States on a Composite System
Recall all the way back to what a quantum state is.
A composite quantum state is a state that lives in a Composite System .
Recall Tensor Product of two Hilbert spaces H A H_A H A H B H_B H B Ket (State) on the individual subsystems.
∣ ψ ⟩ = ∑ i j a i j ∣ v i ⟩ ⊗ ∣ w j ⟩ ∈ H A ⊗ H B \left\lvert \psi \right\rangle=\sum_{ij} a_{ij}\left\lvert v_i \right\rangle\otimes\left\lvert w_j \right\rangle\quad \in H_A\otimes H_B ∣ ψ ⟩ = ij ∑ a ij ∣ v i ⟩ ⊗ ∣ w j ⟩ ∈ H A ⊗ H B Another way to notationally write this is
∣ ψ ( A B ) ⟩ = ∑ i j a i j ∣ v i ⟩ A ⊗ ∣ w j ⟩ B \boxed{\left\lvert \psi^{(AB)} \right\rangle=\sum_{ij} a_{ij}\left\lvert v_i \right\rangle_A\otimes\left\lvert w_j \right\rangle_B} ψ ( A B ) ⟩ = ij ∑ a ij ∣ v i ⟩ A ⊗ ∣ w j ⟩ B The following is a product state. For some ∣ v ∗ ⟩ ∈ H A \left\lvert v^* \right\rangle\in H_A ∣ v ∗ ⟩ ∈ H A ∣ w ∗ ⟩ ∈ H B \left\lvert w^* \right\rangle\in H_B ∣ w ∗ ⟩ ∈ H B
∣ ψ ⟩ = ∣ v ∗ ⟩ ⊗ ∣ w ∗ ⟩ ∣ v ∗ ⟩ ∈ H A , ∣ w ∗ ⟩ ∈ H B \left\lvert \psi \right\rangle=\left\lvert v^* \right\rangle\otimes\left\lvert w^* \right\rangle\quad \left\lvert v^* \right\rangle\in H_A, \left\lvert w^* \right\rangle\in H_B ∣ ψ ⟩ = ∣ v ∗ ⟩ ⊗ ∣ w ∗ ⟩ ∣ v ∗ ⟩ ∈ H A , ∣ w ∗ ⟩ ∈ H B An entangled state is any state ∣ ψ ⟩ \left\lvert \psi \right\rangle ∣ ψ ⟩ Vector to individual subsystems.
Einstein famously called this "Spooky action at a distance."
Example 1
Let dim H A = dim H B = 2 \dim H_A = \dim H_B = 2 dim H A = dim H B = 2
∣ ψ 1 ⟩ = 1 2 ( ∣ 0 , 0 ⟩ + ∣ 0 , 1 ⟩ + ∣ 1 , 0 ⟩ + ∣ 1 , 1 ⟩ ) \left\lvert \psi_1 \right\rangle=\frac{1}{2}(\left\lvert 0,0 \right\rangle+\left\lvert 0,1 \right\rangle+\left\lvert 1,0 \right\rangle+\left\lvert 1,1 \right\rangle) ∣ ψ 1 ⟩ = 2 1 ( ∣ 0 , 0 ⟩ + ∣ 0 , 1 ⟩ + ∣ 1 , 0 ⟩ + ∣ 1 , 1 ⟩ ) Note these are being tensor producted with each other but notation wise, we don’t write ⊗ \otimes ⊗
= 1 2 ( ∣ 0 ⟩ ∣ 0 ⟩ + ∣ 0 ⟩ ∣ 1 ⟩ + ∣ 1 ⟩ ∣ 0 ⟩ + ∣ 1 ⟩ ∣ 1 ⟩ ) = \frac{1}{2}\big(|0\rangle |0\rangle + |0\rangle |1\rangle + |1\rangle |0\rangle + |1\rangle |1\rangle\big) \\ = 2 1 ( ∣0 ⟩ ∣0 ⟩ + ∣0 ⟩ ∣1 ⟩ + ∣1 ⟩ ∣0 ⟩ + ∣1 ⟩ ∣1 ⟩ ) = 1 2 ( ∣ 0 ⟩ ( ∣ 0 ⟩ + ∣ 1 ⟩ ) + ∣ 1 ⟩ ( ∣ 0 ⟩ + ∣ 1 ⟩ ) ) = \frac{1}{2}\big(|0\rangle(|0\rangle + |1\rangle) + |1\rangle(|0\rangle + |1\rangle)\big) = 2 1 ( ∣0 ⟩ ( ∣0 ⟩ + ∣1 ⟩) + ∣1 ⟩ ( ∣0 ⟩ + ∣1 ⟩) ) = 1 2 ( ∣ 0 ⟩ + ∣ 1 ⟩ ) ( ∣ 0 ⟩ + ∣ 1 ⟩ ) = \frac{1}{2}(|0\rangle + |1\rangle)(|0\rangle + |1\rangle) = 2 1 ( ∣0 ⟩ + ∣1 ⟩) ( ∣0 ⟩ + ∣1 ⟩) As ∣ 0 ⟩ + ∣ 1 ⟩ \left\lvert 0 \right\rangle+\left\lvert 1 \right\rangle ∣ 0 ⟩ + ∣ 1 ⟩ H A H_A H A H B H_B H B
∣ ψ ⟩ = ∣ v ∗ ⟩ ∣ w ∗ ⟩ ∣ v ∗ ⟩ ∈ H A , ∣ w ∗ ⟩ ∈ H B \left\lvert \psi \right\rangle=\left\lvert v^* \right\rangle\left\lvert w^* \right\rangle\quad \left\lvert v^* \right\rangle\in H_A, \left\lvert w^* \right\rangle\in H_B ∣ ψ ⟩ = ∣ v ∗ ⟩ ∣ w ∗ ⟩ ∣ v ∗ ⟩ ∈ H A , ∣ w ∗ ⟩ ∈ H B Example 2
Let
∣ ψ 2 ⟩ = 1 2 ( ∣ 0 , 0 ⟩ + ∣ 0 , 1 ⟩ + ∣ 1 , 0 ⟩ − ∣ 1 , 1 ⟩ ) \left\lvert \psi_2 \right\rangle=\frac{1}{2}(\left\lvert 0,0 \right\rangle+\left\lvert 0,1 \right\rangle+\left\lvert 1,0 \right\rangle-\left\lvert 1,1 \right\rangle) ∣ ψ 2 ⟩ = 2 1 ( ∣ 0 , 0 ⟩ + ∣ 0 , 1 ⟩ + ∣ 1 , 0 ⟩ − ∣ 1 , 1 ⟩ ) = 1 2 ( ∣ 0 ⟩ ∣ 0 ⟩ + ∣ 0 ⟩ ∣ 1 ⟩ + ∣ 1 ⟩ ∣ 0 ⟩ − ∣ 1 ⟩ ∣ 1 ⟩ ) = \frac{1}{2}\big(|0\rangle |0\rangle + |0\rangle |1\rangle + |1\rangle |0\rangle - |1\rangle |1\rangle\big) \\ = 2 1 ( ∣0 ⟩ ∣0 ⟩ + ∣0 ⟩ ∣1 ⟩ + ∣1 ⟩ ∣0 ⟩ − ∣1 ⟩ ∣1 ⟩ ) = 1 2 ( ∣ 0 ⟩ ( ∣ 0 ⟩ + ∣ 1 ⟩ ) + ∣ 1 ⟩ ( ∣ 0 ⟩ − ∣ 1 ⟩ ) ) = \frac{1}{2}\big(|0\rangle(|0\rangle + |1\rangle) + |1\rangle(|0\rangle - |1\rangle)\big) = 2 1 ( ∣0 ⟩ ( ∣0 ⟩ + ∣1 ⟩) + ∣1 ⟩ ( ∣0 ⟩ − ∣1 ⟩) ) = 1 2 ( ∣ 0 ⟩ ∣ + ⟩ + ∣ 1 ⟩ ∣ − ⟩ ) = \frac{1}{2}\big(|0\rangle\left\lvert + \right\rangle+\left\lvert 1 \right\rangle\left\lvert - \right\rangle\big) = 2 1 ( ∣0 ⟩ ∣ + ⟩ + ∣ 1 ⟩ ∣ − ⟩ ) This does not follow the form
∣ ψ ⟩ = ∣ v ∗ ⟩ ∣ w ∗ ⟩ ∣ v ∗ ⟩ ∈ H A , ∣ w ∗ ⟩ ∈ H B \left\lvert \psi \right\rangle=\left\lvert v^* \right\rangle\left\lvert w^* \right\rangle\quad \left\lvert v^* \right\rangle\in H_A, \left\lvert w^* \right\rangle\in H_B ∣ ψ ⟩ = ∣ v ∗ ⟩ ∣ w ∗ ⟩ ∣ v ∗ ⟩ ∈ H A , ∣ w ∗ ⟩ ∈ H B This is not written cleanly as a product of two product states in H A H_A H A H B H_B H B
Checking if a state is entangled
To check if a state is entangled, this is an easy way to do it.
Let this be the state that we want to check if it is entangled.
a 11 ∣ v 1 , w 1 ⟩ + a 12 ∣ v 1 , w 2 ⟩ + a 21 ∣ v 2 , w 1 ⟩ + a 22 ∣ v 2 , w 2 ⟩ a_{11}\left\lvert v_1,w_1 \right\rangle+a_{12}\left\lvert v_1, w_2 \right\rangle+a_{21}\left\lvert v_2, w_1 \right\rangle + a_{22}\left\lvert v_2, w_2 \right\rangle a 11 ∣ v 1 , w 1 ⟩ + a 12 ∣ v 1 , w 2 ⟩ + a 21 ∣ v 2 , w 1 ⟩ + a 22 ∣ v 2 , w 2 ⟩ Does it follow in this form
( a 1 ∣ v 1 ⟩ + a 2 ∣ v 2 ⟩ ) ⊗ ( b 1 ∣ w 1 ⟩ + b 2 ∣ w 2 ⟩ ) (a_1\left\lvert v_1 \right\rangle+a_2\left\lvert v_2 \right\rangle)\otimes(b_1\left\lvert w_1 \right\rangle+b_2\left\lvert w_2 \right\rangle) ( a 1 ∣ v 1 ⟩ + a 2 ∣ v 2 ⟩ ) ⊗ ( b 1 ∣ w 1 ⟩ + b 2 ∣ w 2 ⟩ ) To have a solution
a 11 = a 1 b 1 a_{11} = a_1b_1 a 11 = a 1 b 1 a 21 = a 2 b 1 a_{21} = a_2b_1 a 21 = a 2 b 1 a 12 = a 1 b 2 a_{12} = a_1b_2 a 12 = a 1 b 2 a 22 = a 2 b 2 a_{22} = a_2b_2 a 22 = a 2 b 2 a 11 a 22 − a 12 a 21 = a 1 b 1 a 2 b 2 − a 1 a 2 a 2 b 1 = 0 a_{11}a_{22}-a_{12}a_{21}=a_1b_1a_2b_2-a_1a_2a_2b_1=0 a 11 a 22 − a 12 a 21 = a 1 b 1 a 2 b 2 − a 1 a 2 a 2 b 1 = 0 Must all be true.
Hence
product state ⇔ det ( a 11 a 12 a 21 a 22 ) = 0 \text{product state} \Leftrightarrow \det\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}=0 product state ⇔ det ( a 11 a 21 a 12 a 22 ) = 0 Notation
Note that
∣ 0 ( A ) ⟩ ∣ 1 ( B ) ⟩ = ∣ 0 , 1 ⟩ = ∣ 01 ⟩ \ket{0^{(A)}}\ket{1^{(B)}}=\ket{0,1}=\ket{01} ∣ 0 ( A ) ⟩ ∣ 1 ( B ) ⟩ = ∣ 0 , 1 ⟩ = ∣ 01 ⟩ The first bit is implicitly in the basis of the first Qubit , the second bit is implicitly in the basis of the second qubit
Bar-Yossef, Jayram, and Kerenidis (2004)