Definition of the Pauli group and the Clifford groupWhy is the decomposition of a qubit-qutrit Hamiltonian in terms of Pauli and Gell-Mann matrices not unique?Is the Pauli group for $n$-qubits a basis for $mathbbC^2^ntimes 2^n$?Shorthand notation for the sign flip gateIs there a simple rule for the inverse of a Clifford circuit's stabilizer table?How does the stated Pauli decomposition for $operatornameCPcdot Acdot CP$ arise?Isomorphism between the Clifford group and the quaternionsEfficient implementation of the Clifford group for $n$ qubitsFast way to check if two state vectors are equivalent up to Pauli operationsRotation operator on Pauli parity gates $XX$, $YY$ and $ZZ$CRZ and CRY Gates
Articles with professions
Where is this photo of a group of hikers taken? Is it really in the Ural?
Why are there not any MRI machines available in Interstellar?
Historicity doubted by Romans
The 50,000 row query limit is not actually a "per APEX call" as widely believed
Are glider winch launches rarer in the USA than in the rest of the world? Why?
Why do people say "I am broke" instead of "I am broken"?
How could Barty Crouch Jr. have run out of Polyjuice Potion at the end of the Goblet of Fire movie?
Why is DC so, so, so Democratic?
Inadvertently nuked my disk permission structure - why?
My current job follows "worst practices". How can I talk about my experience in an interview without giving off red flags?
Is there a way to shorten this while condition?
Sextortion with actual password not found in leaks
Monty Hall Problem with a Fallible Monty
Other than a swing wing, what types of variable geometry have flown?
How can I deal with someone that wants to kill something that isn't supposed to be killed?
Why did computer video outputs go from digital to analog, then back to digital?
What exactly makes a General Products hull nearly indestructible?
How can I tell if there was a power cut when I was out?
Who has jurisdiction for a crime committed in an embassy?
Short story where a flexible reality hardens to an unchanging one
Character Frequency in a String
What the purpose of the fuel shutoff valve?
High income and difficulty during interviews
Definition of the Pauli group and the Clifford group
Why is the decomposition of a qubit-qutrit Hamiltonian in terms of Pauli and Gell-Mann matrices not unique?Is the Pauli group for $n$-qubits a basis for $mathbbC^2^ntimes 2^n$?Shorthand notation for the sign flip gateIs there a simple rule for the inverse of a Clifford circuit's stabilizer table?How does the stated Pauli decomposition for $operatornameCPcdot Acdot CP$ arise?Isomorphism between the Clifford group and the quaternionsEfficient implementation of the Clifford group for $n$ qubitsFast way to check if two state vectors are equivalent up to Pauli operationsRotation operator on Pauli parity gates $XX$, $YY$ and $ZZ$CRZ and CRY Gates
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
There seem to be two definitions of the Pauli group. In Nielsen and Chuang, the Pauli group on 1 qubit is defined as
beginalign*
mathcalP_1 = pm I, pm iI, pm X, pm iX, pm Y, pm iY, pm Z, pm iZ = langle X, Y, Zrangle.
endalign*
But in other references, such as https://journals.aps.org/pra/abstract/10.1103/PhysRevA.57.127,
beginalign*
mathcalP_1 = pm I, pm X, pm iY, pm Z = langle X, Zrangle.
endalign*
The second definition makes more sense in the stabilizer formalism, since we usually only consider the action of Pauli strings of $X$ and $Z$. What bothers me is that by the second definition, the phase gate $S$, which is an element of the Clifford group, results in
beginalign*
SXS^dagger = Y notin mathcalP_1.
endalign*
I know that the prefactor $pm i$ somehow doesn't matter, but can someone justify it from a more mathematical/group-theoretical perspective?
pauli-gates clifford-group
$endgroup$
add a comment |
$begingroup$
There seem to be two definitions of the Pauli group. In Nielsen and Chuang, the Pauli group on 1 qubit is defined as
beginalign*
mathcalP_1 = pm I, pm iI, pm X, pm iX, pm Y, pm iY, pm Z, pm iZ = langle X, Y, Zrangle.
endalign*
But in other references, such as https://journals.aps.org/pra/abstract/10.1103/PhysRevA.57.127,
beginalign*
mathcalP_1 = pm I, pm X, pm iY, pm Z = langle X, Zrangle.
endalign*
The second definition makes more sense in the stabilizer formalism, since we usually only consider the action of Pauli strings of $X$ and $Z$. What bothers me is that by the second definition, the phase gate $S$, which is an element of the Clifford group, results in
beginalign*
SXS^dagger = Y notin mathcalP_1.
endalign*
I know that the prefactor $pm i$ somehow doesn't matter, but can someone justify it from a more mathematical/group-theoretical perspective?
pauli-gates clifford-group
$endgroup$
add a comment |
$begingroup$
There seem to be two definitions of the Pauli group. In Nielsen and Chuang, the Pauli group on 1 qubit is defined as
beginalign*
mathcalP_1 = pm I, pm iI, pm X, pm iX, pm Y, pm iY, pm Z, pm iZ = langle X, Y, Zrangle.
endalign*
But in other references, such as https://journals.aps.org/pra/abstract/10.1103/PhysRevA.57.127,
beginalign*
mathcalP_1 = pm I, pm X, pm iY, pm Z = langle X, Zrangle.
endalign*
The second definition makes more sense in the stabilizer formalism, since we usually only consider the action of Pauli strings of $X$ and $Z$. What bothers me is that by the second definition, the phase gate $S$, which is an element of the Clifford group, results in
beginalign*
SXS^dagger = Y notin mathcalP_1.
endalign*
I know that the prefactor $pm i$ somehow doesn't matter, but can someone justify it from a more mathematical/group-theoretical perspective?
pauli-gates clifford-group
$endgroup$
There seem to be two definitions of the Pauli group. In Nielsen and Chuang, the Pauli group on 1 qubit is defined as
beginalign*
mathcalP_1 = pm I, pm iI, pm X, pm iX, pm Y, pm iY, pm Z, pm iZ = langle X, Y, Zrangle.
endalign*
But in other references, such as https://journals.aps.org/pra/abstract/10.1103/PhysRevA.57.127,
beginalign*
mathcalP_1 = pm I, pm X, pm iY, pm Z = langle X, Zrangle.
endalign*
The second definition makes more sense in the stabilizer formalism, since we usually only consider the action of Pauli strings of $X$ and $Z$. What bothers me is that by the second definition, the phase gate $S$, which is an element of the Clifford group, results in
beginalign*
SXS^dagger = Y notin mathcalP_1.
endalign*
I know that the prefactor $pm i$ somehow doesn't matter, but can someone justify it from a more mathematical/group-theoretical perspective?
pauli-gates clifford-group
pauli-gates clifford-group
asked Jul 14 at 23:05
snsunxsnsunx
563 bronze badges
563 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of $pm i$ that are missing.
In lieu of a tikz commutative diagram
beginalign
<X,Y,Z> & hookrightarrow & U(2)\
downarrow & & downarrow\
<[X],[Z]> & hookrightarrow & PU(2)
endalign
For both of the inclusion arrows $H hookrightarrow G$, you can form normalizers $N_G (H)$ which fit in between on each row.
$endgroup$
$begingroup$
You mean the Pauli matrices can be thought of as taken from the projective unitary group? In that case should I think of the Clifford group as normalizing $mathcalP_1/U(1)$ rather than normalizing $mathcalP_1$?
$endgroup$
– snsunx
Jul 14 at 23:45
$begingroup$
It would be nicer if could do the commutative diagram with tikz here
$endgroup$
– AHusain
Jul 14 at 23:57
add a comment |
$begingroup$
Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of $pm i Y$ have a +1 eigenspace. That means that you can only describe states which are stabilised by operators with two or more factors of $pm i Y$. This is only enough to simulate real stabiliser circuits, which is what you're noticing.
For that reason, I would use the first definition; and anyone whose work involves taking the Pauli operators first (which consist of $mathbf 1$, $X$, $Z$, and also $Y$) and then generating a group from them will inevitably use the first definition, for the simple reason that this is the group $langle X, Y, Zrangle$. And because the operators $mathbf 1$, $X$, $Y$, and $Z$ form an operator basis for the Hermitian operators on a single qubit, there are often very good theoretical reasons to include the operator $Y$ (in preference to $iY$) in one's analysis.
It must be said that [Phys. Rev. A 57 (127), 1998] is not just any old article which uses the Pauli group, nor will the point I made above have escaped its author. The reason (I believe) why Gottesman didn't make the distinction there, that I am making now, is because his main interest in that article is in describing error correcting codes — in which the scalar factors preceding the Pauli operators are in principle less important.
To start with, many of the best liked and interesting codes — from the 9-qubit code and the Steane $[! [7,1,3]! ]$ code, to the toric code — happen to be CSS codes, whose stabiliser groups are generated by separate X and Z stabilisers; any stabiliser operators involving Y, will indeed have Y operators in pairs. So, in practise, the point I made about $iY$ not having a +1 eigenspace often doesn't come up.
Secondly, if one decides to just do a hand-wave and forget all scalars, then the important thing is not whether a state is a joint +1 eigenstate of the stabilisers, but just that the state is some eigenstate of all the stabilisers. Considering a stabiliser code, the important thing is that all of the states of the code have the same eigenvalues as each other. This second point gives rise to the notion of the Pauli Frame: the eigenvalues which the codespace has for the 'stabilisers', interpreted as a reference frame. The main thing is for the Pauli frame to be well-defined, and for it to change slowly enough to track how it is changing by stabiliser measurements. And if you're not actually responsible for realising a system to do so, but just want to explore the theory of quantum error correction codes itself, specific eigenvalues are largely a distraction.
The thing is, if you want to use the stabiliser formalism for a different purpose — say, to efficiently simulate stabiliser circuits on standard basis states — then you can't be quite so carefree with scalars: a collection of $n$-qubit stabiliser generators define a specific basis in which the state will live, but flip some signs and you change which of the $2^n$ states you are talking about. To perform the correct simulation, signs do matter here. Keeping track of them may be slightly tedious using the usual convention, but in the end it's important to do.
$endgroup$
$begingroup$
But $[pm iY]$ as it's equivalence class still squares to the identity in the projective unitary group. So it may not have $pm 1$ eigenspaces (not being a matrix itself), but that is the corresponding notion.
$endgroup$
– AHusain
Jul 15 at 14:00
$begingroup$
@AHusain: Right, here's a question for you then. I have in mind a single qubit state which is 'stabilised' by $[pm i Y]$. Which state am I thinking of?
$endgroup$
– Niel de Beaudrap
Jul 15 at 16:39
$begingroup$
(Hint: does being an eigenstate of $pm i Y$ characterise a unique single-qubit state?)
$endgroup$
– Niel de Beaudrap
Jul 16 at 15:10
$begingroup$
Travelling last night. Anyway. The action of PU(n) descends to an action on $CP^n-1$. In this case, the equivalence class of $(alpha,beta)$ goes to that of $(beta,-alpha)$. So $(alpha,beta) equiv (beta,-alpha)$ is the condition for fixed points of group action. This translates into $(1,fracbetaalpha)=(1,frac-alphabeta)$. Which becomes $alpha^2=-beta^2$. A homogeneous equation so cuts out a well defined locus in projective space. $alpha = pm i beta$. Rescale solutions and you get two equivalence classes $(1,pm i)$. I never claimed anything about unique solutions.
$endgroup$
– AHusain
Jul 16 at 18:06
$begingroup$
We're just talking about a group. There are well defined groups whether they be matrix groups or in a projective quotient. There are well defined group actions whether they be linear or merely algebraic.
$endgroup$
– AHusain
Jul 16 at 18:08
|
show 1 more comment
Your Answer
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%2f6784%2fdefinition-of-the-pauli-group-and-the-clifford-group%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
$begingroup$
The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of $pm i$ that are missing.
In lieu of a tikz commutative diagram
beginalign
<X,Y,Z> & hookrightarrow & U(2)\
downarrow & & downarrow\
<[X],[Z]> & hookrightarrow & PU(2)
endalign
For both of the inclusion arrows $H hookrightarrow G$, you can form normalizers $N_G (H)$ which fit in between on each row.
$endgroup$
$begingroup$
You mean the Pauli matrices can be thought of as taken from the projective unitary group? In that case should I think of the Clifford group as normalizing $mathcalP_1/U(1)$ rather than normalizing $mathcalP_1$?
$endgroup$
– snsunx
Jul 14 at 23:45
$begingroup$
It would be nicer if could do the commutative diagram with tikz here
$endgroup$
– AHusain
Jul 14 at 23:57
add a comment |
$begingroup$
The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of $pm i$ that are missing.
In lieu of a tikz commutative diagram
beginalign
<X,Y,Z> & hookrightarrow & U(2)\
downarrow & & downarrow\
<[X],[Z]> & hookrightarrow & PU(2)
endalign
For both of the inclusion arrows $H hookrightarrow G$, you can form normalizers $N_G (H)$ which fit in between on each row.
$endgroup$
$begingroup$
You mean the Pauli matrices can be thought of as taken from the projective unitary group? In that case should I think of the Clifford group as normalizing $mathcalP_1/U(1)$ rather than normalizing $mathcalP_1$?
$endgroup$
– snsunx
Jul 14 at 23:45
$begingroup$
It would be nicer if could do the commutative diagram with tikz here
$endgroup$
– AHusain
Jul 14 at 23:57
add a comment |
$begingroup$
The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of $pm i$ that are missing.
In lieu of a tikz commutative diagram
beginalign
<X,Y,Z> & hookrightarrow & U(2)\
downarrow & & downarrow\
<[X],[Z]> & hookrightarrow & PU(2)
endalign
For both of the inclusion arrows $H hookrightarrow G$, you can form normalizers $N_G (H)$ which fit in between on each row.
$endgroup$
The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of $pm i$ that are missing.
In lieu of a tikz commutative diagram
beginalign
<X,Y,Z> & hookrightarrow & U(2)\
downarrow & & downarrow\
<[X],[Z]> & hookrightarrow & PU(2)
endalign
For both of the inclusion arrows $H hookrightarrow G$, you can form normalizers $N_G (H)$ which fit in between on each row.
edited Jul 16 at 18:18
answered Jul 14 at 23:23
AHusainAHusain
2,6952 gold badges4 silver badges13 bronze badges
2,6952 gold badges4 silver badges13 bronze badges
$begingroup$
You mean the Pauli matrices can be thought of as taken from the projective unitary group? In that case should I think of the Clifford group as normalizing $mathcalP_1/U(1)$ rather than normalizing $mathcalP_1$?
$endgroup$
– snsunx
Jul 14 at 23:45
$begingroup$
It would be nicer if could do the commutative diagram with tikz here
$endgroup$
– AHusain
Jul 14 at 23:57
add a comment |
$begingroup$
You mean the Pauli matrices can be thought of as taken from the projective unitary group? In that case should I think of the Clifford group as normalizing $mathcalP_1/U(1)$ rather than normalizing $mathcalP_1$?
$endgroup$
– snsunx
Jul 14 at 23:45
$begingroup$
It would be nicer if could do the commutative diagram with tikz here
$endgroup$
– AHusain
Jul 14 at 23:57
$begingroup$
You mean the Pauli matrices can be thought of as taken from the projective unitary group? In that case should I think of the Clifford group as normalizing $mathcalP_1/U(1)$ rather than normalizing $mathcalP_1$?
$endgroup$
– snsunx
Jul 14 at 23:45
$begingroup$
You mean the Pauli matrices can be thought of as taken from the projective unitary group? In that case should I think of the Clifford group as normalizing $mathcalP_1/U(1)$ rather than normalizing $mathcalP_1$?
$endgroup$
– snsunx
Jul 14 at 23:45
$begingroup$
It would be nicer if could do the commutative diagram with tikz here
$endgroup$
– AHusain
Jul 14 at 23:57
$begingroup$
It would be nicer if could do the commutative diagram with tikz here
$endgroup$
– AHusain
Jul 14 at 23:57
add a comment |
$begingroup$
Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of $pm i Y$ have a +1 eigenspace. That means that you can only describe states which are stabilised by operators with two or more factors of $pm i Y$. This is only enough to simulate real stabiliser circuits, which is what you're noticing.
For that reason, I would use the first definition; and anyone whose work involves taking the Pauli operators first (which consist of $mathbf 1$, $X$, $Z$, and also $Y$) and then generating a group from them will inevitably use the first definition, for the simple reason that this is the group $langle X, Y, Zrangle$. And because the operators $mathbf 1$, $X$, $Y$, and $Z$ form an operator basis for the Hermitian operators on a single qubit, there are often very good theoretical reasons to include the operator $Y$ (in preference to $iY$) in one's analysis.
It must be said that [Phys. Rev. A 57 (127), 1998] is not just any old article which uses the Pauli group, nor will the point I made above have escaped its author. The reason (I believe) why Gottesman didn't make the distinction there, that I am making now, is because his main interest in that article is in describing error correcting codes — in which the scalar factors preceding the Pauli operators are in principle less important.
To start with, many of the best liked and interesting codes — from the 9-qubit code and the Steane $[! [7,1,3]! ]$ code, to the toric code — happen to be CSS codes, whose stabiliser groups are generated by separate X and Z stabilisers; any stabiliser operators involving Y, will indeed have Y operators in pairs. So, in practise, the point I made about $iY$ not having a +1 eigenspace often doesn't come up.
Secondly, if one decides to just do a hand-wave and forget all scalars, then the important thing is not whether a state is a joint +1 eigenstate of the stabilisers, but just that the state is some eigenstate of all the stabilisers. Considering a stabiliser code, the important thing is that all of the states of the code have the same eigenvalues as each other. This second point gives rise to the notion of the Pauli Frame: the eigenvalues which the codespace has for the 'stabilisers', interpreted as a reference frame. The main thing is for the Pauli frame to be well-defined, and for it to change slowly enough to track how it is changing by stabiliser measurements. And if you're not actually responsible for realising a system to do so, but just want to explore the theory of quantum error correction codes itself, specific eigenvalues are largely a distraction.
The thing is, if you want to use the stabiliser formalism for a different purpose — say, to efficiently simulate stabiliser circuits on standard basis states — then you can't be quite so carefree with scalars: a collection of $n$-qubit stabiliser generators define a specific basis in which the state will live, but flip some signs and you change which of the $2^n$ states you are talking about. To perform the correct simulation, signs do matter here. Keeping track of them may be slightly tedious using the usual convention, but in the end it's important to do.
$endgroup$
$begingroup$
But $[pm iY]$ as it's equivalence class still squares to the identity in the projective unitary group. So it may not have $pm 1$ eigenspaces (not being a matrix itself), but that is the corresponding notion.
$endgroup$
– AHusain
Jul 15 at 14:00
$begingroup$
@AHusain: Right, here's a question for you then. I have in mind a single qubit state which is 'stabilised' by $[pm i Y]$. Which state am I thinking of?
$endgroup$
– Niel de Beaudrap
Jul 15 at 16:39
$begingroup$
(Hint: does being an eigenstate of $pm i Y$ characterise a unique single-qubit state?)
$endgroup$
– Niel de Beaudrap
Jul 16 at 15:10
$begingroup$
Travelling last night. Anyway. The action of PU(n) descends to an action on $CP^n-1$. In this case, the equivalence class of $(alpha,beta)$ goes to that of $(beta,-alpha)$. So $(alpha,beta) equiv (beta,-alpha)$ is the condition for fixed points of group action. This translates into $(1,fracbetaalpha)=(1,frac-alphabeta)$. Which becomes $alpha^2=-beta^2$. A homogeneous equation so cuts out a well defined locus in projective space. $alpha = pm i beta$. Rescale solutions and you get two equivalence classes $(1,pm i)$. I never claimed anything about unique solutions.
$endgroup$
– AHusain
Jul 16 at 18:06
$begingroup$
We're just talking about a group. There are well defined groups whether they be matrix groups or in a projective quotient. There are well defined group actions whether they be linear or merely algebraic.
$endgroup$
– AHusain
Jul 16 at 18:08
|
show 1 more comment
$begingroup$
Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of $pm i Y$ have a +1 eigenspace. That means that you can only describe states which are stabilised by operators with two or more factors of $pm i Y$. This is only enough to simulate real stabiliser circuits, which is what you're noticing.
For that reason, I would use the first definition; and anyone whose work involves taking the Pauli operators first (which consist of $mathbf 1$, $X$, $Z$, and also $Y$) and then generating a group from them will inevitably use the first definition, for the simple reason that this is the group $langle X, Y, Zrangle$. And because the operators $mathbf 1$, $X$, $Y$, and $Z$ form an operator basis for the Hermitian operators on a single qubit, there are often very good theoretical reasons to include the operator $Y$ (in preference to $iY$) in one's analysis.
It must be said that [Phys. Rev. A 57 (127), 1998] is not just any old article which uses the Pauli group, nor will the point I made above have escaped its author. The reason (I believe) why Gottesman didn't make the distinction there, that I am making now, is because his main interest in that article is in describing error correcting codes — in which the scalar factors preceding the Pauli operators are in principle less important.
To start with, many of the best liked and interesting codes — from the 9-qubit code and the Steane $[! [7,1,3]! ]$ code, to the toric code — happen to be CSS codes, whose stabiliser groups are generated by separate X and Z stabilisers; any stabiliser operators involving Y, will indeed have Y operators in pairs. So, in practise, the point I made about $iY$ not having a +1 eigenspace often doesn't come up.
Secondly, if one decides to just do a hand-wave and forget all scalars, then the important thing is not whether a state is a joint +1 eigenstate of the stabilisers, but just that the state is some eigenstate of all the stabilisers. Considering a stabiliser code, the important thing is that all of the states of the code have the same eigenvalues as each other. This second point gives rise to the notion of the Pauli Frame: the eigenvalues which the codespace has for the 'stabilisers', interpreted as a reference frame. The main thing is for the Pauli frame to be well-defined, and for it to change slowly enough to track how it is changing by stabiliser measurements. And if you're not actually responsible for realising a system to do so, but just want to explore the theory of quantum error correction codes itself, specific eigenvalues are largely a distraction.
The thing is, if you want to use the stabiliser formalism for a different purpose — say, to efficiently simulate stabiliser circuits on standard basis states — then you can't be quite so carefree with scalars: a collection of $n$-qubit stabiliser generators define a specific basis in which the state will live, but flip some signs and you change which of the $2^n$ states you are talking about. To perform the correct simulation, signs do matter here. Keeping track of them may be slightly tedious using the usual convention, but in the end it's important to do.
$endgroup$
$begingroup$
But $[pm iY]$ as it's equivalence class still squares to the identity in the projective unitary group. So it may not have $pm 1$ eigenspaces (not being a matrix itself), but that is the corresponding notion.
$endgroup$
– AHusain
Jul 15 at 14:00
$begingroup$
@AHusain: Right, here's a question for you then. I have in mind a single qubit state which is 'stabilised' by $[pm i Y]$. Which state am I thinking of?
$endgroup$
– Niel de Beaudrap
Jul 15 at 16:39
$begingroup$
(Hint: does being an eigenstate of $pm i Y$ characterise a unique single-qubit state?)
$endgroup$
– Niel de Beaudrap
Jul 16 at 15:10
$begingroup$
Travelling last night. Anyway. The action of PU(n) descends to an action on $CP^n-1$. In this case, the equivalence class of $(alpha,beta)$ goes to that of $(beta,-alpha)$. So $(alpha,beta) equiv (beta,-alpha)$ is the condition for fixed points of group action. This translates into $(1,fracbetaalpha)=(1,frac-alphabeta)$. Which becomes $alpha^2=-beta^2$. A homogeneous equation so cuts out a well defined locus in projective space. $alpha = pm i beta$. Rescale solutions and you get two equivalence classes $(1,pm i)$. I never claimed anything about unique solutions.
$endgroup$
– AHusain
Jul 16 at 18:06
$begingroup$
We're just talking about a group. There are well defined groups whether they be matrix groups or in a projective quotient. There are well defined group actions whether they be linear or merely algebraic.
$endgroup$
– AHusain
Jul 16 at 18:08
|
show 1 more comment
$begingroup$
Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of $pm i Y$ have a +1 eigenspace. That means that you can only describe states which are stabilised by operators with two or more factors of $pm i Y$. This is only enough to simulate real stabiliser circuits, which is what you're noticing.
For that reason, I would use the first definition; and anyone whose work involves taking the Pauli operators first (which consist of $mathbf 1$, $X$, $Z$, and also $Y$) and then generating a group from them will inevitably use the first definition, for the simple reason that this is the group $langle X, Y, Zrangle$. And because the operators $mathbf 1$, $X$, $Y$, and $Z$ form an operator basis for the Hermitian operators on a single qubit, there are often very good theoretical reasons to include the operator $Y$ (in preference to $iY$) in one's analysis.
It must be said that [Phys. Rev. A 57 (127), 1998] is not just any old article which uses the Pauli group, nor will the point I made above have escaped its author. The reason (I believe) why Gottesman didn't make the distinction there, that I am making now, is because his main interest in that article is in describing error correcting codes — in which the scalar factors preceding the Pauli operators are in principle less important.
To start with, many of the best liked and interesting codes — from the 9-qubit code and the Steane $[! [7,1,3]! ]$ code, to the toric code — happen to be CSS codes, whose stabiliser groups are generated by separate X and Z stabilisers; any stabiliser operators involving Y, will indeed have Y operators in pairs. So, in practise, the point I made about $iY$ not having a +1 eigenspace often doesn't come up.
Secondly, if one decides to just do a hand-wave and forget all scalars, then the important thing is not whether a state is a joint +1 eigenstate of the stabilisers, but just that the state is some eigenstate of all the stabilisers. Considering a stabiliser code, the important thing is that all of the states of the code have the same eigenvalues as each other. This second point gives rise to the notion of the Pauli Frame: the eigenvalues which the codespace has for the 'stabilisers', interpreted as a reference frame. The main thing is for the Pauli frame to be well-defined, and for it to change slowly enough to track how it is changing by stabiliser measurements. And if you're not actually responsible for realising a system to do so, but just want to explore the theory of quantum error correction codes itself, specific eigenvalues are largely a distraction.
The thing is, if you want to use the stabiliser formalism for a different purpose — say, to efficiently simulate stabiliser circuits on standard basis states — then you can't be quite so carefree with scalars: a collection of $n$-qubit stabiliser generators define a specific basis in which the state will live, but flip some signs and you change which of the $2^n$ states you are talking about. To perform the correct simulation, signs do matter here. Keeping track of them may be slightly tedious using the usual convention, but in the end it's important to do.
$endgroup$
Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of $pm i Y$ have a +1 eigenspace. That means that you can only describe states which are stabilised by operators with two or more factors of $pm i Y$. This is only enough to simulate real stabiliser circuits, which is what you're noticing.
For that reason, I would use the first definition; and anyone whose work involves taking the Pauli operators first (which consist of $mathbf 1$, $X$, $Z$, and also $Y$) and then generating a group from them will inevitably use the first definition, for the simple reason that this is the group $langle X, Y, Zrangle$. And because the operators $mathbf 1$, $X$, $Y$, and $Z$ form an operator basis for the Hermitian operators on a single qubit, there are often very good theoretical reasons to include the operator $Y$ (in preference to $iY$) in one's analysis.
It must be said that [Phys. Rev. A 57 (127), 1998] is not just any old article which uses the Pauli group, nor will the point I made above have escaped its author. The reason (I believe) why Gottesman didn't make the distinction there, that I am making now, is because his main interest in that article is in describing error correcting codes — in which the scalar factors preceding the Pauli operators are in principle less important.
To start with, many of the best liked and interesting codes — from the 9-qubit code and the Steane $[! [7,1,3]! ]$ code, to the toric code — happen to be CSS codes, whose stabiliser groups are generated by separate X and Z stabilisers; any stabiliser operators involving Y, will indeed have Y operators in pairs. So, in practise, the point I made about $iY$ not having a +1 eigenspace often doesn't come up.
Secondly, if one decides to just do a hand-wave and forget all scalars, then the important thing is not whether a state is a joint +1 eigenstate of the stabilisers, but just that the state is some eigenstate of all the stabilisers. Considering a stabiliser code, the important thing is that all of the states of the code have the same eigenvalues as each other. This second point gives rise to the notion of the Pauli Frame: the eigenvalues which the codespace has for the 'stabilisers', interpreted as a reference frame. The main thing is for the Pauli frame to be well-defined, and for it to change slowly enough to track how it is changing by stabiliser measurements. And if you're not actually responsible for realising a system to do so, but just want to explore the theory of quantum error correction codes itself, specific eigenvalues are largely a distraction.
The thing is, if you want to use the stabiliser formalism for a different purpose — say, to efficiently simulate stabiliser circuits on standard basis states — then you can't be quite so carefree with scalars: a collection of $n$-qubit stabiliser generators define a specific basis in which the state will live, but flip some signs and you change which of the $2^n$ states you are talking about. To perform the correct simulation, signs do matter here. Keeping track of them may be slightly tedious using the usual convention, but in the end it's important to do.
edited Jul 15 at 13:35
answered Jul 15 at 9:31
Niel de BeaudrapNiel de Beaudrap
7,0341 gold badge12 silver badges40 bronze badges
7,0341 gold badge12 silver badges40 bronze badges
$begingroup$
But $[pm iY]$ as it's equivalence class still squares to the identity in the projective unitary group. So it may not have $pm 1$ eigenspaces (not being a matrix itself), but that is the corresponding notion.
$endgroup$
– AHusain
Jul 15 at 14:00
$begingroup$
@AHusain: Right, here's a question for you then. I have in mind a single qubit state which is 'stabilised' by $[pm i Y]$. Which state am I thinking of?
$endgroup$
– Niel de Beaudrap
Jul 15 at 16:39
$begingroup$
(Hint: does being an eigenstate of $pm i Y$ characterise a unique single-qubit state?)
$endgroup$
– Niel de Beaudrap
Jul 16 at 15:10
$begingroup$
Travelling last night. Anyway. The action of PU(n) descends to an action on $CP^n-1$. In this case, the equivalence class of $(alpha,beta)$ goes to that of $(beta,-alpha)$. So $(alpha,beta) equiv (beta,-alpha)$ is the condition for fixed points of group action. This translates into $(1,fracbetaalpha)=(1,frac-alphabeta)$. Which becomes $alpha^2=-beta^2$. A homogeneous equation so cuts out a well defined locus in projective space. $alpha = pm i beta$. Rescale solutions and you get two equivalence classes $(1,pm i)$. I never claimed anything about unique solutions.
$endgroup$
– AHusain
Jul 16 at 18:06
$begingroup$
We're just talking about a group. There are well defined groups whether they be matrix groups or in a projective quotient. There are well defined group actions whether they be linear or merely algebraic.
$endgroup$
– AHusain
Jul 16 at 18:08
|
show 1 more comment
$begingroup$
But $[pm iY]$ as it's equivalence class still squares to the identity in the projective unitary group. So it may not have $pm 1$ eigenspaces (not being a matrix itself), but that is the corresponding notion.
$endgroup$
– AHusain
Jul 15 at 14:00
$begingroup$
@AHusain: Right, here's a question for you then. I have in mind a single qubit state which is 'stabilised' by $[pm i Y]$. Which state am I thinking of?
$endgroup$
– Niel de Beaudrap
Jul 15 at 16:39
$begingroup$
(Hint: does being an eigenstate of $pm i Y$ characterise a unique single-qubit state?)
$endgroup$
– Niel de Beaudrap
Jul 16 at 15:10
$begingroup$
Travelling last night. Anyway. The action of PU(n) descends to an action on $CP^n-1$. In this case, the equivalence class of $(alpha,beta)$ goes to that of $(beta,-alpha)$. So $(alpha,beta) equiv (beta,-alpha)$ is the condition for fixed points of group action. This translates into $(1,fracbetaalpha)=(1,frac-alphabeta)$. Which becomes $alpha^2=-beta^2$. A homogeneous equation so cuts out a well defined locus in projective space. $alpha = pm i beta$. Rescale solutions and you get two equivalence classes $(1,pm i)$. I never claimed anything about unique solutions.
$endgroup$
– AHusain
Jul 16 at 18:06
$begingroup$
We're just talking about a group. There are well defined groups whether they be matrix groups or in a projective quotient. There are well defined group actions whether they be linear or merely algebraic.
$endgroup$
– AHusain
Jul 16 at 18:08
$begingroup$
But $[pm iY]$ as it's equivalence class still squares to the identity in the projective unitary group. So it may not have $pm 1$ eigenspaces (not being a matrix itself), but that is the corresponding notion.
$endgroup$
– AHusain
Jul 15 at 14:00
$begingroup$
But $[pm iY]$ as it's equivalence class still squares to the identity in the projective unitary group. So it may not have $pm 1$ eigenspaces (not being a matrix itself), but that is the corresponding notion.
$endgroup$
– AHusain
Jul 15 at 14:00
$begingroup$
@AHusain: Right, here's a question for you then. I have in mind a single qubit state which is 'stabilised' by $[pm i Y]$. Which state am I thinking of?
$endgroup$
– Niel de Beaudrap
Jul 15 at 16:39
$begingroup$
@AHusain: Right, here's a question for you then. I have in mind a single qubit state which is 'stabilised' by $[pm i Y]$. Which state am I thinking of?
$endgroup$
– Niel de Beaudrap
Jul 15 at 16:39
$begingroup$
(Hint: does being an eigenstate of $pm i Y$ characterise a unique single-qubit state?)
$endgroup$
– Niel de Beaudrap
Jul 16 at 15:10
$begingroup$
(Hint: does being an eigenstate of $pm i Y$ characterise a unique single-qubit state?)
$endgroup$
– Niel de Beaudrap
Jul 16 at 15:10
$begingroup$
Travelling last night. Anyway. The action of PU(n) descends to an action on $CP^n-1$. In this case, the equivalence class of $(alpha,beta)$ goes to that of $(beta,-alpha)$. So $(alpha,beta) equiv (beta,-alpha)$ is the condition for fixed points of group action. This translates into $(1,fracbetaalpha)=(1,frac-alphabeta)$. Which becomes $alpha^2=-beta^2$. A homogeneous equation so cuts out a well defined locus in projective space. $alpha = pm i beta$. Rescale solutions and you get two equivalence classes $(1,pm i)$. I never claimed anything about unique solutions.
$endgroup$
– AHusain
Jul 16 at 18:06
$begingroup$
Travelling last night. Anyway. The action of PU(n) descends to an action on $CP^n-1$. In this case, the equivalence class of $(alpha,beta)$ goes to that of $(beta,-alpha)$. So $(alpha,beta) equiv (beta,-alpha)$ is the condition for fixed points of group action. This translates into $(1,fracbetaalpha)=(1,frac-alphabeta)$. Which becomes $alpha^2=-beta^2$. A homogeneous equation so cuts out a well defined locus in projective space. $alpha = pm i beta$. Rescale solutions and you get two equivalence classes $(1,pm i)$. I never claimed anything about unique solutions.
$endgroup$
– AHusain
Jul 16 at 18:06
$begingroup$
We're just talking about a group. There are well defined groups whether they be matrix groups or in a projective quotient. There are well defined group actions whether they be linear or merely algebraic.
$endgroup$
– AHusain
Jul 16 at 18:08
$begingroup$
We're just talking about a group. There are well defined groups whether they be matrix groups or in a projective quotient. There are well defined group actions whether they be linear or merely algebraic.
$endgroup$
– AHusain
Jul 16 at 18:08
|
show 1 more 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%2f6784%2fdefinition-of-the-pauli-group-and-the-clifford-group%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