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I have a continuous-variable quantum state which is entangled between two subsystems, that is,

$$\left|\psi_{AB}\right\rangle=\int_{-\infty}^{+\infty} dq_{1} dq_{2}~ \psi(q_{1}, q_{2}) \left|q_{1}, q_{2}\right\rangle, \tag{1} $$

where $\left\lbrace\left|q_{1}, q_{2}\right\rangle \right\rbrace= \left\lbrace \left|q_{1}\right\rangle\left|q_{2}\right\rangle \right\rbrace$ represents a continuous-variable position basis, such that $\left\lbrace q_{1}, q_{2} \right\rbrace \in \mathbb{R}$. Besides, $\left|q_{1}\right\rangle$ is associated with the system $A$ and $\left|q_{2}\right\rangle$ with the system $B$. The state of Eq. (1) is said entangled because it has non-separability; that is, the components of the state of Eq. (1) in the position basis $\left\lbrace\left|q_{1}, q_{2}\right\rangle \right\rbrace$ are $\psi(q_{1}, q_{2})=\left\langle q_{1}, q_{2}\right| \left. \psi_{AB}\right\rangle \neq \psi(q_{1}) \psi(q_{2})$. One of the typical forms to quantify entanglement in bipartite systems is through the von Neumann entropy, that is,

$$E=S(\hat{\rho_{A}})=S(\hat{\rho_{B}}) \tag{2},$$

where $E$ stands for entanglement and $S(\hat{\rho_{A(B)}})$ is the marginal density operator of the system $A(B).$ We can compute $E$, for example, from the marginal density operator of the system $B$; that is,

$$E=-\text{Tr}\left[ \hat{\rho_{B}} \ln \hat{\rho_{B}}\right]. \tag{3}$$

Now, we can derive $E$ in terms of the eigenvalues $\lambda(q_{2})$ associated with the density matrix $\left\langle q_{2}\right|\hat{\rho_{B}}\left|q_{2}' \right\rangle $. To see this argument, we proceed as follows. We can write any continuous-valued hermitean operator $\hat{X}=\int_{-\infty}^{+\infty} dx~ x \left| X\right\rangle \left\langle X \right|$; take into account that $\hat{X}\left| X\right\rangle=x \left| X\right\rangle$. Now, any analytical function $f(\hat{X})$ satisfies $f(\hat{X})=\int_{-\infty}^{+\infty} dx~f(x) \left| X\right\rangle \left\langle X \right|$. Following this argument, we choose $f(\hat{\rho_{B}})=\hat{\rho_{B}} \ln \hat{\rho_{B}}$, to have

$$\hat{\rho_{B}} \ln \hat{\rho_{B}}= \int_{-\infty}^{+\infty} dq_{2} ~\lambda(q_{2}) \ln \lambda(q_{2}) \left| q_{2}\right\rangle \left\langle q_{2} \right|. \tag{4}$$

Taking the trace of the last expression through the basis $\left| x_{2} \right\rangle$

$$\text{Tr}[\hat{\rho_{B}} \ln \hat{\rho_{B}}]=\int_{-\infty}^{+\infty} dx_{2}~\int_{-\infty}^{+\infty} dq_{2} \lambda(q_{2}) \ln \lambda(q_{2}) \left\langle x_{2} \right. \left| q_{2}\right\rangle \left\langle q_{2} \right|\left. x_{2}\right\rangle $$

$$=\int_{-\infty}^{+\infty} dx_{2}~ \lambda(x_{2}) \ln \lambda(x_{2}), \tag{5} $$

using the last equation in Eq. (3), we derive the expression for the von Neumann entropy of a continuous-variable quantum state, like the one of Eq. (1), that is,

$$E=-\text{Tr}[\hat{\rho_{B}} \ln \hat{\rho_{B}}]=-\int_{-\infty}^{+\infty} dx_{2}~ \lambda(x_{2}) \ln \lambda(x_{2}). \tag{6}$$

Then, my question is:

What is the mechanism to obtain the continuous eigenvalues $\lambda(x_{2})$ of the marginal density matrix $\hat{\rho_{B}}$ associated with the state of Eq. (1)?

This is interesting since the marginal density matrix $\left\langle x_{2}\right|\hat{\rho_{B}}\left| x_{2}' \right\rangle = \rho(x_{2}, x_{2}')$ is a continuous matrix; that is, the $x_{2}$ variable can be associate with a horizontal axis (the abscissa) and the $x_{2}'$ with a vertical axis (the ordinate). This is in contrast with the discrete case, where the discrete eigenvalues are obtained by solving the characteristical polynomial associated with the marginal density matrix.

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    $\begingroup$ No, density matrices are by definition trace-class operators, thus compact, and thus admit a discrete spectrum. The partial trace of a density matrix is again a density matrix. Why do you think a density matrix has "continuous" aka generalized eigenstates at all? $\endgroup$ Mar 28, 2023 at 6:08
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    $\begingroup$ I remember from when I looked into this many years ago that there have been indeed attempts to define a continuous entropy along these lines, but I did not find them very satisfying. I'm not so sure that your Eq. 6 works dimensionally; I've seen versions with some kind of unitful regulator added. $\endgroup$
    – Rococo
    Mar 28, 2023 at 6:20
  • $\begingroup$ @Rococo For me even $(4)$ does not make sense already... And, as far as I can see, there is some weird step done in eq. $(5)$, where a $\delta(x_2-q_2)^2$ appears... $\endgroup$ Mar 28, 2023 at 6:25
  • $\begingroup$ To OP: Consider the 1D quantum harmonic oscillator; you can also write, at least formally, $\langle x|H|x^\prime\rangle=H(x,x^\prime)$. Still, $H$ has a discrete spectrum... The Dirac notation just obscure things here, I am afraid... $\endgroup$ Mar 28, 2023 at 6:33
  • $\begingroup$ Seems to me what you want to do is to diagonalize $\psi(q1,q2)$. For that you should be able to do a Schmidt decomposition. It will represent the state as a discrete sum of tensor products. $\endgroup$ Mar 28, 2023 at 12:00

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