In this video, I'm going to present the
rather technical derivation of an
important equation of quantum theory.
The continuity equation for probability,
which both relies on the Borne
interpretation of quantum theory and also
serves as a sort of safety check for it.
Because it establishes the conservation of
probability.
The idea here is that, even though we
cannot see with certainty where exactly
our quantum particle is located, we can
certainly, we can be sure that it's
located somewhere in space so if we
perform a measurement and look for the
particle everywhere, we're going to find
it with a probability equal to one.
And this total probability never becomes
smaller than one and certainly cannot
become larger than one, so it is
conserved.
And this simple conservation law gives
rise to this continuity equation that
we're going to derive.
Now, the form of this equation is actually
not specific to quantum theory, and it
appears in many different fields of
physics.
So raw here is the density of a
conservative quantity.
In our case this is the probability
density, which per the born rule is equal
to the absolute value of the wave function
squared.
And g here is the probability current,
which we actually are going to derive.
So to see that this equation indeed
describes some sort of a conservation
law,.
Let me consider an arbitrary volume in
space.
So let's say this is a volume V, and let
me also denote the surface encircling this
volume as dV.
And let me be interested in the
probability of finding the particle of my
quantum particle inside this volume.
So this probability, let me call it P sub
v, is going to be an integral of the
probability density, which appears here
over this volume.
Now to see to look at the dynamics of this
probability with time, let me integrate
both sides of this equation over the
volume.
Well, the right hand side here is zero,
there is not much to integrate, but the
first term is going to give me just dp
over dt, and the second term is going to
be an integral of the diversions of g.
And everything, the sum of these two terms
is equal to 0.
Now at this stage, I can use the Gauss's
theorem to handle the second term, which
tells me that the integral of a full
diversion, so for a volume, is equal to a
flux of the vector of which I calculated
diversions, through the surface
surrounding my volume.
So this is my dv.
So in ds here is an elementary An
elementary surface element, with the
vector pointing, outwards.
So I'm sure, some of you have already
seen, the Guassian theorem, let's say, in
the theory of electromagnetism.
But if, you forgot about it, or you have
never seen it.
So let me just mention that this Gauss,
theorem is in some sense, similar to the
simple identity that we oftentimes use for
usual integral, so let's see if we have an
integral between a and b or the full
derivative of a function df dx, so we can
write the result in the result as f of b
minus f of So essentially in this case we
have a one-dimensional segment from a to
b, or we, we integrate and of we have a
full derivative we can only focus on the
end points.
So that's the only thing which answer is
the final result.
So likewise if we have more complicated
integral now, an integral over, over three
dimensional volume but of a full
derivative, which it divergence.
E is, so instead of having two boundary
values we have a, a surface integral going
through this surface through this dv.
So in some sense this dv is similar to a
and b, so the volume is similar to the
segment of between a and b and the full
derivative df.
Over the X, is similar to the divergence
of G.
So what we actually have established here
is the derivative, what we call it P dot
of the probability of finding a particle
in the volume V is equal to minus the flux
of the probability current again, we're
going to derive it a little later.
Through the surface surrounding this
volume and let's see if we have a current
say going outwards here luckily, so that
only so that there is forms as small angle
with the ds so then this g.ds is positive
which corresponds to the negative change
in the probability so and it makes sense
because, it means that.
The current carries away the probability,
and the probability of finding a particle
inside decreases.
So if we, on the other hand, have the
current going inside, the probability is
going to increase.
So in order for me to prove the continuity
equation in the form I just formulated,
let me calculate directly the probability
of finding my particle in the volume v
over time using the boron, rule and, the
boron rule basically implies that I
calculate the derivative of this integral
of the volume of the absolute value of my
wave function squared, and the absolute
value can vary in, of course by definition
as a product of the wave function and its
complex integrated everything and degraded
over the volume.
So now if I apply this derivative to this
product, I can write it as psi star dot
psi plus psi star psi dot.
So in order for me to simplify it further
Let me just use the Schrodinger equation
in it's center form, which actually
appears on the logo of our, of our course.
So, and express the psi dot from here as
simply minus i over H bar doing on, acting
on sine.
So I just divided it by H bar, and
multiplied both sides with the imaginary
constant I.
So I can also drive the same equation
with, for psi star dot and the since
Hamiltonian is sort of real cause the
kinetic potential is just going to be plus
I over H bar, H acting on psi.
Now my H here is a combination of kinetic
energy and potential energy.
So kinetic energy is sort of an involved
operator which is the momentum squared and
potential energy is just multiplication,
it just multiplies my wave function.
So if I put everything together, what I
will find is, the following, for P V dot,
so I'm going to have here instead of psi
star dot, I'm going to have this guy, so
I.
H bar h psi, and here we have minus i h
bar psi star h psi.
So if h were a number so these 2 terms
would cancel each other out.
And it does happen, as a matter of fact
for the potential energy draw So if, so
this H can actually be expre-, replaced
with a kinetic energy.
But there's no consolation necessarily for
the kinetic energy because again, this is
an operator which acts on different
functions here and here.
And, this operator is equal to just p
squared over 2m, or minus h squared over
2m Laplacian.
So we should write as delta number
squared.
So again, putting everything together, so
what I have is minus i h bar 2 m, which I
can factor out outside the integral.
And in the brackets I'm going to have the
Laplacian number squared psi start and psi
minus psi star number squared, squared
psi.
So the last step here is to integrate this
expression by parts, and if I do so
essentially by moving this delta from here
to here and from here back to here, so I
see that these terms become identical and
cancel each other out.
And the only term which survives is, the
full derivative of delta up side star.
Psi minus complex conjugated.
So, And this whole thing, well, times this
coefficient, is exactly the current that
we have been looking for because, again,
we ended up with the full derivative of a
vector integrated over the volume.
So we can write it as minus an integral of
sum G over DS over the distribute, this
sort of encircling, encircling our volume.
So finally, we can just collect everything
from this expression and write the final
expression for the current, which I will
arrive in the full length forum, which is,
1/2 side star, p, over m, edging on side,
plus complex country.
So, the reason I can write like this is
because P is an operator, which is equal
to minus I H bar.
Nebla and this minus i h and r, hr nebla
appears here, and so if we look at this 5
expressions, we see that the symmetric
effect makes sort of intuitive sense,
because in classical physics, a classical
current associated with the density, rho,
is simply the density times the velocity
with which it moves.
Now, here in quantum mechanics the
momentum is an operator so velocity which
is momentum divided by m is also an
operator.
And this operator sort of acts on the
density which is which is rho is psi star
psi.
So and so this is the final result which
connects the change in the probability of
finding a particle in the volume v with
the flux of a certain probability current
flowing through the surface.