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The relationship between entanglement of vector states to matrix operations

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The relationship between entanglement of vector states to matrix operations

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1

$begingroup$

I don't understand something which is I believe pretty fundamental. It's said that an operation represented by a matrix A is an entanglement if A can't be written as a tensor product of other matrices. On the other hand, I just learned lately that a vector state too can be determined by these criteria. That is, it's entangled if and only if it can't be written as a tensor product of other vector states.

Here comes the confusion. Suppose you take a CNOT for example. This obviously can't be written as a tensor product of other matrices as it's known to be entangled. What're the consequences of applying a CNOT on a vector? I don't understand it, if you take a vector like (1,0,0,0) or (0,0,10) you will get after CNOT (1,0,0,0) or (0,0,0,1) respectively. Which obviously can be written as the tensor product of other vectors. So what's going on here? Is CNOT supposed to entangle the vector state or not?

Why is it that the gate itself can't be decomposed but the vector which it acted upon can? Or if it isn't necessary then, what's the definition or requirements for entanglement which I am missing?

quantum-gate quantum-state entanglement

share|improve this question

edited 1 hour ago

Blue♦

6,24531355

asked 2 hours ago

bilanushbilanush

976

$endgroup$

add a comment | 

1

$begingroup$

I don't understand something which is I believe pretty fundamental. It's said that an operation represented by a matrix A is an entanglement if A can't be written as a tensor product of other matrices. On the other hand, I just learned lately that a vector state too can be determined by these criteria. That is, it's entangled if and only if it can't be written as a tensor product of other vector states.

Here comes the confusion. Suppose you take a CNOT for example. This obviously can't be written as a tensor product of other matrices as it's known to be entangled. What're the consequences of applying a CNOT on a vector? I don't understand it, if you take a vector like (1,0,0,0) or (0,0,10) you will get after CNOT (1,0,0,0) or (0,0,0,1) respectively. Which obviously can be written as the tensor product of other vectors. So what's going on here? Is CNOT supposed to entangle the vector state or not?

Why is it that the gate itself can't be decomposed but the vector which it acted upon can? Or if it isn't necessary then, what's the definition or requirements for entanglement which I am missing?

quantum-gate quantum-state entanglement

share|improve this question

edited 1 hour ago

Blue♦

6,24531355

asked 2 hours ago

bilanushbilanush

976

$endgroup$

add a comment | 

$begingroup$

I don't understand something which is I believe pretty fundamental. It's said that an operation represented by a matrix A is an entanglement if A can't be written as a tensor product of other matrices. On the other hand, I just learned lately that a vector state too can be determined by these criteria. That is, it's entangled if and only if it can't be written as a tensor product of other vector states.

Here comes the confusion. Suppose you take a CNOT for example. This obviously can't be written as a tensor product of other matrices as it's known to be entangled. What're the consequences of applying a CNOT on a vector? I don't understand it, if you take a vector like (1,0,0,0) or (0,0,10) you will get after CNOT (1,0,0,0) or (0,0,0,1) respectively. Which obviously can be written as the tensor product of other vectors. So what's going on here? Is CNOT supposed to entangle the vector state or not?

Why is it that the gate itself can't be decomposed but the vector which it acted upon can? Or if it isn't necessary then, what's the definition or requirements for entanglement which I am missing?

quantum-gate quantum-state entanglement

share|improve this question

edited 1 hour ago

Blue♦

6,24531355

asked 2 hours ago

bilanushbilanush

976

$endgroup$

share|improve this question

edited 1 hour ago

Blue♦

6,24531355

asked 2 hours ago

bilanushbilanush

976

share|improve this question

edited 1 hour ago

Blue♦

6,24531355

asked 2 hours ago

bilanushbilanush

976

edited 1 hour ago

Blue♦

6,24531355

edited 1 hour ago

Blue♦

6,24531355

asked 2 hours ago

bilanushbilanush

976

asked 2 hours ago

bilanushbilanush

976

1 Answer
1

active

oldest

votes

2

$begingroup$

There are 3 different questions you need to ask yourself. First is the operator a tensor product of smaller matrices. Then is your input state a tensor product of smaller vectors. Finally is the result a tensor product of smaller vectors.

Stick with the two qubit example and CNOT that you had.

First question: CNOT is not able to be written as $A otimes B$ with $A$ and $B$ operating on 1 qubit each.
Second question: The examples you gave were $(1,0,0,0)=|0 0 rangle$ which can be broken up as $|0 rangle otimes |0 rangle$. The other example can as well. It is $|1 rangle otimes |0 rangle$.
Third question: The result states were $(1,0,0,0)$ which was not entangled again. The other was $(0,0,0,1)$ which is $|1 rangle otimes |1 rangle$ also not entangled.

The implication goes differently than you seem to think.

If the operator was not entangled and written as $A otimes B$ and the input state was not entangled and written as $v otimes w$ with the same dimensions as the operator, then you would get $Av otimes Bw$ as the output. So not entangled operator AND not entangled input vector with the same dimensionality imply not entangled on the output.

So the example shows if you only have operator entangled but input not, the output can still be not entangled.

As another example, take the identity operator. It can be written as $I otimes I$ so it is not entangled. Apply it to an entangled input, you get the same state out. That gives the case of unentangled operator and entangled input giving entangled output.

share|improve this answer

edited 1 hour ago

Blue♦

6,24531355

answered 2 hours ago

AHusainAHusain

1,8391311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So why does it matter to call CNOT an entanglement if it doesn't act it on the state? since this is what we mostly care about. I mean, is it possible to say that entanglement operator is the only one which can turn an UNentangled state to an entangled one? Can you show me a state which is unentangled but the CNOT turns it into one? Because otherwise who cares about entangled operation
  $endgroup$
  – bilanush
  8 mins ago
  
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

There are 3 different questions you need to ask yourself. First is the operator a tensor product of smaller matrices. Then is your input state a tensor product of smaller vectors. Finally is the result a tensor product of smaller vectors.

Stick with the two qubit example and CNOT that you had.

First question: CNOT is not able to be written as $A otimes B$ with $A$ and $B$ operating on 1 qubit each.
Second question: The examples you gave were $(1,0,0,0)=|0 0 rangle$ which can be broken up as $|0 rangle otimes |0 rangle$. The other example can as well. It is $|1 rangle otimes |0 rangle$.
Third question: The result states were $(1,0,0,0)$ which was not entangled again. The other was $(0,0,0,1)$ which is $|1 rangle otimes |1 rangle$ also not entangled.

The implication goes differently than you seem to think.

If the operator was not entangled and written as $A otimes B$ and the input state was not entangled and written as $v otimes w$ with the same dimensions as the operator, then you would get $Av otimes Bw$ as the output. So not entangled operator AND not entangled input vector with the same dimensionality imply not entangled on the output.

So the example shows if you only have operator entangled but input not, the output can still be not entangled.

As another example, take the identity operator. It can be written as $I otimes I$ so it is not entangled. Apply it to an entangled input, you get the same state out. That gives the case of unentangled operator and entangled input giving entangled output.

share|improve this answer

edited 1 hour ago

Blue♦

6,24531355

answered 2 hours ago

AHusainAHusain

1,8391311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So why does it matter to call CNOT an entanglement if it doesn't act it on the state? since this is what we mostly care about. I mean, is it possible to say that entanglement operator is the only one which can turn an UNentangled state to an entangled one? Can you show me a state which is unentangled but the CNOT turns it into one? Because otherwise who cares about entangled operation
  $endgroup$
  – bilanush
  8 mins ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

There are 3 different questions you need to ask yourself. First is the operator a tensor product of smaller matrices. Then is your input state a tensor product of smaller vectors. Finally is the result a tensor product of smaller vectors.

Stick with the two qubit example and CNOT that you had.

First question: CNOT is not able to be written as $A otimes B$ with $A$ and $B$ operating on 1 qubit each.
Second question: The examples you gave were $(1,0,0,0)=|0 0 rangle$ which can be broken up as $|0 rangle otimes |0 rangle$. The other example can as well. It is $|1 rangle otimes |0 rangle$.
Third question: The result states were $(1,0,0,0)$ which was not entangled again. The other was $(0,0,0,1)$ which is $|1 rangle otimes |1 rangle$ also not entangled.

The implication goes differently than you seem to think.

If the operator was not entangled and written as $A otimes B$ and the input state was not entangled and written as $v otimes w$ with the same dimensions as the operator, then you would get $Av otimes Bw$ as the output. So not entangled operator AND not entangled input vector with the same dimensionality imply not entangled on the output.

So the example shows if you only have operator entangled but input not, the output can still be not entangled.

As another example, take the identity operator. It can be written as $I otimes I$ so it is not entangled. Apply it to an entangled input, you get the same state out. That gives the case of unentangled operator and entangled input giving entangled output.

share|improve this answer

edited 1 hour ago

Blue♦

6,24531355

answered 2 hours ago

AHusainAHusain

1,8391311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So why does it matter to call CNOT an entanglement if it doesn't act it on the state? since this is what we mostly care about. I mean, is it possible to say that entanglement operator is the only one which can turn an UNentangled state to an entangled one? Can you show me a state which is unentangled but the CNOT turns it into one? Because otherwise who cares about entangled operation
  $endgroup$
  – bilanush
  8 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

There are 3 different questions you need to ask yourself. First is the operator a tensor product of smaller matrices. Then is your input state a tensor product of smaller vectors. Finally is the result a tensor product of smaller vectors.

Stick with the two qubit example and CNOT that you had.

First question: CNOT is not able to be written as $A otimes B$ with $A$ and $B$ operating on 1 qubit each.
Second question: The examples you gave were $(1,0,0,0)=|0 0 rangle$ which can be broken up as $|0 rangle otimes |0 rangle$. The other example can as well. It is $|1 rangle otimes |0 rangle$.
Third question: The result states were $(1,0,0,0)$ which was not entangled again. The other was $(0,0,0,1)$ which is $|1 rangle otimes |1 rangle$ also not entangled.

The implication goes differently than you seem to think.

If the operator was not entangled and written as $A otimes B$ and the input state was not entangled and written as $v otimes w$ with the same dimensions as the operator, then you would get $Av otimes Bw$ as the output. So not entangled operator AND not entangled input vector with the same dimensionality imply not entangled on the output.

So the example shows if you only have operator entangled but input not, the output can still be not entangled.

As another example, take the identity operator. It can be written as $I otimes I$ so it is not entangled. Apply it to an entangled input, you get the same state out. That gives the case of unentangled operator and entangled input giving entangled output.

share|improve this answer

edited 1 hour ago

Blue♦

6,24531355

answered 2 hours ago

AHusainAHusain

1,8391311

$endgroup$

share|improve this answer

edited 1 hour ago

Blue♦

6,24531355

answered 2 hours ago

AHusainAHusain

1,8391311

edited 1 hour ago

Blue♦

6,24531355

edited 1 hour ago

Blue♦

6,24531355

answered 2 hours ago

AHusainAHusain

1,8391311

answered 2 hours ago

AHusainAHusain

1,8391311