The fundamental commutation relations for angular momentum (Equation 4.99) allow for half-integer (as well as integer) eigenvalues. But for orbital angular momentum only the integer values occur. There must be some extra constraint in the specific form$\mathit{L}\mathbf{=}\mathit{r}\mathbf{\times }\mathit{p}$ that excludes half-integer values. Let be some convenient constant with the dimensions of length (the Bohr radius, say, if we're talking about hydrogen), and define the operators

$$\begin{gathered} {\mathit{q}}_{\mathbf{1}}\mathbf{\equiv }\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left[x+\left(a^2/ħ\right)p_y\right]\mathbf{;} \\ {\mathit{p}}_{\mathbf{1}}\mathbf{\equiv }\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{[}{\mathbf{p}}_{\mathbf{x}}\mathbf{-}\left(ħ/a^2\right)\mathbf{y}\mathbf{]}\mathbf{;} \\ {\mathit{q}}_{\mathbf{2}}\mathbf{\equiv }\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{[}\mathbf{x}\mathbf{-}\left(a^2/ħ\right){\mathbf{p}}_{\mathbf{y}}\mathbf{]}\mathbf{;} \\ {\mathit{p}}_{\mathbf{2}}\mathbf{\equiv }\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{[}{\mathbf{p}}_{\mathbf{x}}\mathbf{-}\mathbf{(}\mathbf{ħ}\mathbf{/}{\mathbf{a}}^{\mathbf{2}}\mathbf{)}\mathbf{y}\mathbf{]}\mathbf{;} \end{gathered}$$

(a) Verify that $\left[q_{1,}{q}_2\right]\mathbf{=}\left[p_1,p_2\right]\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{\left[}{\mathbf{q}}_{\mathbf{1}\mathbf{,}}{\mathbf{p}}_{\mathbf{1}}\mathbf{\right]}\mathbf{=}\mathbf{[}{\mathbf{p}}_{\mathbf{2}}\mathbf{,}{\mathbf{q}}_{\mathbf{2}}\mathbf{]}\mathbf{=}\mathit{i}\mathit{ħ}$. Thus the q's and the p's satisfy the canonical commutation relations for position and momentum, and those of index 1are compatible with those of index 2 .

(b) Show that$\mathbf{\left[}{\mathbf{q}}_{\mathbf{1}\mathbf{,}}{\mathbf{q}}_{\mathbf{2}}\mathbf{\right]}\mathbf{=}\mathbf{[}{\mathbf{p}}_{\mathbf{1}}\mathbf{,}{\mathbf{p}}_{\mathbf{2}}\mathbf{]}\mathbf{}\mathbf{}{\mathit{L}}_{\mathbf{z}}\mathbf{=}\frac{\mathbf{ħ}}{\mathbf{2}{\mathbf{a}}^{\mathbf{2}}}\left(q_1^2-q_2^2\right)\mathbf{+}\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}\mathbf{ħ}}\mathbf{(}{\mathbf{p}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{-}{\mathbf{p}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{)}$

(c) Check that , where each is the Hamiltonian for a harmonic oscillator with mass and frequency .

(d) We know that the eigenvalues of the harmonic oscillator Hamiltonian are , where (In the algebraic theory of Section this follows from the form of the Hamiltonian and the canonical commutation relations). Use this to conclude that the eigenvalues of must be integers.

The rotating equivalent of linear momentum is angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

Both the direction and the amplitude of angular momentum are preserved.

(a)

Consider that,

$$\begin{gathered} q_1\equiv \frac{1}{\sqrt{2}}\left[x+\left(a^2/ħ\right)p_y\right]; \\ {q}_1\equiv \frac{1}{\sqrt{2}}\left[px-{\left(ħ/a^2\right)}_y\right]; \\ {p}_2\equiv \frac{1}{\sqrt{2}}\left[x-\left(a^2/ħ\right)p_y\right]; \\ {q}_2\equiv \frac{1}{\sqrt{2}}\left[px+{\left(ħ/a^2\right)}_y\right]; \end{gathered}$$

Now, the commutation relation between $q_1$and $q_{2_{}}$is,

localid="1658311027712" $$\begin{gathered} \left[q_1,q_2\right]=\left[\frac{1}{\sqrt{2}}\left(x+\left(\frac{a^2}{ħ}\right)p_y\right),\frac{1}{\sqrt{2}}\left(x-\left(\frac{a^2}{ħ}\right)p_y\right)\right] \\ =\frac{1}{2}\left[[x,x]+x,\left(-\frac{a^2}{ħ}\right)p_y\right]+\left[\left(\frac{a^2}{ħ}\right)p_y,\left(\frac{a^2}{ħ}\right)p_y\right] \\ =0 \end{gathered}$$

Since, $\left[x,x\right]=\left[p_y,p_y\right]=0$

Also, the commutation relation between $p_1$and $p_2$is,

localid="1658311843930" $$\begin{gathered} \left[p_1,p_2\right]=\left[\frac{1}{\sqrt{2}}\left(p_x-\left(\frac{ħ}{a^2}\right)y\right),\frac{1}{\sqrt{2}}\left(p_x+\left(\frac{ħ}{ħa^2}\right)y\right)\right] \\ =\frac{1}{2}[p_x,p_x]+\frac{1}{2}\left[p_x,\frac{ħ}{a^2}y\right]-\frac{1}{2}\left[\frac{ħ}{a^2}y,p_x\right]-\frac{1}{2}\left[\frac{ħ}{a^2}y,\frac{a^2}{ħ}y\right] \\ As, \\ \left[p_x,p_x\right]=\left[y,y\right]=\left[p_x,y\right]=0 \end{gathered}$$

The commutation relation between $q_1$and $p_1$is,

role="math" localid="1658313061062" $$\begin{gathered} \left[q_1,p_1\right]=\left[\frac{1}{\sqrt{2}}\left(x+\frac{a^2}{ħ}\right),\frac{1}{\sqrt{2}}\left(p_x-\left(\frac{ħ}{a^2}\right)y\right)\right] \\ =\frac{1}{2}\left[x,p_x\right]+\frac{1}{2}\left[\frac{a^2}{ħ}{p}_y,p_x\right]-\frac{1}{2}\left[x,\frac{ħ}{a^2}y\right]-\frac{1}{2}\left[\frac{a^2}{ħ}{p}_y,\frac{ħ}{a^2}\right] \\ =\frac{1}{2}\left[x,p_x\right]-\frac{1}{2}\left[p_y,y\right] \\ =\frac{1}{2}\left[iħ-\left(-iħ\right)\right] \\ =\frac{2iħ}{2} \\ =iħ \end{gathered}$$

And, commutation relation between $q_2$and $p_2$is,

role="math" localid="1658313694232" $$\begin{gathered} \left[q_2,p_2\right]=\left[\frac{1}{\sqrt{2}}\left(x-\frac{a^2}{ħ}{p}_y\right),\frac{1}{\sqrt{2}}\left(p_x+\frac{ħ}{a^2}y\right)\right] \\ =\frac{1}{2}\left[x,p_x\right]+\frac{1}{2}\left[x,\frac{ħ}{a^2}y\right]-\frac{1}{2}\left[\frac{a^2}{ħ}{p}_y,p_x\right]-\frac{1}{2}\left[\frac{a^2}{ħ}{p}_y,\frac{ħ}{a^2}y\right] \\ =\frac{1}{2}\left[x,p_x\right]-\frac{1}{2}\left[p_y,y\right] \\ =\frac{2iħ}{2} \\ =iħ \\ As, \\ \left[x,y\right]=\left[p_y,p_x\right]=0 \\ But \\ \left[x,p_x\right]=iħand\left[p_y,y\right]=-iħ \\ Therefore,theabovefindingsuggeststhatq\text{'}sandp\text{'}sfollowtheposition-momentumquantumcommutationrelations,\left[q_i,p_j\right]=iħ{\delta }_{ij}. \end{gathered}$$

(b)

Evaluate $q_1^2-q_2^2$and $p_1^2-p_2^2$to obtain their relation with the z-component of the angular momentum:

$$\begin{gathered} q_1^2-q_2^2=\frac{1}{2}\left\{x^2+\left(\frac{a^2}{ħ}\right)p_y^2+\left(\frac{a^2}{ħ}\right)\left\{x,p_y\right\}\right\}-\frac{1}{2}\left\{x^2+\left(\frac{a^2}{ħ}\right)p_y^2+\left(\frac{a^2}{ħ}\right)\left\{x,p_y\right\}\right\} \\ =\left(\frac{a^2}{ħ}\right)\left\{x,p_y\right\} \\ =2\left(\frac{a^2}{ħ}\right)x{p}_y \end{gathered}$$

Also,

$$\begin{gathered} p_1^2-p_2^2=\frac{1}{2}\left[p_x^2+{\left(\frac{ħ}{a^2}\right)}^2{y}^2-\frac{ħ}{a^2}\left(p_{x}y+y{p}_x\right)-p_x^2-{\left(\frac{ħ}{a^2}\right)}^2{y}^2-\frac{ħ}{a^2}\left(p_{x}y+y{p}_x\right)\right] \\ =-\frac{2ħ}{a^2}y{p}_x \end{gathered}$$

Now,

$$\begin{gathered} \frac{ħ}{2a^2}\left(q_1^2-q_2^2\right)+\frac{a^2}{2ħ}\left(p_1^2-p_2^2\right)=\frac{ħ}{2a^2}\left(\frac{2a^2}{ħ}x{p}_y\right)+\frac{a^2}{2ħ}\left(\frac{2ħ}{a^2}x{p}_x\right) \\ =x{p}_y-y{p}_x \\ =L_z \end{gathered}$$

Thus, the relation with the z component of angular momentum is established.

(c)

The Hamiltonian for the harmonic oscillator is,

$$\begin{gathered} H=\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2 \\ \mathrm{For}\mathrm{m}=\frac{\mathrm{ħ}}{{\mathrm{a}}^2}\mathrm{and}\mathrm{\omega }=1 \\ \mathrm{H}=\frac{1}{2\mathrm{m}}{\mathrm{P}}^2+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2 \\ =\frac{{\mathrm{a}}^2}{2\mathrm{ħ}}{\mathrm{p}}^2+\frac{\mathrm{ħ}}{2{\mathrm{a}}^2}{\mathrm{x}}^2 \end{gathered}$$

And

$$\begin{gathered} H_1=\frac{a^2}{2ħ}{p}_1^2+\frac{ħ}{2a^2}{q}_1^2 \\ {H}_2=\frac{a^2}{2ħ}{p}_2^2+\frac{ħ}{2a^2}{q}_2^2 \\ {H}_1-H_2=\frac{a^2}{2ħ}\left(p_1^2-p_2^2\right)+\frac{ħ}{2a^2}\left(q_1^2-q_2^2\right) \\ =L_z \end{gathered}$$

Thus, the relationship is established.

(d)

The values of eigenvalues of${\mathrm{L}}_{\mathrm{z}}$are integers. So,

$\mathrm{H}\overline{)\mathrm{\psi }\rangle =\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{ħ\omega }}$

Then,

$$\begin{gathered} \left({\mathrm{H}}_1-{\mathrm{H}}_2\right)\overline{)\mathrm{\psi }\rangle ={\mathrm{L}}_{\mathrm{z}}\overline{)\mathrm{\psi }}}⟩ \\ =\left({\mathrm{n}}_1-\frac{1}{2}-{\mathrm{n}}_2-\frac{1}{2}\right)ħ\omega \overline{)\mathrm{\psi }\rangle } \end{gathered}$$

For$\omega =1$:

$L_2\overline{)\psi \rangle =\left(n_1,n_2\right)}ħ\overline{)\psi ⟩}$

Here $n_1,n_2=0,1,2,3,...$

$L_z\overline{)\psi \rangle =mħ\overline{)\psi }}⟩,$is an integer since .

Hence, the eigenvalues of ${\mathrm{L}}_{\mathrm{z}}$must be integers.