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For a real vector $\mathbf{r}$, the direction is given by: $\hat{\mathbf{n}}=\mathbf{r}/\left|\mathbf{r}\right|$.

The transition dipole moment is a complex vector. How do you define its direction?

The point of this question is that I am trying to understand the meaning of equation 9.29 of the book Charge and Energy Transfer 3rd Ed by May & Kuhn, which reads:

$$J_{mn}=\frac{\left|\mathbf{d}_{m}\right|\left|\mathbf{d}_{n}^{*}\right|}{R_{mn}^3}\left[\mathbf{n}_{m}\cdot\mathbf{n}_{n}-3\left(\mathbf{e}_{mn}\cdot\mathbf{n}_{m}\right)\left(\mathbf{e}_{mn}\cdot\mathbf{n}_{n}\right)\right]$$

In the next paragraph I attempt to explain the notation in this equation. I'll point out the parts that confuse me.

Here we are considering two molecules labeled $m$ and $n$, and we only consider two electronic levels in each, the ground state $g$ and one excited state $e$. $\mathbf{d}_{m}$ is the transition dipole moment for the transition $g\rightarrow e$ of the $m$th molecule, and similarly for $\mathbf{d}_{n}$. $\mathbf{n}_{m}$ is a unit vector pointing in the direction of $\mathbf{d}_{n}$ (I don't understand this), and similarly for $\mathbf{n}_{n}$. Finally, $R_{mn}$ is the distance between the centers of mass of the molecules. The quantity we are calculating here, $J_{mn}$, is the excitonic coupling between the two molecules. This can be seen as the rate at which an exciton at a molecule $n$ will be transfered to a molecule $m$ initially in its ground state. Here we calculate $J_{mn}$ using the dipole-dipole interaction approximation, valid when the molecules are sufficiently far apart.

Note also the factor $\left|\mathbf{d}_{n}^{*}\right|$ in the equation. I don't understand this bit of notation either. I am not sure how to interpret the complex conjugation sign ($*$) as it appears inside absolute-value brackets ($|\square |$).

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2 Answers

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The dipole transition matrix elements $\mathbf{d}_m$ and $\mathbf{d}_n$ are complex-valued vectors which are relatively easy to define. Their "direction" is a mathematical convenience, and it is essentially given by the vector divided by its modulus, for an appropriate interpretation of the latter.

Consider first the case of a single molecule, with a transition from ground $\newcommand{\bra}[1]{\langle#1|}\newcommand{\ket}[1]{|#1\rangle} \ket{g}$ to excited $\ket e$ states. The DTME is defined as the complex-valued vector $$\mathbf{d}=\bra e\hat{\mathbf{d}}\ket g.$$ This is, of course, simply shifting away the burden of definition into defining the vector operator $\hat{\mathbf{d}}$. Such operators are, actually, best understood as working component-by-component: that is, $\mathbf{d}$ is defined as the vector with components $d_j=\bra e\hat{d_j}\ket g$ for $j=1,2,3$, or more abstractly as the unique vector which satisfies $$\mathbf{d}\cdot\mathbf{e}=\bra e\hat{\mathbf{d}}\cdot\mathbf{e}\ket g$$ for all vectors $\mathbf{e}$ - where the 'integrand' $\hat{\mathbf{d}}\cdot\mathbf{e}$ is now a scalar operator and poses no difficulty.

Unfortunately, of course, the components of this vector can now be complex, which poses a challenge when using the rest of our real-oriented vector machinery. If you go to a standard textbook in electrostatics it will tell you that the coupling energy is something more like $$J_{mn}=\frac{1}{R_{mn}^3}\left[\mathbf{d}_{m}\cdot\mathbf{d}_{n}-3\left(\mathbf{e}_{mn}\cdot\mathbf{d}_{m}\right)\left(\mathbf{e}_{mn}\cdot\mathbf{d}_{n}\right)\right]$$ without any fancy jiggamajig. However, if you want to do this properly, what you should be lifting from classical physics is the recipes to make operators, which means that the formula above should be interpreted in an operator sense: $$\hat J_{mn}=\frac{1}{R_{mn}^3}\left[ \hat{\mathbf{d}}_{m}\cdot\hat{\mathbf{d}}_{n}-3\left(\mathbf{e}_{mn}\cdot\hat{\mathbf{d}}_{m}\right)\left(\mathbf{e}_{mn}\cdot\hat{\mathbf{d}}_{n}\right)\right].$$ This makes sense: the dipole moment "in itself" is real, and it is only the states that bring in the complexness. (In a sense, of course!) Thus, what you want is the transition element of this operator in between $\ket e_n\ket g_m$ and $\ket g_n\ket e_m$, and this will get you something like $$\begin{align} J_{mn} &={}_n\bra g{}_m\bra e \hat J_{mn}\ket e_n\ket g_m \\&=\frac{1}{R_{mn}^3}\left[{}_m\bra e \hat{\mathbf{d}}_{m}\ket g_m\cdot{}_n\bra g{}\hat{\mathbf{d}}_{n}\ket e_n-3\left(\mathbf{e}_{mn}\cdot{}_m\bra e \hat{\mathbf{d}}_{m}\ket g_m\right)\left(\mathbf{e}_{mn}\cdot{}_n\bra g{}\hat{\mathbf{d}}_{n}\ket e_n\right)\right] \\&=\frac{1}{R_{mn}^3}\left[\mathbf{d}_{m}\cdot\mathbf{d}_{n}^\ast-3\left(\mathbf{e}_{mn}\cdot\mathbf{d}_{m}\right)\left(\mathbf{e}_{mn}\cdot\mathbf{d}_{n}^\ast\right)\right], \end{align}$$ as I'd defined them above, where I now have to take the complex conjugate, component-by-component, of the transition dipole moment for $n$ because the $e$ and $g$ are switched. This last expression is what your book really wants to define. It is nice, involves very little new/uncomfortable/arbitrary notation, and it arises directly from a standard quantization procedure.

However, this last expression is also a weird mix of the electronic degrees of freedom, which determine the dipole moment components in the molecular frame, together with the molecular orientation, which is very, very important in determining how the coupling will work and whether it will be, e.g., attractive or repulsive. You therefore want a vector that will describe the direction once both effects are taken into account. (Or maybe you don't, really. I'd just leave it like that, to be honest.)

To do this, you simply write $$\mathbf{d}=|\mathbf{d}|\mathbf{n},$$ and take that as a definition for $\mathbf{n}$. You still need to define what you mean by norm, though, and for a complex vector there are two choices that make sense, $$|\mathbf{d}|^2=\sum_j|d_j|^2\ \ \text{ and }\ \ |\mathbf{d}|^2=\sum_jd_j^2.$$ The former is the most common because it is not susceptible to being zero when $\mathbf{d}$ is not, but it will give the slightly unintuitive $|\mathbf{d}|=|\mathbf{d}^\ast|$. Either way, you make a choice of norm, and that makes your definition of the direction vector.

For both choices, though, you will have to take the complex conjugate of $\mathbf{n}$ when you complex-conjugate $\mathbf{d}$: $$\mathbf{d}^\ast=|\mathbf{d}^\ast|\mathbf{n}^\ast,$$ which your book has incorrectly neglected. The direction vector $\mathbf{n}$ is in general complex-valued, and this is inevitable because the real and imaginary parts of $\mathbf{d}$ (both of which are in $\mathbb{R}^3$) need not be parallel. If this is the case, then there is nothing to it but to chug on with a complex direction vector. Usually this indicates nontrivial physics, such as transitions with different magnetic quantum numbers.

It can happen, though, that the real and imaginary parts of $\mathbf{d}$ are indeed parallel. Usually this would be precluded by an appropriate choice of relative phase between $\ket g$ and $\ket e$, but occasionally it goes through for some reason. In this case, you can always write $\mathbf{d}=d\mathbf{n}$, where $d\in\mathbb{C}$ and $\mathbf{n}\in\mathbb{R}^3$, $\mathbf{n}$ has an obvious interpretation as the direction of the dipole, and the formula you quote comes into life. Unfortunately, this case is rather rare.

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+1 Thanks! Very thorough answer. –  becko Nov 26 '13 at 21:07
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Typically the basis wave functions are real, so the transition dipole moment is real too. But if it was complex, then it still can be written as product of magnitude and unit vector n, which then represents its "direction".

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Could you expand this answer a bit more for the benefit of the OP?! –  Michiel Apr 13 '13 at 7:55
    
Agree with @michielm. Can you explain? In what setting are the wave functions real? If it isn't real, the unit vector n is a complex vector. What's the meaning of its "direction"? –  becko Apr 13 '13 at 18:36
    
I think the dipole moment is complex is just because it includes the phase between the states. See here. –  user6048 Apr 22 '13 at 2:37
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