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I Mapping between rotations and operations: sign & handedness

  1. Mar 14, 2016 #1

    Strilanc

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    I have a toy quantum circuit simulator that I work on. I want to visually represent operations in multiple ways: as a Hamiltonian, as a unitary matrix, and as a Bloch sphere rotation. I want to double-check that I haven't flipped anything.

    I'll focus a concrete example: is this animation correct?

    9qVpM6A.gif

    We're in a right-handed coordinate system. Z points up, X points right, and Y points away. I have a qubit; perhaps embodied in the spin of an electron. I begin rotating the qubit around the X axis at 0.5 Hz, so that after a quarter second it's at been rotated by +45 degrees.

    The Hamiltonian for this operation is ##H = X = \begin{bmatrix} 0&1\\1& 0\end{bmatrix}## (I think). The unitary form is:

    ##\begin{align} U(t) &= \exp(-i t H)\\&= \exp(-i t X)\\&= \exp(-i t (1)) \frac{1}{2} \begin{bmatrix} 1&1\\1&1 \end{bmatrix} + \exp(-i t (-1)) \frac{1}{2} \begin{bmatrix} 1&-1\\-1&1 \end{bmatrix} \\&=\frac{1}{2} \begin{bmatrix} e^{-it}+e^{it} & e^{-it}-e^{it} \\ e^{-it}-e^{it} & e^{-it}+e^{it} \end{bmatrix}\\&=\begin{bmatrix} \cos t & -i \sin t \\ -i \sin t & \cos t \end{bmatrix}\end{align}##

    Does that all sound right? Did I make a sign error or a handedness error? The thing I'm most unsure about is the minus sign in ##\exp(-i t H)##. It seems... really unnecessary.
     
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  3. Mar 15, 2016 #2

    haruspex

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    Should det U = 1? If so, you must have a sign error in your last equation.
     
  4. Mar 15, 2016 #3

    Strilanc

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    When I calculate the determinant, I get 1:

    ##\begin{align} \det \begin{bmatrix} \cos t & -i \sin t \\ -i \sin t & \cos t \end{bmatrix} &= (\cos t)^2 - (-i \sin t)^2 \\&= \cos^2 t - (-i)^2 \sin^2 t \\&= \cos^2 t - (-1) \sin^2 t \\&= \cos^2 t + \sin^2 t \\&= 1 \end{align}##
     
  5. Mar 15, 2016 #4

    haruspex

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    Sorry, you're right. I lost track of the number of negations.
     
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