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.