  
  $\begin{array}{l} \hat{J}_{+}|j, m\rangle=\sqrt{j(j+1)-m(m+1)} \hbar|j, m+1\rangle \\ \hat{J}_{-}|j, m\rangle=\sqrt{j(j+1)-m(m-1)} \hbar|j, m-1\rangle \end{array}$
- The result above clearly justifies the names raising and lowering. Note that raising the maximum state (m=j) or lowering the minimum state (m=-j) yield the zero vector, as expected.
Uncertainty: If we have two Hermitian operators such that:

$\comm{\hat{A}}{\hat{B}}=i \hat{C}$

Then, the generalized uncertainty relationship reads:

$\Delta A \Delta B \geq \frac{|\langle C\rangle|}{2}$
Pauli Matrices are the representations of the angular momentum operators of a spin-1/2 particles

$\hat{S}_{x} \rightarrow \frac{\hbar}{2}\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) \quad \hat{S}_{y} \rightarrow \frac{\hbar}{2}\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right) \quad \hat{S}_{z} \rightarrow \frac{\hbar}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$

Warning

How to calculate the matrix representation of angular momenta?

Given the spin j of a system, we wish to represent $\hat{J}_{x}$ , $\hat{J}_{y}$ , $\hat{J}_{z}$ , and $\mathbf{\hat{J}}^2$ in the basis of the eigenstates of $\hat{J}_{z}$ , and $\mathbf{\hat{J}}^2$ . Here is the procedure to follow:

First, we know that for a spin j, there are 2j+1 spin projections; this is the rank of all of the matrices involved.
Second one can easily build the representations of $\hat{J}_{z}$ and $\mathbf{\hat{J}}^2$ in the basis of their eigenstates (the matrices are diagonal with the eigenvalues on the diagonal). For spin-1/2:

$\hat{J}_{z} \rightarrow \frac{\hbar}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$

$\mathbf{\hat{J}}^2 \rightarrow \frac{3\hbar^2}{4}\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)$
How can we determine the other components of the angular momentum? For this we use the raising and lowering operators can be used to obtain $\hat{J}_{x}$ and $\hat{J}_{y}$ .

$\hat{J}_{+} \rightarrow\left(\begin{array}{cc} \bra{1/2,+1/2}\hat{J}_{+}\ket{1/2,+1/2} & \bra{1/2,+1/2}\hat{J}_{+}\ket{1/2,-1/2} \\ \bra{1/2,-1/2}\hat{J}_{+}\ket{1/2,+1/2} & \bra{1/2,-1/2}\hat{J}_{+}\ket{1/2,-1/2} \end{array}\right)=\hbar\left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right)$

$\hat{J}_{-} \rightarrow\left(\begin{array}{cc} \bra{1/2,+1/2}\hat{J}_{-}\ket{1/2,+1/2} & \bra{1/2,+1/2}\hat{J}_{-}\ket{1/2,-1/2} \\ \bra{1/2,-1/2}\hat{J}_{-}\ket{1/2,+1/2} & \bra{1/2,-1/2}\hat{J}_{-}\ket{1/2,-1/2} \end{array}\right)=\hbar\left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right)$
It is now easy to find $\hat{J}_{x}$ and $\hat{J}_{y}$ :

$\begin{array}{l} \hat{J}_{+}=\hat{J}_{x}+i \hat{J}_{y} \\ \hat{J}_{-}=\hat{J}_{x}-i \hat{J}_{y} \end{array}$

and finally:

$\hat{J}_{x} \rightarrow \frac{\hbar}{2}\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)$

and

$\hat{J}_{y} \rightarrow \frac{\hbar}{2}\left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right)$

Test your knowledge

Calculate the following commutator: $\comm{\hat{J}_x}{\hat{J}_y+\hat{J}_z}$
1. $0$
2. $i\hbar(\hat{J}_z+\hat{J}_y)$
3. $i\hbar(\hat{J}_z-\hat{J}_y)$
4. $i\hbar(\hat{J}_x)$
Calculate the following commutator: $\comm{\hat{J}_x}{\hat{J}_y^2}$
1. $i\hbar(\hat{J}_z\hat{J}_y+\hat{J}_y\hat{J}_z)$
2. $i\hbar(\hat{J}_z\hat{J}_y-\hat{J}_y\hat{J}_z)$
3. $\hat{J}_z\hat{J}_y+\hat{J}_y\hat{J}_z$
4. $\hat{J}_z\hat{J}_y-\hat{J}_y\hat{J}_z$
Calculate the following commutator: $\comm{\hat{J}_x^2}{\mathbf{\hat{J}}^2}$
1. $0$
2. $-\hbar^4$
3. $-i\hbar^4$
4. $\hbar^2(\hat{J}_z^2+\hat{J}_y^2)$
You are given the following three matrix representations for a spin-1 particle:

$\hat{J}_{x} \rightarrow \frac{\hbar}{\sqrt{2}}\left(\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right) \quad \hat{J}_{y} \rightarrow \frac{\hbar}{\sqrt{2}}\left(\begin{array}{ccc} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{array}\right) \quad \hat{J}_{z} \rightarrow \hbar\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right).$

Verify that the matrices represent a spin (i.e., calculate the commutation relations)

Use the matrices of the previous question and
1. Calculate $\hat{\mathbf{J}}^{2}=\hat{\mathbf{J}} \cdot \hat{\mathbf{J}}=\hat{J}_{x}^{2}+\hat{J}_{y}^{2}+\hat{J}_{z}^{2}$ .
2. Find a set of eigenvectors that are common to (1) $\hat{J}_{x}$ and $\hat{\mathbf{J}}^2$ and (2) $\hat{J}_{z}$ and $\hat{\mathbf{J}}^2$ .
You know you have a spin 1/2 particle $j=1/2$ . What are the corresponding representations of $\hat{J}_{x}$ , $\hat{J}_{y}$ , and $\hat{J}_{z}$ in the basis of the eigenvectors of $\hat{J}_{z}$ ?
Same question as 6. but for a spin 1 particle (that is: $j=1$ ). It is expected that all students should be able to do this! I’ve worked out this problem completely for you on this video: spin-1.
It is not possible to find a common set of vectors that simultaneously diagonalize the square of the magnitude of the total spin intrinsic angular momentum ( $\hat{J}^2$ ) and its projection along $z$ ( $\hat{J}_z$ ).
1. True
2. False
It is possible to find a common set of vectors that simultaneously diagonalize the projections of the intrinsic spin angular momentum along $z$ ( $\hat{J}_z$ ) and along $x$ ( $\hat{J}_x$ ). Justify your answer briefly.
1. True
2. False
Calculate $\comm{\hat{J}_y}{\hat{J}_-}$ where $\hat{J}_-=\hat{J}_x-i\hat{J}_y$ .
1. 0
2. $i\hbar\hat{J}_z$
3. $-i\hbar\hat{J}_z$
4. None of the other answers is correct.
Calculate $\hat{J}_-^{\dagger}$ where $\hat{J}_-=\hat{J}_x-i\hat{J}_y$ .
1. 0
2. $\hat{J}_{+}$
3. $\hat{J}_{-}$
4. $\hat{J}_x$
5. $\hat{J}_y$
6. None of the other answers is correct.
For a spin-3/2 particle, what is $\bra{\frac{3}{2},\frac{3}{2}}\hat{J}_z\ket{\frac{3}{2},\frac{1}{2}}$ (using the basis of the eigenstates of $\hat{J}_z$ ) ?
1. 0
2. $\hbar\frac{-3}{2}$
3. $\hbar\frac{-1}{2}$
4. $\hbar\frac{1}{2}$
5. $\hbar\frac{3}{2}$
For a spin-3/2 particle, what is $\bra{\frac{3}{2},\frac{3}{2}}\hat{J}_{-}\ket{\frac{3}{2},\frac{3}{2}}$ (using the basis of the eigenstates of $\hat{J}_z$ ) ?
1. 0
2. $\hbar\frac{-3}{2}$
3. $\hbar\frac{-1}{2}$
4. $\hbar\frac{1}{2}$
5. $\hbar\frac{3}{2}$
For a spin-3/2 particle, what is $\bra{\frac{3}{2},\frac{3}{2}}\hat{J}_x\ket{\frac{3}{2},\frac{3}{2}}$ (using the basis of the eigenstates of $\hat{J}_z$ ) ?
1. 0
2. $\hbar\frac{-3}{2}$
3. $\hbar\frac{-1}{2}$
4. $\hbar\frac{1}{2}$
5. $\hbar\frac{3}{2}$

Hint

Find the answer keys on this page: Answers to selected test your knowledge questions. Don’t cheat! Try solving the problems on your own first!

Homework Assignment

Solve the following problems from the textbook: 3.8, 3.9, 3.10, 3.17, 3.20

Recitation Assignment

Solve the following problems from the textbook: 3.1, 3.2, 3.3, and 3.5. In addition, build the matrix representations of the angular momentum operators for spin-2 particles.

Solution to quiz

See here for a short screencast explaining answers to quiz on Chapter 3.