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I want to show that the stabilizer operators ($M_{0}, M_{1}, M_{2}, M_{3}$) for the 5-qubit quantum error correcting code:

If $M_{1} = [XXZIZ]$ and $M_{2} = [XZIZX]$

They commute iff $[M_{1},M_{2}]=0$.

I am given to believe that $[M_{1}, M_{2}] = M_{1}M_{2}-M_{2}M_{1}$

But $M_{1}M_{2} = I \otimes XZ \otimes Z \otimes Z \otimes ZX= I(XZ)ZZ(ZX)$

And $M_{2}M_{1}= I(ZX)ZZ(XZ)$

But, I must be misinterpreting the commutation operation because this only works if we use direct sum:

$M_{1}M_{2} - M_{2}M_{1} = I(-ZX)ZZ(-XZ) - I(ZX)ZZ(XZ) = I(-2ZX)II(-2XZ)$

But this only works if I use $\oplus$ so $I \oplus -2ZX \oplus I \oplus I \oplus -2ZX =0$

So, where did I go wrong with the deifnition of commutation? Why should I be using the direct sum here?

am567
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1 Answers1

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You shouldn't be using the direct sum! You've done most of the calculation, but not followed it through far enough.

You have: $$ M_1M_2=I\otimes XZ\otimes Z\otimes Z\otimes ZX. $$ Now you just need to know that $XZ=-ZX$. Thus, $$ M_1M_2=I\otimes (-ZX)\otimes Z\otimes Z\otimes (-XZ). $$ which is the same as $(-1)(-1)M_2M_1=M_2M_1$.

DaftWullie
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