"E2" Quadrupole decay rate(adsbygoogle = window.adsbygoogle || []).push({});

Problem StatementObtain the angular dependence of the rate for the emission to a single photon (momentum ##\mathbf{k}## and circular polarization ##\lambda##) for the electric quadrupole transition from ##\ell = 2## to ## \ell = 0##. Neglect electron spin.

Setup

(assumed done in the Coulomb gauge/Radiation gauge)

[itex]H^\prime = \frac{1}{2m} \left( -e\mathbf{p} \cdot \mathbf{A} - e \mathbf{A} \cdot \mathbf{p} + e^2A^2 \right) [/itex]

So we wish to claculate

## \langle n_f\, 0\, 0; \gamma\left(\mathbf{k},\, \lambda\right) \mid H' \mid n_i\, 2\, m_i ; 0\rangle e^{i\omega t}##,

where the first three symbols are ##n \ell m## for the hydrogen atom. In the final state, ##\ell = m = 0##. After ';' comes the photon state. In the final state there is a single photon of wave-vector ##\mathbf{k}## and polarization ##\lambda ## which is ##\pm 1##, depending on circular polarization in the plane orthogonal to ##\mathbf{k} ##.

The polarization vector [itex]\mathbf{\epsilon}[/itex] has that the projection along the direction ##\mathbf{k}## is zero, and the chosen basis for the remaining space is circular polarization along the axis of propagation with the state given by [itex]\lambda[/itex]. The vector itself has three components. The ##{}^* ## below doesnotmean conjugate transpose but merely conjugate. If you are curious that came from the definition of the [itex]A[/itex] operator, which connected positive and negative frequency plane waves to ladder operators for a given [itex]\mathbf{k},\, \lambda[/itex] but all still within the same space of polarization vectors. Only the conjugate term survived due to the photon number change in this decay.

I will now submit to you that what we really want to calculate is the following:

## \langle n_f\, 0\, 0\mid \left(i\frac{e}{m}\mathbf{k} \cdot \mathbf{r}\,\boldsymbol{\epsilon}^*\left(\mathbf{k}, \lambda\right)\cdot \mathbf{p}\right)\mid n_i\, 2\, m_i \rangle ##

This came from Taylor expanding ##\mathbf{A}## in powers of ##\mathbf{k} \cdot \mathbf{r}##.

The Conundrum

Now, there are perhaps two competing methods for evaluating this, which is the central point of this post.

I am more familiar with the first method since it was my method of choice. Details of the first method can be found in the attached PDF.

- Brake up the above into symmetric and anti-symmetric parts such that the anti-symmetric operator that is equal to zero when between these two elements, leaving you caring only about [itex]\frac{1}{2}\left[\mathbf{k} \cdot \mathbf{r}\,\boldsymbol{\epsilon}^*\cdot \mathbf{p} + \boldsymbol{\epsilon}^*\cdot \mathbf{r}\,\mathbf{k} \cdot \mathbf{p}\right][/itex]. Then you separate this into a sum of spherical tensor operators, where the coefficients of the operators have an angular dependence on [itex]\mathbf{k}[/itex]. Parameterize the coefficients using spherical coordinates.

- Consider the initial and final total angular momenta and use conservation of angular momentum to constrain the transition. The photon has a superposition of angular momentum states, however if we chose the orbital angular momentum axis to be along [itex]\mathbf{k}[/itex], we are guaranteed [itex]m_{EM, spin} = \pm 1[/itex] and [itex]m_{EM, orbital} = 0[/itex], So the total angular momentum for state with one photon must be a superposition of states all with [itex]m_{EM} = \pm 1[/itex] along the [itex]\mathbf{k}[/itex] axis. Our hydrogen atom was chosen along the z direction however. So the problem becomes finding the component between these angular momentum states directly. The originator of the problem and solution 2 wants the answer to be expressible in terms of a [itex]j = 2[/itex] Wigner d-matrix factor.

It seemed to me when I checked that the resulting angular dependence from either method might agree up to an irrelevant phaseif[itex]\phi = 0[/itex] (the [itex]\phi[/itex] of [itex]\mathbf{k}[/itex] expressed in spherical coordinates). However when [itex]\phi \neq 0[/itex] the results did not seem to agree.

Here's one thing about the second method. We had superposition of different states of the photon angular momentum, which all had some particular projection [itex]m_{EM}[/itex] in the k direction but the terms vary in the total amount of EM angular momentum. We believe the operation of rotating k to be aligned with the conventional axis would would only mix [itex]m_{EM}[/itex] values. Then one might just select one of these terms so that the total angular momentum of of that term matched that of the hydrogen atom before decay.

I am still not sure of the applicability of the Wigner little d matrix. An argument I can make is the following. One can rotate around the axis of angular momentum (z conventionally) and introduce only a phase factor for the individual angular angular momentum states. Then one might suppose the rotation of the particular state that survives [itex]H^\prime[/itex] so that the photon [itex]\mathbf{k}[/itex] of that state was moved to the xz plane would only introduce a phase factor relative to the un-rotated state. Then you would just use the [itex]j = 2[/itex] Wigner d-matrix to give you the correct factor for the [itex]\theta[/itex] between [itex]\mathbf{k}[/itex] and the z axis.

Also I am unsure of dealing with the electromagnetic angular momentum. For example I have read that [itex]\mathbf{L}_{EM}[/itex] and [itex]\mathbf{S}_{EM}[/itex] no longer commute for example, and it's unclear to me what that means for our angular momentum theory or if it means anything important for this problem.

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# Quadrupole Conundrum

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