# What is the physical sense of the transition dipole moment?

So if the states are the same we achieve the expectation value of the dipole moment for a given state. I mean

$\langle \mathbf{\mu} \rangle = \langle \psi \vert \hat{\mathbf{\mu}} \vert \psi \rangle$

But I don't feel the physical sense in the case of transition dipole moment when psi-functions on both sides are different

$\langle \psi_{1} \vert \hat{\mathbf{\mu}} \vert \psi_{2} \rangle$

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This is a good question. I've been lead to believe there isn't really a satisfying answer, but I hope that's not true. Out of curiosity, did you come to this question while taking a class on lasers? –  Colin K Nov 29 '11 at 2:29
While my work in nonlinear optics. It is closely related to lasers, though :) i was reading a chapter about quantum-mechanical theory of the nonlinear optical susceptibility. –  jacksonslsmg4 Nov 29 '11 at 7:04
Optics makes you cool. That's scientific fact. –  Colin K Nov 29 '11 at 21:01
The interpretation of the off-diagonal matrix elements is due to Heisenberg in 1925, who identified their classical limit. The off diagonal elements are the classical Fourier series of the Bohr orbit of the energy levels. This is the birth of quantum mechanics. –  Ron Maimon Nov 30 '11 at 0:06

The dipole transition matrix element has a classical interpretation as the time Fourier series of the classical dipole moment of the Bohr orbit corresponding to one of the energy levels. The interpretation is only exact at high levels, at the correspondence limit, and the m,n matrix element is the m-n-th Fourier series coefficient for either orbit m or orbit n (the difference is higher order in h). This is covered in Wikipedia's page on Matrix Mechanics.

When an operator x(t) is varying in time in a stationary state, that means it has off diagonal matrix elements. On diagonal operators are constant in a stationary state. The electron is orbiting the nucleus, so the position is a function of time x(t), the dipole moment in a certain direction has Fourier components, and these Fourier components are the off diagonal matrix elements. This correspondence was the main tool used by Heisenberg to construct his matrices.

To give a simple example, the x(t) operator in the harmonic oscillator is $a+a^{\dagger}$, so it has off diagonal matrix elements purely between the state of frequency $\omega$ higher and $\omega$ lower, of magnitude about $\sqrt{n}$ (the exact relations are $a|n\rangle = \sqrt{n} |n-1\rangle$, $a^{\dagger}|n\rangle = \sqrt{n+1}|n+1\rangle$. This means that the classical x motion corresponding to the n-th state has exactly two fourier components, one of which is at frequency $\omega$, the other at frequency $-\omega$, both of size $\sqrt{n}$. This means X is sinusoidal with period $2\pi\over \omega$ of size $\sqrt{E}$, and this is indeed the classical harmonic oscillation motion.

The same holds for Rydberg orbits of the H atom, and for all off diagonal matrix elements--- they correspond to the time Fourier series of the classical quantity in the Bohr orbit version of the stationary state. They don't actually make an orbit, because a real orbit has a Fourier series which is multiples of the fundamental frequency, while the quantum system doesn't have exactly equally spaced energy levels, they are only approximately equally spaced at large N in the correspondence limit.

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I never thought of it this way, but off-diagonal components of an operator means that this operator is... Well, non-diagonal. Which means that neither of $| \psi \rangle$ realizes a definite value of this operator.

The dipole operator comes from the term which couples charged particle with electric field. It would be better to write some math here, but I just appeal to quantum approach to Larmor precession. The same situation there (we start from the state which is not the eigenstate of the operator and see how it changes) leads to periodic change of the wavefunction from one state to another where off-diagonal elements are related with the probability of those transitions.

From some other arguments we know that the particle may go down, but can not go up. So, this probability not a probability of some quantum beats, but the probability to go from upper to lower state.

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Normally the transition amplitude is calculated with $e^{i\vec{k}\vec{r}}$. For "small" product $\vec{k}\vec{r}$ one expands the exponential in Taylor series and one leaves only the "dipole" term in the transition amplitude calculation: $\propto\vec{k}\cdot\langle\psi_1|\vec{r}|\psi_2\rangle$. So it is a "dipole" part of the transition amplitude, which in general case is determined with a more complicated formula.

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It may be that there is no satisfying picture for this term in the Copenhagen interpretation, but there most certainly is in the Schroedinger picture. The problem with Copenhagen is we say that the atom is initially in state phi_2, and later it is found in the state phi_1. It gets from 1 to 2 via a mysterious process we are not allowed to talk about, sometimes called the Quantum Leap. The cross-term dipole moment has no clear physical meaning in this picture.

Schroedinger had a different perspective. He said the atom gets from state 2 to state 1 by a continuous transition through intermediate superposition states. So halfway through the transition, the atom is in a state (phi_1 + phi_2). You can think of this in the case of a hydrogen atom as being a superposition of the s_0 and p_z states.

If you evaluate the dipole moment of this state according to your first rule, the one you said makes sense, not the second rule which you call the "transition" dipole moment...according to the first rule you get the sum of four dipole moments: two straight terms and two cross terms. Since the dipole moment of all the eigenstates is zero, the straight terms drop out leaving only the cross terms. Because of normalization (the factor of 1/sqrt(2)) the value of the cross terms is exactly equal to what you already called the transition dipole moment.

In other words, your transition dipole moment is nothing more than the actual dipole moment evaluated for the superposition states. Most importantly, if you recall that the s and p states have different frequencies, it is clear that the dipole moment of the superposition must reverse its direction every half-period. In other words, it is an oscillating dipole. As an oscillating dipole, it radiates energy according to Maxwell's equation. As it loses energy, the amplitude of the p component diminishes in favor of the s component. In other words, the atom makes a smooth transition from the p state to the s state by radiating away the excess energy according to Maxwell's equations.

There is no need for a quantum leap, and there is no need for photons.

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," and there is no need for photons" coming after "As an oscillating dipole, it radiates energy according to Maxwell's equation" is a contradiction. Radiation is photons, period. Do you mean very many soft photons ? i.e. not a quantized transition but a many photon one? –  anna v Nov 29 '11 at 6:43
""Radiation is photons, period."" This is the second or more time You write this. Reality is neither waves or photons (particles). What model You use to describe reality is Your turn. For radiation purpose (that is the radiation under way) waves are much easier to understand. –  Georg Nov 29 '11 at 11:57
@Marty Green: Your explanation is good, but according to Maxwell equations it is a time-dependent current that radiates. It is often approximated with dipole moment time-dependence, but not always. –  Vladimir Kalitvianski Nov 29 '11 at 15:02
@Georg I am pointing out an inconsistency. photons are both particle and wave but in quantum mechanical dimensions they cannot be glossed over with "radiation". In these sizes radiation is photons. –  anna v Nov 29 '11 at 15:19
@Vladimir I'm pretty sure that the approximation is virtually exact in the small-dipole case typified by the s-p transition in Hydrogen. And thanks for the props: I'm not used to people agreeing with me. –  Marty Green Nov 29 '11 at 15:21