- #1
flyusx
- 67
- 10
- Homework Statement
- Find the intensity associated with the transition in the hydrogen atom.
- Relevant Equations
- $$W_{i\to f}=\frac{4\omega_{f\to i}^{3}}{3\hbar c^{3}}\left\vert\boldsymbol{d}_{f\to i}\right\vert^{2}$$
I know that to find the intensity, I must first calculate the transition rate. I started from the transition rate equation for spontaneous emission. $$W_{i\to f}=\frac{4\omega_{f\to i}^{3}}{3\hbar c^{3}}\left\vert\boldsymbol{d}_{f\to i}\right\vert^{2}=\frac{4e^{2}\omega_{f\to i}^{3}}{3\hbar c^{3}}\left\vert\left\langle2,1,m_l\vert\boldsymbol{\epsilon}\cdot e\boldsymbol{r}\vert3,0,0\right\rangle\right\vert^{2}$$
Where ##\boldsymbol{\epsilon}## is the polarisation vector. An equivalent formulation is $$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^{3}}{3\hbar c^{3}}\left\vert\boldsymbol{d}_{f\to i}\right\vert^{2}=\frac{4e^{2}\omega_{f\to i}^{3}}{3\hbar c^{3}}\left\vert\left\langle2,1,m_l\vert r\boldsymbol{\epsilon}\cdot\boldsymbol{r}_{\text{uv}}\vert3,0,0\right\rangle\right\vert^{2}$$
Where I have split ##\boldsymbol{r}=r\boldsymbol{r}_{\text{uv}}## into the product of its magnitude and unit vector. Finding the dipole moment then becomes an issue of computing a radial integral and an integral over the solid angle. The radial integral is straightforward. $$\int_0^\infty \text{d}r r^2 r R_{2,1}^* R_{3,0}=\frac{2^{15}3^{8} a_0^2}{5^{12}}$$
To compute the angular integral, I have used that the unit vector components of the coordinate operator are related to the spherical harmonics in the following way.
$$x_\text{uv}=\sin(\theta)\cos(\phi)=-\sqrt{\frac{2\pi}{3}}\left(Y_{1,1}-Y_{1,-1}\right)$$$$y_\text{uv}=\sin(\theta)\sin(\phi)=i\sqrt{\frac{2\pi}{3}}\left(Y_{1,1}+Y_{1,-1}\right)$$$$z_\text{uv}=\cos(\theta)=\sqrt{\frac{4\pi}{3}}Y_{1,0}$$
Hence $$\boldsymbol{\epsilon}\cdot\boldsymbol{r}_{\text{uv}}=-\sqrt{\frac{2\pi}{3}}\left(Y_{1,1}-Y_{1,-1}\right)\epsilon_x+i\sqrt{\frac{2\pi}{3}}\left(Y_{1,1}+Y_{1,-1}\right)\epsilon_y+\sqrt{\frac{4\pi}{3}}Y_{1,0}\epsilon_z$$ And the angular integral becomes $$\int\text{d}\Omega Y_{1,m_l}^* Y_{0,0}\boldsymbol{\epsilon}\cdot\boldsymbol{r}_{\text{uv}}=\frac{1}{\sqrt{4\pi}}\sqrt{\frac{4\pi}{3}}\int\text{d}\Omega Y_{1,m_l}^* \left(\frac{-\epsilon_x+i\epsilon_y}{\sqrt{2}}Y_{1,1}+\frac{\epsilon_x+i\epsilon_y}{\sqrt{2}}Y_{1,-1}+\epsilon_z Y_{1,0}\right)$$
Spherical harmonics are orthonormal when integrated over the solid angle. This gives $$\int\text{d}\Omega Y_{1,m_l}^* Y_{0,0}\boldsymbol{\epsilon}\cdot\boldsymbol{r}_{\text{uv}}=\frac{1}{\sqrt{3}}\left(\frac{-\epsilon_x+i\epsilon_y}{\sqrt{2}}\delta_{m_l,1}+\frac{\epsilon_x+i\epsilon_y}{\sqrt{2}}\delta_{m_l,-1}+\epsilon_z \delta_{m_l,0}\right)$$
Then $$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^3}{3\hbar c^3}\frac{2^{15} 3^8 a_0^2}{5^{12}}\frac{1}{3}\left(\frac{\epsilon_x^2+\epsilon_y^2}{2} \delta_{m_l,1}+\frac{\epsilon_x^2+\epsilon_y^2}{2}\delta_{m_l,-1}+\epsilon_z^2 \delta_{m_l,0}\right)$$
Where the cross terms in the squared magnitude vanish since they contain the product of ##\delta_{a,b}\delta_{a,c}## which is nonzero only if ##b=c##. The result must be summed over ##m_l=-1,0,1##.
$$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^3}{3\hbar c^3}\frac{2^{15} 3^8 a_0^2}{5^{12}}\frac{1}{3}\left(\frac{\epsilon_x^2+\epsilon_y^2}{2}+\frac{\epsilon_x^2+\epsilon_y^2}{2}+\epsilon_z^2\right)$$
The polarisations satisfy ##\vert\epsilon_x\vert^2+\vert\epsilon_y\vert^2+\vert\epsilon_z\vert^2=1##, leaving $$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^3}{3\hbar c^3}\frac{2^{15} 3^8 a_0^2}{5^{12}}\frac{1}{3}$$.
Note that the initial state has an energy magnitude ##\frac{\mathcal{R}}{9}## while the final state has an energy magnitude ##\frac{\mathcal{R}}{4}##. The difference is ##\frac{5\mathcal{R}}{36}##. Since ##\mathcal{R}=\frac{e^2}{2a_0}##, the energy difference is ##\hbar\omega_{f\to i}=\frac{5e^2}{72a_0}##. Dividing through by ##\hbar## gives ##\omega_{f\to i}##.
Cubing this, substituting into ##W_{i \to f}## and using ##\alpha=\frac{e^2}{\hbar c}## gives $$W_{i \to f}=\frac{2^{17}3^6}{5^{12}}\frac{e^2 a_0^2}{\hbar c^3}\omega_{f\to i}^{3}=\frac{2^{17}3^6}{5^{12}}\frac{e^2 a_0^2}{\hbar c^3}\frac{e^6}{a_{0}^{3}\hbar^{3}}\frac{5^{3}}{2^{9}3^{6}}=\frac{2^8}{5^9}\frac{\alpha^4 c}{a_0}$$
I find that this is off by a factor of three from the well-known value of ##6.31\times 10^{6}## Hz. This is despite the fact that I have summed over the three ##m_l## states that the 3s state can transition into. I suspect that I have actually calculated the transition rate for one photon mode instead of the three photon modes, but I cannot see where specifically I am going wrong.
I am using the method that Zettili uses to obtain the ##2p\to 1s## transition rate in one of his solved problems (of which I have attached). He does it in about a page, and goes through this exact same process. In the end, he derives the correct transition rate.
As a remark, he also introduces (for the ##2p\to 1s## case) the following formula $$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^{3}}{3\hbar c^{3}}\frac{1}{3}\sum_{m_{l}=-1}^{1}\vert\langle2,1,m_{l}\vert\boldsymbol{r}\vert1,0,0\rangle\vert^2$$ where he explicitly sums over the possible ##m_{l}## that the ##2p## state can have and tags along a ##\frac{1}{3}## factor to 'average over the various transitions'. Of course in my case, I would need to replace the initial and final states to apply it. Is the 'averaging' referring to the three ##m_l## states corresponding with ##2p##?
Doing a bit of further digging, I have found that some sources call this the 'Einstein A Coefficient' and state the need to average over the initial degenerate states and sum over the final degenerate states. This would explain the one-third factor from Zettili's workings (##2p## is threefold degenerate and ##1s## is nondegenerate), and it would also explain why I am off by a factor of three (##3s## is nondegenerate but a sum must be done over the three ##2p## degeneracies).
Where ##\boldsymbol{\epsilon}## is the polarisation vector. An equivalent formulation is $$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^{3}}{3\hbar c^{3}}\left\vert\boldsymbol{d}_{f\to i}\right\vert^{2}=\frac{4e^{2}\omega_{f\to i}^{3}}{3\hbar c^{3}}\left\vert\left\langle2,1,m_l\vert r\boldsymbol{\epsilon}\cdot\boldsymbol{r}_{\text{uv}}\vert3,0,0\right\rangle\right\vert^{2}$$
Where I have split ##\boldsymbol{r}=r\boldsymbol{r}_{\text{uv}}## into the product of its magnitude and unit vector. Finding the dipole moment then becomes an issue of computing a radial integral and an integral over the solid angle. The radial integral is straightforward. $$\int_0^\infty \text{d}r r^2 r R_{2,1}^* R_{3,0}=\frac{2^{15}3^{8} a_0^2}{5^{12}}$$
To compute the angular integral, I have used that the unit vector components of the coordinate operator are related to the spherical harmonics in the following way.
$$x_\text{uv}=\sin(\theta)\cos(\phi)=-\sqrt{\frac{2\pi}{3}}\left(Y_{1,1}-Y_{1,-1}\right)$$$$y_\text{uv}=\sin(\theta)\sin(\phi)=i\sqrt{\frac{2\pi}{3}}\left(Y_{1,1}+Y_{1,-1}\right)$$$$z_\text{uv}=\cos(\theta)=\sqrt{\frac{4\pi}{3}}Y_{1,0}$$
Hence $$\boldsymbol{\epsilon}\cdot\boldsymbol{r}_{\text{uv}}=-\sqrt{\frac{2\pi}{3}}\left(Y_{1,1}-Y_{1,-1}\right)\epsilon_x+i\sqrt{\frac{2\pi}{3}}\left(Y_{1,1}+Y_{1,-1}\right)\epsilon_y+\sqrt{\frac{4\pi}{3}}Y_{1,0}\epsilon_z$$ And the angular integral becomes $$\int\text{d}\Omega Y_{1,m_l}^* Y_{0,0}\boldsymbol{\epsilon}\cdot\boldsymbol{r}_{\text{uv}}=\frac{1}{\sqrt{4\pi}}\sqrt{\frac{4\pi}{3}}\int\text{d}\Omega Y_{1,m_l}^* \left(\frac{-\epsilon_x+i\epsilon_y}{\sqrt{2}}Y_{1,1}+\frac{\epsilon_x+i\epsilon_y}{\sqrt{2}}Y_{1,-1}+\epsilon_z Y_{1,0}\right)$$
Spherical harmonics are orthonormal when integrated over the solid angle. This gives $$\int\text{d}\Omega Y_{1,m_l}^* Y_{0,0}\boldsymbol{\epsilon}\cdot\boldsymbol{r}_{\text{uv}}=\frac{1}{\sqrt{3}}\left(\frac{-\epsilon_x+i\epsilon_y}{\sqrt{2}}\delta_{m_l,1}+\frac{\epsilon_x+i\epsilon_y}{\sqrt{2}}\delta_{m_l,-1}+\epsilon_z \delta_{m_l,0}\right)$$
Then $$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^3}{3\hbar c^3}\frac{2^{15} 3^8 a_0^2}{5^{12}}\frac{1}{3}\left(\frac{\epsilon_x^2+\epsilon_y^2}{2} \delta_{m_l,1}+\frac{\epsilon_x^2+\epsilon_y^2}{2}\delta_{m_l,-1}+\epsilon_z^2 \delta_{m_l,0}\right)$$
Where the cross terms in the squared magnitude vanish since they contain the product of ##\delta_{a,b}\delta_{a,c}## which is nonzero only if ##b=c##. The result must be summed over ##m_l=-1,0,1##.
$$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^3}{3\hbar c^3}\frac{2^{15} 3^8 a_0^2}{5^{12}}\frac{1}{3}\left(\frac{\epsilon_x^2+\epsilon_y^2}{2}+\frac{\epsilon_x^2+\epsilon_y^2}{2}+\epsilon_z^2\right)$$
The polarisations satisfy ##\vert\epsilon_x\vert^2+\vert\epsilon_y\vert^2+\vert\epsilon_z\vert^2=1##, leaving $$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^3}{3\hbar c^3}\frac{2^{15} 3^8 a_0^2}{5^{12}}\frac{1}{3}$$.
Note that the initial state has an energy magnitude ##\frac{\mathcal{R}}{9}## while the final state has an energy magnitude ##\frac{\mathcal{R}}{4}##. The difference is ##\frac{5\mathcal{R}}{36}##. Since ##\mathcal{R}=\frac{e^2}{2a_0}##, the energy difference is ##\hbar\omega_{f\to i}=\frac{5e^2}{72a_0}##. Dividing through by ##\hbar## gives ##\omega_{f\to i}##.
Cubing this, substituting into ##W_{i \to f}## and using ##\alpha=\frac{e^2}{\hbar c}## gives $$W_{i \to f}=\frac{2^{17}3^6}{5^{12}}\frac{e^2 a_0^2}{\hbar c^3}\omega_{f\to i}^{3}=\frac{2^{17}3^6}{5^{12}}\frac{e^2 a_0^2}{\hbar c^3}\frac{e^6}{a_{0}^{3}\hbar^{3}}\frac{5^{3}}{2^{9}3^{6}}=\frac{2^8}{5^9}\frac{\alpha^4 c}{a_0}$$
I find that this is off by a factor of three from the well-known value of ##6.31\times 10^{6}## Hz. This is despite the fact that I have summed over the three ##m_l## states that the 3s state can transition into. I suspect that I have actually calculated the transition rate for one photon mode instead of the three photon modes, but I cannot see where specifically I am going wrong.
I am using the method that Zettili uses to obtain the ##2p\to 1s## transition rate in one of his solved problems (of which I have attached). He does it in about a page, and goes through this exact same process. In the end, he derives the correct transition rate.
As a remark, he also introduces (for the ##2p\to 1s## case) the following formula $$W_{i\to f}=\frac{4e^{2}\omega_{f\to i}^{3}}{3\hbar c^{3}}\frac{1}{3}\sum_{m_{l}=-1}^{1}\vert\langle2,1,m_{l}\vert\boldsymbol{r}\vert1,0,0\rangle\vert^2$$ where he explicitly sums over the possible ##m_{l}## that the ##2p## state can have and tags along a ##\frac{1}{3}## factor to 'average over the various transitions'. Of course in my case, I would need to replace the initial and final states to apply it. Is the 'averaging' referring to the three ##m_l## states corresponding with ##2p##?
Doing a bit of further digging, I have found that some sources call this the 'Einstein A Coefficient' and state the need to average over the initial degenerate states and sum over the final degenerate states. This would explain the one-third factor from Zettili's workings (##2p## is threefold degenerate and ##1s## is nondegenerate), and it would also explain why I am off by a factor of three (##3s## is nondegenerate but a sum must be done over the three ##2p## degeneracies).
Last edited: