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