Okay, so in the last video we, we saw how the state of a qubit evolves through a
rotation of the space. So now let's try to understand this a little better. So well,
last time we, we talked about how you know for simplicity we thought of the qubit as
having real amplitude, so we could represent the qubit on the two dimensional
real space. And then we talked about how rotating the space, when you rotate the
space through theta, you can write down a rotation matrix, r sub theta. Which says
how the state of the system evolves. So, for example if the state was psi, then
after the rotation it would become r sub theta times psi. Now, in actual fact, a
qubits sits in a two dimensional complex vector space and so we've got to describe
what, what the rotations of this complex vector space look like. Okay so, so
rotations of this complex vector space are called unitary transformations. Okay and
they are also specified by two by two matrices. So, so our complex rotation is
going to be specified by a matrix like this a, b, c, d where a, b, c, and d are
complex numbers. And what we're, what we're saying is if we start with a state
psi, then under this unitary transformation, it evolves to the state u
psi. Okay so, so now let's, let's try to understand what happens in different
cases. What happens if psi = zero? What is u psi? Well, remember psi, psi = zero
means, well, zero is also another way of writing it in standard vector notation
would be as, as this vector10. And so if you multiply u times this vector, you
would get, get the, get the vector ab which is a0+ b1. And similarly if you
start in the state one, it gets mapped to c0+ d1, okay. Now, saying that u is a
unitary transformation or it's a rotation of the two dimensional complex vector
space, it's equivalent to saying this condition that u, u dagger = to u dagger u
is the identity. Where u dagger, it's the conjugate transpose of u. Meaning first,
we take complex conjugates of all the entries and then we transpose or we do it
in the opposite order, it doesn't matter s o, meaning this one would become a
conjugate, complex conjugate, dcomplex conjugate but these two entries are going
to be exchanged. So this entry would become c conjugate, and this would become
b conjugate. Okay, so now one of the conditions we want from this rigid body
rotation is that angles are preserved. So for example if you do the rigid body
rotation on the standard basis zero and one. What we want is that angles and
lengths are preserved. So we want, we want that, since we started with the unit
vector, this should still be a unit vector. Meaning that we want a^2 + b^2 =
one = c^2 + d^2, right? We also want that these two should be orthogonal because a,
zero, and one were orthogonal. So we want for example ac + bd = zero. So the
inner product between these two is< /i> zero. Okay now, you can see that these
conditions are expressed here in, in, in this because when you look at u dagger u
well, this is. A b c d x a, sorry abcd which is what? Well, this is supposed to
be the identity. But as you can see this entry is just the inner product of ab in
itself. It's just magnitude of a^2 + magnitude of b^2. And similarly this entry
is magnitude of c^2 + magnitude of d^2 where as this entry is the inner product
of ab and cd. And so, so this is saying, this is saying that these two vectors are
orthogonal and this says the same thing and these two say the length of, of the
vectors are preserved. Another property of you know of unitary transformations these
rotations is that they are linear, right? This is what gave us the right to, to
specify the, the rotation by a matrix. So what this means is that if you say that u,
you know u you know if, if I tell you what, what, what the unitary
transformation does on zero, let's say that it maps it to phi zero, which people
calling a0 + b1 and suppose it maps one to phi one which we were calling c0 + d1.
Now, we could ask what does it do? What does u do to the, to the vector a0 + b1?
And the answer is by linearity, it must map it to a phi zero + b phi one. Okay and
you can, you can of course further simplify it and write what, what it looks
like in terms of zero and one. You would just say it's a(a(0) + b(1)) + b ((c0) +
(d1)) which is (aa + bc)(0) + (ab + bd)(1). Okay so, the main point is that we
can specify this unitary transformation by saying what it does on each of the basis
vectors, okay? Not just in two dimensions. Not, not just for qubit but this also
holds, if you were specifying what it does for, you know if you had a, if you had a K
level system then you could specify what the unitary transformation does on the K
basis vectors and you'd have specified the, the unitary transformation
completely. And of course, there's no mystery to this because as we said, the
columns of u are exactly where you map each of these k basis factors.