Understanding this description of teleportationQuantum algorithm for linear systems of equations (HHL09): Step 1 - Confusion regarding the usage of phase estimation algorithmQuantum algorithm for linear systems of equations (HHL09): Step 2 - What is $|Psi_0rangle$?HHL algorithm — problem with the outcome of postselectionQutrit TeleportationGeneral construction of $W_n$-stateCalculating measurement result of quantum swap circuitQuantum teleportation: second classical bit for removing entanglement?Weeding out qubit states with leftmost qubit as 1Clarification of the “Calculations” section of the (HHL09) paperQuantum secret Sharing using GHZ state paper
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Understanding this description of teleportation
Quantum algorithm for linear systems of equations (HHL09): Step 1 - Confusion regarding the usage of phase estimation algorithmQuantum algorithm for linear systems of equations (HHL09): Step 2 - What is $|Psi_0rangle$?HHL algorithm — problem with the outcome of postselectionQutrit TeleportationGeneral construction of $W_n$-stateCalculating measurement result of quantum swap circuitQuantum teleportation: second classical bit for removing entanglement?Weeding out qubit states with leftmost qubit as 1Clarification of the “Calculations” section of the (HHL09) paperQuantum secret Sharing using GHZ state paper
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$begingroup$
In the context of quantum teleportation, my lecturer writes the following (note that I assume the reader is familiar with the circuit):
If the measurement of the first qubit is 0 and the measurement of the second qubit is 0 , then we have the state $left|phi_4right>=c_0left|000right>+c_1left|001right>=left|00right>otimes left(c_0left|0right>+c_1left|1right>right)=left|00right> otimes left|psi 'right>$.
Now to get the final teleported state we have to go through the final two gates $Z, X$.
My lecturer writes this as;
$left|gammaright>=Z^0X^0left|psi 'right>= left|psi'right>= c_0left|0right>+c_1left|1right>$
Here are my questions:
Why is it that the we do not have $left|gammaright>=Z^0X^0left(left|00right>otimes left|psi 'right>right)$? I don't understand why we cut the state $left|phi_4right>$ "in half ", to use bad terminology, at this step.
What does the superscript 0 on the operators refer to?
In the final state again to use bad terminology we only use half of the state $left|phi_4right>$ can we always assume that the final state will be the $left|psi'right>$ part of $left|phi_4right>=left|xyright>otimesleft|psi'right>$ state, and if so what is the significance of the final step.
If my question is unanswerable due to a deep misunderstanding of the math processes here, I'd still really appreciate some clarification on whatever points can be answered and I can edit it to make more sense as I learn.
algorithm quantum-state quantum-information teleportation
edited Apr 15 at 5:24
Sanchayan Dutta
8,0774 gold badges18 silver badges63 bronze badges
asked Apr 14 at 23:22
bhapibhapi
3798 bronze badges
$endgroup$
add a comment |
$begingroup$
In the context of quantum teleportation, my lecturer writes the following (note that I assume the reader is familiar with the circuit):
If the measurement of the first qubit is 0 and the measurement of the second qubit is 0 , then we have the state $left|phi_4right>=c_0left|000right>+c_1left|001right>=left|00right>otimes left(c_0left|0right>+c_1left|1right>right)=left|00right> otimes left|psi 'right>$.
Now to get the final teleported state we have to go through the final two gates $Z, X$.
My lecturer writes this as;
$left|gammaright>=Z^0X^0left|psi 'right>= left|psi'right>= c_0left|0right>+c_1left|1right>$
Here are my questions:
Why is it that the we do not have $left|gammaright>=Z^0X^0left(left|00right>otimes left|psi 'right>right)$? I don't understand why we cut the state $left|phi_4right>$ "in half ", to use bad terminology, at this step.
What does the superscript 0 on the operators refer to?
In the final state again to use bad terminology we only use half of the state $left|phi_4right>$ can we always assume that the final state will be the $left|psi'right>$ part of $left|phi_4right>=left|xyright>otimesleft|psi'right>$ state, and if so what is the significance of the final step.
If my question is unanswerable due to a deep misunderstanding of the math processes here, I'd still really appreciate some clarification on whatever points can be answered and I can edit it to make more sense as I learn.
algorithm quantum-state quantum-information teleportation
edited Apr 15 at 5:24
Sanchayan Dutta
8,0774 gold badges18 silver badges63 bronze badges
asked Apr 14 at 23:22
bhapibhapi
3798 bronze badges
$endgroup$
add a comment |
$begingroup$
In the context of quantum teleportation, my lecturer writes the following (note that I assume the reader is familiar with the circuit):
If the measurement of the first qubit is 0 and the measurement of the second qubit is 0 , then we have the state $left|phi_4right>=c_0left|000right>+c_1left|001right>=left|00right>otimes left(c_0left|0right>+c_1left|1right>right)=left|00right> otimes left|psi 'right>$.
Now to get the final teleported state we have to go through the final two gates $Z, X$.
My lecturer writes this as;
$left|gammaright>=Z^0X^0left|psi 'right>= left|psi'right>= c_0left|0right>+c_1left|1right>$
Here are my questions:
Why is it that the we do not have $left|gammaright>=Z^0X^0left(left|00right>otimes left|psi 'right>right)$? I don't understand why we cut the state $left|phi_4right>$ "in half ", to use bad terminology, at this step.
What does the superscript 0 on the operators refer to?
In the final state again to use bad terminology we only use half of the state $left|phi_4right>$ can we always assume that the final state will be the $left|psi'right>$ part of $left|phi_4right>=left|xyright>otimesleft|psi'right>$ state, and if so what is the significance of the final step.
If my question is unanswerable due to a deep misunderstanding of the math processes here, I'd still really appreciate some clarification on whatever points can be answered and I can edit it to make more sense as I learn.
algorithm quantum-state quantum-information teleportation
edited Apr 15 at 5:24
Sanchayan Dutta
8,0774 gold badges18 silver badges63 bronze badges
asked Apr 14 at 23:22
bhapibhapi
3798 bronze badges
$endgroup$
In the context of quantum teleportation, my lecturer writes the following (note that I assume the reader is familiar with the circuit):
If the measurement of the first qubit is 0 and the measurement of the second qubit is 0 , then we have the state $left|phi_4right>=c_0left|000right>+c_1left|001right>=left|00right>otimes left(c_0left|0right>+c_1left|1right>right)=left|00right> otimes left|psi 'right>$.
Now to get the final teleported state we have to go through the final two gates $Z, X$.
My lecturer writes this as;
$left|gammaright>=Z^0X^0left|psi 'right>= left|psi'right>= c_0left|0right>+c_1left|1right>$
Here are my questions:
Why is it that the we do not have $left|gammaright>=Z^0X^0left(left|00right>otimes left|psi 'right>right)$? I don't understand why we cut the state $left|phi_4right>$ "in half ", to use bad terminology, at this step.
What does the superscript 0 on the operators refer to?
In the final state again to use bad terminology we only use half of the state $left|phi_4right>$ can we always assume that the final state will be the $left|psi'right>$ part of $left|phi_4right>=left|xyright>otimesleft|psi'right>$ state, and if so what is the significance of the final step.
If my question is unanswerable due to a deep misunderstanding of the math processes here, I'd still really appreciate some clarification on whatever points can be answered and I can edit it to make more sense as I learn.
algorithm quantum-state quantum-information teleportation
algorithm quantum-state quantum-information teleportation
edited Apr 15 at 5:24
Sanchayan Dutta
8,0774 gold badges18 silver badges63 bronze badges
asked Apr 14 at 23:22
bhapibhapi
3798 bronze badges
edited Apr 15 at 5:24
Sanchayan Dutta
8,0774 gold badges18 silver badges63 bronze badges
asked Apr 14 at 23:22
bhapibhapi
3798 bronze badges
edited Apr 15 at 5:24
Sanchayan Dutta
8,0774 gold badges18 silver badges63 bronze badges
edited Apr 15 at 5:24
Sanchayan Dutta
8,0774 gold badges18 silver badges63 bronze badges
edited Apr 15 at 5:24
Sanchayan Dutta
8,0774 gold badges18 silver badges63 bronze badges
8,0774 gold badges18 silver badges63 bronze badges
asked Apr 14 at 23:22
bhapibhapi
3798 bronze badges
asked Apr 14 at 23:22
bhapibhapi
3798 bronze badges
asked Apr 14 at 23:22
bhapibhapi
3798 bronze badges
3798 bronze badges
add a comment |
add a comment |
1 Answer
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$begingroup$
The first two qubits stay on the sender's side, and since they are not entangled with the receiver's qubit, it's not necessary to mention them any further — we switched to discussing the receiver's side of the protocol, and all information from the first two qubits is now in measurement results.
The superscripts denote the power of the operator applied: if both measurement results are 0, you don't need to apply X or Z corrections, which is the same as applying these operators raised to power 0 ($X^0 = Z^0 = I$).
After measuring the first two qubits, the state will always be represented as $|xyrangle otimes |textsome staterangle$, but to convert the receiver's qubit from $|textsome staterangle$ to the state that needed to be teleported $|psi'rangle$ you'll need to apply corrections which depend on the measurement results $x$ and $y$. I believe this is what you refer to as the final step, and its goal is fixing the amplitudes of the state for it to match the state that was teleported.
answered Apr 15 at 5:24
Mariia MykhailovaMariia Mykhailova
3,3401 gold badge3 silver badges21 bronze badges
$endgroup$
1
$begingroup$
Minor nitpicks. If you write text within a mathematical expression it's best to use thetextformatting; otherwise, the $spacing space between space letters space is space somewhat space uneven$. Moreover,1.,2.and3.is the only format that works for numbered lists (on SE).1),2)and3)does not work. Cf. help/formatting. I've edited these things; hope you don't mind. Also, it might be useful to clarify the Alice-Bob terminology in the answer itself as the OP doesn't seem to be using it.
$endgroup$
– Sanchayan Dutta
Apr 15 at 5:31
add a comment |
Your Answer
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$begingroup$
The first two qubits stay on the sender's side, and since they are not entangled with the receiver's qubit, it's not necessary to mention them any further — we switched to discussing the receiver's side of the protocol, and all information from the first two qubits is now in measurement results.
The superscripts denote the power of the operator applied: if both measurement results are 0, you don't need to apply X or Z corrections, which is the same as applying these operators raised to power 0 ($X^0 = Z^0 = I$).
After measuring the first two qubits, the state will always be represented as $|xyrangle otimes |textsome staterangle$, but to convert the receiver's qubit from $|textsome staterangle$ to the state that needed to be teleported $|psi'rangle$ you'll need to apply corrections which depend on the measurement results $x$ and $y$. I believe this is what you refer to as the final step, and its goal is fixing the amplitudes of the state for it to match the state that was teleported.
answered Apr 15 at 5:24
Mariia MykhailovaMariia Mykhailova
3,3401 gold badge3 silver badges21 bronze badges
$endgroup$
1
$begingroup$
Minor nitpicks. If you write text within a mathematical expression it's best to use thetextformatting; otherwise, the $spacing space between space letters space is space somewhat space uneven$. Moreover,1.,2.and3.is the only format that works for numbered lists (on SE).1),2)and3)does not work. Cf. help/formatting. I've edited these things; hope you don't mind. Also, it might be useful to clarify the Alice-Bob terminology in the answer itself as the OP doesn't seem to be using it.
$endgroup$
– Sanchayan Dutta
Apr 15 at 5:31
add a comment |
$begingroup$
The first two qubits stay on the sender's side, and since they are not entangled with the receiver's qubit, it's not necessary to mention them any further — we switched to discussing the receiver's side of the protocol, and all information from the first two qubits is now in measurement results.
The superscripts denote the power of the operator applied: if both measurement results are 0, you don't need to apply X or Z corrections, which is the same as applying these operators raised to power 0 ($X^0 = Z^0 = I$).
After measuring the first two qubits, the state will always be represented as $|xyrangle otimes |textsome staterangle$, but to convert the receiver's qubit from $|textsome staterangle$ to the state that needed to be teleported $|psi'rangle$ you'll need to apply corrections which depend on the measurement results $x$ and $y$. I believe this is what you refer to as the final step, and its goal is fixing the amplitudes of the state for it to match the state that was teleported.
answered Apr 15 at 5:24
Mariia MykhailovaMariia Mykhailova
3,3401 gold badge3 silver badges21 bronze badges
$endgroup$
1
$begingroup$
Minor nitpicks. If you write text within a mathematical expression it's best to use thetextformatting; otherwise, the $spacing space between space letters space is space somewhat space uneven$. Moreover,1.,2.and3.is the only format that works for numbered lists (on SE).1),2)and3)does not work. Cf. help/formatting. I've edited these things; hope you don't mind. Also, it might be useful to clarify the Alice-Bob terminology in the answer itself as the OP doesn't seem to be using it.
$endgroup$
– Sanchayan Dutta
Apr 15 at 5:31
add a comment |
$begingroup$
The first two qubits stay on the sender's side, and since they are not entangled with the receiver's qubit, it's not necessary to mention them any further — we switched to discussing the receiver's side of the protocol, and all information from the first two qubits is now in measurement results.
The superscripts denote the power of the operator applied: if both measurement results are 0, you don't need to apply X or Z corrections, which is the same as applying these operators raised to power 0 ($X^0 = Z^0 = I$).
After measuring the first two qubits, the state will always be represented as $|xyrangle otimes |textsome staterangle$, but to convert the receiver's qubit from $|textsome staterangle$ to the state that needed to be teleported $|psi'rangle$ you'll need to apply corrections which depend on the measurement results $x$ and $y$. I believe this is what you refer to as the final step, and its goal is fixing the amplitudes of the state for it to match the state that was teleported.
answered Apr 15 at 5:24
Mariia MykhailovaMariia Mykhailova
3,3401 gold badge3 silver badges21 bronze badges
$endgroup$
The first two qubits stay on the sender's side, and since they are not entangled with the receiver's qubit, it's not necessary to mention them any further — we switched to discussing the receiver's side of the protocol, and all information from the first two qubits is now in measurement results.
The superscripts denote the power of the operator applied: if both measurement results are 0, you don't need to apply X or Z corrections, which is the same as applying these operators raised to power 0 ($X^0 = Z^0 = I$).
After measuring the first two qubits, the state will always be represented as $|xyrangle otimes |textsome staterangle$, but to convert the receiver's qubit from $|textsome staterangle$ to the state that needed to be teleported $|psi'rangle$ you'll need to apply corrections which depend on the measurement results $x$ and $y$. I believe this is what you refer to as the final step, and its goal is fixing the amplitudes of the state for it to match the state that was teleported.
answered Apr 15 at 5:24
Mariia MykhailovaMariia Mykhailova
3,3401 gold badge3 silver badges21 bronze badges
edited Apr 15 at 6:04
answered Apr 15 at 5:24
Mariia MykhailovaMariia Mykhailova
3,3401 gold badge3 silver badges21 bronze badges
answered Apr 15 at 5:24
Mariia MykhailovaMariia Mykhailova
3,3401 gold badge3 silver badges21 bronze badges
answered Apr 15 at 5:24
Mariia MykhailovaMariia Mykhailova
3,3401 gold badge3 silver badges21 bronze badges
3,3401 gold badge3 silver badges21 bronze badges
1
$begingroup$
Minor nitpicks. If you write text within a mathematical expression it's best to use thetextformatting; otherwise, the $spacing space between space letters space is space somewhat space uneven$. Moreover,1.,2.and3.is the only format that works for numbered lists (on SE).1),2)and3)does not work. Cf. help/formatting. I've edited these things; hope you don't mind. Also, it might be useful to clarify the Alice-Bob terminology in the answer itself as the OP doesn't seem to be using it.
$endgroup$
– Sanchayan Dutta
Apr 15 at 5:31
add a comment |
1
$begingroup$
Minor nitpicks. If you write text within a mathematical expression it's best to use thetextformatting; otherwise, the $spacing space between space letters space is space somewhat space uneven$. Moreover,1.,2.and3.is the only format that works for numbered lists (on SE).1),2)and3)does not work. Cf. help/formatting. I've edited these things; hope you don't mind. Also, it might be useful to clarify the Alice-Bob terminology in the answer itself as the OP doesn't seem to be using it.
$endgroup$
– Sanchayan Dutta
Apr 15 at 5:31
1
1
$begingroup$
Minor nitpicks. If you write text within a mathematical expression it's best to use the
text formatting; otherwise, the $spacing space between space letters space is space somewhat space uneven$. Moreover, 1., 2. and 3. is the only format that works for numbered lists (on SE). 1), 2) and 3) does not work. Cf. help/formatting. I've edited these things; hope you don't mind. Also, it might be useful to clarify the Alice-Bob terminology in the answer itself as the OP doesn't seem to be using it.$endgroup$
– Sanchayan Dutta
Apr 15 at 5:31
$begingroup$
Minor nitpicks. If you write text within a mathematical expression it's best to use the
text formatting; otherwise, the $spacing space between space letters space is space somewhat space uneven$. Moreover, 1., 2. and 3. is the only format that works for numbered lists (on SE). 1), 2) and 3) does not work. Cf. help/formatting. I've edited these things; hope you don't mind. Also, it might be useful to clarify the Alice-Bob terminology in the answer itself as the OP doesn't seem to be using it.$endgroup$
– Sanchayan Dutta
Apr 15 at 5:31
add a comment |
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