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I believe that any unit length vector $|\Psi\rangle$, whose components are complex numbers, can be transformed into any other vector $|\Phi\rangle$, of the same dimension, same length, and also composed of complex numbers, by a unitary operator, $\hat{U}$, such that $\hat{U}|\Psi\rangle=|\Phi\rangle$ and $\hat{U^{\dagger}}|\Phi\rangle=|\Psi\rangle$. Is that true and, if so, where can I find a proof of that?

I can't seem to prove it or find such an operator even for simple cases such as $(a_1^2 + a_2^2 + a_3^2 + a_4^2)^{-1/2}\hat{U}\begin{bmatrix} a_1 \\ a_2 \\ a_3 \\ a_4 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}2^{-1/2}$, where each $a_i$ is a complex number and $(a_1^2 + a_2^2 + a_3^2 + a_4^2)^{-1/2}\begin{bmatrix} a_1 \\ a_2 \\ a_2 \\ a_4 \end{bmatrix} = \begin{bmatrix} \alpha_1 \\ \alpha_2 \end{bmatrix} \otimes \begin{bmatrix} \beta_1 \\ \beta_2 \end{bmatrix}$, the tensor product of two other unit length vectors (if the tensor product aspect makes it easier...).

In Quantum Mechanics, the interpretation of the example, above, is that I am looking to prove that any pure state $|a\rangle$ (which we can assume is a tensor product of two systems and so is not entangled) can be transformed into any pure entangled state (for instance the vector in the example that contains two $0$s and two $1s$), by a unitary transformation.

Thanks.

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It suffices to construct a unitary matrix whose first row is an arbitrary unit vector $v$. Take any non-singular matrix with first row $v$ and apply the Gram-Schmidt process to its rows. The vectors one gets form the rows of a unitary matrix with top row $v$.

In more detail. Such a matrix $M$ has $e_1M=v$ where $e_1= \pmatrix{1&0&\cdots&0}$. If you have two unit vectors, $v$ and $w$ there are unitary matrices $M$ and $N$ with $e_1M=v$ and $e_2N=w$. Then $M^{-1}N$ is unitary and $v(M^{-1}N)=w$.

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  • $\begingroup$ @lordshardtheunknown Thank your for your quick response. Could you please tell me if what you said is written somewhere, in a little more detail, that I can get access to? I have looked long and hard on the Web and cannot find a proof of the answer to my question. $\endgroup$ – David Apr 26 '18 at 5:30
  • $\begingroup$ @lordshardtheunknown Is that written up in some lecture notes or book, somewhere that you know of? Also, how do you figure out the rest of the columns of M and N? $\endgroup$ – David Apr 26 '18 at 5:47
  • $\begingroup$ It's the Gram-Schmidt process: see en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process . Although this Wikipedia page concentrates on Euclidean inner products, the procedure works just as well for Hermitian inner products. $\endgroup$ – Lord Shark the Unknown Apr 26 '18 at 5:54
  • $\begingroup$ @lordshardtheunknown After looking up G-S, I came across QR factorization. If, for square matrix $\hat{A}$: $\hat{A} |x\rangle = |b\rangle$, and any $\hat{A} = \hat{Q} \hat{R}$, and the length of $|x\rangle$ and $|b\rangle$ are the same, then the length of $\hat{R} |x\rangle$ must be the same as the length of $ |x\rangle$, so $ \hat{R}$ is Unitary, just as $\hat{Q}$ is, and the product of two unitary operators is Unitary, so $\hat{A}$ is unitary. I think that may prove that the square matrix that carries one vector, of length $l$ into another vector of the same length is unitary. $\endgroup$ – David Apr 26 '18 at 6:51

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