Is there a quantum operation whose output is always orthogonal to the input?

Question

I'm trying to show/convince myself the following statement is correct (I haven't been able to find any similar posts):

"There is no reversible quantum operation that transforms any input state to a state orthogonal to it."

I can see how this could be true based on the operation being unitary and that you could likely find some input state that isn't transformed into an orthogonal state. Is there a simple way to show/prove this? My unsuccesful approach has been:

Let $U$ be some unitary transformation/operation, and $|x\rangle$ some input state decomposed as $$|x\rangle=\lambda_1|0\rangle + \lambda_2|1\rangle + \lambda_1|2\rangle + \cdots + \lambda_n|n\rangle$$ where $$\lambda_1 \lambda_1+\lambda_2 \lambda_2+\cdots+\lambda_n\lambda_n=1,$$ and

$$U|x\rangle = \langle x|U^\dagger$$

I'm trying to show that there exists an input state $x$ such that $\langle x|U^\dagger|x\rangle = 0$ isn't true.

I've tried to make use of the properties of unitary matrices but haven't had much luck.

Any assistance or suggestions on alternative approaches on how to show this would be greatly appreciated.

Answer 1

The existing answers have quite an elegant idea behind them. However, my concern is that they don't seem to allow for the introduction of an ancilla. If we introduced an ancilla in a fixed state, and applied a unitary across the two systems, then it could be that all the eigenvectors are entangled across the two subsystems, and the basic argument wouldn't apply since we'd only be trying to make an argument about a separable input.

The way that I would approach such a calculation (I'll do it only for a single qubit here. That's sufficient to make the necessary argument, but you could extend it to any Hilbert space if you wanted). Let $\Phi$ be the Choi map of our operation. This means that, for an input state $|\psi\rangle$, the output state would be $$ \rho_{\psi}=2\text{Tr}_1(|\psi\rangle\langle\psi|^T\otimes I\cdot\Phi). $$ This is successful if $$ \text{Tr}((I-|\psi\rangle\langle\psi|)\rho_{\psi})=1. $$ Now, we need this to be true for all possible input states $|\psi\rangle=\cos\frac{\theta}{2}|0\rangle+\sin\frac{\theta}{2}e^{i\phi}|0\rangle$, so the average over all possible input states must also be 1. In other words, we want $$ \frac{1}{2\pi}\int_0^{2\pi}d\phi\int_0^{\pi}\sin\theta d\theta\ \text{Tr}\left(|\psi\rangle\langle\psi|^T\otimes(I-|\psi\rangle\langle\psi|)\Phi\right)=1. $$ If we work this out, it means that we require $$ \text{Tr}\left(M\Phi\right)=1 $$ where $$ M=\left(\begin{array}{cccc} \frac{1}{3} & 0 & 0 & -\frac{1}{3} \\ 0 & \frac{2}{3} & 0 & 0 \\ 0 & 0 & \frac{2}{3} & 0 \\ -\frac13 & 0 & 0 & \frac13 \end{array}\right). $$ So, what's the best choice of $\Phi$? It's an eigenvector of maximum eigenvalue of $M$. But the maximum eigenvalue is $\frac23$, meaning that it is impossible to achieve $\text{Tr}(M\Phi)>\frac23$, and hence it is impossible to give an orthogonal output for all possible single-qubit input states.

By way of contrast, if we restrict to just real states (i.e. $\phi=0$), the maximum eigenvalue of $M$ is larger that 1, which leaves open the possibility for creating an orthogonal state, which we know is possible just by applying Pauli-$Y$. This corresponds with $\Phi=(I\otimes Y)|B_{00}\rangle\langle B_{00}|(I\otimes Y)$.

Answer 2

This is easiest to show by contradiction, so lets suppose that there exists an operation TransformToOrthogonal that maps the state of its input qubits to an orthogonal state. Reversible quantum operations can be simulated using unitary operators, so we can simulate TransformToOrthogonal by a matrix $U$ such that $UU^\dagger = 𝟙$. Since $U$ is unitary, $U$ is also normal (that is, $U$ commutes with its transpose, $UU^T - U^T U= 0$), so that we can use the spectral theorem to decompose $U$ into eigenvalues and eigenvectors, $$ U = \sum_i \lambda_i |\psi_i\rangle\langle\psi_i|. $$ At this point, we're almost at the contradiction; consider what happens if you input qubits that are in one of the eigenstates $|\psi_i\rangle$.

Answer 3

Every unitary matrix can be diagonalized as $U=S^\dagger \Lambda S$, where $S$ is unitary and $\Lambda$ is diagonal and contains the eigenvalues ($\lambda_i$) of $U$. So you can consider instead $$\langle x \vert U \vert x \rangle=\langle x \vert S^\dagger \Lambda S \vert x \rangle = 0.$$ Defining $\vert \tilde x \rangle \equiv S \vert x \rangle$, you only have to convince yourself that there always exists a state $\vert \tilde x \rangle$ such that $$\langle \tilde x \vert \Lambda \vert \tilde x \rangle = \sum \limits_i \lambda_{i}\tilde x_i^\ast \tilde x_i \ne 0.$$ That should be pretty easy to see in this form.