Integrating out a variable in quantum mechanics

In the lecture https://www.youtube.com/watch?v=-9G54aCkNH0 (or see https://www.theorie.physik.uni-muenchen.de/activities/schools/archiv/asc_school_14/komargodski_notes.pdf for notes), Zohar Komargodski begins by examining the Hamiltonian
$$H=frac{1}{2}p_x^2+frac{1}{2}p_y^2+frac{1}{2}lambda^2x^2y^2.$$
He argues that for $|x|rightarrowinfty$, the $y$ part of the Hamiltonian behaves like a Harmonic oscillator with a very big mass. Therefore, it will settle in its vacuum, which would have energy $E=frac{1}{2}lambda|x|$. Then the effective Hamiltonian for the $x$ degree of freedom would be $H=frac{1}{2}p_x^2+frac{1}{2}lambda|x|$, after integrating out the degree of freedom of $y$.

I was wondering what this integrating out means quantum mechanically. To respond to this question I wrote the Hamiltonian in the form
$$H=sum_{n=0}^inftyleft(frac{1}{2}p_x^2+left(frac{1}{2}+nright)lambda|x|right)|nrangle_ylangle n|_y,$$
In terms of the usual harmonic oscillator eigenfunctions for the $y$ direction. However, I don’t know how to argue that at large $|x|$ the only important term is the $n=0$ one.

In a thermodynamical setting I guess this would make sense but at no point in the lecture did he mention a low temperature limit.