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This circuit:

Simpler teleportation circuit schematic.

The first (upper) qubit is the one we want to teleport, so could start it any state, call $\alpha|0\rangle+\beta|1\rangle$. Our goal is to teleport it to the third (bottom) bit.

After entangling the second and third bits, we have

$$\frac{1}{\sqrt2}\Big(\alpha(|0\rangle\otimes(|00\rangle+|11\rangle))+\beta(|1\rangle\otimes(|00\rangle+|11\rangle))\Big)$$

After applying CNOT from the first to second bit, we have

$$\frac{1}{\sqrt2}\Big(\alpha(|0\rangle\otimes(|00\rangle+|11\rangle))+\beta(|1\rangle\otimes(|10\rangle+|01\rangle))\Big)$$

At this point, we measure the second qubit. If it's $|0\rangle$, then we end up in the state

$$\alpha(|0\rangle\otimes|00\rangle)+\beta(|1\rangle\otimes|01\rangle)=\alpha|000\rangle+\beta|101\rangle$$

If it's $|1\rangle$, then we end up in

$$\alpha(|0\rangle\otimes|11\rangle)+\beta(|1\rangle\otimes|10\rangle)=\alpha|011\rangle+\beta|110\rangle$$

In this case, we apply X to the third bit, yielding

$$\alpha|010\rangle+\beta|111\rangle$$

In either case, if we have access to the third qubit only, it appears to be in the state $\alpha|0\rangle+\beta|1\rangle$, which is what we want. What's wrong with this scheme? The only thing I can think of is that it "doesn't count" as teleportation because the first and third qubits are still entangled at the end. But I thought the main point of teleportation was just to transmit a state using only classical data and a pre-entangled pair, in which case this does seem to "count".

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    $\begingroup$ the fact that the final state is entangled means the third register does not contain the state you wanted to teleport. The state the third register sees is here $|\alpha|^2 |0\rangle\!\langle0|+|\beta|^2 |1\rangle\!\langle1|$, which is not the same as $\alpha|0\rangle+\beta|1\rangle$ $\endgroup$
    – glS
    Nov 25, 2021 at 22:40
  • $\begingroup$ how can |α|2|0⟩⟨0|+|β|2|1⟩⟨1| be a state? isn't that a 2x2 matrix instead of 2x1? $\endgroup$ Nov 25, 2021 at 22:53
  • $\begingroup$ it's a state represented as a density matrix. Have a look e.g. at quantumcomputing.stackexchange.com/q/2347/55 $\endgroup$
    – glS
    Nov 26, 2021 at 8:34

2 Answers 2

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As others have said, the basic problem here is that qubits 1 and 3 are still entangled. You cannot claim that just looking at the third qubit is like what you had from the input state.

To be more concrete about this, it's supposed to be like Bob has received the initial state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$. If he has, then anything he does to that state will behave exactly as it would when he does it directly to $|\psi\rangle$. So, consider Bob measuring his qubit. Sure, if he measures it in the $Z$ basis, he'll get the same outcomes: 0 with probability $|\alpha|^2$ and 1 with probability $|\beta|^2$. However, Bob could equally well make a different measurement. Let's say he wants to make an $X$ measurement. He should get the answer + with probability $$ \left|\frac{\alpha+\beta}{\sqrt{2}}\right|^2 $$ What does he actually get? Let's say $|\Psi_0\rangle$ was the 3-qubit output state having got the measurement result 0 on the second qubit. The probability of getting the $+$ answer is $$ \langle\Psi_0|\left(I\otimes I\otimes|+\rangle\langle +|\right)|\Psi_0\rangle=\frac{|\alpha|^2+|\beta|^2}{2}=\frac12. $$ This (in general) is different. Bob did not have the state $|\psi\rangle$.

To be even more explicit, imagine that Alice had used $|\psi\rangle=|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$. If Bob receives this correctly, he'll have $|+\rangle$ and if he measures in the $X$ basis, he is guaranteed to get the answer +. However, we can also write $$ |\Psi_0\rangle=\frac{1}{\sqrt{2}}(|+0+\rangle+|-0-\rangle) $$ from which it should be obvious (even if you are less familiar with how to describe a single-qubit measurement on a 3-qubit state) that you will get the - result half the time.

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  • $\begingroup$ Thank you -- working out the projections carefully by hand in each case is illuminating. $\endgroup$ Nov 26, 2021 at 20:26
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@glS is right, if you measure the third qubit, the system will be in the mixed state. Only pure states can be expressed as superposition of basis vectors. In general case (both pure and mixed states), states can be described as density matrices.

For quantum teleportation to work, all the qubits you measure, must not be entangled with the state, you want to teleport.

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  • $\begingroup$ If Alice, holding the third qubit, is completely isolated from Bob, holding the first qubit, she cannot tell whether it is entangled with the third qubit or not, right? So how is that any different than a pure state? Or do we not assume that Bob and Alice are permanently isolated? $\endgroup$ Nov 26, 2021 at 1:16
  • $\begingroup$ She can tell, that her state is not pure, because the impurity will affect probability distributions. $\endgroup$
    – totikom
    Nov 27, 2021 at 8:38
  • $\begingroup$ They are not permanently isolated, as they 1) need to obtain a shared Bell state (the first Hadamard gate + CNOT) 2) exchange two bits of classical information $\endgroup$
    – totikom
    Nov 27, 2021 at 9:02

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