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I am trying to work my way through Dirac-Hartree-Fock. I have reached a point where I am baffled. I will present the problem here, but will have to define a few things since nomenclature is so varied in the area. First the spinor:$$\Psi =\left[ \begin{array}{c} \psi_L^+ \\ \psi_L^- \\ i\psi_S^+ \\ i\psi_S^- \\ \end{array}\right]=\left[ \begin{array}{c} \psi_L \\ i\psi_S \\ \end{array}\right]$$Next, the one-electron operator:$$\left[ \begin{array}{cc} (\varepsilon_n)_2 &c\sigma\cdot\widehat{\mathbf{p}}_c \\c\sigma\cdot\widehat{\mathbf{p}}_c & (-2m_ec^2+\varepsilon_n)_2 \\ \end{array}\right]\left[ \begin{array}{c} \psi_L \\ i\psi_S \\ \end{array}\right]=\varepsilon_1\left[ \begin{array}{c} \psi_L \\ i\psi_S \\ \end{array}\right]$$where$\epsilon_n$ is the nuclear attraction potential. The subscript "2" indicates a diagonal 2 x 2 sub-matrix. The 4 x 4 matrix above is the one-electron operator:$$\widehat{h}_D \Psi = \epsilon_{1}\Psi$$The ``dot product'' of the Pauli matrices and the canonical momentum operator is$$\sigma\cdot\widehat{\mathbf{p}}_c=\left[ \begin{array}{cc} \widehat{p}_z & \widehat{p}_x-i\widehat{p}_y \\ \widehat{p}_x+i\widehat{p}_y & -\widehat{p}_z \\ \end{array}\right]=\left[ \begin{array}{cc} -i\hbar\dfrac{\partial}{\partial z} & -i\hbar\dfrac{\partial}{\partial x}+\hbar\dfrac{\partial}{\partial y} \\ -i\hbar\dfrac{\partial}{\partial x}-\hbar\dfrac{\partial}{\partial y} & i\hbar\dfrac{\partial}{\partial z} \\ \end{array}\right]$$The one-electron operator, in atomic units, is therefore$$\widehat{h}_D=\left[ \begin{array}{cccc} \varepsilon_n & 0&c\widehat{p}_z & c(\widehat{p}_x-i\widehat{p}_y) \\ 0 & \varepsilon_n &c(\widehat{p}_x+i\widehat{p}_y) & -c\widehat{p}_z \\ c\widehat{p}_z & c(\widehat{p}_x-i\widehat{p}_y)& (-2c^2+\varepsilon_n) & 0 \\ c(\widehat{p}_x+i\widehat{p}_y) & -c\widehat{p}_z& 0& (-2c^2+\varepsilon_n) \\ \end{array}\right],$$where$\widehat{p}_z =-i\partial/\partial z$, etc.Because the momentum operators are self-adjoint the conjugate transpose of the matrix leaves it unchange -- so$\widehat{h}_D$ is Hermitian. Thus$$\langle \Psi|\widehat{h}_D|\Psi\rangle =\varepsilon_1,$$where$\epsilon_1$ is the one-electron contribution to the energy --a real number.Now, to my dilemma:Some abbreviations:\begin{align*}m_z&=c\widehat{p}_z=-ic\dfrac{\partial}{\partial z}\\m_{xy}^+ &= c(\widehat{p}_x+i\widehat{p}_y)=\left(-ic\dfrac{\partial}{\partial x}+c\dfrac{\partial}{\partial y} \right)\\m_{xy}^- &= c(\widehat{p}_x-i\widehat{p}_y)=\left(-ic\dfrac{\partial}{\partial x}-c\dfrac{\partial}{\partial y} \right)\\\varepsilon_n^c&=-2c^2+\varepsilon_n.\end{align*}The integrand for the matrix element is then\begin{align*} \Psi^{*T}|\widehat{h}_D|\Psi &= \left[ \begin{array}{cccc} \psi_L^{+*} & \psi_L^{-*} & i\psi_S^{+*} & i\psi_S^{-*} \\ \end{array} \right] \left[ \begin{array}{cccc} \varepsilon_n & 0 & m_z & m_{xy}^+ \\ 0 & \varepsilon_n & m_{xy}^- & -m_z \\ m_z & m_{xy}^+ & \varepsilon_n^c & 0 \\ m_{xy}^- & -m_z & 0 & \varepsilon_n^c \\ \end{array} \right] \left[ \begin{array}{c} \psi_L^+ \\ \psi_L^- \\ i\psi_S^+ \\ i\psi_S^- \\ \end{array}\right]\nonumber\\ &= (\psi_L^{+*}\varepsilon_n \psi_L^+) + (i\psi_L^{+*}m_z\psi_S^+) + (i \psi_L^{+*}m_{xy}^+\psi_S^-)\nonumber\\ &\ \ \ \ +(\psi_L^{-*}\varepsilon_n\psi_L^-)+(i\psi_L^{-*}m_{xy}^-\psi_S^+ )-(i\psi_L^{-*}m_z\psi_S^-)\nonumber\\ &\ \ \ \ +(i\psi_S^{+*}m_z\psi_L^+)+(i\psi_S^{+*}m_{xy}^+\psi_L^- )+(i^2\psi_S^{+*}\varepsilon_n^c\psi_S^+)\nonumber\\ &\ \ \ \ +(i\psi_S^{-*}m_{xy}^-\psi_L^+ )-(i\psi_S^{-*}m_z\psi_L^- )+(i^2\psi_S^{-*}\varepsilon_n^c\psi_S^-). \end{align*}Noting that$m_{xy}^+$ and$m_{xy}^-$ each are the sums of two operators (and hoping that I got my signs right!), the integrand is the sum of sixteen terms, each of which must be integrated. I can see that some of the terms will be real, but how can those with a momentum operator and two different spinor components be real? The most obvious problem is seen for$m_{xy}^+$ or$m_{xy}^-$:$$ \langle \psi_L^{+}|m_{xy}^+|\psi_S^-\rangle =-ic\int_\mathbf{r}\psi_L^{+*}(\mathbf{r})\left|\frac{\partial}{\partial x}\right|\psi_S^-(\mathbf{r})d\mathbf{r} +c\int_\mathbf{r}\psi_L^{+*}(\mathbf{r})\left|\frac{\partial}{\partial y}\right|\psi_S^-(\mathbf{r})d\mathbf{r}.$$How can this possibly be a real number?? Clearly, if one of the expressions in the sum turns out to be real, the other will be complex! I am sure that I am missing something here -- which is my reason for reaching out.
- 1$\begingroup$You know that the result has to be real, and thus whatever intermediate computations become complex, there would be a complex conjugate pair to cancel out the imaginary parts. Why don't you try and compute the complex conjugate of that expression, or whatever it is that will isolate the imaginary component, and show that the imaginary part cancels away?$\endgroup$naturallyInconsistent– naturallyInconsistent2025-11-11 21:30:23 +00:00CommentedNov 11 at 21:30
- 3$\begingroup$You forgot the complex conjugation of the $i$ factors in the row vector ($-i\psi^{\pm*}_S$ instead of $i\psi^{\pm*}_S$). Using this and $m^+_{xy}=(m^-_{xy})^\dagger$, you can see that all complex terms appear together with their conjugates, so the overall expectation value is real, as it should be.$\endgroup$dennismoore94– dennismoore942025-11-12 00:27:12 +00:00CommentedNov 12 at 0:27
- $\begingroup$@naturallyinconsistent Thank you. I made an error in the spinor adjoint.$\endgroup$Dennis W. Bennett– Dennis W. Bennett2025-11-12 04:15:36 +00:00CommentedNov 12 at 4:15
- $\begingroup$You should thank dennismoore instead. I did not even bother to carefully scrutinise your equations.$\endgroup$naturallyInconsistent– naturallyInconsistent2025-11-12 04:19:16 +00:00CommentedNov 12 at 4:19
- $\begingroup$@dennismoore94 Thanks. I was able to do this by changing the explicit imaginary small components to real numbers, based on the literature, where the imaginary components are not always used. I'll go back now and see if it all works!$\endgroup$Dennis W. Bennett– Dennis W. Bennett2025-11-12 04:19:35 +00:00CommentedNov 12 at 4:19
1 Answer1
The most obvious problem is seen for$m_{xy}^+$ or$m_{xy}^-$:$$ \langle \psi_L^{+}|m_{xy}^+|\psi_S^-\rangle =-ic\int_\mathbf{r}\psi_L^{+*}(\mathbf{r})\left|\frac{\partial}{\partial x}\right|\psi_S^-(\mathbf{r})d\mathbf{r} +c\int_\mathbf{r}\psi_L^{+*}(\mathbf{r})\left|\frac{\partial}{\partial y}\right|\psi_S^-(\mathbf{r})d\mathbf{r}.$$How can this possibly be a real number?? Clearly, if one of the expressions in the sum turns out to be real, the other will be complex! I am sure that I am missing something here -- which is my reason for reaching out.
The single term you wrote down is not necessarily real, but it necessarily has a counterpart term that when summed makes the sum real.
Consider a simpler two-dimensional example$$\langle h \rangle \equiv (\alpha^* \beta^*)\left(\begin{matrix}E & m^+ \\m^- & E\end{matrix}\right)\left(\begin{matrix}\alpha \\\beta\end{matrix}\right)\;,\tag{1}$$where$E=E^*$ is real and where$m^{\pm} = (p_x \pm i p_y)$.
We have$$\langle h \rangle = \int dx (|\alpha(x)|^2 E + \alpha(x)^* m^+ \beta(x) + \beta^*(x) m^- \alpha(x) + |\beta(x)|^2 E)\;,\tag{2}$$where the first and fourth terms in Eq. (2) are clearly real. The second and third terms in Eq. (2) are not necessarily real, individually, but their sum is real since$$\int dx\alpha^*(x) m^+ \beta(x) = \left(\int dx \beta^*(x) m^- \alpha(x) \right)^*$$and so letting$$\int dx \alpha^*(x) m^+ \beta(x) \equiv u + i v\;,$$where$u$ and$v$ are real, we have$$\int dx (\alpha^*(x) m^+ \beta(x) + \beta^*(x) m^- \alpha(x)) = 2u\;,$$which is real.
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