Hello, everyone. Today we have a very
interesting lecture, it's straight out of
science fiction so we'll talk about
Quantum Teleportation. But before we get
there, we have to say something about,
quantum circuits. Okay so, lets see.
Remember, last time we talked about
one-qubit gates. So one-qubit gates, we
think of as you know, we have a wire that
carries a qubit of information. So let's
say, α0(0) + α1(1). And now, we have a
gate which, which acts on it and remember
what a gate does is it takes this two
dimensional complex vector space. The
state is a unit vector in that space and
then it takes the space and we rotate it.
Okay, this rotation in the complex vector
space is called the Unitary Transformation
U, which in this case is a two by two
unitary transformation. So U is two by two
unit rate transformation. And, the fact
that this unitary is given by the
condition U times U [inaudible] is, is U
[inaudible] times U is identity. So we saw
several examples of, of, of such gates and
you know, they included x, the nought
gate, Z, the face flip, H, the Hadamard
gate. Okay, so what, what these do is,
they take the input qubit and convert it
into, into some new qubit, alpha nought
prime zero plus alpha one prime one. Now
of course, you remember that you could,
you could describe the you know, the,
this, this rotation, the unitary rotation
by either specifying the unitary matrix,
or you could just say zero gets mapped to
what and one gets mapped to what, right?
So if you, if you specify these, these
four numbers which are that zero gets,
gets mapped to a(0) + b(1) and one gets
back to c(0) + d(1) then you specify it,
what the, what the mapping is because it's
a linear map. So now if you have α(0) +
β(1), what does it go to? Well, it goes
to alpha times a(0) + b(1) + beta times
c(0) + d(1). Also, the unitary matrix
that, that, that describes how these you
know, the mapping is, is just given by a,
b, c, d. Okay, so that's, that's what we
did last time. And now, this time we are
going to talk about two-qubit ga tes. So,
so here is our, here is our picture. So,
we have two wires that are coming in, each
carrying a qubit. And then, this
encounters a gate U and, then you have two
wires coming out, which, which, which are
not carrying two qubits, which are in a
different state. So, lets, what is the
state of our initial qubits? Well remember
that, the state of two-qubits is described
by four complex numbers, the four complex
amplitudes, alpha 00 through alpha eleven.
And output qubits are described by four
new complex numbers, alpha 00 prime
through alpha eleven prime. So now, the
way that this output state is related to
the input state, is, of course, given by
this unitary transformation U, which is a
linear transformation. It's, have what,
what's the dimension of this
transformation. Well, clearly, U is four
dimensional. Right? Meaning, if we write
it out, it would, looked like, this kind
of matrix. Okay. And of course, you
satisfy this condition, uu dagger equal to
u dagger u is the identity. And once
again, you can specify the action of U by
just saying, how does it map 00? How does
it map 01? Etcetera. How does it map
eleven? So, so if you say, how does it map
each of these four states, then you, then
you have specified U as well. Okay? So,
let's, let's say that it maps (00) to
a(00) + b(01) + c(10) + d(11). Then we
know that the first column of this matrix
is going to be a, b, c, d. Right? So check
that you understand why this is, right?
Because 00 is this vector. Okay? So if you
want to know what, what this u map 00 to
u, you multiply this matrix u by this
vector. And what do you get? You get this
column, a, b, c, d, which is exactly what
this was. Okay, so make sure you
understand all this. Okay. So, now, let's
look at an example of, of a two qubit
gate. This is the most basic of the
two-qubit gates. So it takes as input two
qubits, outputs two qubits and it's
denoted by this funny symbol where there's
a dot on the one side align which with
this, with this with this circular round,
which denotes, an XOR. Okay, so if this wa
s a classical gate, so if the inputs a and
b were classical, then outputs would look
like this, the first input is let through
unchanged and the second bit, so the first
bit is let through unchanged and the
second bit is flipped, if and only if the
first is one. Okay, so the first bit, this
is called, the control. And that bit is
called the Target Bit. Okay, so the CNOT
stands for Controlled NOT. Alright. Okay,
so now remember what the rules were that
if you want to describe what the CNOT gate
is, it's sufficient to describe what it
does on input 00, 01, ten, eleven. So
let's see, what, how does the CNOT gate
map 00? So now the control bit is zero so
the target bit remains unchanged so it
gets mapped to 00. How about 01? Well, the
control bit is twenty so the target bit is
unchanged so, 01. How about ten? Well, now
the control bit is a one, so the target
bit flips and so it gets mapped to one,
one. How about one, one, one? Okay, so
this time, okay again the controlled bit
is one, so the target bit flips. So the
output is one, zero. So now you know
exactly how CNOT works because if you are
given any super position of this four
states. So if you are alpha (00). Okay, so
if you had this superposition alpha 00(00)
+ alpha 01(01), + alpha ten(10) + alpha
eleven(11). So what does it get mapped to
under CNOT? Well answer is simple. What,
what it does is zero, zero stays unchaged,
so it goes to alpha 00(00). Zero, one
unchanged so it's 01(01). Now, one, one
gets mapped to one, zero so you would have
alpha eleven(10) and one, zero gets mapped
to one, one so you get alpha ten(11).
Okay. We can also write out what the
matrix is, corresponding to CNOT. And CNOT
is given by this matrix. So, make sure you
see why this matrix is giving you this
transformation. Right? Because this one
gives you the first column, which is one,
zero, zero, zero. This gives you the
second column, this gives you the third
column, where you get a one, in the last
entry, because every one, one. This gives
you the fourth column, where you have one
in the third entry. And of course, its
unitary to check that you, just check that
CNOT times CNOT conjugate transpose is the
identity. In fact, again, what you have is
that the conjugate transpose is itself and
so CNOT is it's own inverse. If you apply
it twice, you get back to where you
started from.