I'm newbie to quantum computing, i read about these two gates on IBM quantum computing docs, but I can't seem to understand what is t-gate and tdg or t-dagger, can someone give me a simple explanation about these two gates? Also S-gate if that is possible!

Thank you so much.

  • 5
    $\begingroup$ That's something you will understand when you'll get older. Now finish your plate $\endgroup$ Jun 1, 2022 at 7:26

1 Answer 1


tdg is the method used to apply $T^\dagger$ (read T dagger). Thus, there are no differences between these two.

For a quantum gate $U$, $U^\dagger$ is the inverse of $U$. That is, if you apply $U$ on a given state $|\psi\rangle$ and then $U^\dagger$, you will be back in the state $|\psi\rangle$ again.

So now, what is $T$? Or rather, why do we care about $T$? You see, there are these gates, $H$, $S$ and the $CNOT$ using which we can do some things. However, we are quite limited with them, there is no way to construct a Toffoli gate (that is, an $X$ gate controlled on two qubits) using them for instance. However, if we do allow ourselves to use these gates and the $T$ gate, then we can construct any quantum gate we'd like.

The $S$ gate is simply the $T$ gate applied twice: $T^2=S$.

On a more "math" level, the $T$ gate is the following matrix: $$\begin{pmatrix}1&0\\0&\mathrm{e}^{\mathrm{i}\frac\pi4}\end{pmatrix}$$ That is, it applies a phase of $\frac\pi4$ to the $|1\rangle$ state and leaves $|0\rangle$ untouched.

I'm not sure that this is a gate you'll have to deal with often. It does appear a lot in the Quantum Circuits you'll build, but I'm not sure that you'll use it directly when defining your own Quantum Gates. For instance, if you use a Toffoli gate in your circuit, under the hood the gates that will be applied to your three qubits are those:

Toffoli gate

Thus, while you will reason without this $T$ gate in most cases, it will be here when decomposing the circuit into the basic gates using which you'll build it.


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