Partial trace on a two-mode continuous-variable quantum state

Question

I have the following continuous-variable entangled state

$$ \left|\psi\right\rangle_{1,2}= \int_{-\infty}^{+\infty} dq_{1} dq_{2}~\psi(q_{1}) f_{G}(q_{2})e^{\frac{i}{2} q_{1}q_{2}}\left|q_{1}\right\rangle \left|q_{2}\right\rangle, \tag{1}$$

where $ f_{G}(q_{2})=\left(\frac{V_{s}^{2}}{2\pi} \right)^{\frac{1}{4}}e^{-V_{s}^2 q_{2}^{2}/4}$ is a Gaussian function with a positive parameter $V_{s}^2 >0$. Besides, $q_{1}$ and $q_{2}$ are position variables; therefore, the $\left\lbrace\left|q_{1}\right \rangle\right\rbrace$ and $\left\lbrace\left|q_{2}\right \rangle\right\rbrace$ are the position basis of systems 1 and 2 respectively. The quantum state of Eq. (1) represents a continuous-variable cluster, as can be verified from Weedbrook, Eq. (175).

Concretely, I need the position and momentum probability distributions of the reduced system 1 of Eq. (1). To compute the position probability distribution, I proceed as follows. First, I calculate the joint density operator associated with the state of Eq. (1); that is,

$$\hat{\rho}_{1,2}=\int_{-\infty}^{+\infty} dq_{1}dq_{1}' dq_{2}dq_{2}'~\psi(q_{1})\psi^{\ast}(q_{1}') f_{G}(q_{2})f_{G}(q_{2}') e^{\frac{i}{2} q_{1}q_{2}}e^{-\frac{i}{2} q_{1}'q_{2}'}\left(\left|q_{1}\right\rangle \left\langle q_{1}'\right| \otimes \left|q_{2}\right\rangle \left\langle q_{2}'\right|\right), \tag{2}$$

where we have used $f_{G}^{\ast}(q_{2})=f_{G}(q_{2})$ since that function is real. Now, we obtain the reduced density operator of system 1 through the partial trace on system 2, for this, I use the basis $\left\lbrace \left|Q_{2}\right\rangle\right\rbrace$; then, I mean

$$\hat{\rho}_{1}=\text{Tr}\left[ \hat{\rho}_{1,2}\right] \\ =\int_{-\infty}^{+\infty} dQ_{2}dq_{1}dq_{1}' dq_{2}dq_{2}'~\psi(q_{1})\psi^{\ast}(q_{1}') f_{G}(q_{2})f_{G}(q_{2}') e^{\frac{i}{2} q_{1}q_{2}}e^{-\frac{i}{2} q_{1}'q_{2}'}\left(\left|q_{1}\right\rangle \left\langle q_{1}'\right| \otimes \left\langle Q_{2}\right.\left|q_{2}\right\rangle \left\langle q_{2}'\right|\left. Q_{2}\right\rangle\right)\\ =\int_{-\infty}^{+\infty} dQ_{2}dq_{1}dq_{1}' ~\psi(q_{1})\psi^{\ast}(q_{1}') \left[f_{G}(Q_{2}) \right]^2 e^{\frac{i}{2} Q_{2}(q_{1}-q_{1}')}\left|q_{1}\right\rangle \left\langle q_{1}'\right| \\ =\int_{-\infty}^{+\infty} dq_{1}dq_{1}' ~\psi(q_{1})\psi^{\ast}(q_{1}') f_{G}(q_{1}-q_{1}') e^{\frac{i}{2} Q_{2}(q_{1}-q_{1}')}\left|q_{1}\right\rangle \left\langle q_{1}'\right|, \tag{3} $$ where in the second line of the last equation, we solve the integrals on $q_{2}$ and $q_{2}'$ by using the Dirac deltas: $\delta(Q_{2}-q_{2})=\left\langle Q_{2}\right.\left|q_{2}\right\rangle $ and $\delta(q_{2}'-Q_{2})=\left\langle q_{2}'\right|\left. Q_{2}\right\rangle$; besides, in the third line we solve the integral on $Q_{2}$ by utilizing the definition of $f_{G}(Q_{2})$; that is,

$$\int_{-\infty}^{+\infty} dQ_{2}~\left[f_{G}(Q_{2}) \right]^2 e^{\frac{i}{2} Q_{2}(q_{1}-q_{1}')}=f_{G}(q_{1}-q_{1}')=e^{\frac{\left(q_{1}-q_{1}'\right)^2}{8V_{s}^2}}. \tag{4}$$

Then, the diagonal elements of $\hat{\rho}_{1}$ provide the position probability distribution $P(q_{1})$ of system 1; then, by utilizing a basis $\left\lbrace \left|Q_{1}\right\rangle\right\rbrace$, I take the diagonal elements of the reduced density operator $\hat{\rho}_{1}$ of Eq. (3); that is,

$$P(Q_{1})=\left\langle Q_{1}\right| \hat{\rho}_{1} \left|Q_{2} \right\rangle=\left| \psi(Q_{1})\right|^2, \tag{5}$$

where I have solved the integrals on $q_{1}$ and $q_{1}'$ by using the Dirac deltas: $\delta(Q_{1}-q_{1})=\left\langle Q_{1}\right.\left|q_{1}\right\rangle $ and $\delta(q_{1}'-Q_{1})=\left\langle q_{1}'\right|\left. Q_{1}\right\rangle$. Hence, Eq. (5) represents the position probability distribution associated with the reduced system 1 of the state of Eq. (1); therefore, my questions are:

How to obtain the momentum probability distribution associated with the reduced system 1 of Eq. (1)?

Can I take the Fourier transform of Eq. (5) in order to obtain the momentum probability distribution of the reduced system 1?

Answer 1

Given a state $\hat{\rho}$, the average position of the state is given by $$ \langle x\rangle = \text{tr}\{\hat{\rho}\hat{X}\} , $$ where $\hat{X}$ is a position operator. In the current context with $|q\rangle$ representing a position basis, the position operator is given by $$ \hat{X} = \int |q\rangle x \langle q|\ dq , $$ so that $\hat{X}|q\rangle=|q\rangle x$.

When we now compute the average position of your final state, we get $$ \langle x\rangle = \text{tr}\{\hat{\rho}_1\hat{X}\} = \int |\psi(q)|^2 q\ dq . $$ So one can interpret the diagonal as the probability distribution for position.

In a similar way, the momentum operator is defined by $$ \hat{P} = \int |p\rangle p \langle p|\ dp , $$ where $$ \langle q|p\rangle = \exp(iqp) . $$ The average momentum is now given by $$ \langle p\rangle = \text{tr}\{\hat{\rho}_1\hat{P}\} = \int p\exp[-ip(q_1-q_1')]\psi(q_1)\psi(q_1')\tilde{f}_G(q_1-q_1')\ dq_1 dq_2 dp , $$ where I changed the OP notation so that $\tilde{f}_G$ is the function obtained in (4). The momentum distribution is now given by the part of the integrant of the integral over $p$ without the factor of $p$.