Quantum Information HT24, Quantum steering

Created: January 16, 2024 | Updated: May 11, 2025 | About these notes | View in context | Study these flashcards

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.
\[\begin{aligned} p _ {AB}(m, n) &= \vert (\langle \alpha _ m \vert \otimes \langle\beta _ n \vert )( \vert \alpha\rangle \otimes \vert \beta\rangle) \vert ^2 \\\\ &= \vert \langle\alpha _ m \vert \alpha\rangle \langle\beta _ n \vert \beta\rangle \vert ^2 \\\\ &= p _ A(m)p _ B(n) \end{aligned}\]

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 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

\[\vert \chi\rangle _ {AB} = \sum _ {m, n} c _ {mn} \vert \alpha _ m\rangle \otimes \vert \beta _ n\rangle\]

where Alice uses the basis

\[\\{ \vert \alpha _ m \rangle _ A \mid m = 0, \cdots, d _ {A} - 1\\}\]

and Bob uses the basis

\[\\{ \vert \beta _ n \rangle _ B \mid n = 0, \cdots, d _ {D} - 1\\}\]

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

\[\vert v _ M \rangle := ({} _ A \langle \alpha _ m \vert \otimes I _ B) \vert \chi \rangle _ {AB}\]

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

\[\frac{ \vert v _ M\rangle}{\big \vert \big \vert \vert v _ M\rangle\big \vert \big \vert }\]

Suppose Alice and Bob share a general quantum state given by

\[\vert \chi\rangle _ {AB} = \sum _ {m, n} c _ {mn} \vert \alpha _ m\rangle \otimes \vert \beta _ n\rangle\]

where Alice uses the basis

\[\\{ \vert \alpha _ m \rangle _ A \mid m = 0, \cdots, d _ {A} - 1\\}\]

and Bob uses the basis

\[\\{ \vert \beta _ n \rangle _ B \mid n = 0, \cdots, d _ {D} - 1\\}\]

Define

\[\vert v _ M \rangle := ({} _ A \langle \alpha _ M \vert \otimes I _ B) \vert \chi \rangle _ {AB}\]

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

\[\frac{ \vert v _ M\rangle}{\big \vert \big \vert \vert v _ M\rangle\big \vert \big \vert }\]

.

First, note that

\[\begin{aligned} \vert v _ M \rangle &:= ({} _ A \langle \alpha _ M \vert \otimes I _ B) \vert \chi \rangle _ {AB} \\\\ &= \sum^{d _ B - 1} _ {n = 0} c _ {Mn} \vert \beta _ n\rangle \end{aligned}\]

so then

\[\big \vert \big \vert \vert v _ M \rangle \big \vert \big \vert ^2 = \sum^{d _ B - 1} _ {n = 0} \vert c _ {Mn} \vert ^2\]

Now calculating the probability of measuring $ \vert \alpha _ M \rangle$:

\[\begin{aligned} p _ A(M) &= \sum^{d _ B - 1} _ {n = 0} p _ {AB}(M, n) \\\\ &= \sum^{d _ B - 1} _ {n = 0} \left \vert (\langle \alpha _ M \vert \otimes \langle \beta _ n \vert ) \sum _ {m', n'} c _ {m'n'} \vert m'\rangle \otimes \vert n'\rangle \right \vert ^2 \\\\ &= \sum^{d _ B - 1} _ {n = 0} \left \vert \sum _ {m', n'} c _ {m'n'} \langle \alpha _ M \vert \alpha _ {m'}\rangle\langle \beta _ n \vert \beta _ {n'}\rangle \right \vert ^2 \\\\ &= \sum^{d _ B - 1} _ {n = 0} \vert c _ {Mn} \vert ^2 \end{aligned}\]

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

\[\vert \langle \beta _ i \vert \varphi\rangle \vert ^2 = p(\text{Bob measures } \vert \beta _ i\rangle \mid \text{Alice measured } \vert \alpha _ M\rangle)\]

Expanding out, we have

\[\begin{aligned} p(\text{Bob measures } \vert \beta _ i\rangle \mid \text{Alice measured } \vert \alpha _ M\rangle) &= \frac{p(\text{Bob measures }\beta _ i, \text{Alice measured } \alpha _ 0)}{p(\text{Alice measured } \vert \alpha _ M\rangle)} \\\\ &= \frac{ \vert (\langle\alpha _ 0 \vert \otimes \langle \beta _ i \vert ) \vert \chi\rangle _ {AB} \vert ^2}{p _ A(M)} \\\\ &= \left \vert \langle \beta _ i \vert \frac{(\langle \alpha _ 0 \vert \otimes I _ B) \vert \chi _ {AB\rangle} }{\sqrt{p _ A(M)} }\right \vert ^2 \\\\ &= \vert \langle\beta _ i \vert \varphi\rangle \vert ^2 \end{aligned}\]

As required.

Suppose we have the state

\[\vert W\rangle = \frac{1}{\sqrt 3} \left( \vert 0\rangle _ A \otimes \vert 0\rangle _ B \otimes \vert 1\rangle _ C + \vert 0\rangle _ A \otimes \vert 1\rangle _ B \otimes \vert 0\rangle _ C + \vert 1\rangle _ A \otimes \vert 0\rangle _ B \otimes \vert 0\rangle _ C \right)\]

Quickly show that after performing a basic measurement corresponding to outcome

\[{} _ C\langle \phi \vert = (\alpha \quad \beta)\]

The updated state on $AB$ is given by

\[\frac{1}{\sqrt{1 + \vert \alpha \vert ^2}\\,} \Big( \vert 0\rangle _ A \otimes (\beta \vert 0\rangle _ B + \alpha \vert 1\rangle _ B) + \alpha \vert 1\rangle _ A \otimes \vert 0\rangle _ B\Big)\]

This is an application of quantum steering. The updated state is given by

\[\frac{ \vert \nu\rangle _ {AB }\\,}{ \vert \vert \\, \vert \nu\rangle _ {AB} \\, \vert \vert }\]

where

\[\vert \nu\rangle _ {AB} = (I _ A \otimes I _ B \otimes {} _ C \langle \phi \vert ) \vert W\rangle\]

After some expanding, this becomes

\[\vert \nu\rangle _ {AB} = \frac{1}{\sqrt 3} \Big( \vert 0\rangle _ A \otimes (\beta \vert 0\rangle _ B + \alpha \vert 1\rangle _ B) + \alpha \vert 1\rangle _ A \otimes \vert 0\rangle _ B\Big)\]

And then

\[\begin{aligned} \vert \vert \\, \vert \nu\rangle _ {AB}\\, \vert \vert &= \sqrt{\frac{ \vert \beta \vert ^2 + \vert \alpha \vert ^2 + \vert \alpha \vert ^2}{3}\\,} \\\\ &= \sqrt{\frac{1 + \vert \alpha \vert ^2}{3}\\,} \end{aligned}\]

Hence

\[\frac{ \vert \nu\rangle _ {AB }\\,}{ \vert \vert \\, \vert \nu\rangle _ {AB} \\, \vert \vert } = \frac{1}{\sqrt{1 + \vert \alpha \vert ^2}\\,} \Big( \vert 0\rangle _ A \otimes (\beta \vert 0\rangle _ B + \alpha \vert 1\rangle _ B) + \alpha \vert 1\rangle _ A \otimes \vert 0\rangle _ B\Big)\]

Proofs



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