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$ \newcommand{\vec}[1]{\mathbf{#1}} \newcommand{\op}[1]{\hat{#1}} \newcommand{\bra}[1]{\langle{#1}\rvert} \newcommand{\ket}[1]{\lvert{#1}\rangle} \newcommand{\braket}[2]{\langle{#1}|{#2}\rangle} \DeclareMathOperator{sgn}{sgn} $ I learned the fact that

dipole-allowed transitions occur only between $N$-electron states differing in one electron that makes a dipole-allowed transition with the other $N-1$ electrons remaining in their initial single particle state.

I tried proving that statement, with some assumptions, following my textbook (Introduction to the Physics of Matter by N. Manini, from which I also pulled that quote) as follows.

The dipole operator driving the transitions is the sum $\op{\vec D}=\sum_{i=1}^N\op{\vec d}^{(i)}$ where I guess that

\begin{equation} \op{\vec d}^{(i)}= \underset{1}{\op{1}} \otimes \dotsb \otimes \underset{i-1}{\op{1}} \otimes \underset{i}{\op{\vec d}} \otimes \underset{i+1}{\op{1}} \otimes \dotsb \otimes \underset{N}{\op{1}} \end{equation} is the single particle operator, in the product space of $N$ electrons. Choosing two sets of $N$ states (picked from a complete orthonormal set) $\{\ket{a_i}\}_{i=1}^N$ for the initial configuration and $\{\ket{b_i}\}_{i=1}^N$ for the final one, we build the appropriate antisymmetrized states \begin{equation} \ket{A}=\frac1{\sqrt{N!}}\sum_{\sigma\in S_N}\sgn(\sigma)\bigotimes_{i=1}^N\ket{a_{\sigma(i)}} \end{equation} as a Slater determinant ($S_N$ is the symmetric group of order $N$), and $\ket{B}$ in a similar fashion. So I have \begin{equation} \begin{split} \bra{B}\op{\vec D}\ket{A}&= \sum_{i=1}^N\bra{B}\op{\vec d}^{(i)}\ket{A}=\\&= \frac1{N!}\sum_{i=1}^N\sum_{\sigma\in S_N}\sum_{\rho\in S_N}\sgn(\sigma\rho)\biggl[\bigotimes_{j=1}^N\bra{b_{\sigma(j)}}\biggr]\op{\vec d}^{(i)}\biggl[\bigotimes_{k=1}^N\ket{a_{\rho(k)}}\biggr]. \end{split} \end{equation} Now, if I approximate the final states as $\ket{b_i}\approx\ket{a_i}$ for each $i$, each braket in \begin{multline} \biggl(\bigotimes_{j=1}^N\bra{\beta_{\sigma(j)}}\biggr)\op{\vec d}^{(i)}\biggl(\bigotimes_{k=1}^N\ket{\alpha_{\rho(k)}}\biggr)=\\= \braket{\beta_{\sigma(1)}}{\alpha_{\rho(1)}}\dotsm\braket{\beta_{\sigma(i-1)}}{\alpha_{\rho(i-1)}}\bra{\beta_{\sigma(i)}}\op{\vec d}\ket{\alpha_{\rho(i)}}\braket{\beta_{\sigma(i+1)}}{\alpha_{\rho(i+1)}}\dotsm\braket{\beta_{\sigma(N)}}{\alpha_{\rho(N)}} \end{multline} approximates a Kronecker delta, i.e. $\braket{b_\mu}{a_\nu}$ is zero if the two states $\ket{a_\nu}$ and $\ket{b_\mu}$ don't coincide (so probably $\braket{b_\mu}{a_\nu}\approx\delta_{\mu,\nu}$?). This has the consequence that $N-1$ terms of the permutations $\rho$ and $\sigma$ must be the same (or else the whole product is zero), thus $\rho=\sigma$ and \begin{equation} \begin{split} \bra{B}\op{\vec D}\ket{A}&= \frac1{N!}\sum_{i=1}^N\sum_{\sigma\in S_N}\bra{b_{\sigma(i)}}\op{\vec d}\ket{a_{\sigma(i)}}\prod_{\substack{j=1\\j\ne i}}^N\braket{b_{\sigma(j)}}{a_{\sigma(j)}}=\\&= \sum_{i=1}^N\bra{b_i}\op{\vec d}\ket{a_i}\prod_{\substack{j=1\\j\ne i}}^N\braket{b_j}{a_j} \end{split} \end{equation} which agrees with the result in the textbook. Now, I can't see why it is the case that just one electron can make the transition while the others must remain in their original state. I would get it if there wasn't a summation in the last line, i.e. it was just $\bra{b_i}\op{\vec d}\ket{a_i}$, but this isn't the case, if I derived it correctly.

Moreover, the book goes on saying that

when $\bra{b_i}\op{\vec d}\ket{a_i}$ vanishes, because it violates the one-electron selection rules, the overall matrix element (i.e. $\bra{B}\op{\vec D}\ket{A}$) vanishes.

How is it so? If $\bra{b_i}\op{\vec d}\ket{a_i}$ vanishes, it is just an addend that is zero, not the whole sum.

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Your math is pretty much correct (including, in particular, your conclusion that this is exclusively due to the fact that the dipole operator is a single-particle operator), and your reduction of the transition matrix element to \begin{align} \bra{B}\op{\vec D}\ket{A} &= \sum_{i=1}^N\bra{b_i}\op{\vec d}\ket{a_i}\prod_{\substack{j=1\\j\ne i}}^N \braket{b_j}{a_j} \end{align} is about right. The reason that the sum doesn't matter is that only one term can ever contribute here. The easiest way to see this is to reduce it to the case of two electrons: \begin{align} \bra{B}\op{\vec D}\ket{A} &= \bra{b_1}\op{\vec d}\ket{a_1}\braket{b_2}{a_2} + \bra{b_2}\op{\vec d}\ket{a_2}\braket{b_1}{a_1}. \end{align} If both $\ket{b_1}$ and $\ket{b_2}$ are different from $\ket{a_1}$ and $\ket{a_2}$, then both terms will vanish because both will have a zero overlap $\braket{b_i}{a_i}$ multiplying the single-particle dipole transition matrix element.

In other words, you're allowed to have one of the $\ket{b_i}$ (say, $\ket{b_k}$) orthogonal to all the $\{\ket{a_j}\}$, but then that means that you'll have $$ \prod_{\substack{j=1\\j\ne i}}^N \braket{b_j}{a_j} = 0 $$ unless $i=k$, in which case your expression reduces to \begin{align} \bra{B}\op{\vec D}\ket{A} &= \sum_{i=1}^N\bra{b_i}\op{\vec d}\ket{a_i}\prod_{\substack{j=1\\j\ne i}}^N \braket{b_j}{a_j} = \sum_{i=1}^N\bra{b_i}\op{\vec d}\ket{a_i}\delta_{ik} = \bra{b_k}\op{\vec d}\ket{a_k}. \end{align}


(A word to the wise, though: it's important to keep in mind that while the orbitals $\ket{a_i}$ are useful mental models for what's going on, they are not unique representations for the Slater determinant, which is the only actual physical object here. For more details see this Q&A, but the gist is that you can apply any linear transformation with determinant 1 (it doesn't even need to be unitary) to the span of the $\ket{a_i}$, and still be left with the same physical state.

That means that many of the $\ket{b_i}$ could be "different" from the $\ket{a_i}$, with $\ket{B}$ looking formally different to $\ket{A}$, but the difference still boiling down to a single-particle excitation. In the end, when we say "only one of the $\ket{b_i}$ can be excited", this requires a mental model where both Slater determinants are drawn from the same orthonormal set of orbitals.)

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  • $\begingroup$ Thank you! But what if I had two pairs of orthogonal states $\langle a_{i_1}|b_{i_1}\rangle=0$ and $\langle a_{i_2}|b_{i_2}\rangle=0$? I think that the whole element would be zero since in every product there would be either one of those two brakets; if there is only one $\langle{a_i|b_i}\rangle=0$, then we would have a term in the sum in which this braket doesn't appear in the product, therefore such term isn't zero. Is my reasoning correct? $\endgroup$ – yellowquark Oct 23 '16 at 9:25
  • $\begingroup$ If I understand your (needlessly muddled) notation correctly, then yes, you are right, that was sort of what I was trying to say. $\endgroup$ – Emilio Pisanty Oct 23 '16 at 12:44

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