# Eigenvalue Quantization

The motivation for exploring eigenvalue quantization comes form wanting to solve the energy eigenvalue problem for a particle in a central potential. It is not possible, in general, to specify and measure more than one component $\hat{\mathbf{n}}\cdot\hat{\mathbf{L}}$ of the orbital angular momentum. It is, however, possible to specify $\hat{\mathbf{L}}^2$ simulataneously with any one component of $\hat{\mathbf{L}},$ since $\hat{\mathbf{L}}^2$ commutes with all of its Cartesian components, as we saw earlier. We typically choose $\hat{L}_z.$ A central potential yields a Hamiltonian that commutes with $\hat{\mathbf{L}},$ and thus the energy eigenstates of the system may be chosen to be eigenvectors of $\hat{\mathbf{L}}^2$ and one component of $\hat{\mathbf{L}},$ usually $\hat{L}_z.$

The quantization of angular momentum follows simply from the commutation relations derived earlier. Recall that $\mathbf{L}^2\!$ is given by

$\hat{\mathbf{L}}^2=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2.$

Let us now define the operators, $\hat{L}_\pm=\hat{L}_x\pm i\hat{L}_y.\!$ Note that $\hat{L}_+$ and $\hat{L}_-$ are Hermitian conjugates of each other.

We choose to work with these operators because, as we will see shortly, the operators $\hat{L}_\pm$ function as raising and lowering operators for eigenvectors of $\hat{L}_z.$

We may use the commutation relations derived earlier to show that

$\hat{L}_+\hat{L}_-=\hat{\mathbf{L}}^2-\hat{L}_z^2+\hbar\hat{L}_z$

and

$\hat{L}_-\hat{L}_+=\hat{\mathbf{L}}^2-\hat{L}_z^2-\hbar\hat{L}_z.$

Therefore,

$[\hat{L}_+,\hat{L}_-]=\hat{L}_+\hat{L}_--\hat{L}_-\hat{L}_+=2\hbar\hat{L}_z.$

We may also show that $[\hat{L}_z,\hat{L}_\pm]=\hat{L}_z\hat{L}_\pm-\hat{L}_\pm\hat{L}_z=\pm\hbar\hat{L}_\pm.$

We may also easily see that $[\hat{\mathbf{L}}^2,\hat{L}_\pm]=0.$

Let $|\beta,m\rangle\!$ be a normalized eigenstate of $\hat{\mathbf{L}}^2$ with eigenvalue $\beta\!$ and of $\hat{L}_z\!$ with eigenvalue $m\hbar.$ Let us first determine what the effects of the operators, $\hat{L}_\pm,$ are. By definition,

$\hat{L}_z|\beta,m\rangle=m\hbar|\beta,m\rangle.$

If we now act on this expression from the left with $\hat{L}_\pm$ and use the above commutation relations, we find that

$\hat{L}_z\hat{L}_\pm|\beta,m\rangle=(m\pm 1)\hbar\hat{L}_\pm|\beta,m\rangle.$

We therefore see that $\hat{L}_\pm|\beta,m\rangle$ is also an eigenvector of $\hat{L}_z,$ but with an eigenvalue of $(m\pm 1)\hbar.$ In other words,

$\hat{L}_\pm|\beta,m\rangle=\alpha_\pm|\beta,m\pm 1\rangle,$

where $\alpha_\pm\!$ is a normalization constant. We see now that, as asserted earlier, the operators $\hat{L}_\pm$ are raising and lowering operators for eigenstates of $\hat{L}_z.$

To find the normalization constant, we will evaluate the norms of these states. The norm for $\hat{L}_-|\beta,m\rangle$ is, using the expressions derived above,

$\langle\beta,m|\hat{L}_+\hat{L}_-|\beta,m\rangle=\beta-m(m-1)\hbar^2.$

The right-hand side of this expression is equal to $|\alpha_-|^2.\!$ If we now take $\alpha_-\!$ to be real, then

$\alpha_-=\sqrt{\beta-m(m-1)\hbar^2}.$

Similarly, the norm for $\hat{L}_+|\beta,m\rangle$ is

$\langle\beta,m|\hat{L}_-\hat{L}_+|\beta,m\rangle=\beta-m(m+1)\hbar^2,$

and thus, again taking $\alpha_+\!$ to be real,

$\alpha_+=\sqrt{\beta-m(m+1)\hbar^2}.$

In summary,

$\hat{L}_\pm|\beta,m\rangle=\sqrt{\beta-m(m\pm 1)\hbar^2}|\beta,m\pm 1\rangle.$

We may now restrict the possible values of $\beta\!$ as follows. Obviously, it is impossible for $\langle\hat{L}_z^2\rangle$ to be larger than $\langle\hat{\mathbf{L}}^2\rangle.$ This means that

$\beta\geq m^2\hbar^2.$

This implies that there must be both a lower bound and an upper bound on the allowed values of $m\!$ for a given value of $\beta.\!$ Let $l\!$ be the upper bound on $m.\!$ Then

$\hat{L}_+|\beta,l\rangle=0.$

This means that

$\sqrt{\beta-l(l+1)\hbar^2}=0,$

or

$\beta=l(l+1)\hbar^2>l^2\hbar^2.$

The value of $\beta\!$ that we obtained is therefore a suitable value, since it is larger than the assumed upper bound on $m^2\hbar^2.$ If we now consider the lowering operator, then, using this value of $\beta,\!$

$\hat{L}_-|\beta,m\rangle=\hbar\sqrt{l(l+1)-m(m-1)}|\beta,m-1\rangle.$

We see by direct substitution that, if there is a lower bound, then it must be equal to $-l.\!$ In order for us to be able to lower a state $|\beta,l\rangle$ to $|\beta,-l\rangle$ by repeated application of the lowering operator, then $l\!$ and $-l\!$ must differ by an integer:

$l=-l+n\!$

This means that the possible values of $l\!$ are quantized in half-integer steps; i.e.,

$l=0,\tfrac{1}{2},1,\tfrac{3}{2},\ldots$

If we chose any other values of $l,\!$ then it would be possible to construct states with arbitrarily low values of $m,\!$ which is impossible. The allowed values of $m\!$ are all integers such that $-l\leq m\leq l\!$ for an integer value of $l,\!$ or all half-integers satisfying the same constraint if $l\!$ is a half-integer.

From this point forward, we will use the notation, $|l,m\rangle,$ for an eigenstate of $\hat{\mathbf{L}}^2$ with eigenvalue $l(l+1)\hbar^2$ and of $\hat{L}_z$ with eigenvalue $m\hbar.$ The quantum number $l\!$ is often called the orbital quantum number and $m\!$ the magnetic quantum number, the latter being so named because it characterizes the energy shift of, say, an atom in the presence of a magnetic field.

## Problem

Evaluate the following expressions:

(a) $\left \langle {l,m\left |{\hat{L}_{x}^{2}} \right |l,m} \right \rangle$

(b) $\left \langle {l,m\left |{\hat{L}_{x}\hat{L}_{y}} \right |l,m} \right \rangle$