I am trying to visualize a Hamiltonian H=$\hat{\sigma_x}$ $$ \hat{\sigma}_{x} = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) $$ acting on the state $| 1 \rangle$.

I can write the state of the qubit at time $t$ by Schrodinger's equation as: $$e^{-iHt} = e^{-i\hat{\sigma_x}t} \, .$$

By Euler's formula, since $H^2 = I$, the general state is $$| \psi (t) \rangle = \left( \begin{array}{cc} \cos t & -i \sin t \\ -i \sin t & \cos t \end{array} \right) \left( \begin{array}{cc} 0 \\ 1 \end{array} \right) \, .$$

Using $$\lvert\Psi\rangle=\cos(\theta / 2) \lvert 0\rangle + e^{i\phi}\sin(\theta / 2) \lvert 1\rangle$$

I write $| \psi \rangle $ as $$| \psi \rangle = \cos( 2t/2 + \pi/2)\left( \begin{array}{cc} 1 \\ 0 \end{array} \right) + e^{-i*0} \sin( 2t/2 + \pi/2)\left( \begin{array}{cc} 0 \\ 1 \end{array} \right) \, .$$

Thus, $ \theta = 2t + \pi$ and $\phi = 0$.

I am struggling with how to represent the angle $\theta$ on the Bloch sphere. Where would $\theta$ point in this case, where we have a $\pi$ term and $2t$ term together? Please give me a graphical answer so I can see the evolution of the qubit $|1\rangle$ and how to extend this visual image to cases of $-\pi/2$, $\pi/4$, $2t$, $4t$, etc.

  • $\begingroup$ From my extensive reading, I did not find an online source that gives a visual representation of such a case in a way that I understood. $\endgroup$
    – NewUser392
    May 30, 2016 at 16:45
  • $\begingroup$ Do you understand the reason that the state is written as $\cos(\theta/2) |0 \rangle + e^{i \phi} \sin(\theta / 2) | 1 \rangle$? Since you say you did extensive research I guess you found the Bloch Sphere, which is the geometrical representation of the state. $\endgroup$
    – DanielSank
    May 30, 2016 at 17:42
  • $\begingroup$ I read this physics.stackexchange.com/q/174562, but it does not exactly answer my question of where an angle like mine should be placed. For instance, the $\theta /2$ on the the green sphere in the image of the linked question would be difficult to see as doubled with a $\pi$ added. Do you say that I just graph $$ sin\theta cos\phi@DanielSank $\endgroup$
    – NewUser392
    May 30, 2016 at 18:34
  • $\begingroup$ I disagree with the last statement in your comment. Doubling the angle and adding $\pi$ can be understood just by drawing it! :) $\endgroup$
    – DanielSank
    May 30, 2016 at 18:39
  • $\begingroup$ Do you say that I just graph $$ sin\theta cos\phi , sin\theta sin\phi, cos\theta$$? $\endgroup$
    – NewUser392
    May 30, 2016 at 18:40

1 Answer 1


$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ The original post leaves out the strength of the Hamiltonian, which is confusing, so I'm going to put it back in. We write $H/\hbar = \Omega \sigma_x$.

Following the original post we have

\begin{align} \exp(-i H t / \hbar) = \exp \left( -i \Omega t \sigma_x \right) &= \left( \begin{array}{cc} \cos(\Omega t) & -i \sin(\Omega t) \\ -i \sin(\Omega t) & \cos(\Omega t) \end{array} \right)\\ &= \cos(\Omega t)1 - i \sin(\Omega t)\sigma_x \, . \end{align} Therefore $$\ket{\psi(t)} = \exp \left( -i \Omega t \sigma_x \right) \ket{1} = \cos(\Omega t)\ket{1} - i \sin(\Omega t) \ket{0} \, . \tag{1}$$ The original post correctly writes $$\ket{\psi} = \cos(\theta/2)\ket{0} + e^{i \phi} \sin(\theta/2) \ket{1} \, \tag{2}$$ where here $\theta$ and $\phi$ are geometric angles on the Bloch sphere. In other words, the 3D position of the state on the Bloch sphere is $$(\sin(\theta) \cos(\phi), \sin(\theta)\sin(\phi), \cos(\theta)) \, .$$

If we want to plot the trajectory of the state under the influence of $H$ we need to solve for the angles in terms of $t$. To do this we have to get rid of the $i$ in front of $\ket{0}$ in $(1)$ and move it to the $\ket{1}$ term. We do this by multiplying the state by a global phase $i$, yielding $$\ket{\psi(t)} = \sin(\Omega t) \ket{0} + i \cos(\Omega t) \ket{1} \, . \tag{3}$$

Matching coefficients in (2) and (3) gives $$\Omega t = \theta/2 + \pi/2 \qquad \text{and} \qquad \phi = 3\pi/2 \, .$$ Inverting the equation with $t$ gives $$\theta = 2(\Omega t - \pi/2)$$ so then the 3D point is at $$ (\sin(2\Omega t - \pi) \cos(3\pi/2), \sin(2\Omega t - \pi) \sin(3\pi/2), \cos(2\Omega t - \pi))$$ which, using some trigonometry, becomes $$(0, \sin(2\Omega t), -\cos(2\Omega t)) \, .$$

  • $\begingroup$ I have a suspicion that the sign of the $y$-component in the final answer is backwards, but I don't see where I would have made a mistake. The $z$-component is right though. $\endgroup$
    – DanielSank
    May 30, 2016 at 19:51
  • $\begingroup$ Ah, found it and fixed it. $\endgroup$
    – DanielSank
    May 30, 2016 at 20:02
  • $\begingroup$ Thank you, I'm working thru this by hand and trying to understand it. I think the key is applying the global phase instead of forcing $i$ to fit the $\psi$ expression. $\endgroup$
    – NewUser392
    May 30, 2016 at 20:13
  • $\begingroup$ You're absolutely right; you have to make sure the coefficient in front of $|0\rangle$ is real in order to match the Bloch sphere representation. To do that, I multiplied by $i$, as indicated in the post. Also, it took me a while to figure out that $\phi = 3\pi/2$ to make all the factors work out. That's another key element. $\endgroup$
    – DanielSank
    May 30, 2016 at 20:19
  • $\begingroup$ One last question on this thread. Would you say that this is a counterclockwise transformation on the Bloch sphere, where $|1\rangle$ goes from $|1 \rangle$ to $|0 \rangle$ to $|1 \rangle$, where its path crosses (some phase of) $|i \rangle$ and $|-i \rangle$? $\endgroup$
    – NewUser392
    May 31, 2016 at 0:19

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