Quantum Entanglement

Bipartite System

Consider a system consists of system $A$ and system $B$. Let $$\left| 0\right>_A,\left| 1\right>_A,...,\left| N\right>_A$$ be the basis kets of system $A$, and $$\left| 0\right>_B,\left| 1\right>_B,...,\left| M\right>_B$$ be the basis kets of system $B$. The state vector of the whole system is $$\left|\Psi\right>_{AB} = \sum^N_{n=0}\sum^M_{m=0} c_{nm} \left| n \right>_A\left| m \right>_B.$$ The state of the system is said to be a separable state if $c_{ij} = f_i g_j$, i.e. $$\left|\Psi\right>_{AB} = \sum^N_{n=0}f_n\left| n\right>_A\sum^N_{m=0}g_m\left| m\right>_B,$$ and an entangeled state if $c_{ij} \neq f_i g_j$. When a system is consisted of more than one system, of which each has their own states, if the wave function of the whole system are not separable (entangeled), the collapse of one of the system will certainly unveil certain states of the remaining systems (or in some cases, completely determine). That system is what we call, entangeled.

Example

Find the state vector of a system defined by the outer product of $\left|\psi_A\right>=\begin{pmatrix}a \\ b\end{pmatrix}$ and $\left|\psi_B\right>=\begin{pmatrix}c \\ d\end{pmatrix}$.

Solution: $$\begin{align} \left|\Psi_{AB}\right> &= \left|\psi_A\right> \otimes \left|\psi_B\right> = \begin{pmatrix}ac \\ ad\\ bc\\bd \end{pmatrix}\\ &= \left(a\left|\uparrow\right>_A + b\left|\downarrow\right>_A\right)\left(c\left|\uparrow\right>_B + d\left|\downarrow\right>_B\right)\\ &= ac\left|\uparrow\right>_A \left|\uparrow\right>_B + ad \left|\uparrow\right>_A \left|\downarrow\right>_B +bc\left|\downarrow\right>_A \left|\uparrow\right>_B + bd\left|\downarrow\right>_A \left|\downarrow\right>_B. \end{align}$$ As $\left|\Psi_{AB}\right>$ can be separated into $\left(a\left|\uparrow\right>_A + b\left|\downarrow\right>_A\right)$ times $\left(c\left|\uparrow\right>_B + d\left|\downarrow\right>_B\right)$, the state is a separable state

Partial Trace and Reduced Density of Matrix

The ket of a system is given by $$\left|\Psi_{AB}\right> = \sum_n\sum_p c_{np} \left| n \right>_A\left| p \right>_B,$$ and its corresponding bar of the state vector is $$\left<\Psi_{AB}\right| = \sum_m\sum_q c^*_{mq} \left< m \right| \left< q \right|.$$ The density matrix is defined as $$\rho_{AB} = \left|\Psi_{AB}\right>\left<\Psi_{AB}\right|,$$ or in terms of the basis kets of system $A$ and $B$ $$\rho_{AB} = \sum_{nm}\sum_{pq}c_{np}c^*_{pq}\left| n \right>_{AA}\left< m \right| \cdot \left| p \right>_{BB}\left< q \right|.$$ The partial trace of the density matrix over the $B$ subsystem is defined as $$Tr_B(\left| n \right>_A\left< m \right| \cdot \left| p \right>_B\left< q \right|) = \left| n \right>_A\left< m \right| \cdot Tr_B(\left| p \right>_B\left< q\right|)$$ The partial trace of the density matrix is then $$\begin{align} \rho_A &= \sum_{nm}\sum_{pq} c_{np}c^*_{mq} \left| n \right>_A\left< m \right| \cdot Tr_B(\left| p \right>_B\left< q \right|)\\ &= \sum_{nm}\sum_{pq}c_{np}c^*_{mq} \left| n \right>_A\left< m \right| \cdot \delta_{pq}\\ &= \sum_{nm}\left(\sum_pc_{np}c^*_{mq}\right)\left| n \right>_A\left< m \right| \end{align}$$

Expectation of a Bipartite System

Consider an observable $M_A$ of the subsystem $A$. The operator of $M_A$ is given by $$M_A = \sum_{ij}F_{ij}\left| i\right>_{AA}\left< j\right|.$$ In the composite system $M_A$ also operates of the subsystem $B$ but has no effect on $B$, so we insert a $I_B = \sum_k\left| k \right>_{BB} \left< k \right|$ into the operator. Thus, the operator of $M_A$ on the composite system becomes $$M_A = I_B \otimes \sum_{ij}F_{ij}\left| i\right>_{AA} = \sum_{ij}F_{ij}\left| i\right>_{AA}\left< j\right|\cdot\sum_k\left| k\right>_{BB}\left< k\right|.$$ The expectation of $M_A$ of the system is then $$\begin{align} \left|\Psi\right> &= \sum_n\sum_m c_{nm}\left| n \right>_A \left| m \right>_B \\ M_A \left|\Psi\right> &= \sum_{ij}F_{ij}\left| i \right>_{AA}\left< j \right|\cdot\sum_k\left| k \right>_{BB}\left< k \right| \sum_n\sum_m c_{nm}\left| n \right>_A \left| m \right>_B\\ &= \sum_{ij}F_{ij}\left| i \right>_{AA}\left< j \right|\cdot\sum_k\sum_n c_{nk}\left| n \right>_A \left| k \right>_B\\ &= \sum_{ijk}F_{ij}c_{jk}\left| i \right>_A \left| k \right>_B. \end{align}$$ The expectation value of the observable $M_A$ is $$\begin{align} \left< \Psi_{AB} \right| M_A \left| \Psi_{AB} \right> &= \sum_{ijk}c^*_{jk}F_{ij}c_{jk} \\ &= \sum_{ij}F_{ij}[\rho_A]_{ji}, \end{align}$$ where $[\rho_A]_ij = \sum_k c_{jk} c^*_{ik}$ are the matrix elements of the reduced density matrix $\rho_A$.

Schmidt Decomposiition

A bipartite state of subsystem $A$ and $B$ with $N$ dimension can be expressed as the form of Schmidt decomposition $$\left|\Psi\right> = \sum_n\sum_m c_{nm}\left| n \right>_A \left| m \right>_B = \sum_k\sqrt{\lambda_k}\left| f_k \right> \left| g_k \right>_B,$$ where $$\begin{align} \left< f_n \middle| f_m \right>_A &= \delta_{nm},\, \sum_n\left| f_n\right>_A\left< f_n \right| = I_A\\ \left< g_n \middle| g_m \right>_B &= \delta_{nm},\, \sum_n\left| g_n\right>_B\left< g_n \right| = I_B. \end{align}$$ Define the entropy of entanglement to be $$S \equiv -\sum_k\lambda_k\log\lambda_k$$ When $S=0$, the state $$\left|\Psi\right> = \left| f \right>_A \left| g \right>_B$$ is a separable state. When $S = \log N$, the state $$\left| \Psi \right> = \sum^N_{k=1}\frac{1}{\sqrt{N}}\left| f_k \right>_A \left| g_k \right>_B$$ is a maximally entangled state.

Example

Express the state $$\left|\psi\right> = \alpha\left| H \right>_A\left| H \right>_B + \alpha\left| V\right>_A\left| V \right>_B + \beta\left| H \right>_A\left| V \right>_B + \beta\left| V \right>_A\left| H \right>_B$$ in the form of Schmidt decomposition.

Solution: \begin{align} \rho_A &= Tr_B(\left|\psi\right>\left<\psi\right|)\\ &= \left(\alpha\left| H \right>_A + \beta \left| V\right>_A\right)\left(\alpha\left< H\right|_A + \beta\left< V \right|_A\right) + \left(\alpha\left| V \right>_A + \beta\left| H \right>_A\right)\left(\alpha\left< V \right|_A + \beta\left< H \right|_A\right)\\ &=(\alpha^2 + \beta^2)\left(\left| H \right>_A\left< H \right| + \left| V \right>_A\left< V \right|\right) + 2\alpha\beta\left(\left| H \right>_A\left< V \right| + \left| V\right>_A\left_A\) are the eigenvectors of $\rho_A\left|\pm \right>_A = \lambda_{\pm}\left|\pm\right>_A$. Similar procedure can be done to obtain $\rho_B$ and hence $\left|\pm\right>_B$. Then, the state is $$\left|\psi\right> = e^{i\theta_+}\sqrt{\lambda_{+}}\left| + \right>_A\left| + \right>_B + e^{i\theta_-}\sqrt{\lambda_-}\left| - \right>_A\left| - \right>_B,$$ where $$e^{i\theta_\pm} = (\left<\pm\right|_A\left<\pm\right|_B)\left|\psi\right>/\sqrt{\lambda_\pm} = (\alpha\pm\beta)/\sqrt{(\alpha\pm\beta)^2}$$

Bell's Inequality

An electron detector and a positron detector are allowed to be rotated independently. The first measurment measures the component of the electron spin in the direction of a unit vector $\vec{a}$. The second measurement measures the spin of the positron along the direction of a unit vector $\vec{b}$. The records of the spin have units of $\hbar/2$. Let $P(\vec{a},\vec{b})$ be the average value of the product of the spins. The average is $$P(\vec{a},\vec{b}) = -\vec{a}\cdot\vec{b}.$$ Suppose there is a hidden variable that characterize the complete state of the electron and positron system. Assume locality is valid, then measure the positron just before the electron measurement is made such that there is not enough time for any kind of information be transferred between two events. Then, the electron measurement is independent of the orientation of the positron detector. let $A(\vec{a},\lambda)$ and $B(\vec{b},\lambda)$ be the functions that give the results of measuring the electron and the positron respectively. Results are +1 for spin up and -1 for spin down. The average of the product of the measurements is $$P(\vec{a},\vec{b}) = \int \rho(\lambda)A(\vec{a},\lambda)B(\vec{b},\lambda)d\lambda.$$ When the detectors are aligned, the results are perfectly (anti)-correlated: $$A(\vec{a},\lambda) = -B(\vec{a},\lambda),$$ for all $\lambda$. So the average of the product is $$P(\vec{a},\vec{b}) = - \int \rho(\lambda)A(\vec{a},\lambda)A(\vec{b},\lambda)d\lambda.$$ Let $\vec{c}$ be another unit vector, $$P(\vec{a},\vec{b}) - P(\vec{a},\vec{c}) = -\int \rho(\lambda)[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]d\lambda.$$ Since $[A(\vec{b},\lambda)]^2=1$, $$P(\vec{a},\vec{b}) - P(\vec{a},\vec{c}) = -\int \rho(\lambda) [1-A(\vec{b},\lambda)A(\vec{c},\lambda)]A(\vec{a},\lambda)A(\vec{b},\lambda)d\lambda.$$ The values of $A(\vec{a},\lambda)A(\vec{b},\lambda)$ is bounded $$-1 \leq [A(\vec{a},\lambda)A(\vec{b},\lambda)] \leq +1,$$ so, we have $$\rho(\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)] \geq 0.$$ Then, $$\left| P(\vec{a},\vec{b}) - P(\vec{a},\vec{c})\right| \leq \int \rho(\lambda) [1 - A(\vec{b},\lambda)A(\vec{c},\lambda)]d\lambda,$$ or $$\bbox[5px,border:2px solid #666] { \left| P(\vec{a},\vec{b}) - P(\vec{a},\vec{c})\right| \leq 1 + P(\vec{b},\vec{c}). }$$ This is known as the Bell's inequality.