Notes - Quantum Information HT24, No-signalling theorem
Flashcards
State the no-signalling theorem, first informally and then formally.
Sharing an entangled state is not enough for Alice and Bob to communicate.
Suppose:
- Alice wants to communicate value of a bit $x$ to Bob, using an entangled state of system $AB$ as the only resource
- To communicate $x = 0$, Alice measures $A$ in ONB $A _ 0 = \{\alpha^0 _ m, m = 0, \cdots, d _ A - 1\}$
- To communicate $x = 1$, Alice measures $A$ in ONB $A _ 1 = \{\alpha^1 _ m, m = 0, \cdots, d _ A - 1\}$
- Bob measures the system in the ONB $B = \{\beta _ m, m = 0, \cdots, d _ B - 1\}$
Then:
- The probability distribution of his outcomes does not depend on $x$.
Quickly prove the no-signalling theorem, i.e. that if:
- Alice wants to communicate value of a bit $x$ to Bob, using an entangled state of system $AB$ as the only resource
- To communicate $x = 0$, Alice measures $A$ in ONB $A _ 0 = \{\alpha^0 _ m, m = 0, \cdots, d _ A - 1\}$
- To communicate $x = 1$, Alice measures $A$ in ONB $A _ 1 = \{\alpha^1 _ m, m = 0, \cdots, d _ A - 1\}$
- Bob measures the system in the ONB $B = \{\beta _ m, m = 0, \cdots, d _ B - 1\}$
then:
- The probability distribution of his outcomes does not depend on $x$.
- To communicate $x = 0$, Alice measures $A$ in ONB $A _ 0 = \{\alpha^0 _ m, m = 0, \cdots, d _ A - 1\}$
- To communicate $x = 1$, Alice measures $A$ in ONB $A _ 1 = \{\alpha^1 _ m, m = 0, \cdots, d _ A - 1\}$
Let $ \vert \Psi\rangle$ be the entangled state of the composite system.
\[\begin{aligned} p_{AB}(m,n) &= \bigg\vert \langle \alpha_m | \otimes \langle\beta_n ||\Psi \rangle \bigg\vert^2 \\\\ &= \bigg[ \langle \Psi| |\alpha_m \rangle\otimes |\beta_n\rangle \bigg]\bigg[\langle \alpha_m | \otimes \langle \beta_n ||\Psi\rangle\bigg] \\\\ &= \langle\Psi| \bigg[ |\alpha_m \rangle\otimes |\beta_n\rangle \bigg]\bigg[\langle \alpha_m | \otimes \langle \beta_n |\bigg] | \Psi\rangle \\\\ &= \langle\Psi| \bigg( |\alpha_m \rangle \langle \alpha_m | \otimes |\beta_n \rangle \langle \beta_n | \bigg) | \Psi\rangle \end{aligned}\]Then
\[\begin{aligned} p_ B(n) &= \sum^{d_ A - 1}_ {m = 0} p_ {AB}(m, n) \\\\ &= \sum^{d_ A - 1}_ {m = 0} \langle\Psi| \bigg( |\alpha_m \rangle \langle \alpha_m | \otimes |\beta_n \rangle \langle \beta_n | \bigg) | \Psi\rangle \\\\ &= \langle\Psi|\Bigg(\sum^{d_ A-1}_ {m = 0} |\alpha_m \rangle \langle \alpha_m | \otimes |\beta_n \rangle \langle \beta_n | \Bigg) | \Psi\rangle \\\\ &=\langle\Psi|\Bigg(\bigg(\sum^{d_ A-1}_ {m = 0} |\alpha_m \rangle \langle \alpha_m |\bigg) \otimes |\beta_n \rangle \langle \beta_n | \Bigg) | \Psi\rangle \\\\ &= \langle \Psi\bigg| I \otimes |\beta_n \rangle \langle \beta_n |\bigg| \Psi\rangle \end{aligned}\]