# Entanglement characterization of a two-level system

### The Questions

The most fundamental object in quantum information theory is qubit, which is essentially a two-level system, whose basis states are denoted by {∣0⟩, ∣1⟩}. The most general (pure) state is represented by *α*∣0⟩ + *β*∣1⟩, the complex numbers satisfy the normalization condition ∣*α*∣2 + ∣*β*∣2 = 1 . Suppose we now have two such qubits. The basis of the whole system becomes {∣00⟩, ∣01⟩, ∣10⟩, ∣11⟩} and the most general (pure) state is of the form:

$$\left| \rm{General} \right\rangle =\alpha \left| 00 \right\rangle +\beta \left| 01 \right\rangle +\gamma \left| 10 \right\rangle +\lambda \left| 11 \right\rangle$$ (most general two-qubit pure state),

where ∣*α*∣2 + ∣*β*∣2 + ∣*γ*∣2 + ∣*λ*∣2 = 1. Here we use a short-hand notation ∣00⟩ ≡ ∣0⟩ ⊗ ∣0⟩ to represent the tensor product. The main questions we ask in this article are :

**Question (1) When is the state entangled?**

**Question (2) How is the entanglement quantified?**

Part of the reasonings in the following is also applicable to more general states for a qubit coupled with any other quantum systems in pure state. However, for pedagogical purposes, we shall focus on two qubits only.

#### Answer to Question (1)

To answer question (1), it may be better to ask when the state is not entangled? We shall define an non-entangled state as **product state**. By definition, a product state is of the form:

$$\left| \rm {Product} \right\rangle =\left( \alpha _{1}\left| 0 \right\rangle +\beta _{2}\left| 1 \right\rangle \right)\otimes \left( \alpha _{2}\left| 0 \right\rangle +\beta _{2}\left| 1 \right\rangle \right)$$.

We then ask under what circumstances (what values of {*α*, *β*, *γ*, *λ*}) the state $\left| \rm General \right\rangle$ becomes $\left| \rm Product \right\rangle$? This can be answered by a mathematical theorem, known as Schmidt decomposition. In our case, it is always possible to write our states as (the Schmidt form)

$$\left| \rm Schmidt \right\rangle =\sqrt{\mu_{1}}\left| p_{1} \right\rangle \otimes \left| p_{2} \right\rangle +\sqrt{\mu_{2}}\left| q_{1} \right\rangle \otimes \left| q_{2} \right\rangle$$, where it is important to note that ⟨*p*1, 2∣*p*1, 2⟩ = ⟨*q*1, 2∣*q*1, 2⟩ = 1 and ⟨*p*1, 2∣*q*1, 2⟩ = 0. The coefficients are assumed to be real and satisfy *μ*1 + *μ*2 = 1.

Thanks to the bi-orthogonal form, the reduced density matrix $\rho _{1,2}=Tr_{2,1}\left| \rm Schmidt \right\rangle \left\langle \rm Schmidt \right|=\mu_{1}\left| p_{1,2} \right\rangle \left\langle p_{1,2} \right|+\mu_{2}\left| q_{1,2} \right\rangle \left\langle q_{1,2} \right|$ can be immediately determined. By comparing the corresponding reduced density matrices taken from $\left| \rm{Product} \right\rangle \left\langle \rm{Product} \right|$, we conclude that

$$\left| \rm{General} \right\rangle$$ is a product state *if and only if* *μ*1 = 1 and *μ*2 = 0. Otherwise, it is an entangled state. Once we find a relationship between {*μ*1, *μ*2} and {*α*, *β*, *γ*, *λ*}, we are able to answer question (1). Here we note that *μ*'s are the same for both *ρ*1 and *ρ*2. Let us consider *ρ*1, which is a 2x2 matrix:

$$\rho _{1}=Tr_{2}\left| \rm General \right\rangle \left\langle \rm General \right|=\left( \begin{matrix}
\left| \alpha \right|^{2}+\left| \beta \right|^{2} & \alpha \gamma ^{*}+\beta \lambda ^{*} \\
\alpha ^{*}\gamma +\beta ^{*}\lambda & \left| \gamma \right|^{2}+\left| \lambda \right|^{2} \\
\end{matrix} \right)$$. The eigenvectors are {∣*p*1⟩, ∣*q*1⟩} and the corresponding eigenvalues are {*μ*1, *μ*2}. The condition *μ*1 = 1 and *μ*2 = 0 is equivalent to say the determinant det(*ρ*1) = 0 vanishes. This gives the above result explicitly (which is surprisingly simple) in terms of the parameters {*α*, *β*, *γ*, *λ*}:

$$\left| \rm{General} \right\rangle$$ is a product state *if and only if* ∣*α**λ* − *β**γ*∣ = 0.

=== Answer to Question (2) ===

The second question we ask is how entanglement can be quantified. Before jumping into the details, perhaps, we should first ask what is the **purpose** of quantifying entanglement? First, entanglement is the essential feature of the quantum theory. This property is also known as non-locality, where the concept was originated from the seminal paper 1 of Einstein, Podolsky and Rosen (EPR) in 1935. In short, non-locality is a property where a local disturbance cannot affect instantaneously, or at a speed faster then light, an distant object. Quantum mechanics, or more precisely, the measurement part, is non-local. For example, if we have a pair electrons in the spin-singlet state $\left( \left| \uparrow \downarrow \right\rangle -\left| \downarrow \uparrow \right\rangle \right)/\sqrt{2}$ and separated far apart, then the measurement on one electron spin will, according to the orthodox interpretation, necessarily cause the wavefunction to "collapse" to a product state e.g. ∣ ↑ ↓ ⟩. However, the idea of wavefunction collapse is purely a theoretical construct. No noticeable signal is transmitted when one of the distant entangled pair is measured. The bottom line here is that quantum entanglement does suggest different results from any local hidden variable theories. This is made possible by John Bell's famous Bell's inequality found in 1964.

Non-locality is an important property of entanglement, should we quantify entanglement based on the degree the Bell's inequality is violated? Although it is known 2 that all bipartite entangled pure state violate Bell's inequality, We would immediately run into trouble if we consider mixed states as well. It is because some mixed entangled states 3 do not violate any Bell's inequality.