I am faced with the problem of teleporting certain gates using modified Bell states. For example, I have solved the problem with the $S$ gate, which is defined as following:

Alice and Bob share a qubit in the state $S|\Phi^+\rangle = (|00⟩ + i|11⟩) / \sqrt2$. Alice has a qubit in the state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ which she wants to send to Bob as $S|\psi\rangle = \alpha|0\rangle + \beta i|1\rangle$. The part that needs to be implemented is Bob's fix-up part, so he ends up with $S|\psi\rangle$. Bob is only allowed to use Pauli and H gates. After the teleportation, Bob has one of the four states $S|\psi\rangle$, $SX|\psi\rangle$, $SZ|\psi\rangle$ or $SZX|\psi\rangle$. We can find the decode circuit by rewriting those as $S|\psi\rangle$, $(SXS^\dagger)S|\psi\rangle$, $(SZS^\dagger)S|\psi\rangle$ and $(SZXS^\dagger)S|\psi\rangle$. From here, we can decode to $S|\psi\rangle$ by finding $SXS^\dagger = Y$ and $SZS^\dagger = Z$.

The problem comes with trying this with the $T$ gate. Alice and Bob share the state $T|\Phi^+\rangle = (|00⟩ + e^{i\pi / 4}|11⟩) / \sqrt2$. Alice wants to send her qubit in the state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ to Bob as $T|\psi\rangle = \alpha|0\rangle + \beta e^{i\pi / 4}|1\rangle$. Bob is only allowed to use Pauli, $H$ and $S$ gates. This gives the problem of finding a Pauli, $H$ and $S$ gate decomposition of $TXT^\dagger$ and $TZT^\dagger$. However, this seems impossible to me, given you can't create a $T$ gate from Pauli, $H$ and $S$ gates. Clearly $TZT^\dagger = Z$, but I can't seem to figure $TXT^\dagger$ out.

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    $\begingroup$ Well, for starters, $T Z T^{-1}$ seems pretty easy to simplify given that Ts commute with Zs. $\endgroup$ – Craig Gidney Jun 10 '19 at 20:43
  • $\begingroup$ @CraigGidney You're right, I figured out $TZT^\dagger = Z$, I'll add it to the post. $\endgroup$ – soud Jun 10 '19 at 20:44
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    $\begingroup$ Try explicitly computing $TXT^{-1}$. What does the matrix look like? $\endgroup$ – Craig Gidney Jun 10 '19 at 23:04

Think about the following sequence: $$ XT^\dagger X=\left(\begin{array}{cc} e^{-i\pi/4} & 0 \\ 0 & 1 \end{array}\right)=e^{-i\pi/4}T. $$ So, that lets us write $$ TXT^\dagger=e^{-i\pi/4}TTX=e^{-i\pi/4}SX $$ Up to some phase, you have the decomposition that you want. In this context, that phase should be an irrelevant global phase that you can ignore.

  • $\begingroup$ Thanks for your reply. I can see how $XT^\dagger X = e^{-i \pi / 4}T$, but I can't see how you then got to $TXT^\dagger = e^{-i \pi / 4}TTX$. I see they're equivalent, but not how you got there in the algebra. $\endgroup$ – soud Jun 11 '19 at 10:48
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    $\begingroup$ I post-multiplied the first equation by X to give $XT^\dagger=e^{-i\pi/4}TX$ and used it to rewrite that $XT^\dagger$ part of $TXT^\dagger$. $\endgroup$ – DaftWullie Jun 11 '19 at 12:10

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