# fine structure of hydrogen atom

\begin{aligned} It turns out that the previous expression is too large, by a factor 2, due to an obscure relativistic effect known as Thomas precession . = -\frac{5}{64}\alpha^2 \ \textrm{Ry}; \\ If you use the reduced mass, you get 656.47 nm for hydrogen and 656.29 nm for deuterium. \hat{H} = \frac{\hat{p}^2}{2m_e} - \frac{Ze^2}{\hat{r}} \\ The interaction energy is that of a magnetic dipole in a magnetic field and takes the form. It is important to know this because, according to Section 1.6, we can only safely apply perturbation theory to the simultaneous eigenstates of the unperturbed and perturbing Hamiltonians. According to Equation ([e12.138]), the fine structure induced energy-shifts of these two states are the same. , and is the total angular momentum quantum number. \end{aligned} Most of these corrections are not difficult to find for general $$n$$, but the number of states to deal with proliferates, so we'll focus on $$n=2$$ to be concrete; our results here are easily extended to higher $$n$$. \begin{aligned} \begin{aligned} However, this is not really the Hamiltonian for the Hydrogen atom. \end{aligned} It is a purely relativistic effect, which arises from the fact that a relativistic quantum particle with spin does not conserve orbital angular momentum when it moves; the result of this is that if we study the evolution of the position operator $$\hat{\vec{r}}(t)$$, we will find a "jittering" of the motion about the classical path. are degenerate. operator. \ev{\hat{W}_D}_{2p} = 0. $\label{e12.121} {\mit\Delta} E_{nlm} = E_n\,\frac{\alpha^{\,2}}{n^{\,2}}\left(\frac{n}{l+1/2}-\frac{3}{4}\right),$ where $\alpha = \frac{e^{\,2}}{4\pi\,\epsilon_0\,\hbar\,c}\simeq \frac{1}{137}$ is the dimensionless fine structure constant. Fine Structure in Hydrogen In this section, we will calculate the fine structure corrections to the Hydrogen spectrum. Note, also, that we are able to employ non-degenerate perturbation theory in the previous calculation, using the $$\psi_{nlm}$$ eigenstates, because the perturbing Hamiltonian commutes with both $$L^2$$ and $$L_z$$. The spin-orbit interaction is also a magnetic interaction, but with the magnetic field generated by the orbital motion of an electron within the atom itself. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. \]. Watch the recordings here on Youtube! \]. \end{aligned} states of a hydrogen atom is illustrated in Fig. \begin{aligned} where is the radial quantum number, V�WLqdlYn�M�:A���J��@㷌��}��%�Ic��wq��*���~ĉ���\$�}�Q�� ����_[��7��:�r�>���w+aR4PfҮ�>��nCs�\f���܅�7p��S� {�{�F&7W2S+�ei���ֵ���ڗ�oj�tݴ����5z�b���AZt��׭[����,MJLa����oj_փ�B|���ĝ6����� A0a6�ǘ�q�,��aV��H�a��H�Y��Y_LNS�]�q��[��d�#0��*yϡ� m����H&B4l�⦘���t�J(�T���Rn��bF�UJ˔�+��J�.���ϵjn &\�O��*\��X^��DJ��7�u��@_es�&�9��y�S5\��}��߆ �R���6����Dw��f���u�Sȥ. as \Rightarrow \bra{\hat{\vec{r}}} \hat{H} \ket{nlm} = -\frac{\hbar^2}{2m_er^2} \left[ \frac{d}{dr} \left( r^2 \frac{d}{dr} \right) - l(l+1) \right] \psi_{nlm}(r) - \frac{Ze^2}{r} \psi_{nlm}(r). \end{aligned} \], Here $$\lambda$$ can be any parameter that appears in the Hamiltonian; the eigenstates and eigenenergies then depend implicitly on $$\lambda$$ as well. I will from here on out set $$Z=1$$, specializing to the hydrogen atom and not just single-electron atoms. Adopting the notation introduced in Section [s11.3], let $$\psi^{(2)}_{l,s;j,m_j}$$ be a simultaneous eigenstate of $$L^2$$, $$S^2$$, $$J^{\,2}$$, and $$J_z$$ corresponding to the eigenvalues \begin{aligned} L^2\,\psi^{(2)}_{l,s;j,m_j} &= l\,(l+1)\,\hbar^{\,2}\,\psi^{(2)}_{l,s;j,m_j},\\[0.5ex] S^2\,\psi^{(2)}_{l,s;j,m_j} &= s\,(s+1)\,\hbar^{\,2}\,\psi^{(2)}_{l,s;j,m_j},\\[0.5ex] J^{\,2}\,\psi^{(2)}_{l,s;j,m_j} &= j\,(j+1)\,\hbar^{\,2}\,\psi^{(2)}_{l,s;j,m_j},\\[0.5ex] J_z\,\psi^{(2)}_{l,s;j,m_j} &= m_j\,\hbar\,\psi^{(2)}_{l,s;j,m_j}.\end{aligned} According to standard first-order perturbation theory, the energy-shift induced in such a state by spin-orbit coupling is given by \begin{aligned} {\mit\Delta} E_{l,1/2;j,m_j} &= \langle l,1/2;j,m_j|H_1|l,1/2;j,m_j\rangle\nonumber\\[0.5ex] &= \frac{e^{\,2}}{16\pi\,\epsilon_0\,m_e^{\,2}\,c^{\,2}}\left\langle 1,1/2;j,m_j\left|\frac{J^{\,2}-L^2-S^2}{r^{\,3}}\right|l,1/2;j,m_j\right\rangle\nonumber\\[0.5ex] &= \frac{e^{\,2}\,\hbar^{\,2}}{16\pi\,\epsilon_0\,m_e^{\,2}\,c^{\,2}}\,\left[j\,(j+1)-l\,(l+1)-3/4\right]\,\left\langle\frac{1}{r^{\,3}}\right\rangle.\end{aligned} Here, we have made use of the fact that $$s=1/2$$ for an electron. \end{aligned} This section discusses the analytical solutions for the hydrogen atom as the problem is analytically solvable and is the base model for energy level calculations in more complex atoms. << \]. However, E_\lambda = \bra{E_{\lambda}} \hat{H}_{\lambda} \ket{E_{\lambda}} \hat{W}_{SO} = \frac{e^2}{2m_e^2 c^2 r^3} \hat{\vec{L}} \cdot \hat{\vec{S}} \Delta E_{2p,j=3/2} = \left( -\frac{7}{192} + \frac{4}{192} \right) \alpha^2\ \textrm{Ry} \\ \end{aligned} \end{aligned} \begin{aligned} Examination of fine structure by saturation spectroscopy, The small splitting of the spectral line is attributed to an interaction between the electron spin S and the orbital angular momentum L. It is called the. Like all the fine structure corrections, this is down by a factor of order from the Hydrogen binding energy. You can find some slightly more rigorous attempts to justify the Darwin term non-relativistically if you look through the literature; in my opinion, you won't gain much from any of them. There are three terms we must consider, which give rise to corrections of roughly the same magnitude: \end{aligned} Relative to the Coulomb potential $$e^2 / r$$, the size of the spin-orbit term can be written, \[, ${\displaystyle \alpha = {\tfrac {1} {4}}Z_ {0}G_ {0}} .$. above calculation, using the eigenstates, because the perturbing Hamiltonian commutes \ev{\hat{W}_{SO}}_{2p,j=1/2} = -\frac{1}{24} \alpha^2\ \textrm{Ry} \\ \begin{aligned} For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org.

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