A simple counter-example to the no-communication theorem?

Question

Let's say Alice and Bob would like to communicate through entangled qubits. They have a machine that generates qubits in the state $$ | \psi \rangle = \alpha | 0 0 \rangle + \beta | 1 1 \rangle . $$ where Alice and Bob both know the values of $\alpha$ and $\beta$ from the start.

From each of these systems, one qubit is sent to Alice and the other to Bob. Now Alice performs some unitary transformation through the matrix $$ U=e^{i \varphi / 2}\left[\begin{array}{cc} e^{i \varphi_1} \cos \theta & e^{i \varphi_2} \sin \theta \\ -e^{-i \varphi_2} \sin \theta & e^{-i \varphi_1} \cos \theta \end{array}\right] , $$ on her qubit such that each entangled pair has the state $$ | \psi^\prime \rangle = e^{i \varphi / 2} \alpha \left( e^{i \varphi_1} \cos \theta | 0 0\rangle - e^{-i \varphi_2} \sin \theta | 0 1 \rangle \right) + e^{i \varphi / 2} \beta \left( e^{-i \varphi_1} \cos \theta | 1 1 \rangle + e^{i \varphi_2} \sin \theta | 1 0 \rangle \right) . $$

Now Bob measures the state of his qubit in each entangled pair and will be able to estimate the probability that he will get a $| 0 \rangle$ or a $| 1 \rangle$. If these probabilities differ from $| \alpha |^2$ and $| \beta |^2$, then he knows Alice has applied the unitary transformation. E.g., the probability now that Bob measures a $| 0 \rangle$ is not $|\alpha|^2$ but $$ |\alpha|^2 \cos^2 \theta + | \beta |^2 \sin^2 \theta . $$

In this way they can communicate through entanglement - Bob can detect if Alice ''Did something'' or ''did nothing''. Furthermore, Alice could tune the probabilities to any value she likes and Bob could read off an unlimited amount of information by figuring out its digits through repeated measurements.

As far as I understand, this contradicts the no-communication theorem of quantum information theory (https://en.wikipedia.org/wiki/No-communication_theorem). It also appears this would allow them to communicate when they are space-like separated and violate causality. What's the resolution?

Answer 1

As others have pointed out, there is a mistake in the derivation. So let us consider the situation from a more general point of view:

If the bipartite system is in the state $\psi$, then the probability for Bob to obtain the measurement outcome corresponding to some orthogonal projection $P$ is given by $$\mathcal P^B_\psi (P) := \mathrm{Tr}\, |\psi\rangle\langle \psi|\,\left( \mathbb I\otimes P\right) \quad ,$$

where $\mathbb I$ denotes the identity operator on $H\cong \mathbb C^2$ and the overall Hilbert space is $H\otimes H$. Your specific example follows by choosing $P=|0\rangle\langle 0|$. Now Alice manipulates her particle through a unitary operation$^\ddagger$, represented by $U$, before Bob performs a measurement, which changes the state as $|\psi\rangle \longrightarrow |\psi^\prime\rangle:= U\otimes \mathbb I\,|\psi\rangle$.

The probability for Bob to obtain the same outcome in this new state reads

$$\mathcal P^B_{\psi^\prime}(P) =\mathrm{Tr}\, |\psi^\prime\rangle\langle \psi^\prime|\, \left(\mathbb I\otimes P\right) \quad . $$

A straightforward calculation shows that the probability is not altered: \begin{align} \mathcal P^B_{\psi^\prime}(P) &=\mathrm{Tr}\, \left(U\otimes \mathbb I\right) |\psi\rangle\langle \psi| \left(U^\dagger\otimes \mathbb I\right) \, \left(\mathbb I\otimes P\right)\\ &=\mathrm{Tr} \, |\psi\rangle\langle \psi| \left(U^\dagger\otimes \mathbb I\right) \, \left(\mathbb I\otimes P\right) \, \left(U\otimes \mathbb I\right) \\ &=\mathrm{Tr}\, |\psi\rangle\langle\psi|\, \left(U^\dagger U\otimes P\right)\\ &=\mathcal P^B_\psi(P) \quad , \end{align}

which we can also write as

$$\mathrm{Tr}^{(B)}\, \rho_B(\psi)\, P = \mathrm{Tr}^{(B)}\, \rho_B(\psi^\prime)\, P \quad ,$$

where $\rho_B(\psi)$ denotes the reduced density matrix of Bob if the overall state is $\psi$. As shown here, this implies that $\rho_B(\psi) =\rho_B(\psi^\prime)$.

Thus, such local unitary operations (here performed by Alice) leave the reduced density matrix of the other party (here Bob) invariant. Bob's state is thus unaltered by the action of Alice and he hence cannot know whether or not she did something at all: All probabilities for the outcomes of Bob's measurements remain the same.

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$^\ddagger$ A more general operation is discussed in the Wikipedia article of the no-communcation theorem.

The above considerations can also be generalized to mixed states and to the case where both sub-Hilbert spaces are different.

Answer 2

The usual convention is that Alice is the first qubit, $|{\color{red}0}0⟩$ and Bob is the second qubit, $|0{\color{red}0}⟩$. In your calculation for $| \psi^\prime \rangle$ you have swapped these, which is likely the error in your calculation which is leading you to erroneous conclusions.

The state after applying your unitary $U$ on Alice's side should read \begin{align} | \psi^\prime \rangle & = e^{i \varphi / 2} \alpha \left( e^{i \varphi_1} \cos \theta | 0 0\rangle - e^{-i \varphi_2} \sin \theta | 1 0 \rangle \right) + e^{i \varphi / 2} \beta \left( e^{-i \varphi_1} \cos \theta | 1 1 \rangle + e^{i \varphi_2} \sin \theta | 0 1 \rangle \right) \\ & = e^{i \varphi / 2} \alpha \left( e^{i \varphi_1} \cos \theta | 0 \rangle - e^{-i \varphi_2} \sin \theta | 1 \rangle \right)| 0 \rangle + e^{i \varphi / 2} \beta \left( e^{-i \varphi_1} \cos \theta | 1 \rangle + e^{i \varphi_2} \sin \theta | 0 \rangle \right) | 1 \rangle \end{align} instead of the expression you gave (which swaps $|10⟩$ for $|01⟩$, i.e., you've effectively applied $U$ followed by a nonlocal SWAP operation).

Once you've corrected this, it's easy to see that the probabilities on Bob's side remain $|\alpha|^2$ and $|\beta|^2$ $-$ your calculation only showed that Alice's probabilities were changed by Alice's application of a unitary, which is not surprising at all.

The more general case of the no-communication theorem has been covered in good detail by Tobias Fünke's existing answer here.

Answer 3

Depending on the order of your tensor product of Alice and Bob's Hilbert spaces, you have either misapplied Alice's unitary or misapplied Bob's measurement.

You have both the unitary and the measurement acting on the second slot in the tensor product. But the unitary should be acting on Alice's slot, while the measurement should be acting on Bob's slot. Assuming that you're using the standard ordering that Alice's slot comes before Bob's in the tensor product, then you misapplied the unitary. If you properly have Alice's unitary and Bob's measurement acting on different slots in the tensor product, then you'll find that Alice's unitary does not affect the probabilities of Bob's measurement outcomes, so they can't use this channel to communicate.

The situation that you have described mathematically is that the same party (Bob, under the standard ordering convention) has both performed a unitary on his half of the system and then measured the impact of his own unitary. So there's no communication there.