Caldeira-Leggett Dissipation: frequency shift due to bath coupling

Question

I am trying to understand the Caldeira-Leggett model. It considers the Lagrangian

$$L = \frac{1}{2} \left(\dot{Q}^2 - \left(\Omega^2-\Delta \Omega^2\right)Q^2\right) - Q \sum_{i} f_iq_i + \sum_{i}\frac{1}{2} \left(\dot{q}^2 - \omega_i^2q^2\right)$$

where $Q$ is the generalised coordinate of the macro variable (an oscillator with natural frequency $\Omega$), $q_i$ are variables related to an array of harmonic oscillators each with natural frequency $\omega_i$.The first term describes the potentail and kinetic energies related to the macro degree of freedom, the second term describes the coupling using constants $f_i$, the third again describes potential and kinetic energies of the array of oscillators,

$$\Delta \Omega^2 = -\sum_i \left( \frac{f_i}{\omega_i} \right)^2$$

is the ad hoc term my first question relates to. The explanation I found goes

the quantity is inserted to cancel the frequency shift $$\Omega^2 \to \Omega^2 - \sum_{i} \left(\frac{f_i}{\omega_i}\right)^2$$ [...] the shift arises because a static Q displaces the bath oscillators so that $$f_i q_i = - \left(f_i^2 / \omega_i^2\right)Q$$ Substituing these values for the $f_i q_i$ into the potential terms shows that the effective potentail seen by $Q$ would have a "shifted" frequency.

I regretfully do not get it. I tried to get to the equation

$$f_i q_i = - \left(f_i^2 / \omega_i^2\right)Q$$

by considering the equation of motion in equilibrium, without success. Why would $\Omega$ be affected by the coupling? Also from the intuitive point of view, I fail to see how a "pre-load" would affect the natural frequency of a harmonic oscillator. Any hint would be so appreciated.

Answer 1

To expand on Adam's answer, let's see the effect of a static $Q$ on the system without the $\Delta \Omega^2$. The Lagrangian becomes $$L = -\frac{1}{2}\Omega^2Q^2 - Q\sum_i f_iq_i + \sum_i \frac{1}{2}(\dot{q}_i^2 - \omega_i^2q_i^2)$$ The equation of motion for each of the $q_i$ is now $$\frac{\partial L}{\partial q_i} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}} \implies -Qf_i -\omega_i^2q_i = \ddot{q}_i$$ To recover a harmonic oscillator we shift the coordinates by defining $q_i' =q_i + Q\frac{f_i}{\omega_i^2}$, and now the EOM is the familiar $\ddot{q'_i} = -\omega_i^2q_i'$. We can immediately identify the static shift as the quantity we subtracted out of the $q_i$, i.e. $$\Delta q_i = -Q\frac{f_i}{\omega_i^2}$$ Multiply both sides by $f_i$ and you recover the equation you want.

Why does this effect the frequency of the macro-oscillator $Q$? Intuitively, you might think that all of the environment oscillators $q_i$ are "dragging" on $Q$. Quantitatively, replace all of the $q_i$ in the Lagrangian with the $\Delta q_i$ above and you'll find the EOM for $Q$ is now $$\ddot Q = -\left(\Omega^2 - \sum_i\frac{f_i^2}{\omega_i^2} \right)Q$$ which gives you the frequency shift $\Delta \Omega^2$.

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Note: Much of this is from Stone & Goldbart's excellent textbook Mathematics for Physics.

Answer 2

If you look at the equation of motions of $Q$ and the $q_i$, and solve them for the bath, then inject this in the equation for $Q$, you will see that the frequency of $Q$ is shifted by the amount quoted in the question.

If you want to define the true frequency of the system (in presence of the bath) by $\Omega^2$, you have to put this shift in the Lagrangian.

PS: because the system is quadratic, solving the classical equations of motion is equivalent to the quantum problem for the purpose of the question.