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Consider a system of two entangled harmonic oscillators. The normalised ground state is denoted by $\psi_0(x_1,x_2)$.

The reduced density matrix of the second oscillator is given by:

$$\rho_2 = \int_{-\infty}^{+\infty} dx_1 \int_{-\infty}^{+\infty} dx_2 \int_{-\infty}^{+\infty} dx_2' \,\psi_0(x_1, x_2)\psi_0^*(x_1, x_2') |x_2\rangle\langle x_2'|$$

So the entries are:

$$\rho_2(x_2,x_2') = \int_{-\infty}^{\infty} dx_1 \psi_0(x_1,x_2) \psi_0^*(x_1,x_2')$$

$x_1$ and $x_2$ are continuous, so I interpreted $\rho_2$ as a matrix of (uncountably) infinite dimensionality: it has an entry for every $(x_2, x_2')$.

Now we want to find the eigenvalues. A paper I'm reading defines these as follows: $$\int_{-\infty}^{+\infty} dx' \rho_2(x, x') f_n(x') = p_n f_n(x)$$

A system of $n$ equations, where $n$ is an integer running from zero to infinity. Now my question is three-fold:

1) Where do these equations come from? Are they simply the generalisation to continuous variables of the discrete case $$\sum_j O_{ij} \psi_j = \lambda \psi_i$$ for each $n$?

2) For a matrix to be diagonalisable, it needs to have a number of eigenvectors equal to its dimensionality. $n$ takes only countably infinite values, while I thought $\rho_2$ had uncountably infinite rows and columns. I suppose my interpretation of $\rho_2$ is incorrect, but I don't see why, since $x$ and $x'$ can clearly take non-integer values.

Now the author 'guesses' the following solution:

$$p_n = (1 - \epsilon)\epsilon^n \\ f_n = H_n(\alpha^{1/2}x) e^{-\alpha x^2 / 2}$$

Where $H_n$ is the (physicist's) Hermite polynomial, $\alpha = (\gamma^2 - \beta^2)^{1/2}$, $\epsilon = \beta / (\gamma + \alpha)$, and $n$ runs from zero to infinity.

In order to define these new quantities, I'll backtrack a bit. The system has Hamiltonian: $$H = \frac{1}{2} [p_1^2 + p_2^2 + k_0(x_1^2 + x_2^2) + k_1(x_1 - x_2)^2]$$

Uncoupling the Hamiltonian, where we use that $x_± = \frac{1}{\sqrt{2}}(x_1 ± x_2)$, $\omega_+ = k_0^{1/2}$ and $\omega_- = (k_0 + 2k_1)^{1/2}$, we obtain the normalised ground state:

$$\psi_0(x_1,x_2) = \pi^{-1/2} (\omega_+ \omega_-)^{1/4} e^{-(\omega_+ x_+^2 + \omega_- x_-^2) / 2}$$

Now we define $\beta = \frac{1}{4} (\omega_+ - \omega_-)^2 / (\omega_+ + \omega_-)$ and $\gamma - \beta = 2 \omega_+ \omega_- / (\omega_+ + \omega_-)$, and write the reduced density matrix entries as:

$$\rho_2 (x_2, x_2') = \pi^{-1/2} (\gamma - \beta)^{1/2} e^{-\gamma (x_2^2 + x_2'^2)/2 + \beta x_2 x_2'}$$

3) I want to check the author's solutions. Using Mathematica I've evaluated the integral to a mess of Gamma and Krummer's fuctions: rewriting that in terms of the eigenstates did not seem doable. Besides, I'd rather do a more explicit check. A second idea was to use the generating fuction:

$$e^{2xt - t^2} = \sum_{n=0}^{\infty} H_n(x) \frac{t^n}{n!}$$

And use orthogonality of Hermite polynomials to get rid of the sum. This seemed doable, since the left-hand side of the eigenvalue equation contains an exponential of this form. I've run into some problems using this method though, the biggest of which is that I've failed to make the arguments of the generated polynomial and the polynomial in the eigenstate equal, and am thus unable to use orthogonality. I've been stuck on this for days, so any help would be greatly appreciated!

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  • $\begingroup$ Perhaps your third question and the background information that comes with it should be split from the rest? $\endgroup$ – Danu Apr 28 '14 at 13:00
  • $\begingroup$ Isn't ρ_2(x_2,x_2′) square integrable in both x_2 and x_2'? Then it defines a Hilbert-Schmidt operator and has at most a countable number of eigenvalues, so the eigenvalue equation makes sense. That x_2 and x_2' run through a continuum of values is irrelevant. $\endgroup$ – Urgje Apr 28 '14 at 13:11
  • $\begingroup$ Danu: I suppose it is a different kind of question, but I didn't think posting twice would be appreciated much. Urgje: I'm not familiar with the Hilbert-Schmidt operator. With regards to question two, I'm kind of looking for an answer that directly shows where my thinking goes wrong. Is it that I try to view $\rho_2$ as a matrix? Is it that I'm comparing it to the discrete case? Or maybe something else entirely. $\endgroup$ – Timsey Apr 28 '14 at 14:30
  • $\begingroup$ I am afraid that your approach makes things rather complicated. Your kernel defines a self-adjoint operator of Hilbert-Schmidt type, see en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_integral_operator. But then it has a countable set of orthogonal eigenvectors with associated real eigenvalues which can only accumulate in 0. $\endgroup$ – Urgje Apr 28 '14 at 21:05
  • $\begingroup$ I'm sure it does, but I would like to know in what way my approach makes it difficult. I do believe that the author's statement is probably correct, it's just that I cannot understand where my argument fails. $\endgroup$ – Timsey Apr 29 '14 at 14:32
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I can answer your question regarding the eigenvalue equation of the reduced density matrix. This is basically Helmholtz integral equation of the second kind. This is an eigenvalue of equation for a continuous variable. You can do the quadrature approximation of the integral and get the matrix eigenvalue equation which is just an approximation due to the discretization of the integral.But If we write the reduced density matrix in a matrix form (Using the Discrete Basis) then you can have the matrix eigenvalue equation. Hence there are two ways of getting the matrix eigenvalue equation. Using the quadrature approximation of the integral equation or using the Discrete basis.

I have done some computation for the bipartite hamiltonian with some gaussian integrals and that might be helpful to you.

You can access the article from here http://link.springer.com/article/10.1140/epjd/e2014-50294-0 or https://www.dropbox.com/s/06xbckd6fyv65et/Sijo_Yue_Sanjuan_EPJD_final.pdf

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