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While reading a paper by Aaronson and Gottesman, I came across a comment as follows:

$$\text{If }|\psi\rangle \neq |\phi\rangle, Stab(|\psi\rangle) \neq Stab(|\phi\rangle),$$

where $|\psi\rangle$ and $|\phi\rangle$ are two arbitrary states, and $Stab(|\psi\rangle)$ is defined to be a set of unitaries that stabilizes $|\psi\rangle$. I tried to prove it by contradiction. However, I was stuck and wonder if anyone could nudge me in the right direction. Thanks in advance.

Following is my argument thus far. Suppose towards contradiction that $Stab(|\psi\rangle) = Stab(|\phi\rangle)$, it means all elements $s \in Stab(|\psi\rangle)$ stabilizes $|\phi\rangle$. That is $s|\phi\rangle = |\phi\rangle$.

Can we complete the proof using linear algebra?

-----------------------------------Follow-up----------------------------------

Thank you everyone for the help and suggestions. Below is the updated proof. If you have any further comments, please let me know! Thank you!

Note that here we are reasoning up to the global phase. This means when $|\psi\rangle$ and $|\phi\rangle$ are colinear (i.e., differ by global phase), $Stab(|\psi\rangle) = Stab(|\phi\rangle)$.

When $|\psi\rangle$ and $|\phi\rangle$ are not colinear and $|\psi\rangle \neq |\phi\rangle$, consider $|\psi\rangle\langle\psi| \in Stab(|\psi\rangle)$. Suppose towards contradiction that $Stab(|\psi\rangle) = Stab(|\phi\rangle)$. Then $|\psi\rangle\langle\psi|$ must stabilize $|\phi\rangle$: $$|\psi\rangle\langle\psi|(|\phi\rangle) = |\phi\rangle.$$

Multiplying both sides of the equation to the left by $\langle \phi|$ yields $$\langle\phi|(|\psi\rangle\langle\psi||\phi\rangle) = \langle\phi|\phi\rangle \Rightarrow \lvert\langle\phi|\psi\rangle\rvert^2 = 1.$$ This means $|\psi\rangle = \pm|\phi\rangle$, a contradiction.

glS
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SML0712
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    To give a watertight answer, you should add the group in which you take the stabilizer of your states. If it's the full unitary group Danylo Y's answer works. In general, the group has to fulfill some minimum requirements, otherwise the statement is obviously false. – Markus Heinrich May 27 '22 at 11:04
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    (In the Aaranson-Gottesman paper, they refer to the full unitary group at this point, so Danylo Y's answer works) – Markus Heinrich May 27 '22 at 11:11

1 Answers1

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You just need to show that there exists a unitary $S$ such that $S|\psi\rangle = |\psi\rangle$, but $S|\phi\rangle \neq |\phi\rangle$.

If $|\psi\rangle, |\phi\rangle$ are not collinear then you can pick any $|\phi'\rangle \neq |\phi\rangle$ such that $\langle \psi|\phi'\rangle = \langle \psi|\phi\rangle$. Since inner products are equal there exists a unitary $S$ such that $S|\psi\rangle = |\psi\rangle$, $S|\phi\rangle = |\phi'\rangle$.

If $|\psi\rangle, |\phi\rangle$ are collinear (that is, differ by global phase) then $Stab(|\psi\rangle) = Stab(|\phi\rangle)$.

Update

You can take $S = 2|\psi\rangle\langle\psi|-I$.

Danylo Y
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  • Ah, thank you! I added your comment along with a short argument, and they together give a contradiction. – SML0712 May 27 '22 at 15:45
  • @SML0712 You've moved to the right direction, however $|\psi\rangle\langle\psi|$ is not unitary. Take unitary $S = 2|\psi\rangle\langle\psi|-I$, then your argument works. Though if $\lvert\langle\phi|\psi\rangle\rvert^2 = 1$ then $|\psi\rangle=e^{i\gamma} |\phi\rangle$. – Danylo Y May 27 '22 at 18:03