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The states $|+\rangle, |-\rangle$ can be mapped to $|0\rangle, |1\rangle$ by a simple rotation.

But if I now have other states ($|\psi_0\rangle, |\psi_1\rangle$) which are not orthogonal, does a unitary transformation of these states to two other states ($|\phi_0\rangle, |\phi_1\rangle$) exist?

And how can I find this unitary transformation?

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3 Answers 3

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There is a unitary that maps $\{|\psi_1\rangle,|\psi_2\rangle\}$ to $\{|\phi_1\rangle,|\phi_2\rangle\}$ if and only if $$ \langle\psi_1|\psi_2\rangle=\langle\phi_1|\phi_2\rangle. $$

Thy "only if" direction is straightforward. Let's assume $U|\psi_i\rangle=|\phi_i\rangle$. Then $$ \langle\phi_1|\phi_2\rangle=\langle\psi_1|U^\dagger U|\psi_2\rangle=\langle\psi_1|\psi_2\rangle. $$

The "if" direction requires a little more care. Let $|\psi^\perp\rangle$ be the component of $|\psi_2\rangle$ that is orthogonal to $|\psi_1\rangle$. Similarly for $|\phi^\perp\rangle$. So, we claim that if we can find a unitary $U$ such that $$ U|\psi_1\rangle=|\phi_1\rangle,\qquad U|\psi^\perp\rangle=|\phi^\perp\rangle, $$ then we are done because $$ U|\psi_2\rangle=U(\cos\theta|\psi_1\rangle+\sin\theta|\psi^\perp\rangle)=(\cos\theta|\phi_1\rangle+\sin\theta|\phi^\perp\rangle)=|\psi_2\rangle $$ where $\cos\theta=|\langle\psi_1|\psi_2\rangle|=|\langle\phi_1|\phi_2\rangle|$.

Now let $\{|\gamma_i\rangle\}$ be any orthonormal basis with $|\gamma_1\rangle=|\psi_1\rangle$ and $|\gamma_2\rangle=|\psi^\perp\rangle$ and let $\{|\lambda_i\rangle\}$ be an orthonormal basis with $|\lambda_1\rangle=|\psi_1\rangle$ and $|\lambda_2\rangle=|\psi^\perp\rangle$. We can define $$ U=\sum_i|\lambda_i\rangle\langle \gamma_i|. $$ This certainly provides the required transformation.

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From your question it appears you know the two sets of states ahead of time. If so and the $|\psi_0\rangle$ and $|\psi_1\rangle$ are not orthogonal, then a unitary transformation that transforms |$\psi_0\rangle$ to $|\phi_0\rangle$ is not guaranteed to transform |$\psi_1\rangle$ to $|\phi_1\rangle$. There could be one, in special cases, but not always.

However, if both sets are orthogonal, then I expect to have a unitary transformation that could do that every time.

Edit: If they are not orthogonal, but the inner products of both the sets are the same i.e. if $\langle \psi_0 | \psi_1 \rangle$ = $\langle \phi_0 | \phi_1 \rangle$ and if $U$ exists where $\phi_0 = U |\psi_0 \rangle$, then it follows $\phi_1 = U |\psi_1 \rangle$.

This is because $\langle \phi_0 | \phi_1 \rangle$ = $\langle \psi_0 | U^{\dagger} U |\psi_1 \rangle$ , and since $U^{\dagger} U = I$ .

And $U = \sum_{j} |\phi_j \rangle \langle \psi_j |$

Ref: Qiskit Universality

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  • $\begingroup$ Yes, such a mapping cannot always exist. If the inner products are different certainly not. But is it always possible to find a mapping when the inner products are the same? And if so, how? $\endgroup$
    – Johny Dow
    Oct 5, 2021 at 1:30
  • $\begingroup$ @JohnyDow , I have edited the answer $\endgroup$
    – codester
    Oct 5, 2021 at 2:51
  • $\begingroup$ Thanks! But looking at the source you provided the defined $U$ works only if the states are an orthonormal basis. $\endgroup$
    – Johny Dow
    Oct 5, 2021 at 3:17
  • $\begingroup$ I believe, the source says it works for orthonormal basis, because the inner products are equal implicitly. $\endgroup$
    – codester
    Oct 5, 2021 at 3:20
  • $\begingroup$ Do you know how to find $U$ if the states are not orthogonal? =) $\endgroup$
    – Johny Dow
    Oct 5, 2021 at 3:22
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A small rewarding of the argument in this other answer:

  1. Given $\{|\psi_1\rangle,|\psi_2\rangle\}$ and $\{|\phi_1\rangle,|\phi_2\rangle\}$ such that $\langle\psi_1|\psi_2\rangle=\langle\phi_1|\phi_2\rangle$, we can always write $$|\psi_2\rangle = \cos\theta |\psi_1\rangle + \sin\theta|\psi_1^\perp\rangle,\\ |\phi_2\rangle = \cos\theta |\phi_1\rangle + \sin\theta|\phi_1^\perp\rangle,$$ for some angle $\theta$ given by $\langle\psi_1|\psi_2\rangle=\cos\theta$, and for some states $|\psi_1^\perp\rangle,|\phi_1^\perp\rangle$. These are defined as $$|\psi_1^\perp\rangle\equiv \frac{|\psi_2\rangle-\cos\theta|\psi_1\rangle}{\sin\theta}, \qquad |\phi_1^\perp\rangle\equiv \frac{|\phi_2\rangle-\cos\theta|\phi_1\rangle}{\sin\theta},$$ and are such that $\langle\psi_1^\perp|\psi_1\rangle=\langle\phi_1^\perp|\phi_1\rangle=0$.

  2. Given any pair of states $|\psi\rangle,|\phi\rangle$, you can always find unitary operators $U$ such that $U|\psi\rangle=|\phi\rangle$. The choice of $U$ is not unique. Let us then assume that $U|\psi_1\rangle=|\phi_1\rangle$. Our goal will be to show that we can find $U$ such that we also have $U|\psi_2\rangle=|\phi_2\rangle$. But given our observation in the first point, this is equivalent to finding $U$ such that $U|\psi_1^\perp\rangle=U|\phi_1^\perp\rangle$.

  3. We thus reduced the problem to finding a unitary $U$ which maps $|\psi_1\rangle\mapsto |\phi_1\rangle$ and $|\psi_1^\perp\rangle\mapsto |\phi_1^\perp\rangle$. But finding a unitary sending orthogonal states into orthogonal states is trivial to do: for example, we could just define $U$ as $$U = |\phi_1\rangle\!\langle\psi_1| + |\phi_1^\perp\rangle\!\langle\psi_1^\perp|,$$ and this can be readily seen to be unitary and implement the correct mapping.

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