Michael Nielsen posted on Twitter about a proof on a circuit for quantum teleportation.

Had some fun this afternoon re-analyzing the circuit for quantum teleportation. Here's a proof I found that the circuit works.

The circuit to be verified: top two qubits are Alice's, including the state psi to be teleported. Bottom qubit is Bob's - want to show it outputs psi.

enter image description here

I would appreciate if someone can explain and elaborate on the steps for the proof? Also what do the $X^x$ and $Z^z$ gates mean?


1 Answer 1


$x$ and $z$ are just numbers $0$ or $1$, so $X^x = I$ or $X$, and $Z^z = I$ or $Z$.

The explanatiton goes as follows

  1. The first step is to note that in our case CNOT operation on the 1st and 2nd qubits has the same effect as CNOT on the 1st and 3rd qubits. That is $$ CNOT_{12} |\psi\rangle(|00\rangle+|11\rangle)\frac{1}{\sqrt{2}} = CNOT_{13} |\psi\rangle(|00\rangle+|11\rangle)\frac{1}{\sqrt{2}} $$ To see why, just define $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ and compute $$ CNOT_{12} |\psi\rangle(|00\rangle+|11\rangle)\frac{1}{\sqrt{2}} = CNOT_{12}(\alpha|0\rangle + \beta|1\rangle)(|00\rangle+|11\rangle)\frac{1}{\sqrt{2}} = $$ $$ =CNOT_{12}(\alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle )\frac{1}{\sqrt{2}} = $$ $$ = (\alpha|000\rangle + \alpha|011\rangle + \beta|110\rangle + \beta|101\rangle )\frac{1}{\sqrt{2}}, $$ $$ CNOT_{13} |\psi\rangle(|00\rangle+|11\rangle)\frac{1}{\sqrt{2}} = CNOT_{13}(\alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle )\frac{1}{\sqrt{2}} = $$ $$ =(\alpha|000\rangle + \alpha|011\rangle + \beta|101\rangle + \beta|110\rangle )\frac{1}{\sqrt{2}} $$ So, we can transform the circuit by switching $CNOT_{12}$ to $CNOT_{13}$. enter image description here
  2. In the second step we note that in our new circuit the second qubit has only one operation that affects it – the measurement. So we can apply it before all other operations. The result of this operation is $x=0$ or $1$, and 2nd and 3rd qubits after this measurement will be $|xx\rangle = |00\rangle$ or $|11\rangle$. enter image description here
  3. Since the 3rd qubit is $|0\rangle$ or $|1\rangle$ at this step, we can swap the order of operations $CNOT_{13}$ and $X^x$ on it. But $X^x|x\rangle = |0\rangle$, so what is left is to analyze the circuit enter image description here
  4. At this step you can directly compute the effect of the CNOT and H gates $$ H\otimes I \cdot CNOT_{12} \cdot |\psi\rangle|0\rangle = H\otimes I \cdot CNOT_{12} \cdot (\alpha|00\rangle + \beta|10\rangle) = $$ $$ = H\otimes I (\alpha|00\rangle + \beta|11\rangle) = \frac{1}{\sqrt{2}}\left( \alpha(|0\rangle+|1\rangle)|0\rangle + \beta(|0\rangle-|1\rangle)|1\rangle \right) = $$ $$ = \frac{1}{\sqrt{2}}\left( \alpha|00\rangle+\alpha|10\rangle + \beta|01\rangle-\beta|11\rangle \right) = \frac{1}{\sqrt{2}}\left( |0\rangle(\alpha|0\rangle+\beta|1\rangle) + |1\rangle(\alpha|0\rangle-\beta|1\rangle) \right) $$
  5. Now if we measure 1st qubit the result will be $z=0$ or $1$, and the state will be $|0\rangle(\alpha|0\rangle+\beta|1\rangle)$ or $|1\rangle(\alpha|0\rangle-\beta|1\rangle)$ correspondingly. So if we apply $Z^z$ on the last qubit it will be in the state $\alpha|0\rangle+\beta|1\rangle = |\psi\rangle$

His second way to prove last circuit (from step 3) is basically by using equality $$ \alpha|00\rangle+\beta|11\rangle = \frac{1}{\sqrt{2}}(|+\rangle(\alpha|0\rangle + \beta|1\rangle) + |-\rangle(\alpha|0\rangle - \beta|1\rangle)) $$

After applying $CNOT$ on $|\psi\rangle|0\rangle$ we will get the state $\alpha|00\rangle+\beta|11\rangle$. Now instead of applying $H$ and measuring the 1st qubit in $|0\rangle, |1\rangle$ basis we can measure in $|+\rangle, |-\rangle$ basis right away. The result will be $z=0$ with the state $|+\rangle(\alpha|0\rangle + \beta|1\rangle)$ or $z=1$ with the state $|-\rangle(\alpha|0\rangle - \beta|1\rangle)$. So $Z^z$ corrects the state of the last qubit.

  • 1
    $\begingroup$ he also gave another way to complete the last step without the full calculation, see here onwards $\endgroup$
    – glS
    May 27, 2019 at 9:27
  • $\begingroup$ Yes, but I think it just hides the neccessary calculation :) I don't see how it is more intuitive. Updated the answer. $\endgroup$
    – Danylo Y
    May 27, 2019 at 10:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.