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Suppose we have a system with a lower state $\left|{\downarrow}\right\rangle$ and and upper state $\left|{\uparrow}\right\rangle$, coupled by a pulsed electromagnetic field. I thought that from an initial state $\left|{\uparrow}\right\rangle$ we can obtain the superposition state $\frac{1}{\sqrt{2}}\left(\left|{\downarrow}\right\rangle + \left|{\uparrow}\right\rangle\right)$ by applying a $\pi/2$ pulse and swap the states $\left|{\uparrow}\right\rangle \longrightarrow \left|{\downarrow}\right\rangle$ by applying a $\pi$ pulse. This would mean that a $2\pi$ pulse brings the system back to its original state. But in [1] I read

[the $2\pi$ pulse] simply reverses the sign of any component in the $\left|{1}\right\rangle\left|{\uparrow}\right\rangle$ state by inducing a complete Rabi cycle from $\left|{1}\right\rangle\left|{\uparrow}\right\rangle \rightarrow \left|{0}\right\rangle\left|\mathrm{aux}\right\rangle \rightarrow -\left|{1}\right\rangle\left|{\uparrow}\right\rangle$,

where $\left|{0}\right\rangle\left|\mathrm{aux}\right\rangle$ is an auxiliary level. Where is my mistake? Or in other words: what is the correct evolution of the state $\left|{\uparrow}\right\rangle$, when applying $1,2,3,\ldots,\pi/2$ pulses?

EDIT: Considering Craig Gidneys answer: in the basis $\left|{\uparrow}\right\rangle = (1, 0)^T, \left|{\downarrow}\right\rangle = (0,1)^T$ would the transformation $$ e^{iXπ/2}=\frac{1}{\sqrt{2}} \left(\begin{matrix} 1 & i \\ i & 1 \end{matrix}\right) $$ represent the action of a $\pi/2$ pulse on a 2-level system? How can this be derived?

  1. Monroe, C., Meekhof, D.M., King, B.E., Itano, W.M. and Wineland, D.J.. Demonstration of a fundamental quantum logic gate. Phys. Rev. Lett. 75(25), p.4714 (1995), NIST eprint.
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The phrase "$r\pi$ pulse" (for $r\in\mathbb R$, usually a multiple of $1/2$) is a specific convention in quantum optics, and particularly it is more specific than the definition you quote,

from an initial state $\left|{\uparrow}\right\rangle$ we can obtain the superposition state $\frac{1}{\sqrt{2}}\left(\left|{\downarrow}\right\rangle + \left|{\uparrow}\right\rangle\right)$ by applying a $\pi/2$ pulse and swap the states $\left|{\uparrow}\right\rangle \longrightarrow \left|{\downarrow}\right\rangle$ by applying a $\pi$ pulse.

Instead, an $r\pi$ pulse usually refers to the specific transformation $$ U(r) = e^{-ir\pi X/2} = \begin{pmatrix} \cos(r\pi/2) & -i\sin(r\pi/2) \\ -i\sin(r\pi/2) & \phantom{-i}\cos(r\pi/2)\end{pmatrix}. $$ It does the state transformations you mention, but in addition it has a very specific convention on the phases of the resulting states: it's not the same to do $\left|{\uparrow}\right\rangle \to \frac{1}{\sqrt{2}}\left(\left|{\downarrow}\right\rangle + \left|{\uparrow}\right\rangle\right)$ as doing $\left|{\uparrow}\right\rangle \to e^{i\theta}\frac{1}{\sqrt{2}}\left(\left|{\downarrow}\right\rangle + \left|{\uparrow}\right\rangle\right)$, even though population-wise they might look the same. The terms $\pi/2$ pulse, $\pi$ pulse, and $2\pi$ pulse are all associated with a specific convention on what phase to take.

Among other things, this implies that a $2\pi$ pulse takes the system from any state $|\psi⟩$ through to $-|\psi⟩$, flipping the phase by $\pi$. There are other transformations that do the same things to the populations, but we just don't use those terms for them.

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  • $\begingroup$ Sorry for coming back to this, but having now come across the derivation of rabi oscillations, i think that the sign of the $U_{21}$ should be negative as well? $\endgroup$ – Schlabonski Jul 26 '16 at 13:49
  • $\begingroup$ @Schlabonski Yes. But the precise formulation doesn't matter, ends up depending on the precise formulation of Pauli matrices you're using, and in any case the $r\pi$ pulses need not be about the $x$ axis to begin with. I.e. that's a nitpick. $\endgroup$ – Emilio Pisanty Jul 26 '16 at 13:59
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Someone please correct me if I'm wrong, but I suspect it might just be a bit ambiguous, like "beam splitter".

The $\pi/2$ part of the name refers to the amount of rotation in the Bloch sphere, but single-qubit quantum operations have an extra degree of freedom on top of the amount of rotation and the axis of rotation: the global phase factor. For example, both $X^{1/2} = (1+i)/2 \begin{bmatrix}1&i\\i&1\end{bmatrix}$ and $e^{i X \pi /2} = \frac{1}{\sqrt 2}\begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix}$ rotate by $\pi/2$ degrees around the X axis, but $(X^{1/2})^2 = X$ while $(e^{i X \pi /2})^2=iX$. And so $(X^{1/2})^4 = I$ while $(e^{i X \pi /2})^4=-I$.

(The global phase factor of an operation usually doesn't matter, but if you start applying the operation conditionally it can become relevant.)

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  • $\begingroup$ Okay, so the latter transformation would correspond to a $\pi/2$ pulse and the overall phase change can be ommited when no other states are involved (which is not the case in the mentioned paper)? Do you know why authors also often ommit the relative phase of $i$ when it comes to the entangled state? I mean it is equivalent when it comes to expectation values but of some importance for the transformation (like above)... $\endgroup$ – Schlabonski Jul 9 '16 at 18:04
  • $\begingroup$ @Schlabonski Unfortunately I don't know enough about the conventions used in physics papers to give a good answer to that. I just know that "do the quantum operation corresponding to this rotation" is slightly ambiguous. $\endgroup$ – Craig Gidney Jul 9 '16 at 18:24
  • $\begingroup$ Hi Craig. You're correct that there's an extra degree of freedom; in fact all quantum systems have an arbitrary global phase. That phase doesn't affect the theory's prediction of measurements at all. Note, however, that if you have two qubits, then what used to be a global phase in the one-qubit system is now no longer global. Therefore, it now has physical consequences. In fact, in several superconducting qubit systems this phase is the essential ingredient of the controlled Z gate. $\endgroup$ – DanielSank Jul 15 '16 at 12:59

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