# If a unitary operator is close to the identity, will it leave any state it acts on unchanged?

I have a question that feels obvious, but I have been having trouble proving it. In words, the question is: If a unitary is close to the identity, will it leave any state it acts on unchanged?

This, to me, would necessarily be the intuitive definition of a unitary "being close to the identity", so that's why I say it "feels obvious".

In physics notation: Consider a unitary, $$\hat{U}$$ which is assumed to be arbitrarily close to identity. In other words: $$$$\hat{U} = \exp(- i \delta \hat{G} ) \approx \hat{\mathbb{I}}-i \delta \hat{G},$$$$ for $$\delta \ll 1$$ and $$\hat{G}$$ Hermitian.

Now, I need to prove that, for any normalized state $$|\psi\rangle$$, the unitary $$\hat{U}$$ will change it at most by order $$\epsilon$$, i.e. $$$$|\langle \psi| \hat{U}|\psi\rangle|^2 = 1 - \epsilon.$$$$

In math (just in case it helps clarify what I'm after): Prove that for every $$\epsilon<<1$$ there exists a $$\delta$$ such that $$$$|\langle \psi| \exp(- i \delta \hat{G})|\psi\rangle|^2 = 1 - \epsilon,$$$$ for every normalized $$|\psi\rangle$$, i.e. $$|\langle\psi|\psi\rangle|=1$$.

I suspect $$\delta$$ should be something like $$\sqrt{\epsilon}$$. Furthermore, we can assume the spectrum of $$\hat{G}$$ is finite and completely positive (or completely negative). Thanks!

My attempts at solving the problem: I've tried the obvious move of expanding $$\hat{U}$$ inside the $$|\langle...\rangle|^2$$, but this leads to an expression of the form $$1+\epsilon$$, which is not what I'm after because we already know this fidelity is upper bounded by 1. I've also tried a few norms on unitaries to induce a distance metric and modify the idea of "close to the identity" such as the Frobenius norm, and the spectral norm. I'm trying to avoid the diamond norm, because it is often unwieldy. Any proofs where we say "assume $$d(\hat{U},\hat{\mathbb{I}})\leq \delta$$ therefore $$|\langle \psi| \hat{U}|\psi\rangle|^2 = 1 - \epsilon$$ would be totally acceptable as well.

• What is wrong with "it will change the state slowly unless it's an Eigenstate of the operator"? Are you trying to do epsilontics on wave functions? You also seem to be moving backwards here because you already seem to know that the exponential form is the solution for the finite operator that derives from an infinitesimal generator. What's wrong with using that to derive the solution? Commented Jul 23 at 1:11
• I am assuming the state is acted on by this unitary, and I want to know if the unitary is close to the identity then, explicitly, how much does the state change-- even if it is a really small amount. I'd rather not make arguments about the state "changing slowly" because I want the explicit form of how much it changes by.
– John
Commented Jul 23 at 14:22
• You already wrote it out explicitly: it changes by the exponential of your infinitesimal. Commented Jul 23 at 14:24
• Sure, changes by the exponential of my infinitesimal. But I am trying to show precisely that it doesn't change more than a small amount. I think it's obvious if I set $\delta=0$ then the overlap $|\langle \psi| U |\psi\rangle|^2$ is $1$. But now I want $0<\delta\ll 1$ and I'd like to lower bound the overlap with $1-\epsilon$. This would use the fact that the state changes by the exponential of my infinitesimal in order to answer how the overlap probability changes.
– John
Commented Jul 23 at 14:30
• But it does change by more than a small amount. The exponential is the formal integral that expresses the finite action of your generator. You are staring right at the concrete solution. Commented Jul 23 at 14:33

I assume that the spectrum of $$G=G^*$$ stays in $$[0,+\infty)$$ and $$\Lambda \leq +\infty$$ is an upper bound of it. $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle+ \langle \psi |(e^{-i\delta G}-I) \psi\rangle =\langle\psi|\psi\rangle + \delta \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\:.$$ If $$\psi \in D(\sqrt{G})$$ (that just means every $$\psi$$ in the Hilbert space if $$\Lambda < +\infty$$), we can estimate $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \int_0^\Lambda \left|\frac{e^{-i\delta \lambda}-1}{\delta}\right| d \langle \psi| P^{(G)}(\lambda)\psi \rangle$$ $$= \int_0^\Lambda \lambda \sqrt{\left(\frac{\cos \delta \lambda -1}{\delta \lambda}\right)^2+ \left(\frac{\sin \delta \lambda }{\delta \lambda}\right)^2} d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \int_0^\Lambda \lambda \sqrt{\sin^2 x + \cos^2 x' } d \langle \psi| P^{(G)}(\lambda)\psi \rangle \leq\sqrt{2} \int_0^\Lambda \lambda d \langle \psi| P^{(G)}(\lambda)\psi \rangle = \sqrt{2} \langle \psi|G\psi\rangle$$ where $$x,x' \in [0, \delta\lambda]$$, so that $$\left| \int_0^\Lambda \frac{e^{-i\delta \lambda}-1}{\delta} d \langle \psi| P^{(G)}(\lambda)\psi \rangle\right|\leq \sqrt{2} \langle \psi| G\psi\rangle\:.$$ In summary, defining $$E^{(G)}(\delta, \psi)$$ implicitly by $$\langle \psi |e^{-i\delta G} \psi\rangle = \langle\psi|\psi\rangle + \delta E^{(G)}(\delta, \psi)\:,\tag{E}$$ we have $$| E^{(G)}(\delta, \psi)|\leq \sqrt{2\langle \psi|G\psi \rangle}\:.$$ Finally, if $$G$$ is bounded, we also have the estimate $$\sqrt{2\langle \psi|G\psi \rangle}\leq ||\psi|| \sqrt{2||G||}\:.$$

Let us fix the $$1-\delta$$ issue.

I explicitly suppose that $$||\psi||=1$$. In this case $$|\langle \psi |e^{-i\delta G} \psi\rangle|\leq ||\psi||\: ||e^{-i\delta G} \psi|| = ||\psi||^2 =1\tag{M}$$ because the operator is unitary. As a consequence, from (E), $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 + \delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))$$ where $$\delta^2 |E^{(G)}(\delta, \psi)|^2 + \delta 2 Re(E^{(G)}(\delta, \psi))\leq 0$$ for every $$\delta>0$$, otherwise (M) is impossible. In conclusion $$|\langle \psi |e^{-i\delta G} \psi\rangle|^2 = 1 - \delta H^{(G)}(\delta, \psi)$$ where, for $$\delta\geq 0$$, $$0\leq H^{(G)}(\delta, \psi) := \left( -\delta |E^{(G)}(\delta, \psi)|^2 - 2 Re(E^{(G)}(\delta, \psi))\right)\leq 2 |E^{(G)}(\delta, \psi)| \leq 2 \sqrt{2\langle \psi|G\psi \rangle}< +\infty\:.$$

• But does this give the $1-\epsilon$ on the RHS the OP seems to want to have (with $\epsilon >0$, I suppose)? Commented Jul 23 at 12:55
• I do not know, I have "interpreted" the question. Commented Jul 23 at 13:08
• I see. I also thought about answering it, and also found some bound similar to yours, IIRC, but refrained from posting because of the "minus sign issue". Perhaps I am missing something obvious, though. Commented Jul 23 at 13:10
• $P^{(G)}(\Delta)$ is the spectral measure of $G$. Yes, in case of point spectrum, $\Delta$ are sets of proper eigenvalues. $E^{(G)}(\delta,\psi)$ is defined implicitly in the text as made clearer now. Commented Jul 23 at 14:35
• @John I am not sure to understand your issue. Now there is no epsilon, only $\delta$, that is $\epsilon=\delta$. Do you think that there is a better estimate? Commented Jul 23 at 16:26

Here is a very simple lower bound for the fidelity, maybe this helps in addition to the existing answer: Let $$U=\exp(-i\delta G)$$, then using Duhamel's formula one can show the inequality $$|| \exp(A) - \exp(B)|| \leq ||A - B||$$ with the operator norm $$||\cdot||$$ (for a proof, see for instance Lemma 1 here). This yields $$||\mathbb{I} - U|| \leq \delta ||G||$$ and finally $$1 - \langle \psi | U | \psi \rangle = \langle \psi | \mathbb{I} - U | \psi \rangle \leq ||\mathbb{I} - U|| \leq \delta ||G||.$$ Rearranging yields a fidelity lower bound $$|\langle \psi | U | \psi \rangle|^2 \geq (1 - \delta ||G||)^2$$. There are known tightness results, whenever $$\delta$$ is small enough (compared to $$||G||$$).

• This answer is really good Commented Jul 24 at 20:07
• Thanks man! . . Commented Jul 25 at 19:36