For example, if an adversary were to get hold of one half of an entangled 2-qubit quantum state, $|\psi \rangle$, travelling along a channel, would they be able to entangle one of their own qubits with that state $|\psi \rangle$, and then re-insert $|\psi \rangle$ back into the channel unaffected? (even though they do not know the value of $|\psi \rangle$)

To further elaborate: A 'sender' $S$, prepare two qubits (denoted as $|\psi_1 \rangle$ and $|\psi_2 \rangle$) entangled with each other as the state $|\psi \rangle$. $S$ sends one of the qubits, $|\psi_2 \rangle$, to a 'receiver' $R$ along some channel. While the qubit, with state $|\psi_2 \rangle$, is travelling through the channel on its way to reach $R$, an 'adversary', call them $A$, entangles one of their own qubits with the qubit $S$ sent, without affecting its state $|\psi_2 \rangle$. Thus, resulting in $3$ qubits all entangled with each other, and the $2$ original qubit states $S$ prepared are maintained. Is this scenario possible??

edit: I didn't frame my question in the way I meant to. It is obvious that the state will change if entangling a third qubit with it: it is now a new state of 3 qubits instead of 2. My question I was really trying to ask was: can an 'adversary' make a third qubit be entangled with an existing 2 qubit entangled pair, in a way such that the 'sender' and 'receiver' are unaware of it. My answer down below explains how this is possible.

  • 1
    $\begingroup$ I don't quite understand the scenario you are thinking about. You want a channel that given $|\psi\rangle$ gives an output bipartite entangled state. But if that's the case, the state $|\psi\rangle$ is not unaffected. For it to be unaffected you'd need something like $\mathcal E(|\psi\rangle\!\langle\psi|)=|\psi\rangle\!\langle\psi|\otimes\rho_\psi$ for some $\rho_\psi$... but then the adversary's state wouldn't be entangled with $|\psi\rangle$ $\endgroup$
    – glS
    Apr 29 at 6:39
  • $\begingroup$ see edits I made $\endgroup$ Apr 30 at 16:08
  • $\begingroup$ well, again, "$A$ entangles one of their own qubit with the one $S$ sent" is in direct contradiction with "without affecting its state". Entangling anything with $|\psi\rangle$ means to change $|\psi\rangle$. You see it e.g. because any channel $\Phi$ entangling some $\rho$ in $H$ with something else, is by definition such that $\Phi(\rho)$ is some entangled state in some $H\otimes H_A$. Thus tracing out the ancillary degrees of freedom $H_A$ from $\Phi(\rho)$ gives a state that is different from $\rho$. $\endgroup$
    – glS
    Apr 30 at 17:03
  • $\begingroup$ you can entangle with a qubit without affecting it. see my comment: taking the state $\left|\Phi^{+}\right\rangle \bigotimes \left|\Phi^{+}\right\rangle$ can give an entangled state with 4 qubits, by apply $I \bigotimes BCNOT \bigotimes I$, where only one out of the 4 qubits has its value modified. $\endgroup$ Apr 30 at 18:07
  • $\begingroup$ $BCNOT$ is like $CNOT$ except it applies the control to the first bit, instead of the second bit. $BCNOT = \left(\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right)$ $\endgroup$ Apr 30 at 18:08

No. The whole point of an entangled state is that it cannot be written in the form $|\psi\rangle|\phi\rangle$. So, if you replace $|\psi\rangle$ with an entangled state, it is no longer in the state $|\psi\rangle$.

In response to the updated question, the answer is still "no". There will always be a test that the sender and receiver could perform that would detect the difference between the state they started from and the state that has become entangled with $A$. Most of the time, if the sender and receiver are performing local tests, the success of the test would only be probabilistic. The adversary might get away with it on a single occasion. But if the protocol is to be repeated many time, the probability of the adversary getting away with it is vanishingly small. For example, if the sender and receiver share a Bell pair $$ (|00\rangle+|11\rangle)/\sqrt{2}, $$ there's a variety of things they can do to test it. For example, they could perform a CHSH test. Or, they could both perform a measurement $\cos\theta X+\sin\theta Z$ (both parties should get identical outcomes, so long as they use the same value of $\theta$).

  • $\begingroup$ see my answer, it may clarify things. $\endgroup$ Apr 30 at 18:43
  • $\begingroup$ what is a CHSH test? did you see the answer I made down below? I think it is possible for bell pairs. $\endgroup$ May 6 at 16:19
  • $\begingroup$ @QuantumGuy123 Yes I saw. No it's not right. For example, if sender and receiver measure in the X basis, with the Bell state they always get the same answer. With two qubits from a GHZ state, they do not always get the same answer. $\endgroup$
    – DaftWullie
    May 6 at 16:55
  • $\begingroup$ What a CHSH tests is is a big topic. You want to go and read about it properly. $\endgroup$
    – DaftWullie
    May 6 at 16:55

No. This is related to the no-cloning theorem. At least, I think it is.

If this were possible, quantum key distribution wouldn't be secure.

One definite issue with the possibility is that it violates monogamy of entanglement: if a qubit is perfectly entangled with another, then it cannot be entangled at all with anything else. So if the quantum state in question is half of a Bell pair, then it is not possible to entangle it with anything else without affecting it.

Therefore it is impossible to do this in general.

  • $\begingroup$ "This is related to the no-cloning theorem." can you elaborate? $\endgroup$ Apr 28 at 22:04
  • $\begingroup$ Is this really correct? I would say that entangling the transmitted qubit with another qubit basically ruins its state. For the receiver who gets the qubit it is no longer in a pure state, but instead is described by a density matrix. What does this have to do with the no-cloning theorem? $\endgroup$ Apr 29 at 6:21
  • $\begingroup$ It is impossible to do of some very specific example (Bell pairs) therefore it is impossible to do in general? Can you add some more details to back up this claim? $\endgroup$
    – Rammus
    Apr 29 at 12:33
  • $\begingroup$ Being able to do it in general means being able to do it no matter the state you have. Therefore if you cannot do it in one specific case it is impossible to do it in general. $\endgroup$ Apr 29 at 14:52
  • $\begingroup$ I see, I thought your claim was that your example implies that there are no instances in which this is possible. $\endgroup$
    – Rammus
    Apr 29 at 16:28

Some fundamentals of quantum mechanics

Consider a composite system $S$, with subsystems $S_1$ and $S_2$. Each subsystem has a description in terms of a Hilbert space, which in quantum computing is just some $N$-dimensional vector space. Denote the space associated with each subsystem as $\mathcal{H}_1$ and $\mathcal{H}_2$, respectively.

Each Hilbert space has a basis that spans all possible system states. This can be represented by vectors in Dirac notation. For instance, a base for the Hilbert space for a qubit is $$\{|0\rangle,|1\rangle\} = \{\begin{bmatrix}1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1\end{bmatrix}\}$$

The composite system has a Hilbert space which is spanned by the tensor product of the basis of each of its subsystems. If $\mathcal{H}$ is the composite Hilbert space, then $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$.

The Hilbert space for a composite system of 2 qubits, for instance, has a basis $$\{|0\rangle_2 \otimes |0\rangle_1, |0\rangle_2 \otimes |1\rangle_1, |1\rangle_2 \otimes |0\rangle_1, |1\rangle_2 \otimes |1\rangle_1\}$$

Which is what you may have seen is denoted as


In vector representation, this is equivalent to

$$\{\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 0 \\ 0 \\ 1\end{bmatrix}\}$$

Quick reminder: Kroenecker or tensor product is defined between matrices $A$ (size $m \times n$) and $B$ (size $p \times q$) by the operation $$A \otimes B = \begin{bmatrix} a_{11}B & \cdots & a_{1n}B \\ \vdots & \ddots & \vdots \\ a_{m1}B & \cdots & a_{mn}B\end{bmatrix}$$

See Wikipedia for more details. And make sure that you understand the relations between Dirac notation, vector spaces and tensor product of Hilbert spaces.

States of a composite system are classified in two groups: separable and entangled. This discrimination is based upon the concept of tensor product.

Measurement of subsystems

Before addressing this issue, imagine a projective measurement is performed on each subsystem. This measurement would be described by Projector operators $\{\hat{P}_m\}$. As their name suggest, they "collpase" a quantum state onto a subspace that has a definite outcome for the desired measurement. For instance, if the bit value of a qubit is measured, the projectors that model this process are

$$\hat{P}_0 = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$$ $$\hat{P}_1 = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$

Each operator corresponds to the cases when a value $0$ is observed, or $1$ is observed, respectively. If, for example, the bit value of a two-qubit system is measured for both subsystems, then the corresponding projectors are

$$\hat{P}_{ab} = \hat{P}_{a} \otimes \hat{P}_{b}$$

For $a,b = 0,1$. It would be instructive for you to get the matrix representation of this projectors. After a measurement, if the $m$-th possible outcome is observed, the state of the composite system is simply

$$|\Psi_{after}\rangle = \hat{P}_m |\Psi_{before}\rangle$$

And the probability that $m$-th outcome is measured (given that system is on state $|\Psi_{before}\rangle$) corresponds to

$$p(m|\Psi_{before}) = \langle\Psi_{before}|P_m|\Psi_{before}\rangle$$

Separable states and independence

Separable states are those that can be factored as a Kroenecker product

$$|\Psi\rangle = |\psi\rangle_2 \otimes |\psi\rangle_1$$

In the case of a system of two qubits, this generic state corresponds to

$$|\Psi\rangle = (\alpha_2|0\rangle_2 + \beta_2|1\rangle_2) \otimes (\alpha_1|0\rangle_1 + \beta_1|1\rangle_1)$$

I think it would be instructive for you to find the vector representation of this state as a column vector. Notice that if a measurement of bit value of this composite system is performed, the probabilities of obtaining values $a$ for the second bit, and $b$ for the first bit is

$$p(a,b) = \langle \Psi | P_{ab} |\Psi\rangle = \langle \Psi | P_a \otimes P_b |\Psi\rangle = (\langle \psi | P_{a} |\Psi\rangle) (\langle \psi | P_{a} |\Psi\rangle) = p(a|\psi_2)p(b|\psi_1)$$

This means that measurements of the two subsystem are independent. There are no correlations between the measurements of each subsystem.

There is no posible measurement that can be performed in one subsystem that reveals correlations with the other. Another way to put it: you would need no information from subsystem $\mathcal{S}_1$ to completely describe subsystem $\mathcal{S}_2$ and viceversa.

Note: Remember that two random variables are independent if their joint distribution is the product of each separate distribution.

Entangles states and measurement

Consider, for instance, that $\mathcal{S}$ is a two qubit system in a state

$$|\Psi\rangle = \alpha|00\rangle + \beta|11\rangle$$

It is easy to show that this cannot be factored as a Kroenecker product. As a result, it is said that it is entangled. The main feature is that, for states that have this property, it is so that

$$p(a,b) \neq p(a)p(b)$$

Unlike before. Try, for instance, computing the probability that chain $00$ is measured, and the individual probabilities that each subsystem is measured on state $|0\rangle$, to convince yourself that this is true.

From probability theory, this means that there are fundamental correlations between the measurements of each subsystems when the composite system is on an entangled state.

Put in other words: you need to obtain some information about subsystem $\mathcal{S}_2$ to completely describe $\mathcal{S}_1$, and viceversa. You can describe each subsystem individually (with the tool of partial trace). However, individualization only yields partial information. If you have two entangled qubits, and measure one of them, yo can see that the correlations affect the other qubit's measurement distribution. As a result, the scenario you picture is impossible.

No cloning theorem

Consider a two-qubit system in a state $|Q\rangle \otimes (\alpha|0\rangle + \beta|1\rangle)$. Now, imagine a linear operation that is able to perform transformations

$$|Q\rangle \otimes |0\rangle \rightarrow |0\rangle \otimes |0\rangle$$ $$|Q\rangle \otimes |1\rangle \rightarrow |1\rangle \otimes |1\rangle$$

Where $Q$ denotes some auxiliar state that allows us to copy some basis states of a qubit. It is easy to see that this transformation maps our initial state $|Q\rangle \otimes (\alpha|0\rangle + \beta|1\rangle)$ to

$$\alpha|00\rangle + \beta|11\rangle$$

As we saw before, this is an entangled state, and the qubits cannot be described independently.

The no cloning theorem states that a linear transformation that performs an operation of the type $$|Q\rangle \otimes |\psi\rangle \rightarrow |\psi\rangle \otimes |\psi\rangle$$ is just physically impossible.


In conclusion, your idea is physically impossible, for a copy machine that would "capture" a qubit's state on a communication channel without disturbing the communicated information between two partners is prohibited by the no-cloning theorem. You can, howeve, device some imperfect copying machine, that would create an entangled pair in which the individual description of each qubit is the same, but in which the copies share information. In this type of attacks, sender-receiver can detect that some information has been obtained by a third party, and eventually abort communication.

It is important to notice that no communication Channel is perfect. There are correlations between the elements of the channel and the surrounding environment. This implies that some information is lost to the environment. This can be profited by a possible adversary. A copy machine can be designed so as to obtain enough information on the message ,but not so much that the two main parties think it is more than environmental noise.

If you are interested on this type of attacks, you can check the BB84 quantum encryption protocol, and its possible vulnerabilities. The work on this area is quite technical, and it might be advisable that you study a bit of quantum information theory. Check Nielsen and Chuang's book, ver even Strini et. al.

  • 1
    $\begingroup$ I feel like the OP was just asking if the adversary could entangle themselves with the state. Not necessarily produce a copy of it. $\endgroup$
    – Rammus
    Apr 29 at 16:30

Yes this is possible, in some sense (for bell-pairs at least, I am not sure about arbitrary states). Take any starting bell-pair

\begin{array}{l} \left|\Psi^{+}\right\rangle=\frac{\left|0 1\right\rangle+\left|1 0\right\rangle}{\sqrt{2}}, \\ \left|\Psi^{-}\right\rangle=\frac{\left|0 1\right\rangle-\left|1 0\right\rangle}{\sqrt{2}}, \\ \left|\Phi^{+}\right\rangle=\frac{\left|0 0\right\rangle+\left|1 1\right\rangle}{\sqrt{2}}, \\ \left|\Phi^{-}\right\rangle=\frac{\left|0 0\right\rangle-\left|1 1\right\rangle}{\sqrt{2}} . \end{array}

and apply the operation $I \otimes CNOT$ to any one of:

\begin{array}{l} \left|\Psi^{+}\right\rangle \otimes \left|0 \right \rangle, \\ \left|\Psi^{-}\right\rangle \otimes \left|0 \right \rangle, \\ \left|\Phi^{+}\right\rangle \otimes \left|0 \right \rangle, \\ \left|\Phi^{-}\right\rangle \otimes \left|0 \right \rangle. \end{array}

This results in a new entagled state between 3 qubits. This is a different state, yes, but from the perspective of the sender and receiver, the two halves of the bell-pair are unaffected. As an example:

$$(I \otimes CNOT) (\left|\Phi^{+}\right\rangle \otimes \left|0 \right \rangle) = \frac{\left|0 0 0\right\rangle+\left|1 1 1\right\rangle}{\sqrt{2}}$$

We can see that when the sender (or receiver) measures their half of the bell-pair (it's not technically a pair, because now there is a third qubit), the same measurement results will arise between both halves of the pair.

  • 1
    $\begingroup$ This does not satisfy the constraint that you gave: the state sent to through the channel should be unaffected. $\endgroup$
    – Rammus
    Apr 30 at 19:04
  • $\begingroup$ perhaps I am not using the right terminology... you are right: the state does change simply because it is now an entangled state with 3 qubits. but from the perspective of the sender and receiver, the state is the same. $\endgroup$ May 6 at 15:53
  • $\begingroup$ I elaborated on my answer more. it should be much more clear now. $\endgroup$ May 6 at 16:00

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