# Unitary transformation of pure states to other pure states

1. May 30, 2014

### m~ray

Is it true that there always exist a unitary matrix that can take a state vector of an arbitrary pure state to another arbitrary pure state ? (of course assuming same hilbert space). If true, how do we prove it ? it look like it is true via geometrical arguments but i have not been able to convince myself completely. I am sure in the case of 1 qubit system; what about multiqubit system ? Thanks.

2. May 30, 2014

### Simon Bridge

Conversely, can you find an example where the proposition is not true?

You would start your proof by defining your terms, then showing what a transformation between states would look like without assuming that the transformation is unitary. Enjoy.

3. May 30, 2014

### atyy

Not exactly what you are asking, but related, I think. Probably the article on reachability is the closes to what you are asking about.

http://arxiv.org/abs/quant-ph/0010031
Complete controllability of quantum systems
S. G. Schirmer, H. Fu, A. I. Solomon

http://arxiv.org/abs/quant-ph/0102017
Complete controllability of finite-level quantum systems
H. Fu, S. G. Schirmer, A. I. Solomon

http://arxiv.org/abs/quant-ph/0110171
Criteria for reachability of quantum states
S. G. Schirmer, A. I. Solomon, J. V. Leahy

4. May 31, 2014

### rubi

Let's say you have $\psi, \phi \in \mathcal H$ with $\lVert\psi\rVert = \lVert\phi\rVert = 1$ and you want $U\psi = \phi$. Let $(v_\alpha)_\alpha$ be an basis for $\mathcal H$. Apply Gram-Schmidt to $(\psi,(v_\alpha)_\alpha)$ to get an ONB $(x_\alpha)_\alpha$ with $x_0=\psi$ and repeat the procedure to get an ONB $(y_\alpha)_\alpha$ with $y_0 = \phi$. Now define $U\sum_\alpha c_\alpha x_\alpha = \sum_\alpha c_\alpha y_\alpha$. This is clearly a surjective (for $\sum_\alpha c_\alpha y_\alpha$ in $\mathcal H$, choose $\sum_\alpha c_\alpha x_\alpha$) linear operator with $U\psi = \phi$ and it is also unitary because $\left<U\sum_\alpha c_\alpha x_\alpha,U\sum_\beta d_\beta x_\beta\right> = \left<\sum_\alpha c_\alpha y_\alpha,\sum_\beta d_\beta y_\beta\right> = \sum_\alpha c_\alpha d_\alpha = \left<\sum_\alpha c_\alpha x_\alpha,\sum_\beta d_\beta x_\beta\right>$ by orthogonality.

5. May 31, 2014

### Simon Bridge

... etc. But m~ray has to do it in order for it to be properly convincing.

6. May 31, 2014

### rubi

Well, of course it would have been better if he had done it himself, but he said that he was unable to do it. I think my proof sketch is pretty clear, so it should be convincing. (If not, don't hesitate to ask, m~ray.)

7. May 31, 2014

### dextercioby

An isometric operator defined on all vectors in a Hilbert space is called a unitary operator. An isometric operator has unit (strong) norm and reverse. Hence if a,b arbitrary vectors of unit norm and U is an endomorphism of H taking a into b, then ||U|| = max ||Ua|| = max ||b|| = 1. , forall ||a|| = 1. If U is unit norm, it's isometric. If it's isometric and maximal on H, it's unitary.

8. Jun 1, 2014

### tom.stoer

Two pure states define a 2-dim. subspace

$$V_{a,a^\prime} = \text{span}\{|a\rangle, |a^\prime\rangle\}$$

of the full Hilbert space $H$. So for the existence of a unitary operator $U$ with

$$U\,|a\rangle = |a^\prime\rangle$$

it's sufficient that $U$ acts as a rotation operator on $V_{a,a^\prime}$ and as the identity on the orthogonal complement $V_\perp$ of $V_{a,a^\prime}$.

$$U = u_{V_{a,a^\prime}} \otimes \text{id}_{V_\perp}$$

From what I said above it should become clear that it's sufficient to proof this for the 2-dim. case, i.e. for $u_{V_{a,a^\prime}}$. And this is just linear algebra

Last edited: Jun 1, 2014