Question about Stark interference

In summary, the conversation discusses a question about a paper that claims to use points A and F for their experiment. The question asks how an electric field in the z direction can connect two states of different m_N values, given that the dipole moment operator doesn't interact with electron or nuclear spins. The conversation also includes equations and references to a paper that discusses a mixture of quantum numbers allowing for a weak, higher-order stark effect.
  • #1
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Hello! I have a question about this paper (@Twigg ?). They claim towards the end of the second page that they use points A and F for their experiment. But for example, at point A the molecular rotation quantum numbers are ##|N=0,m_N=0>## and ##|N=1,m_N=1>##. However, in their experiment the electric field is in the z-direction, which is the direction of the magnetic field, too, which defines the ##m_N##. So if that is the case (and given that the dipole moment operator doesn't interact with the electron or nuclear spins), how can an electric field in the z direction connect 2 states of different ##m_N##? Am I missing something?

Based on my math we should have this:

$$<N=0,m_N=0|\vec{d}\cdot\vec{E}|N=1,m_N=1> = <N=0,m_N=0|d\hat{n}\cdot\vec{E}|N=1,m_N=1>$$
where E is the electric field and ##\hat{n}## is the internuclear axis direction (defined in the frame of the molecule). In general we have:

$$\hat{n} = \sin\theta\cos\phi \hat{x} + \sin\theta\sin\phi \hat{y} + \cos\theta\hat{z}$$
when expressing ##\hat{n}## in the lab frame. From here we get:

$$<N=0,m_N=0|\vec{d}\cdot\vec{E}|N=1,m_N=1> = E_z <N=0,m_N=0|\cos\theta|N=1,m_N=1> $$
We also have that:
$$\cos\theta \propto Y_1^0$$
where ##Y_1^0## is a spherical harmonic and:
$$|N,m_N> \propto Y_N^{m_N}$$
so the above term becomes:

$$<N=0,m_N=0|\vec{d}\cdot\vec{E}|N=1,m_N=1> = E_z \int(Y_0^0\times Y_1^0 \times Y_1^1)$$
where the integral is over ##\theta## and ##\phi##. But that integral is zero (which is a long way of saying that the signed sum of ##m_N## values appearing in the spherical harmonics of that integral is not zero).
 
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  • #2
Sorry for the really slow reply!

Check out this paper: https://arxiv.org/abs/0708.2925
Specifically, the paragraph near the bottom of the left column on page 2 that starts "We calculate C...".

In short, there is some mixture of quantum numbers in either parity state that allows a weak, higher-order stark effect.

Hope that helps!
 
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