Total angular momentum operator $L^2$

Question

Consider a system with a state of fixed total angular momentum $l = 2$. What are the eigenvalues of the following operators

(a)$ L_z$

(b) $3/5L_x −4/5L_y$

(c) $2L_x −6L_y +3L_z$

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My problem is more to do with the definition of the angular momentum operator:

I think the angular momentum operator is $L^2=L_x^2+L_y^2+L_z^2$. I have seen many different eigenvalues this gets when applied to an eigen ket:

• $L^2|\psi\rangle=\hbar^2 k^2|\psi\rangle$
• $L^2|\psi\rangle=\hbar^2 j(j+1 )|\psi\rangle$

along with a few others. I understand that these are sort of equivilent and we are just using numbers to represent the value. However, what is the $l=2$? Is it the $k$, the $j$?

I know what to do from here on, $m$ (the quantum m=number for angular momentum along a given axis) varies from $-j$ to $+j$

Answer 1

I think the trick here is to note that the operator in (b) measures the component of angular momentum along the axis $\hat{n} = (3/5, 4/5, 0)$. It's eigenvalues must be $\{2,1,0,-1,-2\}$, the same as those of $L_z$, because you could have chosen your z-axis to lie along $\hat{n}$.

Similarly, the operator in (c) is 7 times the component of $\vec{L}$ along the normalized axis $\hat{n} = (2/7, -6/7, 3/7)$. It's eigenvalues must therefore be $\{14,7,0,-7,-14\}$, by the same reasoning.

Answer 2

The information you are given, i.e. $l=2$, tells you that the operators $L_x$, $L_y$, $L_z$ can be represented as 5x5 matrices, which operate on a vector space spanned by the five vectors

$$\{|2,-2\rangle,|2,-1\rangle,|2,0\rangle,|2,1\rangle,|2,2\rangle\}$$

which can be represented, for example, by one of the most natural bases:

$$|2,-2\rangle\stackrel{\cdot}{=}\left(\begin{array}{c} 1\\ 0\\ 0\\ 0\\ 0 \end{array}\right),|2,-1\rangle\stackrel{\cdot}{=}\left(\begin{array}{c} 0\\ 1\\ 0\\ 0\\ 0 \end{array}\right),|2,0\rangle\stackrel{\cdot}{=}\left(\begin{array}{c} 0\\ 0\\ 1\\ 0\\ 0 \end{array}\right),|2,1\rangle\stackrel{\cdot}{=}\left(\begin{array}{c} 0\\ 0\\ 0\\ 1\\ 0 \end{array}\right),|2,2\rangle\stackrel{\cdot}{=}\left(\begin{array}{c} 0\\ 0\\ 0\\ 0\\ 1 \end{array}\right)$$

$L_z$ is easy because in this basis is diagonal by definition, and it would be represented by

$$L_z\stackrel{\cdot}{=}\left(\begin{array}{ccccc} -2 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 2\\ \end{array}\right)$$

On the other hand, the two operators $L_+$ and $L_-$, defined as

$$L_\pm=L_x\pm iL_y$$

are then represented by

$$L_-\stackrel{\cdot}{=}\hbar\left(\begin{array}{ccccc} 0 & 2 & 0 & 0 & 0\\ 0 & 0 & \sqrt6 & 0 & 0\\ 0 & 0 & 0 & \sqrt6 & 0\\ 0 & 0 & 0 & 0 & 2\\ 0 & 0 & 0 & 0 & 0\\ \end{array}\right),\quad L_+\stackrel{\cdot}{=}\hbar\left(\begin{array}{ccccc} 0 & 0 & 0 & 0 & 0\\ 2 & 0 & 0 & 0 & 0\\ 0 & \sqrt6 & 0 & 0 & 0\\ 0 & 0 & \sqrt6 & 0 & 0\\ 0 & 0 & 0 & 2 & 0\\ \end{array}\right)$$

So, inverting the definition, $L_x=\frac12(L_++L_-)$ and $L_y=\frac12(L_++L_-)$, one can build the matrices corresponding to

$$\hat O_1 =\frac35 L_x-\frac45 L_y$$

and

$$\hat O_2 = 2L_x-6L_y + 3L_z$$

and calculate the eigenvalues by merely cranking the math, either:

• manually;
• using some software;
• using some trick that I'm not able to see now;

Answer 3

Short answer: the eigenvalue is: $\qquad l\cdot(l+1) \hbar \qquad$ Consequently, you'll get $2\cdot3\hbar=6\hbar$.

$J$ is (sometimes) angular momentum in general. If your concrete angular momentum is $L$, then replace $j$ by $l$.

But this is only when $J$ is used to denote "angular momentum in general". $J$ is usually "total angular momentum" (sum of all angular momenta), which would not be the same anymore.