No Communication Theorem
With access to state
∣ Ψ ( A B ) ⟩ \left\lvert \Psi^{(AB)} \right\rangle Ψ ( A B ) ⟩ which may or may not be entangled, Alice cannot convey information to Bob by
Choosing a basic measurement on A A A
Choosing a unitary evolution operator for A A A
But Alice’s choice of measurement basis cannot affect Bob’s measurement statistics, her measurement outcome CAN affect Bob’s state.
Let ∣ ψ a ⟩ \left\lvert \psi_a \right\rangle ∣ ψ a ⟩ Qubit conditioned on Alice’s outcome a a a
∣ ψ a ⟩ ≜ ⟨ a ( A ) ∣ Ψ ( A B ) ⟩ \left\lvert \psi_a \right\rangle \triangleq \left\langle a^{(A)}|\Psi^{(AB)} \right\rangle ∣ ψ a ⟩ ≜ ⟨ a ( A ) ∣ Ψ ( A B ) ⟩ Let a a a a ∈ { 0 , 1 } a\in\{0,1\} a ∈ { 0 , 1 } a ∈ { + , − } a\in\{+,-\} a ∈ { + , − }
We can decompose the earlier state into
∣ Ψ ( A B ) ⟩ = ∑ a ∣ a ⟩ ⊗ ∣ ψ a ⟩ \left\lvert \Psi^{(AB)} \right\rangle=\sum_a\left\lvert a \right\rangle\otimes \left\lvert \psi_a \right\rangle Ψ ( A B ) ⟩ = a ∑ ∣ a ⟩ ⊗ ∣ ψ a ⟩ What does this mean? This means the basis ∣ a ⟩ \left\lvert a \right\rangle ∣ a ⟩
P ( a , b ) = ∣ ⟨ b ∣ ψ a ⟩ ∣ 2 P(a,b)=|\left\langle b|\psi_a \right\rangle|^2 P ( a , b ) = ∣ ⟨ b ∣ ψ a ⟩ ∣ 2 Via Conditional Probability
P ( b ∣ a ) = 1 P ( a ) ∣ ⟨ b ∣ ψ a ⟩ ∣ 2 = ⟨ ψ a ∣ b ⟩ ⟨ b ∣ ψ a ⟩ ⟨ ψ a ∣ ψ a ⟩ P(b|a)=\frac{1}{P(a)}|\left\langle b|\psi_a \right\rangle|^2=\frac{\left\langle \psi_a|b \right\rangle\left\langle b|\psi_a \right\rangle}{\left\langle \psi_a|\psi_a \right\rangle} P ( b ∣ a ) = P ( a ) 1 ∣ ⟨ b ∣ ψ a ⟩ ∣ 2 = ⟨ ψ a ∣ ψ a ⟩ ⟨ ψ a ∣ b ⟩ ⟨ b ∣ ψ a ⟩ = ∣ ⟨ b ∣ ψ ^ a ⟩ ∣ 2 =|\left\langle b|\hat{\psi}_a \right\rangle|^2 = ∣ ⟨ b ∣ ψ ^ a ⟩ ∣ 2 Recall Magnitude
∣ ψ ^ a ⟩ ≜ 1 P ( a ) ∣ ψ a ⟩ = ∣ ψ a ⟩ ∣ ∣ ∣ ψ a ⟩ ∣ ∣ = ∣ ψ a ⟩ ⟨ ψ a ∣ ψ a ⟩ \left\lvert \hat{\psi}_a \right\rangle\triangleq\frac{1}{\sqrt{P(a)}}\left\lvert \psi_a \right\rangle=\frac{\left\lvert \psi_a \right\rangle}{||\left\lvert \psi_a \right\rangle||}=\frac{\left\lvert \psi_a \right\rangle}{\sqrt{\left\langle \psi_a|\psi_a \right\rangle}} ψ ^ a ⟩ ≜ P ( a ) 1 ∣ ψ a ⟩ = ∣∣ ∣ ψ a ⟩ ∣∣ ∣ ψ a ⟩ = ⟨ ψ a ∣ ψ a ⟩ ∣ ψ a ⟩ Note that generally, conditional state can be obtained via a partial inner product.
⟨ a ( A ) ∣ Ψ ( A B ) ⟩ ≜ ( ⟨ a ∣ ⊗ I ) ∣ Ψ ( A B ) ⟩ \boxed{\left\langle a^{(A)}|\Psi^{(AB)} \right\rangle\triangleq(\left\langle a \right\rvert\otimes I)\left\lvert \Psi^{(AB)} \right\rangle} ⟨ a ( A ) ∣ Ψ ( A B ) ⟩ ≜ ( ⟨ a ∣ ⊗ I ) Ψ ( A B ) ⟩ Note this yields a vector in H B H_B H B ∣ Ψ ( A B ) ⟩ \left\lvert \Psi^{(AB)} \right\rangle Ψ ( A B ) ⟩ H A ⊗ H B H_A \otimes H_B H A ⊗ H B
Example:
This is a Singlet state: the antisymmetric two-spin state
∣ Ψ − ( A B ) ⟩ = 1 2 ( ∣ 0 ( A ) ⟩ ⊗ ∣ 1 ( B ) ⟩ − ∣ 1 ( A ) ⟩ ⊗ ∣ 0 ( B ) ⟩ ) \left\lvert \Psi^{(AB)}_- \right\rangle=\frac{1}{\sqrt{2}}\left(\left\lvert {0}^{(A)} \right\rangle\otimes \left\lvert 1^{(B)} \right\rangle-\left\lvert 1^{(A)} \right\rangle\otimes \left\lvert {0}^{(B)} \right\rangle\right) Ψ − ( A B ) ⟩ = 2 1 ( 0 ( A ) ⟩ ⊗ 1 ( B ) ⟩ − 1 ( A ) ⟩ ⊗ 0 ( B ) ⟩ ) Suppose Alice measures in z basis
∣ Ψ − ( A B ) ⟩ = 1 2 ( ∣ 0 ( A ) ⟩ ⊗ ∣ 1 ( B ) ⟩ − ∣ 1 ( A ) ⟩ ⊗ ∣ 0 ( B ) ⟩ ) \left\lvert \Psi^{(AB)}_- \right\rangle=\frac{1}{\sqrt{2}}\left(\left\lvert {0}^{(A)} \right\rangle\otimes \left\lvert 1^{(B)} \right\rangle-\left\lvert 1^{(A)} \right\rangle\otimes \left\lvert {0}^{(B)} \right\rangle\right) Ψ − ( A B ) ⟩ = 2 1 ( 0 ( A ) ⟩ ⊗ 1 ( B ) ⟩ − 1 ( A ) ⟩ ⊗ 0 ( B ) ⟩ ) Outcome conditional state (unnormalized) a = 0 ∣ ψ 0 ( B ) ⟩ = ⟨ 0 ( A ) ∣ Φ − ( A B ) ⟩ = 1 2 ∣ 1 ( B ) ⟩ p ( a = 0 ) = 1 2 a = 1 ∣ ψ 1 ( B ) ⟩ = ⟨ 1 ( A ) ∣ Φ − ( A B ) ⟩ = − 1 2 ∣ 0 ( B ) ⟩ p ( a = 1 ) = 1 2 \begin{array}{c|c}
\text{Outcome} & \text{conditional state (unnormalized)} \\
\hline
a = 0 & \left\lvert \psi_0^{(B)} \right\rangle = \left\langle {0}^{(A)} | \Phi_-^{(AB)} \right\rangle = \frac{1}{\sqrt{2}} \left\lvert 1^{(B)} \right\rangle \qquad p(a=0) = \tfrac{1}{2} \\[1em]
a = 1 & \left\lvert \psi_1^{(B)} \right\rangle = \left\langle 1^{(A)} | \Phi_-^{(AB)} \right\rangle = -\frac{1}{\sqrt{2}} \left\lvert {0}^{(B)} \right\rangle \qquad p(a=1) = \tfrac{1}{2}
\end{array} Outcome a = 0 a = 1 conditional state (unnormalized) ψ 0 ( B ) ⟩ = ⟨ 0 ( A ) ∣ Φ − ( A B ) ⟩ = 2 1 1 ( B ) ⟩ p ( a = 0 ) = 2 1 ψ 1 ( B ) ⟩ = ⟨ 1 ( A ) ∣ Φ − ( A B ) ⟩ = − 2 1 0 ( B ) ⟩ p ( a = 1 ) = 2 1 We can see that Bob’s qubit is projected onto z eigenstate opposite to Alice’s outcome.
Let’s say that Alice measures in x basis.
Outcome Conditional state (unnormalized) a = + ∣ ψ + ( B ) ⟩ = ⟨ + ( A ) ∣ Φ − ( A B ) ⟩ = ⟨ 0 ( A ) ∣ + ⟨ 1 ( A ) ∣ 2 ⋅ 1 2 ( ∣ 0 ( A ) 1 ( B ) ⟩ − ∣ 1 ( A ) 0 ( B ) ⟩ ) = ∣ 1 ( B ) ⟩ 2 − ∣ 0 ( B ) ⟩ 2 = − ∣ − ( B ) ⟩ 2 p ( a = + ) = 1 2 a = − ∣ ψ − ( B ) ⟩ = ⟨ − ( A ) ∣ Φ − ( A B ) ⟩ = ∣ + ( B ) ⟩ 2 p ( a = − ) = 1 2 \begin{array}{c|l}
\text{Outcome} & \text{Conditional state (unnormalized)} \\
\hline
a = + & \begin{aligned}
\left\lvert \psi_+^{(B)} \right\rangle & = \left\langle +^{(A)} | \Phi_-^{(AB)} \right\rangle \\
& = \frac{\left\langle {0}^{(A)} \right\rvert + \left\langle 1^{(A)} \right\rvert}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \left( \left\lvert {0}^{(A)} 1^{(B)} \right\rangle - \left\lvert 1^{(A)} {0}^{(B)} \right\rangle \right) \\
& = \frac{\left\lvert 1^{(B)} \right\rangle}{2} - \frac{\left\lvert {0}^{(B)} \right\rangle}{2} = -\frac{\left\lvert -^{(B)} \right\rangle}{\sqrt{2}}
\end{aligned} \\
& \qquad p(a=+) = \tfrac{1}{2} \\[1em]
a = - & \left\lvert \psi_-^{(B)} \right\rangle = \left\langle -^{(A)} | \Phi_-^{(AB)} \right\rangle = \frac{\left\lvert +^{(B)} \right\rangle}{\sqrt{2}} \\
& \qquad p(a=-) = \tfrac{1}{2}
\end{array} Outcome a = + a = − Conditional state (unnormalized) ψ + ( B ) ⟩ = ⟨ + ( A ) ∣ Φ − ( A B ) ⟩ = 2 ⟨ 0 ( A ) + ⟨ 1 ( A ) ⋅ 2 1 ( 0 ( A ) 1 ( B ) ⟩ − 1 ( A ) 0 ( B ) ⟩ ) = 2 1 ( B ) ⟩ − 2 0 ( B ) ⟩ = − 2 − ( B ) ⟩ p ( a = + ) = 2 1 ψ − ( B ) ⟩ = ⟨ − ( A ) ∣ Φ − ( A B ) ⟩ = 2 ∣ + ( B ) ⟩ p ( a = − ) = 2 1 Bob’s qubit is projected to opposite basis state as ALice’s measurement
Somehow, Alice’s choice of measurement basis affects Bob’s state.
If Alice measures in z basis, Bob’s conditional state is ∣ 0 ⟩ \left\lvert 0 \right\rangle ∣ 0 ⟩ ∣ 1 ⟩ \left\lvert 1 \right\rangle ∣ 1 ⟩ probability . If Alice measures in x basis, Bob’s conditional state is ∣ + ⟩ \left\lvert + \right\rangle ∣ + ⟩ ∣ − ⟩ \left\lvert - \right\rangle ∣ − ⟩
ρ ^ B = 1 2 ∣ 0 ⟩ ⟨ 0 ∣ + 1 2 ∣ 1 ⟩ ⟨ 1 ∣ = 1 2 ∣ + ⟩ ⟨ + ∣ + 1 2 ∣ − ⟩ ⟨ − ∣ = 1 2 I \hat{\rho}_B = \tfrac{1}{2} \left\lvert 0 \right\rangle\left\langle 0 \right\rvert + \tfrac{1}{2} \left\lvert 1 \right\rangle\left\langle 1 \right\rvert = \tfrac{1}{2} \left\lvert + \right\rangle\left\langle + \right\rvert + \tfrac{1}{2} \left\lvert - \right\rangle\left\langle - \right\rvert = \frac{1}{2}I ρ ^ B = 2 1 ∣ 0 ⟩ ⟨ 0 ∣ + 2 1 ∣ 1 ⟩ ⟨ 1 ∣ = 2 1 ∣ + ⟩ ⟨ + ∣ + 2 1 ∣ − ⟩ ⟨ − ∣ = 2 1 I Observables on a Composite System