# Quantum gate teleportation T gate

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.

• Well, for starters, $T Z T^{-1}$ seems pretty easy to simplify given that Ts commute with Zs. Jun 10 '19 at 20:43
• @CraigGidney You're right, I figured out $TZT^\dagger = Z$, I'll add it to the post.
– soud
Jun 10 '19 at 20:44
• Try explicitly computing $TXT^{-1}$. What does the matrix look like? 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.
• 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.
• 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 $TXT^\dagger$. Jun 11 '19 at 12:10