Skip to main content
edited title
Link
Sanchayan Dutta
  • 17.8k
  • 8
  • 49
  • 111

Hadamard gate as a product of Rx$R_x$, Rz$R_z$ and a phase

edited title
Link
Sanchayan Dutta
  • 17.8k
  • 8
  • 49
  • 111

Hadamard gate as a product of $R_x$Rx, $R_z$Rz and a phase

edited title; edited tags
Source Link
Sanchayan Dutta
  • 17.8k
  • 8
  • 49
  • 111

Writing the Hadamard gate as a product of Rx$R_x$, Rz$R_z$ and a phase

I am having problems with this task.

Since the Hadamard gate rotates a state 180 degrees$180°$ about the $\hat{n} = \frac{\hat{x} + \hat{z}}{\sqrt{2}}$ axis, I imagine the solution can be found the following way:

First rotate $\hat{n}$ so it lies in the z$z$-y$y$ plane: $$R_z(\pi/2)$$ Then rotate $\hat{n}$ so it is parallell with the z$z$-axis: $$R_x(\pi/4)$$ Now do the desired rotation about the z$z$-axis: $$R_z(\pi)$$ Then rotate $\hat{n}$ back to its original position. The Hadamard gate can then be written: $$H = e^{i\lambda}R_z(-\pi/2)R_x(-\pi/4) R_z(\pi) R_x(\pi/4) R_z(\pi/2) $$ However, this doesn't seem to work. Can someone explain where my logic is flawed? Thank you very much

Writing the Hadamard gate as a product of Rx, Rz and a phase

I am having problems with this task.

Since the Hadamard gate rotates a state 180 degrees about the $\hat{n} = \frac{\hat{x} + \hat{z}}{\sqrt{2}}$ axis, I imagine the solution can be found the following way:

First rotate $\hat{n}$ so it lies in the z-y plane: $$R_z(\pi/2)$$ Then rotate $\hat{n}$ so it is parallell with the z-axis: $$R_x(\pi/4)$$ Now do the desired rotation about the z-axis: $$R_z(\pi)$$ Then rotate $\hat{n}$ back to its original position. The Hadamard gate can then be written: $$H = e^{i\lambda}R_z(-\pi/2)R_x(-\pi/4) R_z(\pi) R_x(\pi/4) R_z(\pi/2) $$ However, this doesn't seem to work. Can someone explain where my logic is flawed? Thank you very much

Hadamard gate as a product of $R_x$, $R_z$ and a phase

I am having problems with this task.

Since the Hadamard gate rotates a state $180°$ about the $\hat{n} = \frac{\hat{x} + \hat{z}}{\sqrt{2}}$ axis, I imagine the solution can be found the following way:

First rotate $\hat{n}$ so it lies in the $z$-$y$ plane: $$R_z(\pi/2)$$ Then rotate $\hat{n}$ so it is parallell with the $z$-axis: $$R_x(\pi/4)$$ Now do the desired rotation about the $z$-axis: $$R_z(\pi)$$ Then rotate $\hat{n}$ back to its original position. The Hadamard gate can then be written: $$H = e^{i\lambda}R_z(-\pi/2)R_x(-\pi/4) R_z(\pi) R_x(\pi/4) R_z(\pi/2) $$ However, this doesn't seem to work. Can someone explain where my logic is flawed? Thank you very much

Source Link
QCQCQC
  • 449
  • 3
  • 10
Loading