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Understanding this description of teleportation

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Understanding this description of teleportation

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3

$begingroup$

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:

1. 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.




2. What does the superscript 0 on the operators refer to?




3. 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

share|improve this question

edited 1 hour ago

Sanchayan Dutta♦

6,67641556

asked 7 hours ago

can'tcauchycan'tcauchy

1285

New contributor

can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment | 

3

$begingroup$

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:

1. 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.




2. What does the superscript 0 on the operators refer to?




3. 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

share|improve this question

edited 1 hour ago

Sanchayan Dutta♦

6,67641556

asked 7 hours ago

can'tcauchycan'tcauchy

1285

New contributor

can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

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:

1. 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.




2. What does the superscript 0 on the operators refer to?




3. 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

share|improve this question

edited 1 hour ago

Sanchayan Dutta♦

6,67641556

asked 7 hours ago

can'tcauchycan'tcauchy

1285

New contributor

can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 1 hour ago

Sanchayan Dutta♦

6,67641556

asked 7 hours ago

can'tcauchycan'tcauchy

1285

New contributor

can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 1 hour ago

Sanchayan Dutta♦

6,67641556

asked 7 hours ago

can'tcauchycan'tcauchy

1285

New contributor

can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 1 hour ago

Sanchayan Dutta♦

6,67641556

edited 1 hour ago

Sanchayan Dutta♦

6,67641556

asked 7 hours ago

can'tcauchycan'tcauchy

1285

New contributor

can'tcauchy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 7 hours ago

can'tcauchycan'tcauchy

1285

1 Answer
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2

$begingroup$

1. 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.




2. 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 ($X^0 = Z^0 = I$).




3. After measuring the first two qubits, the state will always be represented as $|xyrangle otimes |text{some state}rangle$, but to convert the receiver's qubit from $|text{some state}rangle$ 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.

share|improve this answer

edited 40 mins ago

answered 1 hour ago

Mariia MykhailovaMariia Mykhailova

1,8651212

$endgroup$

• 
  
  
  

  
  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♦
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

1. 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.




2. 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 ($X^0 = Z^0 = I$).




3. After measuring the first two qubits, the state will always be represented as $|xyrangle otimes |text{some state}rangle$, but to convert the receiver's qubit from $|text{some state}rangle$ 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.

share|improve this answer

edited 40 mins ago

answered 1 hour ago

Mariia MykhailovaMariia Mykhailova

1,8651212

$endgroup$

• 
  
  
  

  
  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♦
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

1. 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.




2. 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 ($X^0 = Z^0 = I$).




3. After measuring the first two qubits, the state will always be represented as $|xyrangle otimes |text{some state}rangle$, but to convert the receiver's qubit from $|text{some state}rangle$ 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.

share|improve this answer

edited 40 mins ago

answered 1 hour ago

Mariia MykhailovaMariia Mykhailova

1,8651212

$endgroup$

• 
  
  
  

  
  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♦
  1 hour ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

1. 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.




2. 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 ($X^0 = Z^0 = I$).




3. After measuring the first two qubits, the state will always be represented as $|xyrangle otimes |text{some state}rangle$, but to convert the receiver's qubit from $|text{some state}rangle$ 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.

share|improve this answer

edited 40 mins ago

answered 1 hour ago

Mariia MykhailovaMariia Mykhailova

1,8651212

$endgroup$

share|improve this answer

edited 40 mins ago

answered 1 hour ago

Mariia MykhailovaMariia Mykhailova

1,8651212

answered 1 hour ago

Mariia MykhailovaMariia Mykhailova

1,8651212

answered 1 hour ago

Mariia MykhailovaMariia Mykhailova

1,8651212