# Notes - Quantum Information HT24, Quantum steering

> Source: https://ollybritton.com/notes/uni/part-a/ht24/quantum-information/notes/quantum-steering/ · Updated: 2025-05-11 · Tags: uni, notes

- [Course - Quantum Information HT24](https://ollybritton.com/notes/uni/part-a/ht24/quantum-information/)
	- 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 $|\alpha \rangle$
- Bob's system is in state $|\beta\rangle$
- Alice measures her system in the orthonormal basis $\{|\alpha_m\rangle, m = 0, \cdots, d_A - 1\}$
- Bob measures his system in the orthonormal basis $\{|\beta_n\rangle, n = 0, \cdots, d_B - 1\}$

Quickly prove that the joint probability $p_{AB}(m, n)$ where probabilities are calculated using the composite system is equal to $p_{A}(m)p_{B}(n)$, i.e. the outcomes are uncorrelated.::

$$
\begin{aligned}
p_{AB}(m, n) &= |(\langle \alpha_m | \otimes \langle\beta_n|)(|\alpha\rangle \otimes |\beta\rangle)|^2 \\
&= |\langle\alpha_m | \alpha\rangle \langle\beta_n | \beta\rangle|^2 \\
&= p_A(m)p_B(n)
\end{aligned}
$$

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 $|\Phi^+\rangle$
- 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 $|0\rangle$, then Bob will also find $|0\rangle$ and if Alice finds $|1\rangle$ then Bob will also find $|1\rangle$. 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 that they end up in the entangled state $|\Phi^+\rangle$
- 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 that they end up in the entangled state $|\Phi^+\rangle$
- 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 $\{|0\rangle, |1\rangle\}$
	- If she wants to send $1$, she measures in the Fourier basis $\{|+\rangle, |-\rangle\}$

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 $\{|0\rangle, |1\rangle\}$ or one of $\{|+\rangle, |-\rangle\}$), 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
$$
\{|\beta_n \rangle_B \mid n = 0, \cdots, d_{B} - 1\}
$$
What is the probability of Alice measuring $|\alpha_M\rangle$, 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 $\big|\big||v_M\rangle\big|\big|^2$ 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
$$
\{|\beta_n \rangle_B \mid n = 0, \cdots, d_{B} - 1\}
$$
Define
$$
|v_
M \rangle := ({}_
A \langle \alpha_
M | \otimes I_
B) | \chi \rangle_
{AB}
$$
Quickly prove that the probability that Alice measures $|\alpha_M\rangle$ is given by $\big|\big||v_M\rangle\big|\big|^2$ 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 $|\alpha_M \rangle$:
$$
\begin{aligned}
p_
A(M) &= \sum^{d_
B - 1}_
{n = 0} p_
{AB}(M, n) \\
&= \sum^{d_
B - 1}_
{n = 0} \left| (\langle \alpha_
M| \otimes \langle \beta_
n |) \sum_
{m', n'} c_
{m'n'} |m'\rangle \otimes |n'\rangle \right|^2 \\
&= \sum^{d_
B - 1}_
{n = 0} \left| \sum_
{m', n'} c_
{m'n'} \langle \alpha_
M | \alpha_
{m'}\rangle\langle \beta_
n | \beta_
{n'}\rangle \right|^2 \\
&= \sum^{d_
B - 1}_
{n = 0} |c_
{Mn}|^2
\end{aligned}
$$

Which is $\big|\big| |v_M \rangle \big|\big|^2$ as above. To show that Bob's system is in state $|\varphi\rangle := \frac{|v_M\rangle}{\big|\big||v_M\rangle\big|\big|}$, we want to find some state such that for every possible measurement against any $|\beta_i\rangle$, 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

$$
\begin{aligned}
p(\text{Bob measures }|\beta_i\rangle \mid \text{Alice measured }|\alpha_M\rangle) &= \frac{p(\text{Bob measures }\beta_i, \text{Alice measured } \alpha_M)}{p(\text{Alice measured } |\alpha_M\rangle)} \\
&= \frac{|(\langle\alpha_M | \otimes \langle \beta_i |) |\chi\rangle_{AB}|^2}{p_A(M)} \\
&= \left|\langle \beta_i | \frac{(\langle \alpha_M | \otimes I_B)|\chi\rangle_{AB}
}{\sqrt{p_A(M)} }\right|^2 \\
&= |\langle\beta_i | \varphi\rangle|^2
\end{aligned}
$$

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

$$
\frac{|\nu\rangle_{AB }\,}{ || \, |\nu\rangle_{AB} \, || } = \frac{1}{\sqrt{1 + |\alpha|^2}\,} \Big(|0\rangle_A \otimes (\beta |0\rangle_B + \alpha |1\rangle_B) + \alpha |1\rangle_A \otimes |0\rangle_B\Big)
$$

### Proofs

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