After setting up the dichotomy between the classical computational category 
and the quantum computational category 
we started to compare the two. This comparison begins with matrices: matrices over 
suffice to encode both objects and morphisms in , whereas matrices over 
are needed to encode objects and morphisms in 
but the encoding works the same way. Change of ordering in the one-hot encoding of a finite set 
is implemented by left multiplication by permutation matrices, whereas change of orthonormal basis in a finite-dimensional Hilbert space is implemented by left multiplication by unitary matrices; in this sense, unitary matrices are the quantum generalization of permutation matrices. Change of ordering in the source and target sets of 
corresponds to left and right multiplying the matrix encoding of morphisms by permutation matrices, while change of ordered orthonormal bases in the source and target spaces of 
corresponds to left and right multiplying the matrix encoding of morphisms by unitary matrices.
A more interesting line of investigation is to consider the extent to which can be enriched beyond 
i.e. made into something more than a set. First, the existence of a vector space structure on is very straightforward, and in fact a special case of the fact that the set of functions from any set into a vector space is again a vector space under the pointwise operations.
It is also true that is finite-dimensional. Let 
and 
be orthonormal bases, which is the same thing as saying 
is isomorphic to 
and 
is isomorphic to 
What we showed is that is isomorphic to 
Indeed, for each 
define a corresponding elementary transformation 
by
We proved that 
is a basis of 
and indeed that for any 
we have
This is a good way to present the information contained in the matrix of 
relative to the orthonormal bases and without having to choose orderings just to write things in an array.
The elementary basis 
of relative to a pair of orthonormal bases and also gives us a convenient way to analyze topological properties of arbitrary morphisms 
Indeed, for any 
we have that
where the inequality is Cauchy-Schwarz and we used the fact that 
and 
are unit vectors. Thus, each morphism in the elementary basis 
of is a contractive mapping. Thus for an arbitrary morphism 
the triangle inequality gives us
where
is a norm on , specifically the 
-norm relative to the elementary basis of , which is itself defined relative to the orthonormal bases and 
This points us to a much better norm on since the above tells us that every is not just continuous but Lipschitz, with Lipschitz constant bounded by 
Definition 3.1. The operator norm 
of is by definition its Lipshitz constant.
One of the many virtues of the operator norm is that it is basis independent: to define it we only need to know that all morphisms in 
are Lipschitz functions. Definition 3.1 does not care that we used orthonormal bases of and to establish Lipschitz continuity of linear transformations. On the other hand, one could argue that a demerit of the operator norm is that it is not induced by a scalar product.
Problem 3.1. Prove that there is no scalar product on such that 
coincides with the operator norm.
On the other hand, it is completely straightforward to promote to a Hilbert space: we simply equip it with the scalar product in which forms an orthonormal basis. This is the Frobenius scalar product on , and it is given explicitly as follows: for any 
we have
Problem 3.2. Show that for any we have
The norm 
on obtained from the Frobenius scalar product is called the Frobenius norm, and denoted
The problem with the above is that we don’t know whether we have built the Frobenius scalar product or a Frobenius scalar product: if we had performed this construction using two different orthonormal bases 
and 
would we have arrived at the same Hermitian form on ? We would like to know whether 
is basis independent, like the operator norm 
, or whether it is basis dependent, like 
The answer is that the Frobenius scalar product is basis independent. A morphism 
such that 
for every is called an isometry. Polarization shows that norm preservation is the same as scalar product preservation: is an isometry if and only if
Furthermore, every isometry is injective, so that an isometry in 
is automatically an automorphism. Isometric automorphisms of are also referred to as unitary operators. That’s right, there are two names for the same thing. While it might be better practice to never ever override general categorical terminology in category-specific settings, this is just one of many things about the world that could be improved. Even worse, if and are possibly different finite-dimensional Hilbert spaces with 
then again an isometry is automatically an isomorphism, and such isometric isomorphisms are also often called unitary transformations. Saints preserve us. The reason for this abuse of terminology is that the matrix of an isometric isomorphism is a unitary matrix even when the spaces 
are not the same, just like the matrix of set isomorphism 
is a permutation matrix even when and 
are distinct sets.
Problem 3.3. Show that 
is a unitary operator if and only if the image 
of an orthonormal basis under 
is an orthonormal basis of .
Problem 3.2 already removes the dependence of the Frobenius scalar product on a choice of orthonormal basis in the target space. Thus what is left to check is that for any orthonormal basis and any unitary operator we have
In other words, thanks to polarization, we only need to check that the Frobenius norm is unitarily invariant. We have
Thus,
and to finish it remains only to establish the following, which is a consequence of Problem 3.3.
Problem 3.4. For any unitary operator and any orthonormal basis 
we have
In the lectures I presented a different proof of unitary invariance of the Frobenius scalar product on which was inductive in the dimension of 
The reason for doing this was that in the base case, 
we are dealing with the dual space of and this leads to a discussion of vectors and covectors (Riesz duality). However, I did not do a very good job with this discussion and therefore have decided to leave it out of the exam topics.
