The commutativity of $I$ and $Y$ in a stabilizer code

Question

Let $P_1 = \lbrace I, -I, iI, -iI, X, -X, iX, -iX, Y, -Y, iY, -iY, Z, -Z, iZ, -iZ\rbrace$. Let $P_n$ be the $n$-tensor fold of $P_1$. It is said that two operators either commute if $AB = BA$ or anti-commute if $AB = -BA$ for all $A,B \in P_n$.

Let us have $n=1$ and $A=I$ and $B=Y$, then we have:

\begin{align*} IY &\stackrel{\text{true}}{=} YI,\\ IY &\stackrel{\text{true}}{=} -YI. \end{align*}

In other words, $I$ and $Y$ both commute and anti-commute. I have also added a matlab code snippet for completeness.

I = [1 0; 0 1];
Y = [0 -i;i  0];
if isequal(I*Y,Y*I)
   disp('commute') 
end
if isequal(I*Y,-Y*I)
   disp('ANTI-commute') 
end

I am referring to Daniel Gottesman's PhD thesis.

What am I missing?

Edit: My Matlab code should run only one of the if statements (not both). It seems that my machine was caching the answers from previously. A restart of the program has fixed the issue.

Did you run the code? What does it show? (I did, and it showed what it should.) — Norbert Schuch, Aug 31 '20 at 13:41
Slightly off-topic, but you really should not include any $i$'s or $-i$'s in these kind of sets if you're talking about stabilizer codes. And definitely not $-I$! Also, note that the concept of commutativity is not only defined for the Pauli group, but for any pair of operators (of course, the concept even extends beyond operators). — JSdJ, Aug 31 '20 at 15:36
There are two groups at play : the "normal" Pauli group with center ${\pm I,\pm \imath}$ and the "real" version with center ${\pm I}$. Authors switch between them without warning sometimes. "Two operators commute or anti-commute" applies to the real case only. In general the commutator is in the center of the group in both cases. — unknown, Aug 31 '20 at 17:02
@NorbertSchuch, the output is that they commute and anti-commute. My concern that it should be one or the other, not both. — M. Al Jumaily, Aug 31 '20 at 22:33
@NorbertSchuch, I realized that the output I was receiving was a cached output, not the one corresponding to the current script I had. I closed Matlab and reopened it. Indeed, only the first if statement executed showing they do commute :). — M. Al Jumaily, Sep 01 '20 at 06:03

Davit Khachatryan · Accepted Answer · 2020-09-01T10:15:21.740

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It should be $IY \ne - YI$ instead of $IY = - YI$. $$IY \ne - YI \\ \begin{pmatrix} 1&0\\ 0& 1 \end{pmatrix} \begin{pmatrix} 0&-i\\ i& 0 \end{pmatrix} \ne -\begin{pmatrix} 0&-i\\ i& 0 \end{pmatrix} \begin{pmatrix} 1&0\\ 0& 1 \end{pmatrix} \\ Y = \begin{pmatrix} 0&-i\\ i& 0 \end{pmatrix} \ne -\begin{pmatrix} 0&-i\\ i& 0 \end{pmatrix} = -Y $$

So $I$ and $Y$ don't anti-commute. Moreover $I$ anti-commutes only with the matrix with all $0$ entries. Also, $I$ commutes with all matrices, because from definition $IM = M = MI$, where $M$ is a random matrix.

On the contrary all pairs from $\{X, Y, Z \}$ anti-commute with each other. As an example:

$$\{X Z \} = XZ + ZX = 0$$

because $XZ = -ZX$. Also, we have thess relations: $XY = -YX$ and $YZ = -ZY$ from which we can conclude $\{XY\} = \{ZY\} = 0$.

edited Sep 01 '20 at 10:15

answered Aug 31 '20 at 09:27

Davit Khachatryan

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In the thesis referred to in the question in equation 1.4, the anti-commuter function is defined as: $\lbrace \sigma_i,\sigma_j\rbrace = \sigma_i\sigma_j + \sigma_j\sigma_i = 0 $ for $i, j \in \lbrace X,Y,Z\rbrace$. It means that $\lbrace \sigma_i,\sigma_j\rbrace = \sigma_i\sigma_j = -\sigma_j\sigma_i $. Am I missing something? – M. Al Jumaily Aug 31 '20 at 09:39
@M.AlJumaily, you have written in the comment $i,j \in {X, Y, Z }$...note that $I$ is not included. Please add the reference link to the question. – Davit Khachatryan Aug 31 '20 at 09:48
Another resource that shows $I$ is also included, not just $X,Y,Z$ is Page 34 eq 7.115 – M. Al Jumaily Aug 31 '20 at 09:55
@M.AlJumaily, In your first reference, I didn't find a statement that $Y$ and $I$ anti-commute. In your second reference eq 7.115 (7.116) is a Pauli group that has the following property mentioned in the next page "(iv) Any to elements $M, N \in G_n$ either commute or anti-commute: $MN = \pm NM$", so still I didn't find a statement that $Y$ and $I$ anti-commute. Why do you think that they anti-commute? Note that in the answer I have proved that they don't anti-commute. – Davit Khachatryan Aug 31 '20 at 10:07
I chose $Y$ and $I$ arbitrarily as an example with $n = 1$, i.e., dealing with $2 \times 2$ $I,X,Y,Z$ matrices. Thanks for your clarification that probably lead me to figure-out the issue. The $Y$ gate is written as $ZX$ and the $i$ factor is eliminated because it is an global phase which is not observable (I think this is what has occurred in the second reference eq 7.114). Now, using the definition of anti-commutativity in the question, everything follows through. Do you think this is a valid conclusion? – M. Al Jumaily Aug 31 '20 at 10:19
@M.AlJumaily, ok, note that $Y$ with or without $i$ factor don't anti-commute with $I$. – Davit Khachatryan Aug 31 '20 at 10:24
@M.AlJumaily, also I am not sure if we can talk about global phases when we try to figure out commutation relations. – Davit Khachatryan Aug 31 '20 at 10:27
If you have the matrix of $Y$ to be $\begin{matrix}0 & -1\1 & 0\end{matrix}$, then $I$ and $Y$ will commute (which they should). However, keeping $Y$ is $\begin{matrix}0 & -i\i & 0\end{matrix}$ will not follow through and will commute and anti-commute, which shouldn't happen. – M. Al Jumaily Aug 31 '20 at 10:31
@M.AlJumaily, I don't agree that $Y =\begin{pmatrix}0& -i \ i &0 \end{pmatrix}$ anti-commutes with $I$ and I have proved that in the answer. – Davit Khachatryan Aug 31 '20 at 10:36
I think the definition which is given by both references is if $IY=−YI$ is satisfied, then both anti-commute. Furthermore, if we go into your definition, which is, $IY \neq −YI$, then what can you conclude from $YI \neq −IY$? – M. Al Jumaily Aug 31 '20 at 10:42
@M.AlJumaily, ${I, Y } = IY + YI = 2Y \ne 0$, so I conclude that $I$ and $Y$ don't anti-commute. – Davit Khachatryan Aug 31 '20 at 10:46
@M.AlJumaily, I didn't define anything and my statements are not in disagreement with your references. – Davit Khachatryan Aug 31 '20 at 10:47
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Alright, I have been working for long hours now and I need to sleep (it is 7am now). I will go over all of this and see what happens. Thank you for the help :)! – M. Al Jumaily Aug 31 '20 at 10:50
Don't drop the "global" phase when talking about the identity $Y = iZX$. $Z$ and $X$ actually do anti-commute, and this only works because $ZX = -iY$ but $XZ = +iY$, so that $ZX = -XZ$ and they thus anti-commute. – JSdJ Aug 31 '20 at 15:43
@JSdJ, just clarifying, my answer is about commutation relation between $Y$ and $I$...that they don't anti-commute (in the question, it was stated the contrary) and the answer is not about $X$ and $Z$. I haven't dropped the global phase. I have just said that $I$ and $Y$ don't anti-commute no matter if we have a global phase added to $Y$ or not. Even more... the only gate that anti-commutes with $I$ is the matrix with all $0$ entries. – Davit Khachatryan Aug 31 '20 at 16:00
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This was addressed towards M.AlJumaily, not towards you:) They dropped the $i$ somewhere in the comments, it was just a reminder to not do that:) – JSdJ Aug 31 '20 at 16:22
@JSdJ, thank you for the suggestion, I will surely keep it in mind. Also, the problem was a cache issue with Matlab where it kept running cached versions instead of the current script I had. A simple restart fixed the problem. The conclusion is that yes, the answer is they do commute. The second condition (which is written correctly) $IY = - YI$ did not execute after restarting. – M. Al Jumaily Sep 01 '20 at 05:50
@DavitKhachatryan, suppose that we are finding the relation between $X$ and $Z$. We know they anti-commute (end of page 3). How would you use your approach, which is $XZ \neq −ZX$, to show this? – M. Al Jumaily Sep 01 '20 at 07:36
@M.AlJumaily, my calculations are only about $I$ and $Y$, not about $X$ and $Z$. You have claimed that $I$ and $Y$ anti-commute...my answer and comments are about this claim...that they don't anti-commmute: the claim is wrong. BTW, $X$ and $Z$ anti-commute: ${ X, Z} = XZ + ZX = 0$, because $XZ = -ZX$. – Davit Khachatryan Sep 01 '20 at 08:56
@M.AlJumaily, I never said that $XZ \ne -ZX$ and my calculations didn't assume (were not about) that! – Davit Khachatryan Sep 01 '20 at 08:57
@DavitKhachatryan, Okay, so... we both agree with the definition of the anti-commutativity and your answer was targeting to prove that $I$ and $Y$ don't anti-commute (i.e., commute), correct? – M. Al Jumaily Sep 01 '20 at 09:11
@M.AlJumaily, the target is only that they don't anti-commute. The prove that they commute should be done separately. – Davit Khachatryan Sep 01 '20 at 09:14
@DavitKhachatryan, yes, so we both agree on the definition of anti-commutativity and that your answer was to only prove that the claim they anti-commute is false, correct? – M. Al Jumaily Sep 01 '20 at 09:18
@M.AlJumaily, if by anti-commutativity you mean ${A, B } = AB + BA =0$, yes I agree with this definition and yes my answer is only to prove that $IY \ne -YI$ (in the question, it was written the opposite). – Davit Khachatryan Sep 01 '20 at 09:23
@DavitKhachatryan, thanks for the clarification. May I suggest to kindly add that you are proving the statement $IY \ne -YI$ and that it is not a general rule of thumb (as I thought with the $X$ and $Z$ relation) for the next person to see the complete answer :). – M. Al Jumaily Sep 01 '20 at 09:53
@M.AlJumaily, I have added some details to my answer, I hope they are helpful :) – Davit Khachatryan Sep 01 '20 at 10:16
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@DavitKhachatryan, thank you for all of the help, the edit and the time :)! – M. Al Jumaily Sep 01 '20 at 10:22
@M.AlJumaily, you are welcome :) – Davit Khachatryan Sep 01 '20 at 10:26

The commutativity of $I$ and $Y$ in a stabilizer code

1 Answers1