Notes  Quantum Information HT24, Quantum steering

[[Course  Quantum Information HT24]]^{U}
 The lecture notes contain a proof of the general rule for quantum steering in the appendix of Chapter 18
Flashcards
Suppose:
 $A$, $B$ are quantum systems, looked after by Alice and Bob respectively
 Alice’s system is in state $ \vert \alpha \rangle$
 Bob’s system is in state $ \vert \beta\rangle$
 Alice measures her system in the orthonormal basis $\{ \vert \alpha _ m\rangle, m = 0, \cdots, d _ A  1\}$
 Bob measures his system in the orthonormal basis $\{ \vert \beta _ n\rangle, n = 0, \cdots, d _ B  1\}$
Quickly prove that the joint probabilty $p _ {AB}(m, n)$ where probabilities are calculated using the composite system is equal to $p _ {A}(m)p _ {B}(n)$, i.e. the outcomes are uncorrelated.
Suppose:
 $A$, $B$ are quantum systems, looked after by Alice and Bob respectively
 Charlie lets system $A$ and $B$ interact, so that they end up in the entangled state $ \vert \Phi^+\rangle$
 Charlie sends qubit $A$ to Alice and qubit $B$ to Bob
 Alice measures her qubit in the computational basis
 Bob measures his qubit in the computational basis
Given that it can be shown
 $p _ {AB}(0, 0) = \frac 1 2$
 $p _ {AB}(0, 1) = 0$
 $p _ {AB}(1, 0) = 0$
 $p _ {AB}(1, 1) = \frac 1 2$
Explain the problem of “spooky action at a distance”.
Assume Alice measures first. After the measurement, if Alice finds $ \vert 0\rangle$, then Bob will also find $ \vert 0\rangle$ and if Alice finds $ \vert 1\rangle$ then Bob will also find $ \vert 1\rangle$. Hence, regardless of distance, Alice’s measurement has an instantaneous effect on Bob’s system.
Suppose:
 $A$, $B$ are quantum systems, looked after by Alice and Bob respectively
 Charlie lets system $A$ and $B$ interact, so so that they end up in the entangled state $ \vert \Phi^+\rangle$
 Charlie sends qubit $A$ to Alice and qubit $B$ to Bob
 Alice makes a measurement
What is the phenomenon of “quantum steering”?
Alice’s choice of measurement determines the types of states that Bob’s system can jump into.
Suppose:
 $A$, $B$ are quantum systems, looked after by Alice and Bob respectively
 Charlie lets system $A$ and $B$ interact, so so that they end up in the entangled state $ \vert \Phi^+\rangle$
 Charlie sends qubit $A$ to Alice and qubit $B$ to Bob
 Alice wants to send a bit to Bob.
 If she wants to send $0$, she measures in the computational basis $\{ \vert 0\rangle, \vert 1\rangle\}$
 If she wants to send $1$, she measures in the Fourier basis $\{ \vert +\rangle, \vert \rangle\}$
What assumption, that turns out to be wrong, would then allow for fasterthanlight communication, and why?
 If she wants to send $0$, she measures in the computational basis $\{ \vert 0\rangle, \vert 1\rangle\}$
 If she wants to send $1$, she measures in the Fourier basis $\{ \vert +\rangle, \vert \rangle\}$
If Bob can determine which basis his state is in (i.e. is it one of $\{ \vert 0\rangle, \vert 1\rangle\}$ or one of $\{ \vert +\rangle, \vert \rangle\}$), then this allows for faster than light communication. This is because Alice’s measurement makes Bob’s system jump.
Suppose Alice and Bob share a general quantum state given by
\[\chi\rangle_{AB} = \sum_{m, n} c_{mn} \alpha_m\rangle \otimes \beta_n\rangle\]
where Alice uses the basis
\[\\{\alpha_m \rangle_A \mid m = 0, \cdots, d_{A}  1\\}\]
and Bob uses the basis
\[\\{\beta_n \rangle_B \mid n = 0, \cdots, d_{D}  1\\}\]
What is the probability of Alice measuring $ \vert \alpha _ M\rangle$, and if this is the resulting measurement, what state is Bob’s system now in?
Define
\[v_M \rangle := ({}_A \langle \alpha_m  \otimes I_B)  \chi \rangle_{AB}\]Then the probability that Alice measures $\alpha _ M$ is given by $\big \vert \big \vert \vert v _ M\rangle\big \vert \big \vert ^2$ and Bob’s system is now in state
\[\frac{v_M\rangle}{\big\bigv_M\rangle\big\big}\]Suppose Alice and Bob share a general quantum state given by
\[\chi\rangle_{AB} = \sum_{m, n} c_{mn} \alpha_m\rangle \otimes \beta_n\rangle\]
where Alice uses the basis
\[\\{\alpha_m \rangle_A \mid m = 0, \cdots, d_{A}  1\\}\]
and Bob uses the basis
\[\\{\beta_n \rangle_B \mid n = 0, \cdots, d_{D}  1\\}\]
Define
\[v_
M \rangle := ({}_
A \langle \alpha_
M  \otimes I_
B)  \chi \rangle_
{AB}\]
Quickly prove that the probability that Alice measures $ \vert \alpha _ M\rangle$ is given by $\big \vert \big \vert \vert v _ M\rangle\big \vert \big \vert ^2$ and if this is the resulting measurement then Bob’s system is now in state
\[\frac{v_M\rangle}{\big\bigv_M\rangle\big\big}\]
.
First, note that
\[\begin{aligned} v_ M \rangle &:= ({}_ A \langle \alpha_ M  \otimes I_ B)  \chi \rangle_ {AB} \\\\ &= \sum^{d_ B  1}_ {n = 0} c_ {Mn} \beta_ n\rangle \end{aligned}\]so then
\[\big\big v_ M \rangle \big\big^2 = \sum^{d_ B  1}_ {n = 0} c_ {Mn}^2\]Now calculating the probability of measuring $ \vert \alpha _ M \rangle$:
\[\begin{aligned} p_ A(M) &= \sum^{d_ B  1}_ {n = 0} p_ {AB}(M, n) \\\\ &= \sum^{d_ B  1}_ {n = 0} \left (\langle \alpha_ M \otimes \langle \beta_ n ) \sum_ {m', n'} c_ {m'n'} m'\rangle \otimes n'\rangle \right^2 \\\\ &= \sum^{d_ B  1}_ {n = 0} \left \sum_ {m', n'} c_ {m'n'} \langle \alpha_ M  \alpha_ {m'}\rangle\langle \beta_ n  \beta_ {n'}\rangle \right^2 \\\\ &= \sum^{d_ B  1}_ {n = 0} c_ {Mn}^2 \end{aligned}\]Which is $\big \vert \big \vert \vert v _ M \rangle \big \vert \big \vert ^2$ as above. To show that Bob’s system is in state $ \vert \varphi\rangle := \frac{ \vert v _ M\rangle}{\big \vert \big \vert \vert v _ M\rangle\big \vert \big \vert }$, we want to find some state such that for every possible measurement against any $ \vert \beta _ i\rangle$, we have
\[\langle \beta_i  \varphi\rangle^2 = p(\text{Bob measures }\beta_i\rangle \mid \text{Alice measured } \alpha_M\rangle)\]Expanding out, we have
\[\begin{aligned} p(\text{Bob measures }\beta_i\rangle \mid \text{Alice measured }\alpha_M\rangle) &= \frac{p(\text{Bob measures }\beta_i, \text{Alice measured } \alpha_0)}{p(\text{Alice measured } \alpha_M\rangle)} \\\\ &= \frac{(\langle\alpha_0  \otimes \langle \beta_i ) \chi\rangle_{AB}^2}{p_A(M)} \\\\ &= \left\langle \beta_i  \frac{(\langle \alpha_0  \otimes I_B)\chi_{AB\rangle} }{\sqrt{p_A(M)}\\,}\right^2 \\\\ &= \langle\beta_i  \varphi\rangle^2 \end{aligned}\]As required.
Suppose we have the state
\[W\rangle = \frac{1}{\sqrt 3} \left(0\rangle_A \otimes 0\rangle_B \otimes 1\rangle_C + 0\rangle_A \otimes 1\rangle_B \otimes 0\rangle_C + 1\rangle_A \otimes 0\rangle_B \otimes 0\rangle_C \right)\]
Quickly show that after performing a basic measurement corresponding to outcome
\[{}_C\langle \phi  = (\alpha \quad \beta)\]
The updated state on $AB$ is given by
\[\frac{1}{\sqrt{1 + \alpha^2}\\,} \Big(0\rangle_A \otimes (\beta 0\rangle_B + \alpha 1\rangle_B) + \alpha 1\rangle_A \otimes 0\rangle_B\Big)\]
This is an application of quantum steering. The updated state is given by
\[\frac{\nu\rangle_{AB }\\,}{  \\, \nu\rangle_{AB} \\,  }\]where
\[\nu\rangle_{AB} = (I_A \otimes I_B \otimes {}_C \langle \phi) W\rangle\]After some expanding, this becomes
\[\nu\rangle_{AB} = \frac{1}{\sqrt 3} \Big(0\rangle_A \otimes (\beta 0\rangle_B + \alpha 1\rangle_B) + \alpha 1\rangle_A \otimes 0\rangle_B\Big)\]And then
\[\begin{aligned} \\,\nu\rangle_{AB}\\, &= \sqrt{\frac{\beta^2 + \alpha^2 + \alpha^2}{3}\\,} \\\\ &= \sqrt{\frac{1 + \alpha^2}{3}\\,} \end{aligned}\]Hence
\[\frac{\nu\rangle_{AB }\\,}{  \\, \nu\rangle_{AB} \\,  } = \frac{1}{\sqrt{1 + \alpha^2}\\,} \Big(0\rangle_A \otimes (\beta 0\rangle_B + \alpha 1\rangle_B) + \alpha 1\rangle_A \otimes 0\rangle_B\Big)\]