91影视

When two equal and opposite charges are separated by a distance in space, is called a dipole. The product of the distance between them and their magnitude is called the dipole moment of that dipole.

As you know that dipole moment will be

$$\begin{gathered} p=-eRe\left(e^{i鈭�Et/h}鈭�r{{蠄}^{*}}_f\left(\stackrel{炉}{r}\right){蠄}_i\left(\stackrel{炉}{r}\right)dV\right) \\ {{蠄}^{*}}_f\left(\stackrel{炉}{r}\right){蠄}_i\left(\stackrel{炉}{r}\right) \\ r=x\hat{x}+y\hat{y}+z\hat{z}......\left(1\right) \end{gathered}$$

By solving the product ${{蠄}^{*}}_f\left(\stackrel{炉}{r}\right){蠄}_i\left(\stackrel{炉}{r}\right)$and writing out the r vector in its Cartesian component, you get

$$\begin{gathered} r=x\hat{x}+y\hat{y}+z\hat{z} \\ r=r\sin 胃\cos 蠒\hat{x}+r\sin 胃\sin 蠒\hat{y}+r\cos 胃\hat{z}......\left(2\right) \end{gathered}$$

Now, by using equation (2) in equation (1) and leaving R(r) and general,

Where, r is the distance, $胃$ is the colatitude, $蠒$ is the azimuth, and $蠄$wave function.

Therefore,

$鈭�\left(r\sin 胃\cos 蠒\hat{x}+r\sin 胃\sin 蠒\hat{y}+r\cos 胃\hat{z}\right)=R_f\left(\stackrel{鈫�}{r}\right){鈯�}_f\left(胃\right)e^{-i{m}_l{l}_f}{R}_i\left(\stackrel{鈫�}{r}\right){鈯�}_i\left(胃\right)e^{-i{m}_l{l}_l}{r}^2\sin 胃drd胃d蠒$

The x-component involves the integral,

$l_x={鈭�}_0^{2x}\cos 蠒e^{+i(m{l}_l+m{l}_l)蠒}d蠒$

In the above integral, if$鈭�m_l=0$:

The term will be the cosine integral from 0 to $2\mathrm{蟺}$ which will give zero.

Where, is the magnetic quantum number.

If $鈭�m_l=卤1$

$$\begin{gathered} l_x={鈭�}_0^{2\mathrm{蟺}}\cos 蠒\left(\cos 蠒卤i\sin 蠒\right)d蠒 \\ ={鈭�}_0^{2\mathrm{蟺}}{\cos }^2蠒d蠒卤i{鈭�}_0^{2\mathrm{蟺}}\cos 蠒\sin 蠒d蠒 \\ ={\left[\frac{1}{4}\sin 2蠒+\frac{蠒}{2}\right]}_0^{2\mathrm{蟺}}+i{\left[-\frac{1}{2}{\cos }^2蠒\right]}_0^{2\mathrm{蟺}} \\ =\left[\frac{1}{4}\sin \left(4\mathrm{蟺}\right)+\frac{2\mathrm{蟺}}{2}-\frac{1}{4}\sin \left(0\right)+0\right]+i\left[-\frac{1}{2}{\cos }^2\left(2\mathrm{蟺}\right)+\frac{1}{2}{\cos }^2\left(0\right)\right] \\ {l}_x=\left[\mathrm{蟺}\right]+i\left[-\frac{1}{2}+\frac{1}{2}\right] \\ =\mathrm{蟺} \end{gathered}$$

The y-component involves the integral,

$l_y={鈭�}_0^{2\mathrm{蟺}}\sin 蠒e^{+i{(m{l}_i+m{l}_f)}^{蠒}}d蠒$

In the above integral,

If $鈭�m_l=0$,

The term will be the sine integral from 0 to $2\mathrm{蟺}$ which will give zero.

If $鈭�m_l=卤1$,

$$\begin{gathered} l_y={鈭�}_0^{2\mathrm{蟺}}\sin 蠒\left(\cos 蠒卤i\sin 蠒\right)d蠒 \\ ={鈭�}_0^{2\mathrm{蟺}}\cos 蠒\sin 蠒d蠒卤{鈭�}_0^{2\mathrm{蟺}}{\sin }^2蠒d蠒 \\ =卤\mathrm{蟺} \end{gathered}$$

The z-component involves the integral,

$l_z={鈭�}_0^{2\mathrm{蟺}}{e}^{+i\left(m{l}_i+m{l}_f\right)蠒}d蠒$

In the above integral,

If$鈭�m_l=0$,

The term will give$2\mathrm{蟺}$

If$鈭�m_l=卤1$,

It will give $\cos 蠒卤i\sin 蠒$ integrated over one period, i.e., zero.

From the steps 2, 3 and 4 given above, you get, $鈭�m_l=卤1$corresponds to a dipole moment in the xy-plane, while $鈭�m_l=0$corresponds to a moment along the z-axis.