# The minimum time for a quantum state to evolve to an orthogonal state

I'm reading this paper by Margolus and Levitin

The maximum speed of dynamical evolution: http://arxiv.org/abs/quant-ph/9710043

about the so called Margolus-Levitin theorem. For the main result, basically, they take a state $$\left|\psi_t\right\rangle=\sum_nc_n\,e^{-i\frac{E_n}{\hbar}t}\left|E_n\right\rangle$$ and find that the smallest value of $t$ such that $S(t)=\left\langle\psi_0\big|\psi_t\right\rangle=0$, where $\left|\psi_0\right\rangle=\sum_nc_n\left|E_n\right\rangle$, is $$t=\frac{h}{4E}$$ where $E=\sum_n\left|c_n\right|^2E_n$ is the average energy.

They get this using the inequality $$\mathrm{Re}(S)-\frac{2}{\pi}\mathrm{Im}(S)\geq1-\frac{2E}{\pi\hbar}t$$ (there's actually a sign typo in the last line of their expression (8)) which was obtained using $$\cos{x}+\frac{2}{\pi}\sin{x}\geq1-\frac{2}{\pi}x$$ valid for $x\geq0$, and then they just take $t$ such that $S(t)=0$.

What is special about this inequality? It just seems to have come out of nowhere. It's reasonable that it should include both the real and the imaginary parts of $S$, but why not just take $\mathrm{Re}(S)+\mathrm{Im}(S)$ on the LHS, for example, or why not any other linear combination whatsoever; also, what is special about the RHS?

So, aside of this details, I would like to know how to obtain this result in a more systematic way, or in any case, how is the procedure by Margolus and Levitin justified or explained with more detail.

• Someone else probably already knows, so perhaps they'll answer, but I suspect that the factor of $2/\pi$ is the smallest such factor that makes the inequality work with the right-hand side being linear. I also suspect that you want the right-hand side linear so that you can directly solve for $t$. Perhaps you can get a better bound by taking the right-hand side out to higher-order terms, but in that case, it makes it harder to solve for $t$, I imagine. – march Nov 17 '15 at 18:19
• See this Mathematica gif. I am plotting $\cos(x) + \alpha \sin(x)$ and $1 - \alpha x$ together, and exactly when $\alpha = 2/\pi$, the inequality holds. – march Nov 17 '15 at 18:26
• Look at this!. There is a sense in which the $2/\pi$ gives you the best bound, in the sense that the slope on the linear function is the smallest at that value. I'm still not sure exactly why this is important for the calculation, but perhaps you can go from there! I'm going to continue thinking about it as well. – march Nov 17 '15 at 22:38
• The more I think about it, the less I get it. Supposing $b=1$, we can actually make $a$ arbitrarily large, which allows us to chase $c$ arbitrarily large, which in turn allows us to choose $t$ arbitrarily small. Now, the inequality in a sense gets worse, as the gif shows, so maybe there's something to the idea that you need to be as close as possible. I'm just missing something, so I don't feel comfortable writing up an answer. – march Nov 19 '15 at 6:56
• There's no sign typo because $Im(e^{-iEt/\hbar})=-\sin(Et/\hbar)$. – OON Nov 20 '15 at 17:32

It took me a while to figure it out, even though it was obvious from the start that it should be ridiculously simple. Reminded me of the (in)famous question about the plane and the conveyor belt in that sense. The longest part was to figure out why we need to maximize time even though we're supposed to look for the minimum time. I guess that's because I'm not used to the concept of a tight inequality.

Let's consider the inequality $\cos x \ge b - cx - a\sin x$, where $x=\frac{E}{\hbar}t$. What's important here is that $x$ is proportional to $t$ so we can forget about $t$ for a while and concentrate on $x$ to simplify our formulas.

Of course, as March noted in the comment, we can just take some arbitrarily large value of $c$ and get, say, $\cos x \ge 1 - 1000 x - \sin x$, and it follows that $x \ge \frac{1}{1000}$ then. Or we could set $c=10000$ and get even smaller $x$. But that is not our objective here! We aren't looking for an inequality of the form $x \ge x_0$, where $x_0$ is some arbitrarily small number. If that was the case, we wouldn't even have to worry about writing complicated inequalities. We could just say $x \ge 0$ and be done with it. But that certainly doesn't mean that $t=0$ is the earliest possible time when the state $\left | \psi_0 \right\rangle$ can evolve into an orthogonal state. It would only mean that the time $t_0$ when it happens satisfies the $t_0\ge0$ inequality, which is trivial.

The same goes for other small values of $x_0$. In fact, it's pretty obvious that if we prove that $x \ge x_0$ for some $x_0$, then it would certainly be true that $x \ge x_0^{\prime}$ for any $x_0^{\prime}<x_0$. On the other hand, we can't generally state that if we take some number $x_0^{\prime\prime}>x_0$, then $x\ge x_0^{\prime\prime}$ will still be true.

It follows that we're looking for the largest possible $x_0$ so that $x \ge x_0$ is still true. Or we can say that we're looking for the most restrictive inequality for $x$, or simply a tight inequality. The logic here is that it's the inequality that contains the maximum possible information. What is the radius of Earth? If we don't know exactly we can say it's more than 1 millimeter or more than 1 meter. Which of these statements contain more information? Obviously the latter. If we say it's more than 6000 km, we'll be giving even more information.

That essentially means that the inequality should turn into the equality for some values of $x$ because once we get to that point, we can't get closer without breaking it at some points. Moreover, since both $\sin x$ and $\cos x$ disappear when we substitute the real and imaginary parts, what we'll get in the end will be $0 \ge b-cx$ or $x \ge \frac{b}{c}$. That means we should try to maximize $\frac{b}{c}$ while keeping the inequality tight at the same time. Therefore, we should look for the largest possible $b$ and the smallest possible $c$.

Let's substitute $x=0$ into the inequality

$$\cos x \ge b - cx - a \sin x.$$

We have $b\le 1$, all other parameters disappear. Obviously the inequality will still hold true for any $b<1$, but we're looking for the largest $b$, and it's obvious that the inequality will get violated at $x=0$ for any $b>1$, so $b=1$ is our best result. Note that the inequality turns into the equality at that point.

Now we're looking for $a$ and $c$. And we're trying to minimize $c$, remember! But we don't actually care about $a$ as long as the inequality holds true. This leads us to thinking about other values of $x$ where $a$ doesn't matter. We've already checked $x=0$, so let's consider $x=\pi$ now:

$$-1 \ge 1 - c \pi.$$

This gives us $c \ge \frac{2}{\pi}$. Obviously the minimum such $c$ is $\frac{2}{\pi}$. Note that again, the inequality turns into the equality at that point too.

Now we only have to pick any value for $a$ such that the inequality still holds true for every $x \ge 0$. Since the inequality turns into equality for $x=\pi$, where $\sin x$ goes from positive to negative, it follows that we must make sure that $\cos x + a \sin x$ doesn't cross $1 - \frac{2}{\pi} x$ around $x=\pi$, that is, $1 - \frac{2}{\pi} x$ is the tangent to $\cos x + a \sin x$ at that point. By differentiating them, we get

$$-\sin x + a \cos x = -\frac{2}{\pi}.$$

For $x=\pi$, this gives us $a=\frac{2}{\pi}$. The following plot illustrates how deviating from this value breaks things:

To sum it up, we were looking for the largest $\frac{b}{c}$ such that the inequality holds true. If we took $b$ larger than 1, it would break at $x=0$, so that's one restriction. Then we prove that for $x=\pi$ the largest possible $c$ is $\frac{2}{\pi}$ or else it would break at that point. And finally, that gives us the value of $a$ we can't deviate from without crossing the $1-\frac{2}{\pi}$ line.

• This sort of reasoning feels right, and it was the avenue I was going to go down (and it was what I was trying to show with those gifs in the comments), but I still don't have a good understanding of why that bound needs to be the most restrictive. (By the way, if you want to embed the gifs in your answer, feel free to do so.) – march Nov 24 '15 at 4:36
• @march, simply because it's the objective. Suppose you found that $t \ge 1$ for a particular case. OK, but how much information does this fact give us? It could very well be that $t$ never gets smaller than, say, 100. If that's the case, the inequality $t \ge 1$ is still correct, only is not that useful. So we're looking for $t_0$ such that $t \ge t_0$, but we can't find WLOG another $t_1$ such that $t \ge t_1$ is still true. That means, we're looking for the largest such $t_0$. I believe that's what you call a tight inequality. – Sergei Tachenov Nov 24 '15 at 15:30
• @march, as for the gifs, they lack a legend so it's hard to figure out quickly what is plotted. And I don't have a useful tool at hand to produce plots with legends. Maybe I'll look for some and update my answer later. – Sergei Tachenov Nov 24 '15 at 15:32