Okay. So, in today's lecture, I'm going to introduce some of the quantities we saw
last time, a little more formally. So, we'll talk about observables with respect
to continuous quantum states. I'll go over Schrodinger's equation for, for free
particle on a, on a line more precisely. And then, then we are going to step on
and, and talk about particle in a box. Okay, which will give us a toy model, you
know, a very simplified model in which we can actually talk about atomic qubits you
know, this simple model is surprisingly accurate and, and it will allow you to
picture what a qubit might look like. Okay, so, so, so, let's start, let's you
know, so, r, what we, what we want to do is we want to model the particle on a line
and to begin with, we think of, you know, we think of the particle which can be in
one of a discrete number of states on the line say, say, it's in one of these K
different locations. And then, of course, we'd write, write the, state of this
particle as a superposition sum over, sum over j of alpha sub j, j where j goes from
zero to K-1. And we might also wish to wish to, create an observable which gives
us the position of the system. So, let's say, we have a position, observable M. So,
M is going to be her mission matrix. So, M = M conjugate transpose. And what, what
and it, it therefore it has an orthonormal set of eigenvectors with real eigenvalues.
S, o what are, what are the eigenvectors of this observable M going to be? Well, if
it's a position observable, then the eigenvectors should, should clearly be
the, be the states of definite positions which is j. So, so, its a standard basis
states and, and M, of course, will be diagonal in the standard basis and then
what we want, want the position value to be. So, well, we wanted to be, of course,
this matrix, right? So Mj, if you have the definite state j. And it gives us this.
It, it diagonally was, is also j. Right, and, and so, that, that's what that' s
what it gives us. So, for, for, for example you know if we happen to be in the
state, one / square root three, zero + square root two / 3,1. This is not a state
which, you know, the, here, the, the position of the particle is not definite.
So, if you have, if you have a, a thousand such, such particles, each prepared in the
same state and we measure each of these 1,000 particles, then roughly one / three
of them, you know, roughly 333, will, will, will end, you know, when we measure
them, they will be, they will be at position zero. So, the, so, the outcome
will be zero and for roughly 666, that position would be one, right, ? O course
if you want a definite outcome maybe we should be in one of the eigenstates, which
is, which is either zero or one or two, etc. Okay, so, so that's, that's a quick
recap of, of, of, of the discrete state of the events. And now, in the, in the case
of you know, in the case that we don't want this discrete states but we want the
particle to be anywhere on the line then we replace this. This are, are the state
vector psi, which are finite dimensional vector, k-dimensional vector by a wave
function, psi (x), right? So, think of psi (x) as the amplitude being at x. So, psi
(x) is, is a function from the rear line to the complex numbers, okay. So now, we
have you know, it's as though we are replacing this, this k-dimensional vector
psi, right, often through alpha k - one, b this infinite dimensional function psi
(x). So, so, think of k is becoming very, very large and continuous and that's what,
that's what this function is. And, of course, we must have an normalization
condition and that is given by integral for minus infinity to infinity, of psi (x)
magnitude ^two dx = 0.We also have a notion of inner product, so we have an
inner product. So, if you have two different wave functions, psi (x) and phi
(x), then the inner product, is just an integral from minus infinity to infinity
of phi (x) conjugates psi (x) dx. So, the way, let me point out something that , I
guess, if alpha is a complex number, a + ib, then we, we've been using the notation
alpha star to denote a - ib, it's a, it's a conjugate of alpha. And, I also
interchangeably use the notation alpha bar. So, depend, you know, sometimes I use
alpha star, sometimes alpha bar, depending upon which is convenient to write, and so,
you should interpret them to be exactly the same thing. Okay, alright, so now,
let's, let's look at the fact that observable M was Hermitian, which means
that M = M conjugate transverse. Another way of writing that is, this is the same
thing as saying that, the ij entry of alright? You know, make, make sure you,
you understand why this is the ith and jth entry of M, is equal to, if you look at
the jth, ith entry of M, it's a conjugate of that. Now, it's can also, this, this is
also called self-adjoint. And of course, this, this was not just for basis vectors
i and j, but it's actually holds for any two states, phi and psi. So, so, M is
self-adjoint to Hermitian, if and only if this holds for any two states psi and phi.
Okay? So, convince yourself that this, this really is trivially equivalent to, to
the previous, previous assertion. Okay. So, so, what's, what corresponds to
Hermitian, to, to a self-adjoint to an observable in the, in the, in the finite
dimensional case? Well, it's just, you know, you know, it's, it's just going to
be some linear mapping on wave functions, right? So, an observable is, is a linear
function, it's self-adjoint Mapping M that maps wave functions to wave functions So,
for example, there's, there's a position observable, x hat, and the way x hat works
is x hat, when it's applied to a wave function psi (x), it gives us some wave
function phi (x), where, phi (x) is just x phi (x). Okay. So, what, where did this
come from? So, remember, what, what, what, we had for the position, you know, if you
had a position observable in the finite dimensional case, we wrote it down like
this. Write it, it was, so, where, where, this was the position of the particle. It
was at position zero, one or etc. And when we, when we applied it to some, to some
state, alpha zero, so on, to alpha K-1, we just got the, a new state which was zero
times alpha zero, one times alpha 1e, K - one times alpha K - one, right. So, so,
this corresponds to psi (x), right. So, at, at x psi (x) is the amplitude for
being at that point. What does phi (x) tell us? Well, phi (x) is the observable
applied to psi (x), and what should it tell us? Well it should, it should
multiply that, that amplitude by the position, which is the eigenvalue of that,
of, right? So, so, these are the eigenvectors. So, so they're, so the, the
position eigenvectors are, are x, itself, write zero or one , etc, right? So, the
position eigenvectors are, are, in, in this case, are, are, of the form j, and
so, so, j is, so, so, whatever is in this jf location, the delta function of j, it
gets multiplied by, by j, alright? So, you get j alpha jM. And that's exactly what's
happening with the position. So, and of course, you can see that this is, you
know, this is self-adjoint. And, you know, submission, etc. Okay. So, the other
interesting operator is the momentum operator. It is denoted by p hat and it's
i h bar d / dx. Okay, so, what, what this means is that when you, when you apply
this, two of a function psi. So, so, p hat when applied to psi (x) is i H bar d / dx
of psi (x). Okay, so, let's try to wrap our heads around it and see why is this a
Hermitian operator? So, the easiest thing to do is to just think about it in terms
of the discrete analog. Som let's, let's try to understand the discrete analog. So,
again, remember we, we are thinking about it like this, we have a particle on the
line. It's, you know, its at one of a discreet, we have a lot, you know, we have
these evenly spaced points and it's in one of these evenly spaced points, a finite
side of them. And so, so, we write our state vector as, alpha NOT, alpha one
through alpha K - one. And now, how should we write the derivative? You know, d / dx.
Well, d / dx has the following interpretation. So, to, to get the, the
derivative at this point j, one way I can figure out the derivative at, at this
point, if, if I have some function is at, at x, I can look at the point to the right
of it, x + delta x and look at the function value there, I can't look at the
point to the left of it, x - delta x and look at the value there. Take the
difference and divide by two delta x. Okay, so, what does this correspond to?
Well, what it corresponds to, I claim, is, if you have ones on the off diagonal, zero
on the diagonal, and -one below the diagonal. Okay, so, why is that? Well, you
see because, because if you have, if you have let's say, alpha j, alpha j - one,
alpha j + one, and now, now imagine, what's the, what's it going to look like
when you multiply by the, what's the jth row going to give us? Well, you take this
vector and slide it along the, along the jth row and , of course, alpha j
multiplied with zero, alpha j - one with -one, and alpha j + one with +one. So,
you'll get (alpha j + one) - (alpha J - one), which is exactly what we wanted
here, that's the first derivative. So, this matrix corresponds to the first
derivative. But it is Hermitian. Well, no, because if you take it's conjugate, well,
it's real sub-conjugate does nothing and when you take transpose, you get negative
of what it used to be. Okay, so, so now, how do we make it ission? Well, it's easy,
we just multiply by i. So, when, when we multiply by i, these gets, get changed to
i's and these gets changed to -i's, okay? And now, what happens when we take the
conjugate? Well, when, when we get the con, take the conjugate, the i gets
replaced by -i, -i.i. And then, when we take the transpose, we get back what we
started from. So, so, indeed, this matrix is the, is, is self-adjoint to Hermitian.
And so, similarly, this, this momentum operator is self-adjoint. It's something
that you can verify. Okay, so, so now, having done that, we can start trying to
piece together what, what changed equation for, for particle on a l ine should look
like. Okay. So, so, remember, we had a, a momentum operator [unknown] which is i h
bar d / dx and now, we want to, we want to understand what Schrodinger equation
should look like for a particle on a line. So, we have a free particle on a line and
okay, so, what Schrodinger equation tells us is this that i h bar d psi / dt = h
psi, right? Okay, so, to, to understand what Schrodinger equation says for this
particle on the line, all we have to do is understand what H is. Well, what is H? H
is the Hamiltonian. It's the energy operator. So, what does the energy of a
particle on a line? Well, the particle, you know, classically, the energy would
be, would be the potential energy plus the kinetic energy. In our case, the, the
particle is a free particle so we have a free particle which means, by definition,
that there's no potential energy. What about the kinetic energy? Well,
classically, the kinetic energy is p^2 / 2M, where M is the mass and p for
momentum. Okay, so now, this is realistic way of guessing what the, what the
Hamiltonian is going to be for in the, in the quantum case. So, what you do is as
you do something which is which sounds a bit ad hoc, which is you, you, you look at
the classical case and you replace your momentum by the momentum operator, okay.
And you know, to make it sound very official and make it sound correct
physicists have invented a name for this procedure. They call it the correspondence
principle. Okay. So, so, so using this correspondence principle, which is a nice
way of guessing what the, what, what, what this basic equation of motion should be.
We, we get that i h bar d psi / dt = b hat squared / 2M psi. So, what's P hat
squared? Well you just apply P hat twice. So, it's i h bar d / dx of i h bar d / dx
one / 2M psi. So, that's really, if you're applying, if you take the derivative
twice, it's the second derivative. I^2 is -one so it's -h bar square / 2M d^2 / dx^2
of psi. And this is exactly the form that we intuitively came up with last time. S
o, last time we said, let's not worry about the constants and we, we said,
Schrodinger's equation for free particles should look in form, something like i d
psi / dt is d^2 psi / dx^2. And that's what this other way of looking at
Schrodinger's equation also tells us.