 ### Lei Mao

Machine Learning, Artificial Intelligence, Computer Science.

Twitter Facebook LinkedIn GitHub   G. Scholar E-Mail RSS

# Quantum Entanglement

### Introduction

Quantum entanglement has attracted some attention from the public that this mysterious process might be useful for lots of future applications, such as quantum computing and teleportation.

In this blog post, I would like to discuss quantum entanglement superficially using mathematics.

### Quantum Entanglement

Assume we have two independent quantum systems $Q$ and $Q^{\prime}$, represented respectively by the vector spaces $\mathbb{V}$ and $\mathbb{V}^{\prime}$. The quantum system obtained by merging $Q$ and $Q^{\prime}$ will have the tensor product $\mathbb{V} \otimes \mathbb{V}^{\prime}$ as a state space.

If $\mathbb{V} \in \mathbb{C}^n$ with basic states $| x_0 \rangle, | x_1 \rangle, \cdots, | x_{n-1} \rangle$ and $\mathbb{V}^{\prime} \in \mathbb{C}^m$ with basic states $| y_0 \rangle, | y_1 \rangle, \cdots, | y_{m-1} \rangle$, the tensor product state space would be $\mathbb{V} \otimes \mathbb{V}^{\prime} \in \mathbb{C}^{n \times m}$ with basic states $| x_i \rangle \otimes | y_j \rangle$ for $0 \leq i \leq n-1$ and $0 \leq j \leq m-1$. Here, the basic state $| x_i \rangle \otimes | y_j \rangle$ means the first quantum systems $Q$ is in the state $| x_i \rangle$ and the second quantum system $Q^{\prime}$ is in the state $| y_j \rangle$.

More independent quantum systems $Q_0, Q_1, \cdots, Q_{k-1}$ could be merged together and the state space of the merged quantum system would be $\mathbb{V}_0 \otimes \mathbb{V}_1 \otimes \cdots\ \otimes \mathbb{V}_{k-1}$.

The system state of the merged quantum system from two independent quantum systems $Q$ and $Q^{\prime}$ could be expressed using basic states.

$| \psi \rangle = \sum_{i=0}^{n-1} \sum_{j=0}^{m-1} c_{i,j} | x_i \rangle \otimes | y_j \rangle$

The probability of the system being in the state $| x_i \rangle \otimes | y_j \rangle$ after measurement, according to quantum mechanics, would be

$P(| x_i \rangle \otimes | y_j \rangle) = \frac{|c_{i,j}|^2}{\sum_{s=0}^{n-1} \sum_{t=0}^{m-1} |c_{s,t}|^2 }$

If $c_{i,j} = 0$, then $P(| x_i \rangle \otimes | y_j \rangle) = 0$

We further consider the merged system whose $n = 2$ and $m = 2$.

$| \psi \rangle = c_{0,0} | x_0 \rangle \otimes | y_0 \rangle + c_{0,1} | x_0 \rangle \otimes | y_1 \rangle + c_{1,0} | x_1 \rangle \otimes | y_0 \rangle + c_{1,1} | x_1 \rangle \otimes | y_1 \rangle$

Suppose $c_{0,0} = 0$, $c_{0,1} = 1$, $c_{1,0} = 1$, $c_{1,1} = 0$, i.e.,

$| \psi \rangle = | x_0 \rangle \otimes | y_1 \rangle + | x_1 \rangle \otimes | y_0 \rangle$

When we measure the state of the first quantum system $Q$, there are two possible outcomes, either the system was in the state $| x_0 \rangle$ or $| x_1 \rangle$. Similarly, when we measure the state of the first quantum system $Q^{\prime}$, there are two possible outcomes, either the system was in the state $| y_0 \rangle$ or $| y_1 \rangle$.

If we saw the system state of $Q$ is $| x_0 \rangle$, the system state of $Q^{\prime}$ must be $| y_1 \rangle$ even without measurement. Similarly, if we saw the system state of $Q$ is $| x_1 \rangle$, the system state of $Q^{\prime}$ must be $| y_0 \rangle$ even without measurement. This is because the probability of observing $| x_0 \rangle \otimes | y_0 \rangle$ or $| x_1 \rangle \otimes | y_1 \rangle$ is 0.

This is called quantum entanglement.

In this case, no matter how far the two subsystems in the entangled system is, we know the system states of both subsystems by just observing one single subsystem.

The quantum entanglement is simultaneous and instantaneous. If we describe this process using “velocity”, we would find the “velocity” is infinitely large, faster than light speed, which is the fastest “velocity” we are currently using for information transmission. However, we prefer not to describe this process using velocity, since it is not a conventional physical “travel” process.

Formally, if a merged system state $| \psi \rangle$ can be broken into the tensor product of states from the constituent subsystems, i.e.,

\begin{align} | \psi \rangle &= \sum_{i=0}^{n-1} \sum_{j=0}^{m-1} c_{i,j} | x_i \rangle \otimes | y_j \rangle \\ &= \Big(\sum_{i=0}^{n-1} c_{x, i} | x_i \rangle \Big) \otimes \Big(\sum_{j=0}^{m-1} c_{y, j} | y_j \rangle \Big) \\ &= \sum_{i=0}^{n-1} \sum_{j=0}^{m-1} c_{x, i} c_{y, j} | x_i \rangle \otimes | y_j \rangle \\ \end{align}

We call this merged system state $| \psi \rangle$ separable state. Otherwise we call it entangled state, as we would find there is no solution to solve the above equation for an entangled state.

### Conclusions

Some people who did not study physics sometimes described the process of quantum entanglement as if there is a causation relationship, which means that it is the state of one subsystem affecting the other subsystem. However, from our mathematics analysis, we know this causation relationship does not exist.

### FAQs

#### How to Understand the “Information Transimission Velocity” is Greater than the Light Speed?

According to Einstein’s theory that matter cannot travel faster than light, anything that has mass could not travel faster than light speed. The “instance information transimission velocity” in the entangled quantum system does not vilate Einstein’s theory, in my opinion, because it is not transimission at all and should not be considered as velocity.

For example, suppose we have two pre-synchronized accurate clock in Beijing, China and San Francisco, USA, respectively. At 0:00 AM PST according to the two clocks, two people at Beijing and San Francisco send a light beam to a satellite simutaneously. The satellite will receive the two light beams almost simutaneously. Without knowing what is happening on the earth, can the satellite draw conclusion that because the two people from Beijing and San Francisco were sending light beams simutaneously, these two people must have communicated with each other before sending the light beams, and the communication velocity is greater than light speed? The answer is defnitely no.