Vjylku

Why does this relation fail symmetry and transitivity properties?

Solving the linear first order differential equation?

Crack the bank account's password!

Are all power cords made equal?

Taking an academic pseudonym?

Count repetitions of an array

Homeostasis logic/math problem

Growth of Mordell-Weil Rank of Elliptic Curves over Field Extensions

Calculating the strength of an ionic bond that contains poly-atomic ions

Is the fingering of thirds flexible or do I have to follow the rules?

Why is it that Bernie Sanders is always called a "socialist"?

Isn't a semicolon (';') needed after a function declaration in C++?

Identical projects by students at two different colleges: still plagiarism?

How do I avoid the "chosen hero" feeling?

Bug in VectorFieldPlot[] with InterpolatingFunction[]?

Planting on shale filled patch with weed underlay

Trying to detect if any checked values contains a specific string

Why did Luke use his left hand to shoot?

Insecure private-key encryption

Critique vs nitpicking

Dealing with an internal ScriptKiddie

How to fly a direct entry holding pattern when approaching from an awkward angle?

Fraction within another fraction

No option to ask a question in https://developer.salesforce.com discussion forums

The relationship between entanglement of vector states to matrix operations

How to keep track of entanglements when emulating quantum computation?Entanglement distillation by local operations and post-selection using one entanglement pairHow is a single qubit fundamentally different from a classical coin spinning in the air?Matrix representation and CX gateRelation between quantum entanglement and quantum state complexityHow to analyze highly entangled quantum circuits?Composing the CNOT gate as a tensor product of two level matricesIs it correct to say that we need controlled gates because unitary matrices are reversible?Decomposition of any 2-level matrix into single qubit and CNOT gatesWhat is the meaning of the state $|1rangle-|1rangle$?

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 is 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 state 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
  18 mins ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "694"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f5547%2fthe-relationship-between-entanglement-of-vector-states-to-matrix-operations%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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 is 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 state 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
  18 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 is 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 state 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
  18 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