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In a diatomic molecule, the nuclear potential in the ground and excited states can take different shapes. The coordinates might be displaced and the curvatures might not be the same. For example, we may have

$$ H_g=\frac{p^2_g}{2m_g} + \frac{1}{2}m_g\omega_g^2x_g^2 \\ H_e=\frac{p^2_e}{2m_e} + \frac{1}{2}m_e\omega_e^2x_e^2 $$

The coordinates are usually related by $x_g = x_e - x_0$ where $x_0$ is a constant and since momentum is just derivative of position, it follows that $p_g=p_e$. We can then proceed by second quantization:

$$ x_g=\sqrt{\frac{\hbar}{2m_g\omega_g}}(b_g^{\dagger}+b_g) \\ x_e=\sqrt{\frac{\hbar}{2m_e\omega_e}}(b_e^{\dagger}+b_e) $$

and the number state of ground can be written as $|n_g\rangle$ while for excited state $|n_e\rangle$.

The Franck-Condon factor $\langle n_g|n_e \rangle$ was derived analytically and algebraically in many papers. However, why is one allowed to calculate $\langle n_g|n_e \rangle$ in the first place? Because from my understanding, these are not the same harmonic oscillators since they have different shapes and we second quantized them using different operators. Doesn't this mean that $|n_g\rangle$ and $|n_e\rangle$ live in different Hilbert space and thus $\langle n_g|n_e \rangle=0$ no matter what?

I know I must be wrong here because there are plenty existing works on this but I just don't see how we are allowed to calculate overlap integral between $|n_g\rangle$ and $|n_e\rangle$. I would really appreciate if someone could help clarify.

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    $\begingroup$ You have already given the relation between them: $x_e=x_g-x_0, p_e=p_g$, so $H_e$ is just a displaced $H_e$. $|n_g\rangle$ and $|n_e\rangle$ are two different states in the same Hilbert space and there is an overlap between them. $\endgroup$
    – Meng Cheng
    Commented May 29, 2022 at 14:01
  • $\begingroup$ Thank you for your help @MengCheng , I understand two harmonic oscillators that are displaced with respect to each other can be related by a displacement operator. However, is this case the frequency is also different. $\endgroup$ Commented May 30, 2022 at 0:43
  • $\begingroup$ Sure, that's not an issue. Any two states in the same Hilbert space (square-integrable functions on $\mathbb{R}$) have an overlap. $\endgroup$
    – Meng Cheng
    Commented May 30, 2022 at 2:29
  • $\begingroup$ I still don't fully understand this. Could you also have a look at another question that I posted: link? I explained my question in more details there with graphics $\endgroup$ Commented May 30, 2022 at 15:17

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Your assumption that each oscillator acts on a different Hilbert space is incorrect. Two different operators can act on the same Hilbert space.

In this case, the two operators defining the two different harmonic oscillators act on the same Hilbert space. The complete set of eigenfunctions of any of these operators, form a complete basis for this space. Any state belonging to this Hilbert space can be expanded into either basis set. The basis functions themselves are also part of this Hilbert space and thus one set can be expanded into the other set. The expansion coefficients and change of basis transformation can be found by calculating $\langle n_g|n_e\rangle$.

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