Recall
Definition A State $\rho$ is a self-adjoint operator on a complex Hilbert space (e.g., $\mathbb{C}^d, L^2(\mathbb{R})$) where $d < \infty$, satisfying $\rho \ge 0$ and $\mathrm{Tr}(\rho) = 1$. For this note, the Hilbert space will be $\mathbb{C}^d$.
Definition A Pure State is a rank-1 projection, i.e., $\rho = |\psi\rangle\langle\psi|$, where $\langle\psi|\psi\rangle = 1$ and $|\psi\rangle \in \mathbb{C}^d$.
Definition A Mixed State is a convex combination of pure states, $\rho = \sum_i \lambda_i |\psi_i\rangle\langle\psi_i|$.
Entanglement Testing
- Classical notion of entanglement (dependence)
- Entanglement of pure states
- Entanglement of mixed states
- Tensor norms
- Story of norms
- Graphical tensor notation in Quantum Computing
Classical Notion of entanglement (dependence)
Consider $X, Y$ discrete random variables with finite support.
Definition "$X, Y$ are Independent" if for $x_i \in \text{Range}(X)$, $y_j \in \text{Range}(Y)$: $$ P(X=x_i, Y=y_j) = P(X=x_i) P(Y=y_j) $$ "$X, Y$ are Dependent" if not independent.
Goal
From the matrix $M_{ij} = P(X=x_i, Y=y_j)$, decide whether $X, Y$ are independent.
Observation
If $X, Y$ are independent, then $M = \mathbf{u}\mathbf{v}^T$ where
$\mathbf{u} = (P(X=x_1), P(X=x_2), \dots)^T$
$\mathbf{v} = (P(Y=y_1), P(Y=y_2), \dots)^T$.
$M$ is of rank 1.
Theorem $X, Y$ are independent $\iff M$ has rank 1.
($\implies$) Done (by Observation 2.2).
($\impliedby$) If $M$ has rank 1, then $M = \mathbf{u}\mathbf{v}^T$. Since $M$ represents a joint probability distribution with marginals $P(X=\cdot)$ and $P(Y=\cdot)$, we get that
- $u_iv_j\geq 0$
- $\sum_i\sum_j u_iv_j = 1$
- $u_i\sum_jv_j = P(X=x_i)$
- $v_j\sum_iu_i = P(Y=y_j)$
Entanglement of Pure States
Let $\mathcal{H}_X = \mathrm{span}\{|x_1\rangle, \dots, |x_m\rangle\}$ and $\mathcal{H}_Y = \mathrm{span}\{|y_1\rangle, \dots, |y_n\rangle\}$. We can map the classical probability matrix $M$ to a quantum state vector: $$ |\psi\rangle = \sum_{i,j} \sqrt{P(X=x_i, Y=y_j)} |x_i\rangle \otimes |y_j\rangle $$
Proposition $M=uv^T$ iff $|\psi\rangle = |\psi_X\rangle \otimes |\psi_Y\rangle$ for some state vectors $|\psi_X\rangle \in \mathcal{H}_X, |\psi_Y\rangle \in \mathcal{H}_Y$.
This means that $X$ and $Y$ are independent iff the state vector $|\psi\rangle$ is a pure tensor.
Definition (Tensor Rank Decomposition) Given Hilbert Spaces $\mathcal{H}_1, \dots, \mathcal{H}_k$, a state vector $|\psi\rangle \in \mathcal{H}_1 \otimes \dots \otimes \mathcal{H}_k$ can be written as: $$ |\psi\rangle = \sum_{a=1}^r |\varphi_1^{(a)}\rangle \otimes |\varphi_2^{(a)}\rangle \otimes \dots \otimes |\varphi_k^{(a)}\rangle $$
Definition (Tensor Rank) $R(|\psi\rangle) = \min \{ r \ge 1 \mid |\psi\rangle \text{ has a decomposition into } r \text{ pure tensors} \}$.
- $R(|\psi\rangle)=1$ "$|\psi\rangle$ is a separable pure state"
- $R(|\psi\rangle)>1$ "$|\psi\rangle$ is an entangled pure state"
Examples (Computation of Tensor Rank)
① Bipartite Quantum States ($k=2$)
If $|\psi\rangle = \sum_{ij} \Psi_{ij} |x_i\rangle \otimes |y_j\rangle$, then its tensor rank $R(|\psi\rangle)$ is equal to the matrix rank of the coefficient matrix $\Psi = (\Psi_{ij})$. This is also known as the Schmidt rank.
Caution In our classical-to-quantum mapping, $R(|\psi\rangle) = \mathrm{rank}((\sqrt{M_{ij}}))$, which is not equal to $\mathrm{rank}(M)$ in general. However, one of them has rank 1 iff the other does. Hence, either can be used for testing entanglement.
In practice, we can compute the rank of a matrix using Singular Value Decomposition (SVD). Let's recall its definition as we will need it later as well.
Definition (Singular Value Decomposition) For a bounded linear operator $M$ on $\mathcal{H}^d$, there exist orthonormal bases $\{|\alpha_i\rangle\}, \{|\beta_i\rangle\}$ and singular values $\sigma_1 \ge \dots \ge \sigma_r > 0$ ($r=\mathrm{rank}(M)$) such that: $$ M = \sum_{i=1}^r \sigma_i |\alpha_i\rangle\langle\beta_i| $$ The singular values are unique upto permutation.
② The GHZ state - $|\text{GHZ}\rangle = \frac{1}{\sqrt{2}}|000\rangle + \frac{1}{\sqrt{2}}|111\rangle$
The decomposition shows $R(|\text{GHZ}\rangle) \le 2$. To show it is entangled, we assume it is separable for contradiction, i.e., $R(|\text{GHZ}\rangle) = 1$. Then $|\text{GHZ}\rangle = |\phi_A\rangle \otimes |\phi_B\rangle \otimes |\phi_C\rangle$. The density operator would be $\rho_{\text{GHZ}} = \rho_A \otimes \rho_B \otimes \rho_C$. Note that $$\rho_{\text{GHZ}} = \frac{1}{2}(|000\rangle\langle000| + |111\rangle\langle111| + |000\rangle\langle111| + |111\rangle\langle000|)$$ while $\rho_A \otimes \rho_B \otimes \rho_C = |\phi_A\phi_B\phi_C\rangle\langle \phi_A\phi_B\phi_C|$. Now it's tempting to apply the uniqueness of SVD to conclude that the LHS has non zero singular values $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}$, while the RHS has non zero singular value $1$, giving a contradiction. However, note that the vectors $|000\rangle, |111\rangle, |000\rangle, |111\rangle$ are not orthogonal, so we cannot apply the uniqueness of SVD directly.
However, we can "cheat" a bit by rearranging the last two qubits between the input and output, as follows: $$\frac{1}{2}|000\rangle\langle 000| + \frac{1}{2}|111\rangle \langle 111|+ \frac{1}{2}|011\rangle\langle 100| + \frac{1}{2}|100\rangle\langle 011| = | \phi_A\phi_B\phi_C\rangle\langle \phi_A\phi_B\phi_C |$$ Now we can safely apply the uniqueness of SVD, as the vectors $|000\rangle, |111\rangle, |011\rangle, |100\rangle$ are orthogonal. This gives us a contradiction, hence $R(|\text{GHZ}\rangle) > 1$.
The trick we used above is called the partial transpose, which we will come back to later while discussing entanglement testers.
Remark Tensor rank is called a discrete norm. It satisfies some but not all properties of a norm: $R(|\psi\rangle) \ge 0$, $R(|\psi\rangle)=0 \iff |\psi\rangle=0$, and $R(|\psi\rangle + |\phi\rangle) \le R(|\psi\rangle) + R(|\phi\rangle)$. However, it fails positive homogeneity, as $R(\lambda|\psi\rangle) = R(|\psi\rangle)$ for $\lambda \neq 0$.
Entanglement of Mixed Quantum States
We now consider general density operators $\rho$.
Definition (Separable Mixed State) A density operator $\rho \in \mathcal{M}_{d_1} \otimes \dots \otimes \mathcal{M}_{d_k}$ is separable if it can be written as a convex combination of product density operators: $$ \rho = \sum_i p_i (\rho_i^{(1)} \otimes \rho_i^{(2)} \otimes \dots \otimes \rho_i^{(k)}) $$ where $p_i > 0, \sum p_i = 1$, and each $\rho_i^{(j)}$ is a density operator.
Clarifying Notation (Digression) The notation $\rho_i^{(1)} \otimes \rho_i^{(2)}$ means the tensor product of operators. If $\rho_i^{(1)} = |\psi_i^{(1)}\rangle\langle\psi_i^{(1)}|$ and $\rho_i^{(2)} = |\psi_i^{(2)}\rangle\langle\psi_i^{(2)}|$, then their product is $|\psi_i^{(1)}\rangle\langle\psi_i^{(1)}|\otimes |\psi_i^{(2)}\rangle\langle\psi_i^{(2)}|$, which lives in $\mathbb{C}^{d_1}\otimes (\mathbb{C}^{d_1})^{*}\otimes \mathbb{C}^{d_2}\otimes (\mathbb{C}^{d_2})^{*}$. But it can also be viewed as an element of $\mathbb{C}^{d_1d_2}\otimes (\mathbb{C}^{d_1d_2})^{*}$ using the notation $|\psi_i^{(1)}\psi_i^{(2)}\rangle\langle\psi_i^{(1)}\psi_i^{(2)}|$, or as an element of $\mathbb{C}^{d_1d_2}\otimes \mathbb{C}^{d_1d_2}$ using the notation $|\psi_i^{(1)}\psi_i^{(2)}\rangle\otimes|\psi_i^{(1)}\psi_i^{(2)}\rangle$.
Question: How useful is (matrix) rank for mixed quantum states?
| Example State ($\rho$) | Rank of $\rho$ |
|---|---|
| $|00\rangle\langle00|$ | 1 |
| $|01\rangle\langle01|$ | 1 |
| $|10\rangle\langle 10|$ | 1 |
| $|11\rangle\langle 11|$ | 1 |
| $\frac{1}{4}(|00\rangle\langle00| + |01\rangle\langle01| + |10\rangle\langle 10| + |11\rangle\langle 11|)$ | 4 |
But the last state is separable by definition. This demonstrates that rank is not a good measure of entanglement for mixed states: it does not behave well with respect to taking convex combinations.
Criteria for Entanglement
Desirable Properties of an Entanglement Measure
We seek a function $\Phi$ that can test for entanglement. A key property is convexity. If $\rho = \sum_i p_i \rho_i$, we want $\Phi(\rho) \le \sum_i p_i \Phi(\rho_i)$. If $\Phi$ is convex and normalized such that $\Phi(\rho_{prod}) \le C$ for any product state, then for any separable state $\rho_{sep}$: $$ \Phi(\rho_{sep}) = \Phi\left(\sum_a p_a \rho_{prod, a}\right) \le \sum_a p_a \Phi(\rho_{prod, a}) \le \sum_a p_a \cdot C = C $$ Thus, if we find $\Phi(\rho) > C$, the state $\rho$ must be entangled. This motivates the use of norms.
Definition (Convex Envelope) The convex envelope of a function $f$ on a convex set $C$ is the largest convex function $g$ such that $g(x) \le f(x)$ for all $x \in C$.
Theorem The convex envelope of the rank function $R(A)$ (on the set of matrices with operator norm $\|A\|_{S_\infty} \le 1$) is the nuclear norm $\|A\|_{S_1} = \sum \sigma_i(A)$. (Sometimes, denoted by $\|A\|_*$)
Lemma (Variational characterization of Nuclear Norm) For a bounded linear operator $A$ on $\mathcal{H}^d$: $$ \|A\|_{S_1} = \max_{U \text{ unitary}} \lvert\mathrm{Tr}(AU)\rvert $$
- First, recall that $A = \sum_{i=1}^r \sigma_i |\alpha_i\rangle\langle\beta_i|$ for some orthonormal bases. Then $\lvert\mathrm{Tr}(AU)\rvert = \lvert\sum_{i=1}^r \sigma_i \langle\beta_i|U\alpha_i\rangle\rvert \leq \sum_{i=1}^r \sigma_i \|\beta_i\|\|U\alpha_i\|$ by Cauchy-Schwarz.
- Then for $U$ unitary, the RHS is just $ \sum_{i=1}^r \sigma_i \|\beta_i\|\|\alpha_i\| = \sum_{i}^r \sigma_i$
- Finally, equality is achieved by taking $U = A (A^{*}A)^{-1/2}$
Theorem (Nuclear Norm is a Norm) The nuclear norm $\|\cdot\|_{S_1}$ satisfies all properties of a norm.
The PPT Entanglement Test
Now we come to the reason why we introduced the nuclear norm: entanglement testing.Theorem (Entanglement Test using Partial Transpose) Let $\rho$ be a $d_A \times d_B$ dimensional bipartite density operator. If $\|\rho^{\Gamma_B}\|_{S_1} > 1$, then $\rho$ is entangled.