Eigenvalue Quantization

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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

Solution

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