# Quantum Entanglement

**Quantum entanglement** occurs when a system of multiple particles in quantum mechanics interact in such a way so that the particles cannot be described as independent systems but only as one system as a whole. Measurement (e.g. of the spin of an entangled electron) may instantaneously affect another electron's spin at an arbitrarily distant location, apparently (but not actually) faster than the lightspeed limit of special relativity. The fact that electron spin measurements can be highly correlated, violating Bell's inequality, is one of the cornerstone experimental results in the modern theory and interpretation of quantum mechanics.

The properties of quantum entanglement may engender quantum teleportation, where the state of one entangled particle is sent from one location to another without moving the particle. This phenomenon may prove extremely useful in the nascent field of quantum computing, where manipulating quantum states without losing information by exposing them to the environment is highly valued.

#### Contents

## EPR Paradox and Stern-Gerlach Experiments

The original Stern-Gerlach experiment examined the behavior of silver atoms in magnetic fields. Magnetic dipoles couple to magnetic fields; they feel a force $F = \nabla (\vec{\mu} \cdot \vec{B})$ where $\mu$ is the magnetic dipole moment and $B$ is the magnetic field vector. Since silver atoms only have one valence electron with zero orbital angular momentum (a $5s$ electron), any coupling of silver atoms to magnetic fields is due to an **intrinsic magnetic dipole moment** or **spin** of the electron. Since only the behavior of the valence electron matters, the rest of this article will refer to electrons where silver atoms were actually used originally.

In the Stern-Gerlach experiment, the magnetic field gradient $\nabla B$ is aligned along the z-axis and set to some constant $C$ so that electrons are deflected in the magnetic field according to the magnetic dipole moment: $F = C\mu$. If the electrons are spin-up in the z-direction, they are deflected upwards, whereas if the electrons are spin-down in the z-direction, they are deflected downwards. This experiment provided the first evidence for the quantization of spin: on a detecting screen behind the magnets, only two peaks appear rather than a continuous spectrum, corresponding to the two possible quantized values of spin rather than a continuous distribution of possible magnetic dipole moments.

The entirety of a Stern-Gerlach experiment can be thought of as a black box which is able to measure whether an entering electron is spin-up or spin down.

In the EPR thought experiment, Einstein, Podolsky, and Rosen considered a system of two electrons which are sent in opposite directions through two Stern-Gerlach apparatuses, measuring each of their spins. If the electrons are **entangled**, then when electron 1 is measured to be spin-up, electron 2 is measured to be spin-down a large percentage of the time, and vice versa. The precise meaning of "entangled" in quantum mechanics is discussed below mathematically. The result seemed to imply that at one Stern-Gerlach apparatus, the measurement result was being communicated instantly to the other apparatus. Although EPR did not perform the experiment themselves, actual experimental results are consistent with their mathematics. The **nonlocality** of the EPR paradox would be in violation of the principles of relativity, since it seems like information is being communicated to a distant location faster than the speed of light.

EPR tried to avoid this conclusion by asserting that there must have been some **local hidden variables**: at the electron source, some variable that cannot be measured was determining the outcomes of spin measurements when the entangled electrons were produced. They took this to mean that the theory of quantum mechanics was not complete, since the spin and spatial wavefunctions of the electrons did not include the hidden variables and so could not account for all the physics that they would observe. However, in the 1960s, Bell's theorem showed that this was not the case: no local hidden variables theory could possibly account for the degree of correlation present in measurement of spin-entangled electrons.

Entanglement usually originates on the subatomic scale via processes that produce two electrons simultaneously in a correlated way, although complicated systems of fiber optics can be used to produce e.g. systems of two photons with entangled polarizations.

## Measurement of Electron Spin

A single electron spin in quantum mechanics is described by a vector in a two-dimensional vector space. The basis vectors are taken to be the eigenvectors of the $\hat{S}_z$ operator giving the spin of the electron along the $z$ axis, and are labeled $|0\rangle$ and $|1\rangle$, $|+\rangle$ and $|-\rangle$, or $|\uparrow\rangle$ and $|\downarrow\rangle$ depending on context. In this article, the latter notation will be used. In general, an arbitrary electron spin state can be a superposition of the "spin-up" and "spin-down" states:

$|\Psi\rangle = c_1 |\uparrow\rangle + c_2 |\downarrow\rangle,$

where $c_1$ and $c_2$ are some complex constants normalized so that $|c_1|^2+|c_2|^2 = 1$. This notation suggests that a single-spin state can be written as a vector:

$|\Psi \rangle = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}.$

The spin states are defined so that the **spin operator** in the z-direction,

$\hat{S}_z = \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},$

has eigenvalues $\pm\frac{\hbar}{2}$ corresponding to the eigenvectors $|\uparrow\rangle$ and $|\downarrow\rangle$. This operator acts on the vector representation of spin states by simple matrix multiplication.

A electron spin in quantum mechanics is given by the superposition state:

$\large |\Psi\rangle = \frac{i}{2} |\uparrow\rangle + \dfrac{\sqrt{3}}{2} |\downarrow\rangle.$

What is the probability of measuring the spin in the $z$-direction to have value equal to $-\dfrac{\hbar}{2}$?

## Entanglement and Tensor Products

In a two-particle system, the spins of two electrons can be described jointly using **tensor product spaces**. The tensor product of two electron spin states lives in a *four-dimensional* vector space, with basis vectors labeled:

$|\uparrow\rangle\otimes|\uparrow\rangle, \quad |\uparrow\rangle\otimes|\downarrow\rangle, \quad |\downarrow\rangle\otimes|\uparrow\rangle, \quad |\downarrow\rangle\otimes|\downarrow\rangle.$

The tensor product of two states $|x\rangle$ and $|y\rangle$ obeys the following rules:

1) Scaling the tensor product is equivalent to scaling either state:

$c(|x\rangle \otimes |y\rangle) = (c|x\rangle) \otimes |y\rangle = |x\rangle \otimes (c|y\rangle).$

2) The tensor product is distributive in both the first and second slots:

$\begin{aligned} &|x\rangle \otimes (|y_1\rangle + |y_2\rangle) = |x\rangle \otimes |y_1\rangle + |x\rangle\otimes |y_2\rangle \\ &(|x_1\rangle + |x_2\rangle) \otimes |y\rangle = |x_1\rangle \otimes |y\rangle + |x_2\rangle\otimes |y\rangle \end{aligned}$

An arbitrary tensor product state for a two-spin system can be written out in the four-dimensional basis as:

$|\Psi\rangle = c_{11} |\uparrow\rangle\otimes|\uparrow\rangle + c_{12} |\uparrow\rangle\otimes|\downarrow\rangle + c_{21} |\downarrow\rangle\otimes|\uparrow\rangle + c_{22} |\downarrow\rangle\otimes|\downarrow\rangle.$

Two spin state are given by $|\Psi_1\rangle = |\uparrow\rangle$ and $|\Psi_2 \rangle = \frac{1}{\sqrt{2}} (|\uparrow\rangle + |\downarrow\rangle)$.

Which gives the correct tensor product state $|\Psi_1 \rangle \otimes |\Psi_2\rangle$?

**Note**: The notation $|\uparrow\downarrow\rangle = |\uparrow\rangle \otimes |\downarrow\rangle$ is a common shorthand for tensor products of spin states.

This notation suggests that it is more convenient to represent a two-spin state as a matrix, in analogue to the vector representation of a one-spin state:

$|\Psi \rangle = \begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix}.$

To perform a measurement on a two-spin system, one needs to specify the operators acting on each independent space. For instance, to measure the spin of just the first particle, use the operator written as $\hat{S}_z \otimes I$, which measures the spin of the first particle in the z-direction and acts on the second space as the identity. To measure the spins in the z-direction of both particles simultaneously, use $\hat{S}_z \otimes \hat{S}_z$ and so on.

Having defined this notation, it is now possible to state an explicit mathematical criterion of entanglement that is useful for proving things in quantum information theory.

A two-electron spin state is a

product stateif it can be written as the tensor product of exactly two independent single-spin states.

Some states are obviously product states: $|\uparrow\rangle \otimes |\downarrow\rangle$, for instance, is clearly the product of $|\uparrow\rangle$ and $|\downarrow\rangle$. This state, therefore, can be described just by taking each independent single-spin state independently and ignoring the other. A state such as $|\uparrow\rangle \otimes |\downarrow\rangle + |\downarrow\rangle \otimes |\uparrow\rangle$, however, is clearly not a product state. Not all cases are so obvious, however, which is why formulating a precise mathematical criterion is important.

The intuition above suggests the definition of entanglement:

A two-electron spin state is

entangledif it is not a product state, that is, if the matrix representation $M$ of the state does not satisfy $\det M = 0$.

For states of more than two electrons, this condition is actually too strong -- if $\det M \neq 0$, then *all* of the electron spins are entangled. But it's enough for just two of the electrons to be entangled in a multiparticle state for the whole state to be considered entangled. The more accurate condition for a general multiparticle state is that $\text{rank } M \geq 2$, since this is equivalent to entanglement in a two-dimensional subspace. Note that this condition reduces to $\det M \neq 0$ for a two-electron spin state.

## Challenge problem: prove the determinant condition for entanglement described above

Is the following state $|\Psi\rangle$ entangled?

$|\Psi\rangle = \frac{1}{\sqrt{2}} (|\downarrow\downarrow\rangle + |\downarrow\uparrow\rangle).$

**Note**: The notation $|\uparrow\downarrow\rangle = |\uparrow\rangle \otimes |\downarrow\rangle$ is a common shorthand for tensor products of spin states.

$|\Psi\rangle = \frac{1}{\sqrt{5}}|\uparrow\uparrow\rangle +\frac{\sqrt{2}}{\sqrt{5}} |\downarrow\uparrow\rangle - \frac{\sqrt{2}}{\sqrt{5}} |\downarrow\downarrow\rangle$

Is the state $|\Psi\rangle$ above entangled?

**Note**: The notation $|\uparrow\downarrow\rangle = |\uparrow\rangle \otimes |\downarrow\rangle$ is a common shorthand for tensor products of spin states.

The reason for this definition of entanglement is as described above. If a two-spin state can be written as a product state, it is as if each spin is independent. Measurements on each spin do not affect the other. But if a two-spin state is entangled, the fact that the the state is not a product state means that collapsing to one eigenvector in one space causes collapse in the other space as well. Measurements on one spin thus affect any entangled spins instantaneously by influencing what measurements of the other spin are possible, regardless of the spatial location of the other electron! It is important to note, however, that this is not in violation of special relativity: information *cannot* be sent faster than lightspeed via entanglement; the effects of entanglement can only be seen after the fact when two observers meet up to compare spin measurements.

## Consider the entangled state given by:

$|\Psi_E \rangle =\frac{1}{\sqrt{2}} ( |\uparrow\rangle \otimes |\downarrow\rangle - |\downarrow\rangle \otimes |\uparrow\rangle),$

and the non-entangled state given by

$|\Psi_N \rangle = \frac{1}{\sqrt{2}}( |\uparrow\rangle \otimes |\uparrow\rangle + |\uparrow\rangle \otimes |\downarrow\rangle).$

Show that measuring the spin of the first particle for $|\Psi_E\rangle$ affects the probabilities of finding the second particle in each possible state, while the same measurement for $|\Psi_N\rangle$ does not affect the results.

Solution:

For the entangled state, before measurement the second particle has an equal chance to be found in either the spin-up or spin-down states. After measurement of the first particle's spin with the operator $\hat{S}_z \otimes I$, if the first electron is found in the spin-up state, $|\Psi_E\rangle$ collapses to $|\uparrow \rangle \otimes |\downarrow\rangle$ and the second electron

mustbe found in the spin-down state. If the first electron is found in the spin-down state the reverse is true; thus, measurement of the state of the first electron has affected the overall state of the system.For the non-entangled state, the probability of finding the first particle in the spin-up state is one. After measurement, the state does not change, and the second particle is still in an equal superposition of both spin states. Note that there also exist non-entangled states for which measuring the first particle would change the overall spin wavefunction without changing the state of the second particle.

**Cite as:**Quantum Entanglement.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/quantum-entanglement/