Parity of photon in nuclear transitions

1. Jul 4, 2015

Yaste

Let's say we have a transition from $J^P = \dfrac{1}{2}^+$ to something like $J^P = \dfrac{5}{2}^+$. It radiates a photon with some energy $E_\gamma$. How does one know the parity of said photon? How does conservation of parity work here?

Last edited: Jul 4, 2015
2. Jul 4, 2015

bcrowell

Staff Emeritus
It's parity that's conserved, not polarity. (The word "polarity" isn't used in this context.)

You can't infer the multipolarity of a photon from the initial and final spin-parity of the nucleus. What you can do is rule out certain multipolarities.

Transitions with multipolarity M1, M3, M5, ... and E2, E4, ... have even parity, and therefore when they are emitted, the nucleus doesn't change parity, as in your example. Transitions with the other multipolarities have odd parity, and their emission means the nucleus has changed its parity, so in your example those multipolarities are ruled out.

In your example the nuclear spin is changing by 2 units, so by the triangle inequality for vector addition, the photon's angular momentum must be at least 2. This means that M1 is ruled out for your example.

So the transition in your example is going to be a mixture of E2, M3, E4, ... In most cases the photon's wavefunction is dominated by the lowest multipolarity, so your photon would probably be almost pure E2 in character.

3. Jul 4, 2015

Yaste

@bcrowell, sorry, I actually meant parity, not polarity. I'll edit the post. I've gone long without sleeping... In my example, what is the parity of the photon and why? I don't know what those M and E's stand for..

EDIT: I think I understand it now. the parity, in my example, is positive. I was adding the parities but they're a multiplicative quantity. So, according to conservation of parity, the parity of the initial state, must be the same as the parity of the final state, that is, $P_i = P_f*P_\gamma$, so $+1=+1*+1$..

Last edited: Jul 4, 2015
4. Jul 14, 2015

vanhees71

The $E_l$ are the elctric and the $M_l$ the magnetic multipole modes of the electromagnetic field.

The free electromagnetic field always has two independent real field-degrees of freedom. For different applications different choices are convenient. Here you observe radiation from some distance of a small source (in your case the nucleus radiating due to electromagnetic transitions). This calls for a description in spherical coordinates with the source at rest in the center. It turns out that in this case the most convenient two field components are the Debye potentials in terms of which the magnetic field is given as
$$\vec{B}=\vec{r} \times \vec{\nabla} \psi + \vec{\nabla} \times (\vec{r} \times \vec{\nabla} \chi).$$
For a wave with harmonic time dependence $\propto \exp(-\mathrm{i} \omega t)$ you have from the Ampere Maxwell law (with $c=1$)
$$\vec{\nabla} \times \vec{B}=-\dot{\vec{E}}=\mathrm{i} \omega \vec{E} \; \Rightarrow \; \vec{E}=-\frac{\mathrm{i}}{\omega} \vec{\nabla} \times \vec{B}.$$
Using the Helmholtz equation, valid for the Debye potentials, as can be shown from the Maxwell equations, one finds
$$\vec{E}=\mathrm{i}{\omega} \vec{\nabla} \times (\vec{r} \times \vec{\nabla} \psi)+\mathrm{i} \omega \vec{r} \times \vec{\nabla} \chi.$$
For the first piece, involving $\psi$, only the electric field can have a component in radial direction, and that's why this piece is called a electric-field or E mode (or sometimes transverse magnetic, because the corresponding piece of the magnetic field has only components perpendicular to the direction of $\vec{r}$), while for the 2nd piece, involving $\chi$, only the magnetic field has a component in radial direction, and that's why this piece of the field is called a magnetic or M mode (or transverse electric mode).

Since the Debye potentials obey the free Helmholtz equation $(\Delta+\omega^2)\psi=(\Delta + \omega^2) \chi=0$, these can be expanded in terms of spherical harmonics with the spherical Hankel functions (or Bessel functions) as the radial piece. This leads to an expansion of the electromagnetic field (or photon field) in terms of multipole fields with field modes in terms of defined angular momentum (corresponding to a number $\ell \in \{1,2,\ldots \}$) and magnetic (corresponding to a number $m \in \{-\ell,-\ell+1,\ldots, \ell \}$. The expansion starts from $\ell=1$, because there's no monopole radiation, i.e., the piece corresponding to $\ell=0$ describes a static electric field of a radially symmetric charge distribution outside of this charge distribution. The constant coefficients in this series expansion are the multipole moments of the radiation field.

See also the quite nice Wikipedia page (the only drawback is that they don't make use of the Debye potentials, which help a lot to make sense of these rather complicated formal math)