T-dagger gate
| Identifier |
Operator |
Example statement |
| Tdag |
\(T^\dagger\) |
Tdag q[0] |
Description
The T-dagger gate, i.e., the complex conjugate (inverse) of the T gate,
is an anti-clockwise rotation of \(-\pi / 4\) [rad] about the \(\hat{\mathbf{z}}\)-axis and
a global phase of \(-\pi / 8\) [rad].
It is equal to half the \(S^{\dagger}\) rotation and a quarter of the inverse of the \(Z\) rotation:
\(T = (S^{\dagger})^{1/2} = S^{-1/2} = Z^{-1/4}\).
Representation
\[\begin{align}
T^\dagger &= \left(\begin{matrix}
1 & 0 \\
0 & \frac{1 - i}{\sqrt{2}}
\end{matrix}\right)
\end{align}\]
Any single-qubit operation in \(U(2)\) (including global phase) can be expressed by 5 parameters in the
canonical representation \(R_\hat{\mathbf{n}}\)
\[R_\hat{\mathbf{n}}\left([n_x, n_y, n_z]^T, \theta, \phi\right) = e^{i\phi} \cdot e^{-i\frac{\theta}{2}\left(n_x\cdot\sigma_x + n_y\cdot\sigma_y + n_z\cdot\sigma_z\right)},\]
where \(\hat{\mathbf{n}}=[n_x, n_y, n_z]^T\) denotes the axis of rotation, \(\theta\in(-\pi, \pi]\) the angle of rotation [rad], and \(\phi\in[0,2\pi)\) the global phase angle [rad].
The T-dagger gate is given by:
\[\begin{align}
T^\dagger &= R_\hat{\mathbf{n}}\left([0, 0, 1]^T, -\frac{\pi}{4}, -\frac{\pi}{8}\right) = e^{-i\frac{\pi}{8}} \cdot e^{i\frac{\pi}{8}\sigma_z}, \\
\\
T^\dagger &= \left(\begin{matrix}
1 & 0 \\
0 & \frac{1 - i}{\sqrt{2}}
\end{matrix}\right).
\end{align}\]
In the Hadamard basis \(\{|+\rangle, |-\rangle\}\), the T-dagger gate \(T^\dagger_H\) is given by:
\[T^\dagger_H = HT^\dagger H = \frac{1}{2\sqrt{2}}\left(\begin{matrix}
\sqrt{2} + 1 - i & \sqrt{2} - 1 + i \\
\sqrt{2} - 1 + i & \sqrt{2} + 1 - i
\end{matrix}\right).\]
Operation examples
Standard basis
\[\begin{align}
T^\dagger\,|0\rangle &= |0\rangle \\
\\
T^\dagger\,|1\rangle &= \frac{1 - i}{\sqrt{2}} |1\rangle \\
\end{align}\]
Hadamard basis
\[\begin{align}
T^\dagger\,|+\rangle &= \frac{\sqrt{2} + 1 - i}{2\sqrt{2}}|+\rangle + \frac{\sqrt{2} - 1 + i}{2\sqrt{2}}|-\rangle \\
\\
T^\dagger\,|-\rangle &= \frac{\sqrt{2} - 1 + i}{2\sqrt{2}}|+\rangle + \frac{\sqrt{2} + 1 - i}{2\sqrt{2}}|-\rangle \\
\end{align}\]