I recently read the (excellent) online resource Quantum Computing for the Very Curious by Andy Matuschak and Michael Nielsen. Upon reading the proof that all length-preserving matrices are unitary and trying it out myself, I came to believe that there is an error in the proof as written, specifically with trying to show that off-diagonal entries in are zero if is length-preserving.
Using the identity , a suitable choice of with , and the fact that is length-preserving, Nielsen first shows that for .
He then goes on to write “But what if we’d done something slightly different, and instead of using we’d used ? … I won’t explicitly go through the steps – you can do that yourself – but if you do go through them you end up with the equation: .”
I was an undergraduate physics and math major, but either I never worked with bra-ket notation and Hermitian conjugates or I’ve forgotten whatever I knew about them. In any case in working through this I could not get the same result as Nielsen; I simply ended up once again proving that .
After some thought and experimentation I concluded that the key is to choose . Below is my (possibly mistaken!) attempt at a correct proof that all length-preserving matrices are unitary.
Proof: Let be a length-preserving matrix such that for any vector we have . We wish to show that is unitary, i.e., .
We first show that the diagonal elements of , or , are equal to 1.
To do this we start with the unit vectors and with 1 in positions and respectively, and 0 otherwise. The product is then the th column of , and is the th entry of or .
From the general identity we also have . But since is length-preserving we have since is a unit vector.
We thus have . So all diagonal entries of are 1.
We next show that the non-diagonal elements of , or with , are equal to zero.
Let with . Since is length-preserving we have
We also have where . From the definition of the dagger operation and the fact that the nonzero entries of and have no imaginary parts we have .
We then have
since we previously showed that all diagonal entries of are 1.
Since and also we thus have for .
Now let with . Again we have since is length-preserving, so that
Since has an imaginary part for its (single) nonzero entry, in performing the dagger operation and taking complex conjugates we obtain . We thus have
We also have
Since we have or so that .
But we showed above that . Adding the two equations the terms for cancel out and we get for . So all nondiagonal entries of are equal to zero.
Since all diagonal entries of are equal to 1 and all nondiagonal entries of are equal to zero, we have and thus the matrix is unitary.
Since we assumed was a length-preserving matrix we have thus shown that all length-preserving matrices are unitary.