3
$\begingroup$

Suppose we have an operator $U$, and a register $|\lambda\rangle$ in an eigenstate of $U$ with eigenvalue $\lambda=1$. Repeatedly applying $U$ to $|\lambda\rangle$ does not affect $|\lambda\rangle$ - that is, $\langle\lambda|U^r|\lambda\rangle=1$ for all $r\in \mathbb{Z}$.

However suppose we have another register $|\mu\rangle$, which is only "close to" the eigenstate $|\lambda\rangle$. That is, suppose we have a state $|\mu\rangle$ with $\langle\lambda|\mu\rangle=1-\delta$, for some small $\delta$. I believe repeated applications of $U$ will eventually scramble $|\mu\rangle$ beyond recognition. But, with respect to $\delta$, how quickly does this scrambling take place?

How much do we have to know about $U$ and $\langle\lambda|\mu\rangle$ to be able to say anything about how big $r$ has to be before $\langle\mu|U^r|\mu\rangle\lt 1/3$?

I think this is akin to asking for more spectral properties of $U$. Do we need to know about other eigenvalues of $U$ to answer how big $\delta$ has to be before $\langle\mu|U^r|\mu\rangle\lt 1/3$?

If we are able to prepare two separate registers, both initially "close to" an eigenstate, we can apply $U$ many times to the second register, and do a SWAP test between the first register $|\mu\rangle$ and the second register $U^r|\mu\rangle$. If the SWAP test fails then we know that $r$ was large enough to scramble $U$.

I think this is a natural question that comes about from thinking of the BBBV theorem, but I'm not sure, and my only intuition is that if $U$ is more "connected" then $U^r$ acts rapidly to scramble $|\mu\rangle$.

$\endgroup$
2
$\begingroup$

You certainly need to know all the other eigenvectors and eigenvalues. It's also important to know $|\mu\rangle$ and the nature of the operator $U$. Say, if $U$ is unitary, then the eigenvectors form an orthonormal set and their eigenvalues have modulus $1$. It's easy to resolve $|\mu\rangle$ along these orthonormal vectors.

To be concrete, suppose the orthonormal eigenvectors are $|\lambda_1\rangle, |\lambda_2\rangle, \ldots, |\lambda_n\rangle$, with eigenvalues $c_1, c_2, \ldots, c_n$. Now consider $\langle\lambda_1|\mu\rangle = 1 - \delta$ (where $\delta \in \mathbb R^+$). Contrary to your intuition, if $c_1 = 1$ and $|c_2|, \ldots, |c_n| = 1$, then after each application of the operator $U$, $\langle\lambda_1|\mu\rangle$ remains constant i.e, $1-\delta$. Repeated applications of $U$ would not necessarily scramble it.

Note that, in general, $\langle \lambda_1|U^r|\mu\rangle$ is not necessarily real and it doesn't make sense to directly compare it with a real number like $\frac{1}{3}$. Though, $|\langle \lambda_1|U^r|\mu\rangle|$ will always remain constant for a unitary transformation $U$ ($\forall r \in \Bbb N$), as all the eigenvalues of $U$ have a magnitude of $1$.

As $|\mu\rangle = a_1|\lambda_1\rangle + a_2|\lambda_2\rangle + \ldots a_n|\lambda_n\rangle$ (where $a_i \in \Bbb C$), the value of $\langle \mu|U^r|\mu\rangle$ would very much depend on the exact values of $c_1, c_2, \ldots, c_n$.


Example:

The phase shift ($R_\phi$) gate has eigenvalue $e^{i\phi}$ corresponding to eigenvector $|\lambda\rangle = |1\rangle$.

Say $|\mu\rangle = \frac{1|0\rangle + 99|1\rangle}{\sqrt{1^2+99^2}}$ and $|\langle 1|\mu\rangle| = \frac{99}{\sqrt {1^2+99^2}}$.

$$\therefore U^r|\mu\rangle = \frac{1|0\rangle + 99e^{ir\phi}|1\rangle}{\sqrt{1^2+99^2}}$$ $$\implies \langle 1|U^r|\mu\rangle = \frac{99 e^{ir\phi}}{\sqrt {1^2+99^2}}$$ $$\implies |\langle 1|U^r|\mu\rangle| = \frac{99 }{\sqrt {1^2+99^2}}.$$

So the magnitude of $\langle \lambda|U^r|\mu\rangle$ clearly remains constant, irrespective of $r$. Though, it certainly might pick up a phase factor upon each application of $U$.

Now, $$\langle \mu |U^r|\mu\rangle = \frac{1^2+99^2e^{ir\phi}}{1^2+99^2}.$$

Unless you take the magnitude it's not possible to compare this with $\frac{1}{3}$, and even then the value very much depends on $\phi$. Notice that if $\phi = 0$, $\langle \mu |U^r|\mu\rangle$ remains $1$ irrespective of how large $r$ is. If $\phi = \pi$, the magnitude keeps flipping between $1$ and $|\frac{1^2-99^2}{1^2+99^2}|$ as $r$ increases, without ever reaching below $\frac{1}{3}$.

$\endgroup$
  • $\begingroup$ Ahh! Wow my intuition was pretty wrong. How do I square this with Grover's algorithm? Certainly the operators there are unitary, and if there's only one marked state eventually we get $|\langle\mu|U^r|\mu\rangle|\lt1/3$ (but it swings back up to $1−\delta$ again). Does it always swing back? I.e. for all unitaries $U$ and all states $|\mu\rangle$ must there be a positive integer $r\gt 1$ such that $|\langle\mu|U^r|\mu\rangle|=|\langle\mu|\mu\rangle|$? The special unitary group is continuous. $\endgroup$ – Mark S Oct 30 at 12:12
  • $\begingroup$ In Grover's algorithm the initial state $|\mu\rangle$ is a uniform superposition of all strings. Could you reframe your question in that context (and ask it as a new question)? In that case $|\langle\mu|\mu\rangle|$ is definitely $1$, and your question boils down to whether $U^r = I$ for some $r$. The answer is yes, iff $U$ is of finite order. $\endgroup$ – Sanchayan Dutta Oct 30 at 13:02
  • $\begingroup$ @MarkS It appears that Householder transformations are involuntary, so it's posssible that such a $r$ always exists. Though, I'm not sure whether product of two involuntary matrices i.e., $U_sU_\omega$ necessarily has finite order. In any case, I think you should ask that as a new question to allow for more elaborate answers. $\endgroup$ – Sanchayan Dutta Oct 30 at 14:21
  • $\begingroup$ Cool! involutary ($U^2=I$). But if I recall, in Grover's algorithm, there is not one fixed $r\gt 0$ that necessarily brings $(U_sU_\omega)^r|\mu\rangle$ back exactly to $|\mu\rangle$; rather, I think there's an $r$ that gets pretty close to $|\mu\rangle$. I'll ponder your answer here, and find a way to ask a question closer to the applications that I'm thinking of. $\endgroup$ – Mark S Oct 30 at 15:25
  • 2
    $\begingroup$ @MarkS about the question on Grover's algorithm, you can actually see pretty easily when there is such an $r$ when you write the evolution as an SU(2) matrix in the target/initial state subspace, which you can do as I show here. Using the notation there, to go full circle you need $\pi k/\alpha\in\mathbb N$ for some $k\in\mathbb N$, where $\sin\alpha$ is the overlap between initial and target states. If this is true, than you go back to the starting point after $r=k\pi/\alpha$ iterations $\endgroup$ – glS Nov 5 at 17:12
2
$\begingroup$

Given the constraints of the problem, we have for $U$ unitary, $$ \newcommand{\ketbra}[1]{\lvert#1\rangle\!\langle #1\rvert} \newcommand{\ket}[1]{\lvert#1\rangle} \newcommand{\bra}[1]{\langle#1\rvert} \newcommand{\sqoverlap}[2]{|\langle #1|#2\rangle|^2} U=\ketbra{\lambda}+\sum_k e^{i\varphi_k}\ketbra{\lambda_k},$$ for some orthonormal basis $\{\ket\lambda\}\cup\{\ket{\lambda_k}\}_k$ and phases $\varphi_k\in\mathbb R$. Given a target $\ket\mu$, the expectation value of $U^r$ equals

$$ \bra\mu U^r\ket\mu=\sqoverlap{\mu}{\lambda} + \underbrace{ \sum_k e^{ir\varphi_k} \sqoverlap{\mu}{\lambda_k} }_{\equiv F_r}. $$

Note also that the normalisation of $\ket\mu$ implies $$ \sqoverlap\mu\lambda + \sum_k |\langle\mu|\lambda_k\rangle|^2 = 1 $$ The problem here is clearly in how to handle $F_r\in\mathbb C$, which has a pretty nasty behaviour: as a function of $r$, it oscillates wildly in the complex plane. Without any constraint on the possible $\ket{\lambda_k}$ and $\varphi_k$, or on the total dimension of $U$, pretty much any shape is possible (and here I cannot not link this beautiful video by 3Blue1Brown). Just to give a mental picture of what I mean, here are two examples of what you can get when you plot $r\mapsto F_r$ for randomly sampled phases and overlaps, for a six-dimensional $U$:

enter image description here enter image description here

The black dots correspond to the integer values of $r$, the blue line gives the value of $F_r$ for the real values in-between, and the red circle has radius $1-\sqoverlap{\mu}{\lambda}$. Only values of $r$ between $1$ and $100$ are shown.

Plotting even more points (all $r$ between $1$ and $10^3$), we see that the circle is essentially filled:

enter image description here

This is telling us that, in the general case, all we can say about $F_r$ is that it satisfies $$ |F_r|\le1-\sqoverlap\mu\lambda. $$ Now, to get the probability of finding the initial state $\ket\mu$ unchanged after $r$ applications of $U$, we need to study the quantity $|\langle\mu|U^r|\mu\rangle|=\big|\sqoverlap\mu\lambda+F_r\big|$. When $\sqoverlap\mu\lambda\ge1/2$ we know that $|F_r|\le\sqoverlap\mu\lambda$, and we thus have the bound $$ (2\sqoverlap{\mu}{\lambda}-1)^2 \le |\langle\mu|U^r|\mu\rangle|^2 \le 1,\quad \forall r. $$ On the other hand, when the initial squared overlap is smaller than $1/2$, we can have $F_r=-\sqoverlap\mu\lambda$, and thus we get the trivial bound $|\langle\mu|U^r|\mu\rangle|\in[0,1]$.

Putting the two cases together, if we plot the possible values of $|\langle\mu|U^r|\mu\rangle|^2$ as a function of $p\equiv\sqoverlap\mu\lambda$ we get the following:

enter image description here

The interesting conclusion is that we can say something about the possible probabilities of a state remaining unchanged after a number of applications of the same unitary $U$, provided that the initial overlap between the state and one of the eigenvectors of $U$ is big enough. Let me remark a few other things here:

  • There is no loss of generality in our initial assumption that the eigenvalue of some $\ket\lambda$ is $1$. We can repeat the same identical argument by just taking an arbitrary eigenvector of $U$ with eigenvalue $e^{i\varphi}$, by only replacing each instance of $U$ with $U e^{-i\varphi}$. Most notably, this does not affect the final result on $|\langle\mu|U^r|\mu\rangle|$.

  • If $\sqoverlap\mu\lambda=1-\delta$ with $\delta\ge0$ small, then the bound can be simplified to $|\langle\mu|U^r|\mu\rangle|^2\ge 1-4\delta$.

  • Because we are not putting any constraint on $U$ or $r$ other than the initial overlap $\sqoverlap\mu\lambda$, we can also understand this result as telling us about the probability of $\ket\mu$ being unchanged upon a single application of a unitary $U$, or more generally about the probability of $\ket\mu$ being unchanged upon arbitrary applications of arbitrary unitaries all having an eigenvector close enough to $\ket\mu$.


Mathematica code to generate the plots of $F_r$:

dim = 5;
probs = 0.8 * # / Total @ # & @ RandomReal[{0, 1}, dim];
phases = RandomReal[{0, 2 Pi}, dim];
otherFactor[probs_, phases_, r_] := Total[Exp[I r phases] probs];
Show[
 ParametricPlot[ReIm@otherFactor[probs, phases, r], {r, 0, 100}, 
  PlotPoints -> 60, MaxRecursion -> 10, 
  PlotRange -> ConstantArray[{-.9, .9}, 2]],
 Graphics[{ 
   Point@Table[ReIm@otherFactor[probs, phases, r], {r, 1, 100}],
   Thick, Red, Circle[{0, 0}, 0.8]
   }]
 ]
$\endgroup$
  • $\begingroup$ on second thought, this means that there is some scrambling, as the probability of the state remaining $|\mu\rangle$ is the square of $|\langle\mu|U^r|\mu\rangle^2$, which is then lower bounded by the square of the initial probabilities of finding $|\mu\rangle$ as $|\lambda\rangle$. I'll think more about it later $\endgroup$ – glS Nov 5 at 19:37
  • $\begingroup$ Thanks! This is a really pretty derivation, that I'm surprised I understand as much as I do. But I'm still struggling to square this with my intuition about Grover's algorithm. In that case I think of $U$ as a combination of rotate-satisfying-assignment and invert-about-the-mean. Clearly there's an $r$ that will have "scrambled" the state far enough away from the uniform distribution. With Grover's algorithm, is it the combination of the $Z$-rotation of the satisfying assignments along with the inversion about the mean, that gets us far away from the initial state? $\endgroup$ – Mark S Nov 5 at 19:38
  • $\begingroup$ @MarkS I agree that the connection with Grover's algorithm is interesting. As per my previous comment, the answer might lie in the fact that there might be some scrambling after all, because if you ask for the probability of $|\mu\rangle$ remaining itself, you have to square $\langle \mu|U^r|\mu\rangle$, which then gives you as lower bound a quantity that is the square of the initial one. This might correspond to a "maximum amount of scrambling" allowed by $U$, but I need to think a bit more about it $\endgroup$ – glS Nov 5 at 19:46
  • $\begingroup$ Thanks so much! My application is to characterize the "mixing" properties of a unitary $U$ if we have a way to prepare a state "perturbed" from an eigenstate of $U$ by an amount $\delta$, as in this question. What is the relationship, if any, between $\delta$, $r$, and the (TVD) mixing time? $\endgroup$ – Mark S Nov 5 at 19:52
  • $\begingroup$ @MarkS going back to this with a fresh mind I realised there were a few problems. See edits. The new bound makes much more sense, but is still interesting I think. The gist is that if a state is close enough to an eigenstate of $U$, it will not go too far from it, but if there isn't an eigenstate that is clearly privileged (i.e. corresponds to prob $\ge1/2$) then nothing can be said. Grover most notably falls into the latter category, and thus these arguments tell us nothing about it $\endgroup$ – glS Nov 6 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.