6
$\begingroup$

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?

$\endgroup$
1
8
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ 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 '21 at 1:30
  • $\begingroup$ @JohnyDow , I have edited the answer $\endgroup$
    – codester
    Oct 5 '21 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 '21 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 '21 at 3:20
  • $\begingroup$ Do you know how to find $U$ if the states are not orthogonal? =) $\endgroup$
    – Johny Dow
    Oct 5 '21 at 3:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.