8 Entanglement. longer examples

8.1 Entanglement swapping

In this subsection, we check a calculation that would be relatively difficult to check by hand. Consider a pair of entangled qubits $A$ and $B$, and another entangled pair $C$ and $D$. By performing a joint measurement on, say $B$ and $C$, we can put $A$ and $B$ in an entangled state although they may be widely separated. We begin by considering the most general projective measure on $B$ and $C$, and calculate the reduced density matrix for a single qubit and the probability of outcome. In this example we calculate these quantities two ways-- one, directly from the density matrix formalism, and two, via formulas taking advantage of the particulars of this problem. To do the first calculation by hand would be extremely unpleasant, as it involves multiplying $16\times16$ matrices with several factors in a single element. Carrying it out below with Maxima is a concise exercise. At present this example does not continue by discussing the measurements that maximize the resulting entanglement of $A$ and $D$.

Qubits $A$ and $B$ are in the state

$${\lvert\alpha \rangle}=\sqrt{\alpha}{\lvert 00 \rangle}+ \sqrt{1-\alpha}{\lvert 11 \rangle},$$

$C$ and $D$ are in the state

$${\lvert\beta \rangle}=\sqrt{\beta}{\lvert 00 \rangle}+ \sqrt{1-\beta}{\lvert 11 \rangle},$$

with the Schmidt coefficients satisfying $\alpha,\beta >1/2$. For now, we only want to tell Maxima that the coefficients of the kets are real.

We consider the projective measurement $\{E_m\}$, that is $E_m={\lvert u_m \rangle}{\langle u_m \rvert}$ and $\sum_m E_m=\mathbb{1}_4$. We consider only a single basis vector here, so we don't use the subscript $m$ for Maxima vector name. We need to use Maxima's declare^$\dagger$ to declare that the components are complex. The state ${\lvert u_m \rangle}$ is normalized, but we don't need to impose that condition in Maxima at this point.

The initial joint state ${\lvert\alpha\beta \rangle}{\langle\alpha\beta \rvert}$ is pure and remains so after the measurement applying ${\lvert u_m \rangle}{\langle u_m \rvert}$ to qubits $B$ and $C$. But we write the density operator because we will examine the reduced states, which are mixed. In the case that $B$ and $C$ are projected onto ${\lvert u_m \rangle}$, the state of the entire system of four qubits after the measurement is given by

$$\rho = \left( (\mathbb{1}_2 \otimes {\lvert u_m \rangle}{\l... ...\lvert\alpha\beta \rangle}{\langle\alpha\beta \rvert}\right),$$ (8)

with

where conjsimp (supplied via the Maxima listserv by Barton Willis) replaces $xx^*$ with $\vert x\vert^2$, and ${\bf ident}($n $)\index{ident@{\bf ident}($n$)}$ is the $n\times n$ identity matrix. The output was suppressed with the trailing dollar sign because the $\rho$ is a $16\times16$ matrix with large expressions for entries. The reduced state of qubits $A$ and $D$ is obtained by tracing out components $2$ and $3$ corresponding to qubits $B$ and $C$, ie $\rho_{AD} =\mbox{Tr}_{BC}\rho$.

Likewise, the reduced state of just qubit $D$ is $\rho_{D} =\mbox{Tr}_{ABC}\rho$.

The second method of calculating $\rho_{D}$ is as follows. Considering the following map from $\mathbb{C}^2\otimes \mathbb{C}^2$ to $M(\mathbb{C},2)$:

$${\lvert a \rangle}=\sum_{i,j=0}^1 a_{ij}{\lvert ij \rangle}... ...t{a}= \pmatrix{ a_{00} & a_{01} \cr a_{10} & a_{11} \cr },$$ (9)

one can show that $\rho_{D}$ is equal to $X_m^{\dagger}X_m$, with $X_m = \widehat{\alpha} \widehat{u}_m \widehat{\beta}$. The Maxima function implementing the mapping (9) is

Then the second calculation of $\rho_D$, which we call rho_4a is given by the following lines.

We compare (%o9) and (%o13), to see that the two methods of calculating the reduced state for qubit $D$ after the measurement give the same result

Now we compute the probability $p_m=\mbox{Tr}(\rho)=\mbox{Tr}(\rho_D)$ that the state is in fact projected onto ${\lvert u_m \rangle}$.

Finally, we compare this to the trace computed by hand from the expression following (9), which is given by

$$p_m=\sum^1_{i,j=0} {\alpha}_{i} {\beta}_{j} \vert\widehat{u}_{m,ij}\vert^2,$$ (10)

where $\alpha_0=\alpha, \alpha_1=1-\alpha, \beta_0=\beta, \beta_1=1-\beta$,

Here we have used Maxima's apply^$\dagger$ and create_list^$\dagger$ functions. Once again we compare the probabilities computed via the two methods

and see that they give the same result.