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I have been studying this paper for the past few days but really get stuck on one specific derivation that seems easy but I can't comprehend it. It's about quantum collapse models, so quite niche.

We have the following evolution of a density operator $\rho$ (equation 12 of the paper), which is a coherent Hamiltonian evolution and a dissipation term derived from Lindblad dynamics: $$\frac{d}{dt}\rho = -2\eta(\hat{q}^L\hat{q}^L\hat{\rho} - 2\hat{q}^L\hat{\rho}\hat{q}^L + \hat{\rho}\hat{q}^L\hat{q}^L) - 2\eta(\hat{q}^R\hat{q}^R\hat{\rho} - 2\hat{q}^R\hat{\rho}\hat{q}^R + \hat{\rho}\hat{q}^R\hat{q}^R)$$

Here, $\hat{q}^L$ and $\hat{q}^R$ are the position operators of a harmonic oscillator. It is clear how one arrives at this equation to me and it is also clear how to get the matrix elements of this operator in, let's say, a Fock basis. Now, the authors say that

Let us take the solutions of the above equation. Then, the density matrix in the position representation evolves as $\rho = \rho_0 e^{-4\eta(\Delta z)^2t}$.

I really don't understand this step. First of all, they don't specify what $\rho_0$ is. Then, they don't specify what basis elements they are using for their position representation. If I use basis elements $\{q^L, q^R\}$ and do the projection $\langle q^L q^R |\rho|q'^L q'^R \rangle$, I not nearly arrive at the given result. Next, I don't understand if the above equation refers to one matrix element or the the total operator, which would mean that all elements decay on the same timescale.

In summary, I stared at this stuff for three days now and didn't advance at all, so any help is greatly appreciated. Also my supervisor for this project doesn't know how to continue.

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  • $\begingroup$ Maybe I'm missing something, but looking into the paper, the authors specify the basis (eq. (14)) and $\rho_0$ (eq. (18)), before arriving at the line you don't understand (eq. (28)). Also, it is said that they neglect contributions from $H$. Then, eq. (28) follows pretty directly from eq. (15). $\endgroup$
    – Cream
    Commented Apr 29, 2022 at 14:20
  • $\begingroup$ @Cream Well, what I am confused about is that they mention the position representation. How is the Fock basis (eq. (14)) a position basis? $\endgroup$ Commented Apr 29, 2022 at 14:29
  • $\begingroup$ @Cream Also, how does eq. (28) follow pretty directly from eq. (15)? I use Mathematica and don't get the result. $\endgroup$ Commented Apr 29, 2022 at 15:08
  • $\begingroup$ @Cream So I explicitly checked your steps. If I solve equation (15) and neglect contributions from $H$ using Mathematica's DSolve tool, I get a non-uniform decay for different elements $\rho_{ij}$. For example, $\rho_{11} = \frac{1}{4}(1 - e^{-2\Lambda t})$ whereas $\rho_{22} = \frac{1}{4}(1 + e^{-2\Lambda t})$. $\endgroup$ Commented Apr 30, 2022 at 7:05

1 Answer 1

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Okay, let me try to address some points:

Well, what I am confused about is that they mention the position representation. How is the Fock basis (eq. (14)) a position basis?

They write in the paper:
"We restrict the analysis to the four-dimensional subspace,tensor product of the two two-dimensional “left” and “right” subspaces generated by the vacuum and the single-phononstates."
So, they do not consider full Fock space, but the 4D tensor product space $\text{span}\{|0_L\rangle \otimes |0_R\rangle,~ |0_L\rangle \otimes |1_R\rangle ,~ |1_L\rangle \otimes |0_R\rangle,~ |1_L\rangle \otimes |1_R\rangle\}$.
I don't understand why they write of $\rho_t$ in position representation, maybe because $L$ and $R$ correspond to different positions in the system?

So I explicitly checked your steps. If I solve equation (15) and neglect contributions from H using Mathematica's DSolve tool

Yes, they also do that in the paper (between eq. (18) and (19)) and obtain the same results as you do. There, they even include $H$ (which does not seem to change the result for $\rho_{11}$ and $\rho_{22}$). I was wrong that eq. (28) follows directly from eq. (15). They say, it follows from the solutions of eq. (15) (which are between eq. (18) and (19) again).

As far as I understand, eq. (26) is an approximation in the regime specified by eq. (27). It is evaluated at physical values, e.g. at $T=250$ fs.
In this "small time regime", $\rho_t = \rho_0 e^{-\Lambda t}$ seems to be a good enough approximation to the actual solutions.
At least for the off-diagonal elements $\rho_{23}$ and $\rho_{32}$ which correspond to the breakdown of entanglement and are of interest here (because the approximation is bad/wrong for $\rho_{11}$ and $\rho_{22}$), since $$ \rho_{23}(t) = \frac{1}{2} e^{-\Lambda t} \frac{\Lambda^2\cosh(\Omega t) - 4 \omega^2}{\Lambda^2 - 4 \omega^2} = e^{-\Lambda t} (1 + O(t^2)),~ \text{since}~ \cosh(t) = 1 + O(t^2). $$

If you have more specific questions, I will try to answer modify the answer accordingly.

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  • $\begingroup$ Thanks a lot for your answer. What do you mean by "evaluated at physical values"? $\endgroup$ Commented May 5, 2022 at 15:29
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    $\begingroup$ The computation is done with an experiment in mind which is cited at the beginning of the paper. I think they are inserting the paramters of the experiment into their equation. $\endgroup$
    – Cream
    Commented May 5, 2022 at 16:28
  • $\begingroup$ Thanks. Why can they neglect the contribution of the Hamiltonian at all? I guess it doesn't change the approximation they make? $\endgroup$ Commented May 6, 2022 at 7:50
  • $\begingroup$ Also: When I solve for $\rho_{23}$ without the coherent part, I get $\rho_{32} = \rho_{23} = \frac{1}{4} e^{-2\Lambda t}(1 + e^{2 \Lambda t})$. Is my reasoning correct? We assume $\Lambda T \ll 1 \Rightarrow 1 + e^{2 \Lambda t} \approx 2 \Rightarrow \rho_{23} \approx \frac{1}{2}e^{-2 \Lambda t}$? $\endgroup$ Commented May 6, 2022 at 8:07
  • $\begingroup$ So I am pretty sure this is incorrect, at least my calculation. First of all, they have $e^{-\Lambda T}$ while I get an extra factor of 2 in the exponent: $e^{-2 \Lambda T}$. Then, I don't see how $\rho_{32} = \frac{1}{4}(e^{-2 \Lambda t} + 1)$ should ever reduce to expression (26) $\endgroup$ Commented May 6, 2022 at 8:30

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