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.