What is coherence in quantum mechanics?

What are coherence and quantum entanglement? Does it mean that two particles are the same?

I read this in a book called Physics of the Impossible by Michio Kaku. He says that two particles behave in the same way even if they are separated. He also says that this is helpful in teleportation. How can this be possible? Could somebody please explain?

Answer

Coherent (or pure) state

Consider 2 basic states $\lvert0\rangle$ and $\lvert1\rangle$. (If you never heard about states, treat them as ordinary complex vectors.) Here we suppose that $\lvert0\rangle$ and $\lvert1\rangle$ are orthogonal ($\langle 0\lvert1\rangle =0$).

Now, consider $\lvert c\rangle = \frac{1}{\sqrt{2}} (\lvert0\rangle + e ^{i\phi}\lvert1\rangle)$.

$\lvert c\rangle$ is a (normalized) coherent state, because the phase $\phi$ is constant.

We may look at the density matrix, defined as $\rho = \lvert c\rangle \langle c\lvert$ (that is $\langle i\lvert\rho\lvert j\rangle=\rho_{ij} = c_i^* c_j$, with $c_1 = \frac{1}{\sqrt{2}}, c_2 = \frac{1}{\sqrt{2}}e ^{i\phi}$). We have:

$$\rho =\frac{1}{2} \begin{pmatrix} 1&e ^{i\phi}\\e ^{-i\phi}&1 \end{pmatrix}. \tag{1}$$

This density matrix describes a coherent state. You may verify that $\rho^2 = \rho$, that is $\rho$ is a projector onto the coherent state $\lvert c\rangle$.

Now, suppose the phase $\phi$ is random (that is: the phase difference between $c_1$ and $c_2$ is random), so the mean expectation value of $e ^{i\phi}$ is just zero, and we have a density matrix:

$$\rho' =\frac{1}{2} \begin{pmatrix} 1&0\\0&1 \end{pmatrix}. \tag{2}$$

The density matrix $\rho'$ does not represent no more a coherent state, this is simply a classical statistical probability law. The off-diagonal elements of the matrix $\rho$ have disappeared.

In the 2 cases, we are dealing with only one particle, and the probabilities to find the particle in the $\lvert0\rangle$ state or the $\lvert1\rangle$ state, are the same, and are $\frac{1}{2}$.

$$\langle 0\lvert\rho\lvert0\rangle= \langle 1\lvert\rho\lvert1\rangle= \langle 0\lvert\rho'\lvert0\rangle= \langle 1\lvert\rho'\lvert1\rangle = \frac{1}{2}. \tag{3}$$

Entanglement

Entanglement is about at least $2$ particles, for instance the pure state

$$ \frac{1}{\sqrt{2}}(\lvert0\rangle\lvert0\rangle+\lvert1\rangle\lvert1\rangle )\tag{4}$$is a maximally 2-particles entangled state.

From a given entanged state, you may calculate the correlations for the joint measurement of the 2 particles.

It appears that these entangled quantum correlations are stronger than the classical statistical correlations.

Correlations do not mean that you may exchange instantaneous information; see also this previous answer.