# Distinguishing Quantum States

1. Sep 21, 2009

### Kreizhn

Hey,

I've been looking into different aspects of distinguishing two pure quantum states. I've ended up reading a lot of books/papers covering things like "accessible information", but there haven't been too many explanations on how to find optimal measurements.

The book by (Kaye, Laflamme, Mosca) outlines a simple procedure in their appendix, which requires sending the two states $| \Psi_Y \rangle, |\Psi_x \rangle$ to the states
$$\cos(\theta) | 0 \rangle + \sin(\theta) |1 \rangle$$
$$\sin(\theta) |0 \rangle + \cos(\theta) |1 \rangle$$
where $0 \leq \theta \leq \frac \pi4$ and $\langle \Psi_y | \Psi_x \rangle = \sin(2\theta)$. However, I'm uncertain as to how to even construct a unitary operator that does the associated mapping.

Other sources consider the states $| \Psi_x \rangle |\Psi_x \rangle, | \Psi_y \rangle |\Psi_y \rangle$, or just talk about accessible information rather than optimal measurements.

If anyone could shed some light on how to construct the Unitary Map necessary for the procedure in Kaye et al. or any other insight into this, it would be much appreciated.

2. Sep 21, 2009

### Avodyne

First you have to express $| \Psi_x \rangle, |\Psi_y \rangle$ in terms of $| 0 \rangle, |1 \rangle$, eg, $| \Psi_x \rangle = a| 0 \rangle+b |1 \rangle, |\Psi_y \rangle= c| 0 \rangle+d |1 \rangle$, with $|a|^2+|b|^2=1, |c|^2+|d|^2=1, a^*c+b^*d=\sin(2\theta)$. Then you want to find a 2x2 unitary matrix that takes the column vector $(a,b)$ to $(\cos\theta,\sin\theta)$ and the column vector $(c,d)$ to $(\sin\theta,\cos\theta)$. This is four equations involving the elements of U. Since a 2x2 unitary matrix has four parameters, this should allow you to determine all four. Looks kind of messy to actually do it, though.

EDIT: one specific case is pretty easy: let $| \Psi_x \rangle = | 0 \rangle, |\Psi_y \rangle= \sin(2\theta)| 0 \rangle+\cos(2\theta) |1 \rangle$. Then $U=((\cos\theta,-\sin\theta),(\sin\theta,\cos\theta))$.

Last edited: Sep 21, 2009
3. Sep 21, 2009

### Kreizhn

Yes, this is the method I had originally tried. I haven't followed the math all the way through, but I don't believe it's as well-defined as you say. The problem is we infact have more than 4 equations to determine the unknowns components of the unitary transformation. These extra equations come from the fact that the mapping is unitary. Hence we may have an over-determined system and the solution may not exist. Is there something in how we've defined $\theta$ that ensures this doesn't happen?

4. Sep 21, 2009

### Avodyne

The four equations are linear, and so have a unique solution for the four elements of U. As long as the norms and the inner product of the states are unchanged, the result should be consistent with unitarity.

5. Sep 22, 2009

### kanato

Are |0> and |1> orthogonal? If they are, then there is no unitary transformation which will produce |x> and |y>, because a unitary transformation will preserve orthogonality.

6. Sep 22, 2009

### Kreizhn

Kanato, yes |0> and |1> are orthogonal, but this doesn't change anything. In the even that the two states I'm trying to distinguish are orthogonal, then theta =0 and so the unitary transformation is trivial. However, no statement as to the orthogonality of Psi_x and Psi_y were made; we want to map them to the points (cos theta, sin theta) and (sin theta, cos theta) on the space spanned by |0> and |1>. Furthermore, the theta we've chosen is such that the inner-product is preserved as necessary for a unitary mapping.

7. Sep 27, 2009

### kanato

Maybe I'm not understanding what you're trying to do. Are you looking for an operator that behaves like $$|\psi_x\rangle = U |0\rangle$$, $$|\psi_y\rangle = U |1\rangle$$? If so, then any unitary transformation will necessarily result in $$\langle \psi_x | \psi_y \rangle = \langle 0 | 1 \rangle$$. If that's not the type of transformation you're looking for, then I don't understand the original question.