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Is there a known formula for the reduced Green's function for the (1D) harmonic oscillator?

As far as I am aware, the reduced Green's function could be used as a tool for working with time-independent perturbation theory of discrete quantum mechanical systems, for example, to express the first order perturbative correction to the wavefunction using the reduced resolvent for the n-th state:

$$ |n^{(1)} \rangle = \hat{R}_n \hat{V} |n^{(0)} \rangle $$

The representation of the reduced resolvent operator on a given basis (eg. spatial basis) is the reduced Green's function for that state. It can be expressed for example as a sum-over-states:

$$G_n(x,y)=\sum_{m\neq n}\frac{\Psi(x)\Psi^*(y)}{E_n-E_m}$$

And from this definition, it can be seen that the reduced Green's function for the $n$th state is given from the full Green's function, evaluated at the energy of the state in question, by substracting the non-local density matrix of that state.

A question which is related to the applicability of reduced Green's functions in time-independent perturbation theory (vs. the propagator methods) is for example this one. However, it does not answer my question.

I am aware that an expression of the full Green's function for a simple harmonic oscillator exists and can be given by the Mehler kernel or just using Hermite polynomials. I am also aware that, in principle, the reduced Green's function could be found from the full Green's function for example by the procedure I've described above. However, the algebra is way too formidable for me to find anything meaningful. I also believe that a simple model system of the 1D harmonic oscillator has to had been studied.

Therefore, my question is anyone aware of a time-independent reduced Green's function for the simple 1D quantum harmonic oscillator?

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  • $\begingroup$ Maybe I'm missing something, but if you take the full GF at $\omega=E_n$ and subtract a delta function doesn't it suppose to give you the reduced GF? $\endgroup$ – user245141 May 4 at 15:28
  • $\begingroup$ @yu-v An immediate problem of that approach is that the Green's function has a pole at $\omega = E_n$ $\endgroup$ – Ezze May 4 at 15:33
  • $\begingroup$ That's what you remove with the delta function. The retarded full GF is shifted by an $i 0^+$, which means that the imaginary part has a delta function when taking the limit. You can take the real part of the retarded GF, which should be finite at $\omega=E_n$. But maybe I'm missing something basic here. $\endgroup$ – user245141 May 4 at 15:36
  • $\begingroup$ Or maybe I am the one missing it. Would you mind elaborating it in a comment, if you have the time for it? $\endgroup$ – Ezze May 4 at 15:41
  • $\begingroup$ it was too long for a comment, and it also started to veer into an answer so I wrote it as such, but I think that I might still be missing something fundamental about the expression you are looking for, or have, when you say Green function $\endgroup$ – user245141 May 4 at 15:57
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Maybe I'm missing something, but the retarded GF can be written as $$ G^r(x,y ; \omega) = \sum_n \frac{\psi_n(x)\psi^*_n(y)}{\omega+i0^+-E_n}$$,

which we can expand as

$$ G^r(x,y ; \omega) = \sum_n \frac{\psi_n(x)\psi^*_n(y)}{\omega+i0^+-E_n} = \sum_n \left[\frac{\omega-E_n}{(\omega-E_n)^2+0^2}-i\pi\delta(\omega-E_n) \right]\psi_n(x)\psi^*_n(y)$$

which gives the reduced GF as

$$G^{red}_n (x,y) = {\rm{Re}}\{G^r(x,y ; E_n)\}$$

now I am not familiar with the expression for the full GF for a quantum harmonic oscillator so I don't know how if taking the real part of it makes sense, but if you have it then checking numerically if this expression works should be straight-forward.

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  • $\begingroup$ The full propagator is the Mehler kernel, but the OP is avoiding it on account of a messy FT? $\endgroup$ – Cosmas Zachos May 4 at 16:41
  • $\begingroup$ Well the real reason I wanted to avoid it is that I strictly wanted to exclude any time, or frequency, from my treatment, and only work with two spatial coordinates. The Mehler kernel is only defined for finite time difference/frequency as far as I can tell. $\endgroup$ – Ezze May 4 at 17:48
  • $\begingroup$ The above answer is just the Fourier Transform of the Mehler kernel, with the frequency set equal to the selected $E_n$. $\endgroup$ – Cosmas Zachos May 4 at 22:11

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