Notes - Quantum Information HT24, Quantum steering
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[[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 faster-than-light 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
\[\frac{|v_M\rangle}{\big|\big||v_M\rangle\big|\big|}\]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
\[\frac{|v_M\rangle}{\big|\big||v_M\rangle\big|\big|}\]
.
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
\[\frac{|\nu\rangle_{AB }\\,}{ || \\, |\nu\rangle_{AB} \\, || } = \frac{1}{\sqrt{1 + |\alpha|^2}\\,} \Big(|0\rangle_A \otimes (\beta |0\rangle_B + \alpha |1\rangle_B) + \alpha |1\rangle_A \otimes |0\rangle_B\Big)\]