No signalling and no cloning

Intuition

If measuring $A$ at once changes what can be predicted about $B$ , can $A$ send $B$ a message? No. Whatever $A$ does — measure along any axis, turn her spin, or nothing — $B$ ’s reduced state is unchanged, and everything $B$ can observe depends only on it. $A$ ’s choice changes how $B$ ’s ensemble is split, not the ensemble: after $A$ measures along $z$ , $B$ is $\uparrow$ or $\downarrow$ , half the time each; after $A$ measures along $x$ , he is $+x$ or $-x$ , half the time each; either way he holds $\tfrac12\cdot1$ . The correlations show only when the two lists of results are brought together by ordinary means. A second limit backs this up: an unknown quantum state cannot be copied. If it could, $B$ would copy his spin many times and learn which ensemble he held.

Matching sealed envelopes cannot carry a message: opening mine tells me about yours but changes nothing you see. Entanglement is stronger than envelopes, and just as silent.

Spin $B$ of a singlet, on a cut through its Bloch sphere. If $A$ measures along $z$ , $B$ is left at $\ket{\uparrow}$ or ; if along , at or — half the time each. Averaged over ’s unknown result, sits at the centre either way, so nothing measures reveals ’s choice.

Whatever is done to $A$ alone — a unitary $\hat{U}_{A}$ , or a measurement with projectors $\hat{P}_{k}$ whose result is not sent — leaves with

\sum_{k}\operatorname{tr}_{A}\bigl[(\hat{P}_{k}\otimes\hat{1})\hat{\rho}(\hat{P}_{k}\otimes\hat{1})\bigr]=\operatorname{tr}_{A}\hat{\rho}=\hat{\rho}_{B}

Properties

The identity holds because $\sum_{k}\hat{P}_{k}=\hat{1}$ and the trace over lets move around the cycle; for a unitary, does the same.

No cloning

\begin{aligned}&\hat{U}\ket{\psi}\ket{e}=\ket{\psi}\ket{\psi},\ \ \hat{U}\ket{\varphi}\ket{e}=\ket{\varphi}\ket{\varphi}\\ &\Longrightarrow\ \braket{\varphi|\psi}\in\{0,1\}\end{aligned}

A unitary keeps inner products. Compute the inner product of the two inputs, which is $\braket{\varphi|\psi}$ because the blank is normalised, and of the two outputs, which is its square. A number equal to its own square is 0 or 1: the two states are orthogonal or the same.

Proof steps

\hat{U}\ket{\psi}\ket{e}=\ket{\psi}\ket{\psi},\qquad \hat{U}\ket{\varphi}\ket{e}=\ket{\varphi}\ket{\varphi}

Suppose one device copies two states onto a blank $\ket{e}$ .

\bigl(\bra{\varphi}\bra{e}\bigr)\hat{U}^{\dagger}\hat{U}\bigl(\ket{\psi}\ket{e}\bigr)=\braket{\varphi|\psi}\braket{e|e}=\braket{\varphi|\psi}

$\hat{U}^{\dagger}\hat{U}=\hat{1}$ and .

\bigl(\bra{\varphi}\bra{\varphi}\bigr)\bigl(\ket{\psi}\ket{\psi}\bigr)=\braket{\varphi|\psi}^{2}

The same inner product, from the outputs.

\braket{\varphi|\psi}=\braket{\varphi|\psi}^{2}\ \Longrightarrow\ \braket{\varphi|\psi}\in\{0,1\}

Equate the two.

\begin{aligned}&\text{one device copies only states}\\ &\text{that are equal or orthogonal}\end{aligned}

So no device copies every state.

Applications

Practice

B Sees Nothing

Averaged over $A$ ’s results, which $B$ does not know, $B$ ’s reduced state is the same whatever $A$ did. So are all of $B$ ’s statistics.

Try it

$A$ measures her spin of a singlet along $x$ instead of $z$ . What changes in $B$ ’s statistics, before any comparison of results?