1
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ That's something you will understand when you'll get older. Now finish your plate $\endgroup$ Jun 1 at 7:26

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.