Subsections


D.36 Photon wave function derivations

The rules of engagement are as follows:


D.36.1 Rewriting the energy integral

As given in the text, the energy in an electromagnetic field in free space that satisfies the Coulomb-Lorenz gauge is, writing out the square magnitudes and individual components,

\begin{displaymath}
E = {\textstyle\frac{1}{2}}\epsilon_0 \int
\Bigg(\left\v...
...gg(A_{i,t}^* A_{i,t} + c^2\sum_{j=1}^3 A_{i,j}^*A_{i,j}\bigg)
\end{displaymath}

However, a bit more general expression is desirable. If only the Lorenz condition is satisfied, there may also be an electrostatic potential $\varphi$. In that case, a more general expression for the energy is:
$\parbox{400pt}{\hfill$\displaystyle
E = {\textstyle\frac{1}{2}}\epsilon_0 \bi...
..._t^* \varphi_t
+ \sum_{j=1}^3 \varphi_j^* \varphi_j\bigg)\bigg]
$\hfill(1)}$
The minus sign for the $\varphi$ terms appears because this is really a dot product of relativistic four-vectors. The zeroth components in such a dot product acquire a minus sign, chapter 1.2.4 and 1.3.2. In derivation {D.32} it was shown that each of the four integrals is constant. That is because each component satisfies the Klein-Gordon equation. So their sum is constant too.

The claim to verify now is that the same energy can be obtained from integrating the electric and magnetic fields as

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
E = {\textstyle\frac{1}{2}}\...
...
\bigg({\cal E}_i^*{\cal E}_i + c^2{\cal B}_i^*{\cal B}_i\bigg)
$\hfill(2)}$
Since $\skew3\vec{\cal E}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\partial\skew3\vec A/\partial{t}-\nabla\varphi$ and $\skew2\vec{\cal B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\nabla$ $\times$ $\skew3\vec A$:

\begin{displaymath}
{\cal E}_i = - A_{i,t} -\varphi_i
\qquad
{\cal B}_i = ...
...h}}} - A_{{\overline{\imath}},{\overline{\overline{\imath}}}}
\end{displaymath}

From now on, it will be understood that there is a summation over $i$ and $j$ and that everything has a ${\textstyle\frac{1}{2}}\epsilon_0$. Therefore these will no longer be shown.

Start with the electric field integral. It is, using the above expressions and multiplying out,

\begin{displaymath}
\int A_{i,t}^* A_{i,t} +
A_{i,t}^* \varphi_i + \varphi_i^* A_{i,t} + \varphi_i^* \varphi_i
\end{displaymath}

The first term already gives the vector-potential time derivatives in (1). That leaves the final three terms. Perform an integration by parts on the first two. It will always be assumed that the potentials vanish at infinity or that the system is in a periodic box. In that case there are no boundary terms in an integration by parts. So the three terms become

\begin{displaymath}
\int {} - A_{i,it}^* \varphi - \varphi^* A_{i,it} + \varphi_i^* \varphi_i
\end{displaymath}

However, the divergence $A_{i,i}$ is according to the Lorenz condition equal to $-\varphi_t$$\raisebox{.5pt}{$/$}$$c^2$, so

\begin{displaymath}
\int \frac{1}{c^2} \varphi_{tt}^* \varphi
+ \frac{1}{c^2} \varphi^* \varphi_{tt} + \varphi_i^* \varphi_i
\end{displaymath}

Using the Klein-Gordon equation, $\varphi_{tt}$$\raisebox{.5pt}{$/$}$$c^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varphi_{ii}$, and then another integration by parts on the first two terms and renotating $i$ by $j$ gives the $\varphi_j^*\varphi_j$ terms in (1).

Now consider the integral of $\vert\skew2\vec{\cal B}\vert^2$ in (2). You get, multiplying out,

\begin{displaymath}
c^2 \int \bigg(A_{{\overline{\overline{\imath}}},{\overlin...
... A_{{\overline{\imath}},{\overline{\overline{\imath}}}}\bigg)
\end{displaymath}

Now the first and last terms in the right hand side summed over $i$ produce all terms $A_{i,j}^*A_{i,j}$ in (1) in which $i$ and $j$ are different. That leaves the middle terms. An integration by parts yields

\begin{displaymath}
c^2 \int \bigg(A_{{\overline{\overline{\imath}}},{\overlin...
...verline{\imath}}}}^* A_{{\overline{\overline{\imath}}}}\bigg)
\end{displaymath}

Renotate the indices cyclically to get

\begin{displaymath}
c^2 \int \bigg(A_{{\overline{\imath}},{\overline{\imath}}i...
...line{\imath}}},{\overline{\overline{\imath}}}i}^* A_{i}\bigg)
\end{displaymath}

(If you want, you can check that this is the same by writing out all three terms in the sum.) This is equivalent to

\begin{displaymath}
c^2 \int \bigg(A_{i,i}^* + A_{{\overline{\imath}},{\overli...
...ne{\overline{\imath}}}}^*\bigg)_i A_{i}
- A_{i,i i}^* A_{i}
\end{displaymath}

as you can see from differentiating and multiplying out. The final term gives after integration by parts the $A_{i,j}^*A_{i,j}$ terms in (1) in which $i$ and $j$ are equal. That leaves the first part. The term in parentheses is the divergence $-\varphi_t$$\raisebox{.5pt}{$/$}$$c^2$, so the first part is

\begin{displaymath}
\int - \varphi_{it}^* A_{i}
\end{displaymath}

Perform an integration by parts

\begin{displaymath}
\int \varphi_t^* A_{i,i}
\end{displaymath}

Recognizing once more the divergence, this gives the final $-\varphi_t^*\varphi_t$$\raisebox{.5pt}{$/$}$$c^2$ term in (1)


D.36.2 Angular momentum states

The rules of engagement listed at the start of this section apply. In addition:


D.36.2.1 About the scalar modes

The scalar modes are the $jY$.

It will be assumed that the $j$ are zero at the large radius $r_{\rm {max}}$ at which the domain is assumed to terminate. That makes the scalar modes a complete set; any scalar function $f$ can be written as a combination of them. (That is because they are the eigenfunctions of the Laplacian inside the sphere, and the zero boundary condition on the sphere surface $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_{\rm {max}}$ makes the Laplacian Hermitian. This will not be explicitly proved since it is very standard.)

The Bessel function $j$ of the scalar modes satisfy the ordinary differential equation, {A.6}

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
r^2 j'' + 2 r j' = l(l+1) j - k^2r^2 j
$\hfill(3)}$
The following integral is needed (note that $j$ is real):
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\int_0^{r_{\rm max}} j^2r^2{\,\rm d}r \sim \frac{r_{\rm{max}}}{2k^2}
$\hfill(4)}$
This is valid for large $kr_{\rm {max}}$, which applies since $r_{\rm {max}}$ is large and the $k$ values of interest are finite. The above result comes from the integral of the square two-di­men­sion­al Bessel functions $J$, and a recurrence relation, [40, 27.18,88], using $j_l(kr)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $J_{l+\frac12}(kr)\sqrt{\pi/2kr}$, [1, p 437, 10.1.1], and the asymptotic behavior of the Bessel function you get from {A.6} (A.19). To get the leading asymptotic term, each time you have to differentiate the trigonometric function. And where the trigonometric function in $j_l$ is zero at $r_{\rm {max}}$ because of the boundary condition, the one in $j_{l+1}$ has magnitude 1.

The spherical harmonics are orthonormal on the unit sphere, {D.14.4}

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\int \bar Y Y{\,\rm d}\Omega = \delta_{\bar{l}l}\delta_{\bar{m}m}
$\hfill(5)}$
In other words, the integral is only 1 if $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar{l}$ and $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\bar{m}$ and otherwise it is zero. Further
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\int \left(\bar Y_\theta Y_\...
...ight) {\,\rm d}\Omega = l(l+1) \delta_{\bar{l}l}\delta_{\bar{m}m}
$\hfill(6)}$


D.36.2.2 Basic observations and eigenvalue problem

For any function $f$

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
{\skew0\vec r}\times \nabla f = - \nabla \times ({\skew0\vec r}f)
$\hfill(7)}$
This follows from writing out the right hand side

\begin{displaymath}
- (r_{\overline{\overline{\imath}}}f)_{\overline{\imath}}+...
...math}}+ r_{\overline{\imath}}f_{\overline{\overline{\imath}}}
\end{displaymath}

the latter since ${\overline{\imath}}$ and ${\overline{\overline{\imath}}}$ are different indices.

The electric modes

\begin{displaymath}
\nabla\times{\skew0\vec r}\times\nabla f
\end{displaymath}

are solenoidal because $\nabla\cdot\nabla$ $\times$ $\ldots$ gives zero. The magnetic modes

\begin{displaymath}
{\skew0\vec r}\times\nabla f
\end{displaymath}

are solenoidal for the same reason, after noting (7) above.

The Laplacian commutes with the operators in front of the scalar functions in the electric and magnetic modes. That can be seen for the magnetic ones from

\begin{displaymath}
(r_{\overline{\imath}}f_{\overline{\overline{\imath}}}- r_...
...}}} - 2 f_{{\overline{\overline{\imath}}}{\overline{\imath}}}
\end{displaymath}

and the final two terms cancel. And the Laplacian also commutes with the additional $\nabla\times$ in the electric modes since differentiations commute.

From this it follows that the energy eigenvalue problem is satisfied because by definition of the scalar modes $-\nabla^2jY$ $\vphantom0\raisebox{1.5pt}{$=$}$ $k^2jY$. In addition,

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\nabla\times\nabla\times{\skew0\vec r}\times\nabla f
= k^2 {\skew0\vec r}\times\nabla f
$\hfill(8)}$
because $\nabla$ $\times$ $\nabla\times$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$\nabla^2$ for a solenoidal function, (D.1).


D.36.2.3 Spherical form and net angular momentum

In spherical coordinates, the magnetic mode is

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\skew3\vec A^{\rm M} = {\ske...
...\theta - {\hat\imath}_\theta
\frac{1}{\sin\theta} Y_\phi\right]
$\hfill(9)}$
and then the electric mode is
$\parbox{400pt}{\hspace{15pt}\hfill$\displaystyle
\skew3\vec A^{\rm E} = \nabl...
..._\theta + {\hat\imath}_\phi
\frac{1}{\sin\theta} Y_\phi\right]
$\hfill(10)}$
from [40, 20.74,76,82] and for the $r$ component of $\skew3\vec A^{\rm {E}}$ the eigenvalue problem of chapter 4.2.3.

Now note that the $\phi$ dependence of $Y$ is through a simple factor $e^{{{\rm i}}m\phi}$, chapter 4.2.3. Therefore it is seen that if the coordinate system is rotated over an angle $\gamma$ around the $z$-​axis, it produces a factor $e^{{{\rm i}}m\gamma}$ in the vectors. First of all that means that the azimuthal quantum number of net angular momentum is $m$, {A.19}. But it also means that, {A.19},

\begin{displaymath}
{\widehat J}_z {\skew0\vec r}\times\nabla f = {\skew0\vec ...
...\times\nabla f = \nabla\times{\skew0\vec r}\times\nabla L_z f
\end{displaymath}

because either way the vector gets multiplied by $m\hbar$ for the modes. And if it is true for all the modes, then it is true for any function $f$. Since the $z$-​axis is not special for general $f$, the same must hold for the $x$ and $y$ angular momentum operators. From that it follows that the modes are also eigenfunctions of net square angular momentum, with azimuthal quantum number $l$.

At the cut-off $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_{\rm {max}}$, $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, which gives:

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\mbox{At $r_{\rm{max}}$:} \q...
..._\theta + {\hat\imath}_\phi
\frac{1}{\sin\theta} Y_\phi\right]
$\hfill(11)}$
Also needed is, differentiating (10):
$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\mbox{At $r_{\rm{max}}$:} \q...
..._\theta + {\hat\imath}_\phi
\frac{1}{\sin\theta} Y_\phi\right]
$\hfill(12)}$
which used (3) to get rid of the second order derivative of $j$.


D.36.2.4 Orthogonality and normalization

Whether the modes are orthogonal, and whether the Laplacian is Hermitian, is not obvious because of the weird boundary conditions at $r_{\rm {max}}$.

In general the important relations here

$\parbox{400pt}{\hspace{11pt}\hfill$\displaystyle
\begin{array}{l}
\displays...
...i,j})
\frac{\partial r_j}{\partial r} {\,\rm d}S
\end{array}
$\hfill(13)}$
where $S$ is the surface of the sphere $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_{\rm {max}}$. The second last line can be verified by differentiating out and the last line is the divergence theorem.

The first and second line in (13) show that the Laplacian is Hermitian if all unequal modes are orthogonal (or have equal $k$ values, but orthogonality should be shown anyway.). For unequal $k$ values orthogonality may be shown by showing that the final surface integral is zero.

It is convenient to show right away that the electric and magnetic modes are always mutually orthogonal:

\begin{displaymath}
\int (r_{\overline{\imath}}\bar f_{\overline{\overline{\im...
...\overline{\imath}}}\bar f A_{i,{\overline{\imath}}}^{\rm {E}}
\end{displaymath}

The first two terms in the right hand side can be integrated in the ${\overline{\overline{\imath}}}$, respectively ${\overline{\imath}}$ direction and are then zero because $\bar{f}$ is zero on the spherical surface $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_{\rm {max}}$. The final two terms summed over $i$ can be renotated by shifting the summation index one unit down, respectively up in the cyclic sequence to give

\begin{displaymath}
\sum_i - \int \bar f r_i(A_{{\overline{\overline{\imath}}}...
...\int \bar f {\skew0\vec r}\cdot {\skew0\vec r}\times \nabla f
\end{displaymath}

the latter because of the form of $\skew3\vec A^{\rm {E}}$, the fact that $\nabla$ $\times$ $\nabla\times$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$\nabla^2$ for a solenoidal vector, and the energy eigenvalue problem established for $\skew3\vec A^{\rm {M}}$. The final term is zero because ${\skew0\vec r}\cdot{\skew0\vec r}\times$ is.

Next consider the orthogonality of the magnetic modes for different quantum numbers. For $\bar{l}$ $\raisebox{.2pt}{$\ne$}$ $l$ or $\bar{m}$ $\raisebox{.2pt}{$\ne$}$ $m$, the orthogonality follows from (9) and (6). For $\bar{k}$ $\raisebox{.2pt}{$\ne$}$ $k$, the orthogonality follows from the final line in (13) since the magnetic modes are zero at $r_{\rm {max}}$, (11).

Finally the electric modes. For $\bar{l}$ $\raisebox{.2pt}{$\ne$}$ $l$ or $\bar{m}$ $\raisebox{.2pt}{$\ne$}$ $m$, the orthogonality follows from (10), (5), and (6). For $\bar{k}$ $\raisebox{.2pt}{$\ne$}$ $k$, the orthogonality follows from the final line in (13). To see that, recognize that $A_{i,j}\partial{r}_j$$\raisebox{.5pt}{$/$}$$\partial{r}$ is the radial derivative of $\skew3\vec A$; therefore using (11) and (12), the integrand vanishes.

The integral of the absolute square integral of a magnetic mode is, using (9), (6), and (4),

\begin{displaymath}
\int \skew3\vec A^{{\rm {M}}*}\cdot \skew3\vec A^{\rm {M}}
= l(l+1) \frac{r_{\rm {max}}}{2k^2}
\end{displaymath}

The integral of the absolute square integral of an electric mode is, using (10), (5), and (6),

\begin{displaymath}
l^2(l+1)^2 \int_0^{r_{\rm max}} j^2 {\,\rm d}r
+ l(l+1) \int_0^{r_{\rm max}} (rj)'(rj)' {\,\rm d}r
\end{displaymath}

Apply an integration by parts on the second integral,

\begin{displaymath}
l^2(l+1)^2 \int_0^{r_{\rm max}} j^2 {\,\rm d}r
- l(l+1) \int_0^{r_{\rm max}} jr(rj)'' {\,\rm d}r
\end{displaymath}

and then use (3) to get

\begin{displaymath}
\int \skew3\vec A^{{\rm {M}}*}\cdot \skew3\vec A^{\rm {M}}
= k^2 l(l+1) \frac{r_{\rm {max}}}{2k^2}
\end{displaymath}

The normalizations given in the text follow.


D.36.2.5 Completeness

Because of the condition $\nabla\cdot\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, you would generally speaking expect two different types of modes described by scalar functions. The electric and magnetic modes seem to fit that bill. But that does not mean that there could not be say a few more special modes. What is needed is to show completeness. That means to show that any smooth vector field satisfying $\nabla\cdot\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 can be written as a sum of the electric and magnetic modes, and nothing else.

This author does not know any simple way to do that. It would be automatic without the solenoidal condition; you would just take each Cartesian component to be a combination of the scalar modes $jY$ satisfying a zero boundary condition at $r_{\rm {max}}$. Then completeness would follow from the fact that they are eigenfunctions of the Hermitian Laplacian. Or from more rigorous arguments that you can find in mathematical books on partial differential equations. But how to do something similar here is not obvious, at least not to this author.

What will be done is show that any reasonable solenoidal vector can be written in the form

\begin{displaymath}
\skew3\vec A= {\skew0\vec r}\times\nabla f + \nabla\times{\skew0\vec r}\times\nabla g
\end{displaymath}

where $f$ and $g$ are scalar functions. Completeness then follows since the modes $jY$ provide a complete description of any arbitrary function $f$ and $g$.

But to show the above does not seem easy either, so what will be actually shown is that any vector without radial component can be written in the form

\begin{displaymath}
\vec v = {\skew0\vec r}\times\nabla f + {\hat\imath}_r\times{\skew0\vec r}\times\nabla g
\end{displaymath}

That is sufficient because the Fourier transform of $\skew3\vec A$ does not have a radial component, so it will be of this form. And the inverse Fourier transform of $\vec{v}$ is of the form $\skew3\vec A$, compare any book on Fourier transforms and (7).

The proof that $\vec{v}$ must be of the stated form is by construction. Note that automatically, the radial component of the two terms is zero. Writing out the gradients in spherical coordinates, [40, 20.74,82], multiplying out the cross products and equating components gives at any arbitrary radius $r$

\begin{displaymath}
- \frac{\partial f}{\partial\phi}
- \sin\theta \frac{\pa...
...eta}
- \frac{\partial g}{\partial\phi}
= v_\phi\sin\theta
\end{displaymath}

Now decompose this in Fourier modes $e^{{{\rm i}}m\phi}$ in the $\phi$ direction:

\begin{displaymath}
- {\rm i}m f_m
- \sin\theta \frac{\partial g_m}{\partial...
..._m}{\partial\theta}
- {\rm i}m g_m
= v_{\phi m}\sin\theta
\end{displaymath}

For $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, $f_0$ and $g_0$ follow by integration. Note that the integrands are periodic of period $2\pi$ and antisymmetric about the $z$-​axis. That makes $f$ and $g$ periodic of period $2\pi$ too,

For $m$ $\raisebox{.2pt}{$\ne$}$ 0 make a coordinate transform from $\theta$ to

\begin{displaymath}
t = \int{\rm d}\theta/\sin\theta = \ln\tan{\textstyle\frac{1}{2}}\theta
\end{displaymath}

Note that $\vphantom0\raisebox{1.5pt}{$-$}$$\infty$ $\raisebox{.3pt}{$<$}$ $t$ $\raisebox{.3pt}{$<$}$ $\infty$. If anybody is actually reading this, send me an email. The system becomes after cleaning up

\begin{displaymath}
- {\rm i}m f_m
- \frac{\partial g_m}{\partial t}
= v_{...
...tial f_m}{\partial t}
- {\rm i}m g_m
= v_{\phi m}/\cosh t
\end{displaymath}

It is now easiest to solve the above equations for each of the two right hand sides separately. Here the first right hand side will be done, the second right hand side goes similarly.

From the two equations it is seen that $f_m$ must satisfy

\begin{displaymath}
\frac{\partial^2 f_m}{\partial t^2} - m^2f_m
= \frac{-2m{\rm i}v_{\theta m}}{e^t+e^{-t}}
\end{displaymath}

and ${{\rm i}}mg_m$ must the derivative of $f_m$. The solution satisfying the required regularity at $\pm\infty$ is, [40, 19.8],

\begin{displaymath}
f_m =
\int_t^{\infty} {\rm i}v_{\theta m}\frac{e^{-m(\ta...
...{\theta m}\frac{e^{m(\tau-t)}}{e^\tau+e^{-\tau}}{\,\rm d}\tau
\end{displaymath}

That finishes the construction, but you may wonder about potential nonexponential terms in the first integral at $\vphantom0\raisebox{1.5pt}{$-$}$$\infty$ and the second integral at $\infty$. Those would produce weak logarithmic singularities in the physical $f$ and $g$. You could simply guess that the two right hand sides will combine so that these terms drop out. After all, there is nothing special about the chosen direction of the $z$-​axis. If you choose a different axis, it will show no singularities at the old $z$-​axis, and the solution is unique.

For more confidence, you can check the cancellation explicitly for the leading order, $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 terms. But there is a better way. If the right hand sides are zero within a nonzero angular distance $\Delta\theta_1$ from the $z$-​axis, there are no singularities. And it is easy to split off a part of $\vec{v}$ that is zero within $\Delta\theta_1$ of the axis and then changes smoothly to the correct $\vec{v}$ in an angular range from $\Delta\theta_1$ to $\Delta\theta_2$ from the axis. The remainder of $\vec{v}$ can than be handled by using say the $x$-​axis as the axis of the spherical coordinate system.


D.36.2.6 Density of states

The spherical Bessel function is for large arguments proportional to $\sin(kr)$$\raisebox{.5pt}{$/$}$$r$ or $\cos(kr)$$\raisebox{.5pt}{$/$}$$r$. Either way, the zeros are spaced $\Delta{k}\,r_{\rm {max}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$ apart. So there is one state $\Delta{N}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 in an interval $\Delta{E}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\Delta{k}c$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar{\pi}c$$\raisebox{.5pt}{$/$}$$r_{\rm {max}}$. The ratio gives the stated density of states.


D.36.2.7 Parity

Parity is what happens to the sign of the wave function under a parity transformation. A parity transformation inverts the positive direction of all three Cartesian axes, replacing any position vector ${\skew0\vec r}$ by $\vphantom0\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. The parity of something is 1 or even if it does not change, and $\vphantom0\raisebox{1.5pt}{$-$}$1 or odd if it changes sign. Under a parity transformation, the operators ${\skew0\vec r}\times$ and $\nabla\times$ flip over the parity of what they act on. On the other hand, $\nabla{j}_jY_j^{m_j}$ has the same parity as $j_jY_j^{m_j}$; the spatial components flip over, but so do the unit vectors that multiply them. And the parity of $j_jY_j^{m_j}$ is even if $j$ is even and odd if $j$ is odd. The stated parities follow.


D.36.2.8 Orbital angular momentum of the states

In principle a state of definite net angular momentum $j$ and definite spin 1 may involve orbital angular momentum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j-1$, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$ and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j-1$, chapter 7.4.2. But states of definite parity restrict that to either only odd values or only even values, {A.20}. To get the stated parities, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j$ for magnetic states and $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j-1$ or $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $j+1$ for electric ones.

woof.