# When can pairs of states be transformed into other pairs of states via unitary mapping?

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?

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.

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

• 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? Oct 5, 2021 at 1:30
• @JohnyDow , I have edited the answer Oct 5, 2021 at 2:51
• Thanks! But looking at the source you provided the defined $U$ works only if the states are an orthonormal basis. Oct 5, 2021 at 3:17
• I believe, the source says it works for orthonormal basis, because the inner products are equal implicitly. Oct 5, 2021 at 3:20
• Do you know how to find $U$ if the states are not orthogonal? =) Oct 5, 2021 at 3:22

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.