Why do we have to uncompute rather than simply set registers to zero?
$begingroup$
In implementing a quantum subroutine it is important to uncompute temporary registers after use, to ensure the output state of the subroutine is not entangled with them (which would affect its behaviour).
Why is it necessary to perform this through uncomputation rather than simply setting temporary registers to zero afterwards?
quantum-state entanglement
asked 5 hours ago
Sideshow BobSideshow Bob
1325
$endgroup$
add a comment |
$begingroup$
In implementing a quantum subroutine it is important to uncompute temporary registers after use, to ensure the output state of the subroutine is not entangled with them (which would affect its behaviour).
Why is it necessary to perform this through uncomputation rather than simply setting temporary registers to zero afterwards?
quantum-state entanglement
asked 5 hours ago
Sideshow BobSideshow Bob
1325
$endgroup$
$begingroup$
a related question on cstheory
$endgroup$
– glS
2 hours ago
add a comment |
$begingroup$
In implementing a quantum subroutine it is important to uncompute temporary registers after use, to ensure the output state of the subroutine is not entangled with them (which would affect its behaviour).
Why is it necessary to perform this through uncomputation rather than simply setting temporary registers to zero afterwards?
quantum-state entanglement
asked 5 hours ago
Sideshow BobSideshow Bob
1325
$endgroup$
In implementing a quantum subroutine it is important to uncompute temporary registers after use, to ensure the output state of the subroutine is not entangled with them (which would affect its behaviour).
Why is it necessary to perform this through uncomputation rather than simply setting temporary registers to zero afterwards?
quantum-state entanglement
quantum-state entanglement
asked 5 hours ago
Sideshow BobSideshow Bob
1325
asked 5 hours ago
Sideshow BobSideshow Bob
1325
asked 5 hours ago
Sideshow BobSideshow Bob
1325
asked 5 hours ago
Sideshow BobSideshow Bob
1325
asked 5 hours ago
Sideshow BobSideshow Bob
1325
1325
$begingroup$
a related question on cstheory
$endgroup$
– glS
2 hours ago
add a comment |
$begingroup$
a related question on cstheory
$endgroup$
– glS
2 hours ago
$begingroup$
a related question on cstheory
$endgroup$
– glS
2 hours ago
$begingroup$
a related question on cstheory
$endgroup$
– glS
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because all operations you carry out (besides measurement) need to be unitary and this one would not.
answered 4 hours ago
cnadacnada
1,982212
$endgroup$
1
$begingroup$
There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
$endgroup$
– Niel de Beaudrap
2 hours ago
$begingroup$
@NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
$endgroup$
– cnada
2 hours ago
$begingroup$
My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
$endgroup$
– Niel de Beaudrap
57 mins ago
add a comment |
$begingroup$
Uncomputing temporary registers seems to me to be the most efficient way to set temporary registers back to zero.
You can't measure these and classically flip them. 1) because a measurement of 1 doesn't necessarily mean the the state of the temporary register is |1>. and 2) Measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.
And somehow how implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.
answered 2 hours ago
Malcolm ReganMalcolm Regan
1697
$endgroup$
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f5222%2fwhy-do-we-have-to-uncompute-rather-than-simply-set-registers-to-zero%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because all operations you carry out (besides measurement) need to be unitary and this one would not.
answered 4 hours ago
cnadacnada
1,982212
$endgroup$
1
$begingroup$
There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
$endgroup$
– Niel de Beaudrap
2 hours ago
$begingroup$
@NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
$endgroup$
– cnada
2 hours ago
$begingroup$
My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
$endgroup$
– Niel de Beaudrap
57 mins ago
add a comment |
$begingroup$
Because all operations you carry out (besides measurement) need to be unitary and this one would not.
answered 4 hours ago
cnadacnada
1,982212
$endgroup$
1
$begingroup$
There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
$endgroup$
– Niel de Beaudrap
2 hours ago
$begingroup$
@NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
$endgroup$
– cnada
2 hours ago
$begingroup$
My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
$endgroup$
– Niel de Beaudrap
57 mins ago
add a comment |
$begingroup$
Because all operations you carry out (besides measurement) need to be unitary and this one would not.
answered 4 hours ago
cnadacnada
1,982212
$endgroup$
Because all operations you carry out (besides measurement) need to be unitary and this one would not.
answered 4 hours ago
cnadacnada
1,982212
answered 4 hours ago
cnadacnada
1,982212
answered 4 hours ago
cnadacnada
1,982212
answered 4 hours ago
cnadacnada
1,982212
1,982212
1
$begingroup$
There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
$endgroup$
– Niel de Beaudrap
2 hours ago
$begingroup$
@NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
$endgroup$
– cnada
2 hours ago
$begingroup$
My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
$endgroup$
– Niel de Beaudrap
57 mins ago
add a comment |
1
$begingroup$
There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
$endgroup$
– Niel de Beaudrap
2 hours ago
$begingroup$
@NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
$endgroup$
– cnada
2 hours ago
$begingroup$
My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
$endgroup$
– Niel de Beaudrap
57 mins ago
1
1
$begingroup$
There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
$endgroup$
– Niel de Beaudrap
2 hours ago
$begingroup$
There is one thing you don't address in your answer --- why not use a measurement then, and then flip the bit if the outcome is 1? That would effectively reset the qubit. Not that this is actually a good idea, but could you elaborate your answer to describe why this is not a good idea?
$endgroup$
– Niel de Beaudrap
2 hours ago
$begingroup$
@NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
$endgroup$
– cnada
2 hours ago
$begingroup$
@NieldeBeaudrap If you measure you collapse into one possibility. But say you want to continue applying operations on this output state (in the register of interest)... you see why this is not a good idea?
$endgroup$
– cnada
2 hours ago
$begingroup$
My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
$endgroup$
– Niel de Beaudrap
57 mins ago
$begingroup$
My comment was that it is something you don't address, which is conspicuous given that 'resetting' the state is exactly what the OP asked about. While I know why it isn't a good idea, I'm not the one who asked the original question.
$endgroup$
– Niel de Beaudrap
57 mins ago
add a comment |
$begingroup$
Uncomputing temporary registers seems to me to be the most efficient way to set temporary registers back to zero.
You can't measure these and classically flip them. 1) because a measurement of 1 doesn't necessarily mean the the state of the temporary register is |1>. and 2) Measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.
And somehow how implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.
answered 2 hours ago
Malcolm ReganMalcolm Regan
1697
$endgroup$
add a comment |
$begingroup$
Uncomputing temporary registers seems to me to be the most efficient way to set temporary registers back to zero.
You can't measure these and classically flip them. 1) because a measurement of 1 doesn't necessarily mean the the state of the temporary register is |1>. and 2) Measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.
And somehow how implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.
answered 2 hours ago
Malcolm ReganMalcolm Regan
1697
$endgroup$
add a comment |
$begingroup$
Uncomputing temporary registers seems to me to be the most efficient way to set temporary registers back to zero.
You can't measure these and classically flip them. 1) because a measurement of 1 doesn't necessarily mean the the state of the temporary register is |1>. and 2) Measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.
And somehow how implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.
answered 2 hours ago
Malcolm ReganMalcolm Regan
1697
$endgroup$
Uncomputing temporary registers seems to me to be the most efficient way to set temporary registers back to zero.
You can't measure these and classically flip them. 1) because a measurement of 1 doesn't necessarily mean the the state of the temporary register is |1>. and 2) Measuring would cause entangled qubits to collapse into a corresponding state so that would defeat the purpose of uncomputing temporary registers that you mentioned in your question.
And somehow how implementing a circuit that detected a temporary register's state moved it back to 0 would be much more difficult than simply uncomputing the temporary register.
answered 2 hours ago
Malcolm ReganMalcolm Regan
1697
edited 2 hours ago
answered 2 hours ago
Malcolm ReganMalcolm Regan
1697
answered 2 hours ago
Malcolm ReganMalcolm Regan
1697
answered 2 hours ago
Malcolm ReganMalcolm Regan
1697
1697
add a comment |
add a comment |
Thanks for contributing an answer to Quantum Computing Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fquantumcomputing.stackexchange.com%2fquestions%2f5222%2fwhy-do-we-have-to-uncompute-rather-than-simply-set-registers-to-zero%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
a related question on cstheory
$endgroup$
– glS
2 hours ago