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

Question

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: \begin{equation} \hat{U} = \exp(- i \delta \hat{G} ) \approx \hat{\mathbb{I}}-i \delta \hat{G}, \end{equation} 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. \begin{equation} |\langle \psi| \hat{U}|\psi\rangle|^2 = 1 - \epsilon. \end{equation}

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

Answer 1

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 $\delta\geq 0$, $$0\leq H^{(G)}(\delta, \psi) := \left( -\delta |E^{(G)}(\delta, \psi)|^2 - 2 Re(E^{(G)}(\delta, \psi))\right)\leq 2 |E^{(G)}(\delta, \psi)| \leq 2 \sqrt{2\langle \psi|G\psi \rangle}< +\infty\:.$$

Answer 2

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 $||G||$).