# Parity of photon in nuclear transitions

• Yaste
In summary, the parity of a photon emitted from a nucleus undergoing a transition from ##J^P = \dfrac{1}{2}^+## to ##J^P = \dfrac{5}{2}^+## is positive due to the conservation of parity between the initial and final states. The photon is a mixture of E2, M3, E4, and other multipole modes, with the lowest multipolarity dominating. The electromagnetic field can be described in terms of Debye potentials, which can be expanded in terms of multipole fields with defined angular momentum and magnetic quantum numbers. The expansion starts from ##\ell=1##, as there is no monopole radiation.
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?

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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.

@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##..

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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)

## 1. What is the concept of parity in nuclear transitions?

The concept of parity in nuclear transitions refers to the intrinsic property of particles, such as photons, to have a specific orientation or spin. In simple terms, it describes whether a particle is symmetrical or asymmetrical in its behavior. In the context of nuclear transitions, parity is used to determine the selection rules for the emission or absorption of photons during a transition between nuclear energy levels.

## 2. How is parity of photons determined in nuclear transitions?

The parity of a photon in a nuclear transition can be determined by analyzing its angular momentum and spin. Photons with odd angular momentum and spin have opposite parity, while those with even angular momentum and spin have the same parity. Additionally, the parity of a photon in a nuclear transition can also be determined by its direction of propagation, which is either parallel or antiparallel to its spin direction.

## 3. What is the significance of parity in nuclear transitions?

The concept of parity is crucial in nuclear transitions as it helps to predict the type of radiation emitted or absorbed by a nucleus during a transition. It also helps to determine the probability of a nuclear transition occurring and the allowed energy levels for the emitted or absorbed photons. In addition, parity conservation is an important principle in nuclear physics and plays a role in understanding the fundamental interactions between particles.

## 4. Can the parity of a photon change in a nuclear transition?

Yes, the parity of a photon can change in a nuclear transition. This can occur if the nucleus undergoes a change in its angular momentum or spin, which can result in a change in the direction of propagation of the emitted or absorbed photon. However, the total parity of the nuclear system must remain conserved, meaning that the sum of the parities of all particles involved in the transition must remain the same before and after the transition.

## 5. How is parity violation observed in nuclear transitions?

Parity violation in nuclear transitions can be observed by studying the properties of the emitted or absorbed photons, such as their energy levels and polarization. If the observed properties do not follow the expected selection rules based on parity conservation, it indicates that parity has been violated in the nuclear transition. This phenomenon has been observed in certain types of weak interactions, such as beta decay, and has played a significant role in our understanding of the fundamental forces of nature.

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