Chapter 4 - Entanglement

Entanglement

Quantum entanglement is the most uniquely quantum mechanical property of quantum systems. It is also believed to be responsible for the advantages of quantum computing systems have over classical computing systems. Entangled states puzzled the founders of quantum theory, including Einstein. For these reasons, it is not surprising that they have become a central part of many investigations into quantum theory and especially quantum information theory.

There are many open problems in this area of research. Some of the most basic and fundamental questions about entangled states are still unanswered. For example, given a mixed-state density matrix for a quantum system, we still do not know how to tell if the systems being described by that matrix are entangled or not. Also, with few exceptions, we do not know how to quantify the entanglement in a system of particles.

What we do understand and what we can explain, is the entanglement between systems of particles which are describable by pure quantum states. Thus this section is concerned with pure entangled states of qubits. Extensions of these ideas will be discussed in this section.

Pure Entangled States

Let us consider two particles, one will be called particle A and the other particle B which are in pure states. (It is also possible to separate a system into two subsystems and call the subsystems A and B.) If the particles are independent and have never interacted, then the state of the system of the composite system of the two particles can be written as

	$\left\vert \Phi \right\rangle =\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B},$	(4.1)

where $\left\vert \psi \right\rangle _{A}\,\!$ is the state of particle A, and $\left\vert \phi \right\rangle _{B}\,\!$ is the state of particle B. This is sometimes stated as a postulate of quantum mechanics \cite{Nielsen/Chuang:book}.

For two particles (or systems) to become entangled, they must interact with each other. They could then be taken far apart and still share their entanglement. This entanglement cannot increase by acting on an individual particle, or even both particles separately. Only joint measurements on both, or interactions between the two, can increase entanglement. Actions on an individual particle, without involving the other, are called local actions or local operations. Local operations on individual particles can be written as

	$L=A\otimes B,$	(4.2)

so that

	$L\left\vert \Phi \right\rangle =A\left\vert \psi \right\rangle _{A}\otimes B\left\vert \phi \right\rangle _{B}.$	(4.3)

A common example, and one often used, is a local unitary transformation

	${\mathcal {U}}_{L}\left\vert \Phi \right\rangle =U_{A}\left\vert \psi \right\rangle _{A}\otimes U_{B}\left\vert \phi \right\rangle _{B}.$	(4.4)

Local unitary transformations will not change the entanglement of a system.

Notice that the density matrix for the composite system in Eq.~(\ref{eq:simpsep}) is

	${\begin{aligned}\rho _{ss}&=\left\vert \Phi \right\rangle \left\langle \Phi \right\vert \\&=(\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B})(_{B}\left\langle \psi \right\vert \otimes _{A}\!\!\left\langle \phi \right\vert )\\&=\left\vert \psi \right\rangle _{A}\left\langle \psi \right\vert \otimes \left\vert \phi \right\rangle _{B}\left\langle \phi \right\vert \\&=\rho _{A}\otimes \rho _{B},\end{aligned}}\,\!$	(4.5)

where $\rho _{A}=\left\vert \psi \right\rangle _{A}\left\langle \psi \right\vert \,\!$ and $\rho _{B}=\left\vert \phi \right\rangle _{B}\left\langle \phi \right\vert \,\!$ so the density operator of a product state is a product of density operators.

Bell States

The simplest examples of entangled states are the entangled states of two two-state systems. There are four different versions of what the "maximally entangled state" of two qubits. The "maximally" will be explained below. These four different versions are called Bell states and are

	${\begin{aligned}\left\vert \psi _{+}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 01\right\rangle +\left\vert 10\right\rangle ),\\\left\vert \psi _{-}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 01\right\rangle -\left\vert 10\right\rangle ),\\\left\vert \phi _{+}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle +\left\vert 11\right\rangle ),\\\left\vert \phi _{-}\right\rangle &={\frac {1}{\sqrt {2}}}(\left\vert 00\right\rangle -\left\vert 11\right\rangle ).\end{aligned}}$	(4.6)

These are an orthonormal ste of states and are all able to be obtained from each other by acting on one particle alone or both individual particles, i.e. acting with local unitary transformations. For example consider the local unitary transformation $\mathbb {I} \otimes \sigma _{3}\,\!$ acting on $\left\vert \phi _{-}\right\rangle \,\!$. The result is $\left\vert \phi _{+}\right\rangle \,\!$. Acting with $\sigma _{1}\otimes \mathbb {I} \,\!$ on $\left\vert \psi _{+}\right\rangle \,\!$ gives $\left\vert \phi _{+}\right\rangle \,\!$, etc.

These states certainly cannot be written in the form

$\left\vert \Phi \right\rangle =\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B}.\,\!$

If they could, then, letting $\left\vert \psi \right\rangle _{A}=\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle \,\!$ and $\left\vert \psi \right\rangle _{B}=\beta _{0}\left\vert 0\right\rangle +\beta _{1}\left\vert 1\right\rangle \,\!$, and notice that the general form is

	${\begin{aligned}\left\vert \psi \right\rangle _{A}\otimes \left\vert \phi \right\rangle _{B}&=(\alpha _{0}\left\vert 0\right\rangle +\alpha _{1}\left\vert 1\right\rangle )\otimes (\beta _{0}\left\vert 0\right\rangle +\beta _{1}\left\vert 1\right\rangle ),\\&=\alpha _{0}\beta _{0}\left\vert 00\right\rangle +\alpha _{1}\beta _{0}\left\vert 10\right\rangle +\alpha _{0}\beta _{1}\left\vert 01\right\rangle +\alpha _{1}\beta _{1}\left\vert 11\right\rangle ,\end{aligned}}$	(4.7)

so the coefficient of $\left\vert 00\right\rangle \,\!$ times the coefficient of $\left\vert 11\right\rangle \,\!$ minus the coefficient of $\left\vert 01\right\rangle \,\!$ times the coefficient of $\left\vert 10\right\rangle \,\!$ is zero. This is not true for any of the Bell states, therefore they cannot be written as a tensor product of two 1-particle states. So, for some 2-particle state,

	$\left\vert \Psi \right\rangle =\alpha _{00}\left\vert 00\right\rangle +\alpha _{01}\left\vert 01\right\rangle +\alpha _{10}\left\vert 10\right\rangle +\alpha _{11}\left\vert 11\right\rangle ,$	(4.8)

the state is separable, or unentangled, if $\alpha _{00}\alpha _{11}-\alpha _{01}\alpha _{10}=0\,\!$. Otherwise it is entangled.

Entangled Mixed States

The state, Eq.~(\ref{eq:simpsep}) is not entangled, so it is called separable. More precisely it is referred to as a simply separable state. In general, a state is separable if its density matrix can be written in the form

	$\rho _{s}=\sum _{i}p_{i}\rho _{A}^{(i)}\otimes \rho _{B}^{(i)},$	(4.9)

where $\rho _{A}^{(i)}\,\!$ $(\rho _{B}^{(i)})\,\!$ is a valid density matrix for subsystem $A\,\!$ $(B)\,\!$, and $\sum _{i}p_{i}=1,\;p_{i}\geq 0\,\!$. An entangled state is one that cannot be written in the form of Eq.~(\ref{eq:defsepstate}).

For a pure state, the situation is simpler. A pure state is entangled if and only if it cannot be written in the form of Eq.~(\ref{eq:simpsep}). In other words a pure state is entangled if it cannot be written as the product of two states of the individual subsystems.

Reduced Density Operators and the Partial Trace

The Bell states are maximally entangled states. To understand this, one may consider the fact that these states are pure states, but that information about the individual particles in the system is lacking. In this section, a more precise meaning of this statement is given.

Let us first consider a tool which will be useful, the partial trace. The partial trace is the trace over one of the subsystems (particle states) in a composite system. Let us suppose that the density matrix for a composite system is given by

	$\rho _{ss}=\rho _{A}\otimes \rho _{B}.\,\!$	(4.10)

The partial trace is the trace over one of the subsystems. For example the trace over subsystem $B\,\!$ is

	${\mbox{Tr}}_{B}(\rho _{ss})=\rho _{A}{\mbox{Tr}}(\rho _{B})=\rho _{A},\,\!$	(4.11)

since ${\mbox{Tr}}(A\otimes B)={\mbox{Tr}}(A){\mbox{Tr}}(B)\,\!$ and the trace of a density matrix is one. The matrix $\rho _{A}\,\!$ is called the reduced density operator, or reduced density matrix. However, this is a special case. The density matrix for a composite system of two (or more) subsystems cannot be written in this form unless the two subsystems are not entangled.

For the cases where the two subsystems are entangled, there are at least two ways to calculate the partial trace. One is to write the matrix form of the state in terms of sum of tensor products of Pauli matrices. (See Appendix {app:polvec}, Sec. {sec:2qdmat}.) The other is to realize that the trace can be calculated by summing the projections onto the diagonal elements of the subsystem over which you are tracing. For example, for a general $2\times 2\,\!$ density matrix $\rho _{2}\,\!$, the trace is

	${\mbox{Tr}}(\rho _{2})={\mbox{Tr}}(\left\vert 0\right\rangle \left\langle 0\right\vert \rho _{2}+\left\vert 1\right\rangle \left\langle 1\right\vert \rho _{2})=\left\langle 0\right\vert \rho _{2}\left\vert 0\right\rangle +\left\langle 1\right\vert \rho _{2}\left\vert 1\right\rangle .$	(4.12)

For the general case, let us consider a density matrix for a bipartite system, $\rho \,\!$. Let the subsystem $A\,\!$ have Greek letters as indices and let the subsystem $B\,\!$ have Latin indices

	${\begin{aligned}\rho &=\sum _{\alpha ,\beta ,i,j}\rho _{\alpha \beta ,ij}\left\vert \alpha i\right\rangle \left\langle \beta j\right\vert \\&=\sum _{\alpha ,\beta ,i,j}\rho _{\alpha \beta ,ij}\left|\alpha \right\rangle \left\langle \beta \right|\otimes \left|i\right\rangle \left\langle j\right|.\end{aligned}}$	(4.13)

To calculate the reduced density matrix for subsystem $B\,\!$, the trace over $A\,\!$ is taken

	${\begin{aligned}\rho _{A}&={\mbox{Tr}}_{B}(\rho )\\&=\sum _{k}\left\langle k\right\vert \rho \left\vert k\right\rangle \\&=\sum _{k}\sum _{\alpha ,\beta ,i,j}\left\langle k\right\vert \left(\rho _{\alpha \beta ,ij}\left\vert \alpha \right\rangle \left\langle \beta \right\vert \otimes \left\vert i\right\rangle \left\langle j\right\vert \right)\left\vert k\right\rangle \\&=\sum _{k}\sum _{\alpha ,\beta ,i,j}\rho _{\alpha \beta ,ij}\left\vert \alpha \right\rangle \left\langle \beta \right\vert \delta _{ki}\delta _{jk}\\&=\sum _{\alpha ,\beta ,k}\rho _{\alpha \beta ,kk}\left\vert \alpha \right\rangle \left\langle \beta \right\vert \\\end{aligned}}$	(4.14)

For the partial trace of a $4\times 4\,\!$ density matrix $\rho \,\!$ over the subsystem $B\,\!$,

	${\mbox{Tr}}_{B}(\rho )={}_{B}\left\langle 0\right\vert \rho \left\vert 0\right\rangle _{B}+{}_{B}\left\langle 1\right\vert \rho \left\vert 1\right\rangle _{B},$	(4.15)

which leaves the part of the matrix corresponding to $A\,\!$ alone, and projects the $B\,\!$ part onto the two diagonal elements and then adds those. Now let us calculate the partial trace of a Bell state, for example $\left\vert \psi _{+}\right\rangle \,\!$,

	${\begin{aligned}{\mbox{Tr}}_{B}(\left\vert \psi _{+}\right\rangle \left\langle \psi _{+}\right\vert )&={}_{B}\left\langle 0\right\vert (\left\vert \psi _{+}\right\rangle \left\langle \psi _{+}\right\vert )\left\vert 0\right\rangle _{B}+{}_{B}\left\langle 1\right\vert (\left\vert \psi _{+}\right\rangle \left\langle \psi _{+}\right\vert )\left\vert 1\right\rangle _{B},\\&={}_{B}\left\langle 0\right\vert \left({\frac {1}{2}}(\left\vert 01\right\rangle +\left\vert 10\right\rangle )(\left\langle 01\right\vert +\left\langle 10\right\vert )\right)\left\vert 0\right\rangle _{B}\\&\;\;+{}_{B}\left\langle 1\right\vert \left({\frac {1}{2}}\left(\left\vert 01\right\rangle +\left\vert 10\right\rangle \right)(\left\langle 01\right\vert +\left\langle 10\right\vert )\right)\left\vert 1\right\rangle _{B}\\&={\frac {1}{2}}(\left\vert 1\right\rangle _{A}\left\langle 1\right\vert )+{\frac {1}{2}}(\left\vert 0\right\rangle _{A}\left\langle 0\right\vert ),\end{aligned}}$	(4.16)

which can be written simply as

	${\mbox{Tr}}_{B}(\left\vert \psi _{+}\right\rangle \left\langle \psi _{+}\right\vert )=\rho _{A}=\left({\begin{array}{cc}1/2&0\\0&1/2\end{array}}\right).$	(4.17)

This is quite an interesting and significant find. The density matrix for the whole system of two qubits is in a pure state indicating maximal knowledge. However, the reduced density matrix, representing our knowledge of one of the individual particles, is completely, or maximally mixed, indicating minimal knowledge. They way this is interpreted is that the two particles or subsystems, taken together, are in a definite state, a pure state, whereas taken separately, they contain as little information as possible. This indicates entanglement since the two states together make up a definite pure state, but separately the two states contain as little information as possible.

It is important to note that for a pure state, the trace over subsystem $A\,\!$ produces the same result as the trace over subsystem $B\,\!$. In other words, for a pure state $\rho _{p}\,\!$,

	${\mbox{Tr}}_{A}(\rho _{p})={\mbox{Tr}}_{B}(\rho _{p}).\,\!$	(4.18)

How Entangled Is It?

Due to the previous discussion, there is a definite notion of maximal entanglement for pure states. From the determinant condition, there is a method for identifying unentangled pure states. One question is, how entangled is it if it is not separable, nor maximally entangled? There are now many ways of defining measures of entanglement which will be explored in a later section and an appendix. Here two very common ways of measuring the entanglement for pure states are given. One based on each of the two properties identified in this section: the partial trace and the determinant of the pure state's coefficients.