All length-preserving matrices are unitary

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 M^\dagger M are zero if M is length-preserving.

Using the identity || M \left|\psi\right> ||^2 = \left<\psi\right| M^\dagger M \left|\psi\right>, a suitable choice of \left|\psi\right> = \left|e_j\right> + \left|e_k\right> with j \ne k, and the fact that M is length-preserving, Nielsen first shows that (M^\dagger M)_{jk} + (M^\dagger M)_{kj} = 0 for j \ne k.

He then goes on to write “But what if we’d done something slightly different, and instead of using \left|\psi\right> = \left|e_j\right> + \left|e_k\right> we’d used \left|\psi\right> = \left|e_j\right> - \left|e_k\right>? … 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: (M^\dagger M)_{jk} - (M^\dagger M)_{kj} = 0.”

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 (M^\dagger M)_{jk} + (M^\dagger M)_{kj} = 0.

After some thought and experimentation I concluded that the key is to choose \left|\psi\right> = \left|e_j\right> + i\left|e_k\right>. Below is my (possibly mistaken!) attempt at a correct proof that all length-preserving matrices are unitary.

Proof: Let M be a length-preserving matrix such that for any vector \left|\psi\right> we have || M \left|\psi\right> || = || \left|\psi\right> ||. We wish to show that M is unitary, i.e., M^\dagger M = I.

We first show that the diagonal elements of M^\dagger M, or (M^\dagger M)_{jj}, are equal to 1.

To do this we start with the unit vectors \left|e_j\right> and \left|e_k\right> with 1 in positions j and k respectively, and 0 otherwise. The product M^\dagger M \left|e_k\right> is then the kth column of M^\dagger M, and \left<e_j\right| M^\dagger M \left|e_k\right> is the jkth entry of M^\dagger M or (M^\dagger M)_{jk}.

From the general identity \left<\psi\right| M^\dagger M \left|\psi\right> = || M \left|\psi\right> ||^2 we also have \left<e_j\right| M^\dagger M \left|e_j\right> = || M \left|e_j\right> ||^2. But since M is length-preserving we have || M \left|e_j\right> ||^2 = || \left|e_j\right> ||^2 = 1^2 = 1 since \left|e_j\right> is a unit vector.

We thus have (M^\dagger M)_{jj} = \left<e_j\right| M^\dagger M \left|e_j\right> = || M \left|e_j\right> ||^2 =  1. So all diagonal entries of M^\dagger M are 1.

We next show that the non-diagonal elements of M^\dagger M, or (M^\dagger M)_{jk} with j \ne k, are equal to zero.

Let \left|\psi\right> = \left|e_j\right> + \left|e_k\right> with j \ne k. Since M is length-preserving we have

|| M \left|\psi\right> ||^2 = || \left|\psi\right> ||^2 = || \left|e_j\right> + \left|e_k\right> ||^2 = 1^2 + 1^2 = 2

We also have || M \left|\psi\right> ||^2 = \left<\psi\right| M^\dagger M \left|\psi\right> where \left<\psi\right| = \left|\psi\right>^\dagger = (\left|e_j\right> + \left|e_k\right>)^\dagger. From the definition of the dagger operation and the fact that the nonzero entries of \left|e_j\right> and \left|e_k\right> have no imaginary parts we have (\left|e_j\right> + \left|e_k\right>)^\dagger = \left<e_j\right| + \left<e_k\right|.

We then have

|| M \left|\psi\right> ||^2 = \left<\psi\right| M^\dagger M \left|\psi\right>

= \left|\psi\right>^\dagger M^\dagger M \left|\psi\right>

= (\left|e_j\right> + \left|e_k\right>)^\dagger M^\dagger M (\left|e_j\right> + \left|e_k\right>)

= (\left<e_j\right| + \left<e_k\right|) M^\dagger M (\left|e_j\right> + \left|e_k\right>)

= \left<e_j\right| M^\dagger M \left|e_j\right> + \left<e_j\right| M^\dagger M \left|e_k\right> + \left<e_k\right| M^\dagger M \left|e_j\right> + \left<e_k\right| M^\dagger M \left|e_k\right>

= (M^\dagger M)_{jj} + (M^\dagger M)_{jk} + (M^\dagger M)_{kj} + (M^\dagger M)_{kk}

= 2 + (M^\dagger M)_{jk} + (M^\dagger M)_{kj}

since we previously showed that all diagonal entries of M^\dagger M are 1.

Since || M \left|\psi\right> ||^2 = 2 and also || M \left|\psi\right> ||^2 = 2 + (M^\dagger M)_{jk} + (M^\dagger M)_{kj} we thus have (M^\dagger M)_{jk} + (M^\dagger M)_{kj} = 0 for j \ne k.

Now let \left|\psi\right> = \left|e_j\right> + i\left|e_k\right> with j \ne k. Again we have || M \left|\psi\right> ||^2 = || \left|\psi\right> ||^2 since M is length-preserving, so that

|| M \left|\psi\right> ||^2 = || \left|\psi\right> ||^2 = || \left|e_j\right> + i\left|e_k\right> ||^2

= (\left|e_j\right> + i\left|e_k\right>)^\dagger (\left|e_j\right> + i\left|e_k\right>)

Since i\left|e_k\right> has an imaginary part for its (single) nonzero entry, in performing the dagger operation and taking complex conjugates we obtain (\left|e_j\right> + i\left|e_k\right>)^\dagger = \left<e_j\right| - i\left<e_k\right|. We thus have

|| M \left|\psi\right> ||^2 = (\left|e_j\right> + i\left|e_k\right>)^\dagger (\left|e_j\right> + i\left|e_k\right>)

= (\left<e_j\right| - i\left<e_k\right|)(\left|e_j\right> + i\left|e_k\right>)

= \left<e_j\right| \left|e_j\right> + \left<e_j\right| i \left|e_k\right> - i \left<e_k\right| \left|e_j\right> - i \left<e_k\right| i \left|e_k\right>

= \left<e_j|e_j\right> + i\left<e_j|e_k\right> - i \left<e_k|e_j\right> - i^2\left<e_k|e_k\right>

= \left<e_j|e_j\right> + i\left<e_j|e_k\right> - i\left<e_k|e_j\right> + \left<e_k|e_k\right>

= 1^2 + i\cdot 0 - i\cdot 0 + 1^2 = 2

We also have

|| M \left|\psi\right> ||^2 = \left<\psi\right| M^\dagger M \left|\psi\right>

= \left|\psi\right>^\dagger M^\dagger M \left|\psi\right>

= (\left|e_j\right> + i\left|e_k\right>)^\dagger M^\dagger M (\left|e_j\right> + i\left|e_k\right>)

= (\left<e_j\right| - i\left<e_k\right|) M^\dagger M (\left|e_j\right> + i\left|e_k\right>)

= \left<e_j\right| M^\dagger M \left|e_j\right> + \left<e_j\right| M^\dagger M i\left|e_k\right> - i\left<e_k\right| M^\dagger M \left|e_j\right> - i\left<e_k\right| M^\dagger M i\left|e_k\right>

= \left<e_j\right| M^\dagger M \left|e_j\right> + i\left<e_j\right| M^\dagger M i\left|e_k\right> - i\left<e_k\right| M^\dagger M \left|e_j\right> - i^2\left<e_k\right| M^\dagger M \left|e_k\right>

= (M^\dagger M)_{jj} + i(M^\dagger M)_{jk} - i(M^\dagger M)_{kj} + (M^\dagger M)_{kk}

= 2 + i\left((M^\dagger M)_{jk} - (M^\dagger M)_{kj}\right)

Since || M \left|\psi\right> ||^2 = 2 we have 2 = 2 + i\left((M^\dagger M)_{jk} - (M^\dagger M)_{kj}\right) or 0 = i\left((M^\dagger M)_{jk} - (M^\dagger M)_{kj}\right) so that (M^\dagger M)_{jk} - (M^\dagger M)_{kj} = 0.

But we showed above that (M^\dagger M)_{jk} + (M^\dagger M)_{kj} = 0. Adding the two equations the terms for (M^\dagger M)_{kj} cancel out and we get (M^\dagger M)_{jk} = 0 for j \ne k. So all nondiagonal entries of M^\dagger M are equal to zero.

Since all diagonal entries of M^\dagger M are equal to 1 and all nondiagonal entries of M^\dagger M are equal to zero, we have M^\dagger M = I and thus the matrix M is unitary.

Since we assumed M was a length-preserving matrix we have thus shown that all length-preserving matrices are unitary.

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