For the past few tutorials we have been looking at the concept of the quantum harmonic oscillator.
With the Schrödinger equation solved we are now ready to interpret our results
and apply this information to gain some knowledge about the quantum world.
Let’s start by looking at the wavefunctions. We will use the probability density function,
which we are familiar with from earlier tutorials, because this will be the easiest to understand.
Once again, this function tells us the probability of finding the particle at a certain position.
So let’s use this in conjunction with the ground state wavefunction
which we derived in the previous tutorial. We found earlier that H naught of X equals one.
Now for the probability density function, we need psi star times psi, and since there are no complex
numbers in this expression, psi star is the same as psi, so we are just squaring this expression.
That will double this exponent operating on pi, and it will double this exponent operating on e
as well, which means we just lose the two in the denominator, and we get this.
Now this will take on the well-known bell shape, or Gaussian curve, which we can see quite clearly
if we plot position against the probability density function. This is centered around zero,
and because there is only one bump we can call this an odd function. We can do the same thing
for other eigenstates, though in the interest of time we will not calculate them, but rather simply
show them graphically so we can make general conclusions. So jumping up to n equals one,
we have the first excited state, and the probability density function is represented
by P1. Rather interestingly, the particle can’t be found at x equals zero. This is a rather striking
example of quantum behavior. Using a previous analogy regarding a tennis ball in a bowl,
this would be like never finding the tennis ball at the precise bottom of the bowl, which is rather
unintuitive. Also notice that this function has two bumps, so we can call this an even function.
The second excited state is an odd function again, and that will look like this.
Then we can look at the third excited state, which becomes even again.
This is reminiscent of the particle in the box, where we also found both odd and even solutions.
This is a very common result for symmetric systems such as the harmonic oscillator,
which are mirror images of themselves with respect to a central axis.
To get an idea of how these states are distributed, this is how the ground state would
look if we plot P + E, that is, the probability density function plus the eigenenergy of a
particular eigenstate, together with the harmonic potential. This is a common way to visualize the
eigenstates distributed along the potential where they live. So to be clear, the gray line
represents the eigenenergy of the ground state, and the blue line is the harmonic potential,
V of X. So we are understanding that the energy of the ground state is not zero, as it is not at the
bottom of the parabola. This should no longer be surprising after the last few quantum problems
we have discussed. Then we plot the probability density function, P of X, for the ground state,
precisely where the boundary conditions dictate. The boundary conditions are given
by the blue curve, and that’s why most of the Gaussian-shaped red curve is within this region.
So the gray line essentially indicates the bottom, or minimum of that wavefunction. So to summarize,
here we see the ground state, and it is clear that the particle can’t be found outside of the
harmonic trap, as beyond these boundaries the probability density function equals zero, and
the particle is primarily localized at the center. We can plot additional eigenstates the same way.
Here we see the first five excited states going higher in energy, and we can see the
particle can be found in different regions of the harmonic potential, but rarely outside,
given that the probability density function drops to zero in each case. Now we should point out that
even though the likelihood of finding the particle outside V of x is extremely low, it is not zero,
since the tails of the Gaussian function extend to infinity. It is quite puzzling to think that,
precisely because of these infinite tails, a quantum particle has a non-zero probability
of being in London even if the harmonic potential is in New York.
Of course, there are many reasons why no one would ever physically measure such a situation,
but it’s fun to think about regardless. Now let’s go ahead and reiterate a few things.
We know that a particle in a quantum harmonic oscillator possesses a set of quantized energies,
meaning it can’t have any energy from a continuous spectrum, and that these quantized
energies are evenly spaced. Additionally, the probability density functions tell us that the
particle will be localized in certain positions along the x axis, depending on the eigenenergy
that is possessed, and it will have a very low probability of being found in other locations.
To finish things off, let’s perform a quantitative comparison between the classical result that we
found in the first tutorial, and the quantum result we just found here. Since the easiest
way to visualize and understand the quantum result is by looking at graphical representations
of the probability density function, we will make use of these here. To do this, we first
need to calculate the classical correspondent of the probability density function. We will
call this “classical probability density”, and we will represent it with PCL as opposed to simply P,
which will continue to denote the quantum probability density. To intuitively grasp
the results that we are about to find, let’s think about how a classical particle behaves
in a harmonic oscillator. If we imagine a ball in a harmonic “trap”, so to speak, as we have done
before, there are several results that we already know about from simple classical mechanics. First,
there exist two turning points. These are the points located at x equals plus or minus x,
where the particle changes directions. Second, there is an equilibrium point located at x equals
zero. This is the point of lowest potential energy, such as the case where there is zero
tension in a spring, as opposed to the tension that results from stretching or compressing the
spring in one direction or the other. This is also the point where kinetic energy is maximized
as the particle moves along its path, as total energy must be conserved. So at x equals zero,
kinetic energy T is at its maximum, and therefore so is the particle’s velocity.
By contrast at x equals plus or minus x, or the boundaries,
the particle will possess its maximum potential energy and will for an instant be at rest.
Because the particle travels more quickly in the middle and more slowly at the boundaries,
if we make random observations as time progresses, the most likely observations occur
at the turning points, statistically speaking. We will define PCL of x dx as the probability of
finding a particle in a certain region of space, for instance x plus dx. This quantity corresponds
to the fraction of time that the particle takes to travel from x to x plus dx. To put it another way,
we know that a period, T, is the time it takes for a particle to travel between
turning points. So this applies to any periodic or cyclical phenomenon.
With the harmonic oscillator, a classical particle that begins with a certain kinetic energy, E, will
bounce in the harmonic trap from one side of the harmonic potential to the other, indefinitely. So
it will move between negative x and x, always in the same amount of time, or with the same period.
But the particle is not always moving at the same speed. At the bottom of the parabola it
moves much faster. So even though it always takes the same amount of time to traverse a full period,
it doesn’t spend the same amount of time in all the x values it travels through. PCL deals with
precisely this, how long did the particle spend in a specific infinitesimal amount of space,
x + dx, compared with the full period? So mathematically, that all reads like this,
as being equal to 2dt over T, where T represents the period. Recall that the period is defined
as two pi over omega. We can understand this as the fraction of time, dt, of one period T, that
the particle spends in the dx interval. Remember that we are calculating PCL of x, so now we want
to work through the right side of the equation. First, notice that v equals dx over dt, so we
can rearrange this a bit and say that dt equals dx over v. Let’s plug this expression in for dt,
and we will see that we have dx on both sides, so we can get rid of those, and PCL of x
will equal two over vT. Now we need to write velocity as a function of x. So we can’t use
the equation that we found before for x of t, since this is the position with respect to time.
Let’s instead find this from the following equation representing total energy, which is
the sum of the kinetic and potential energies, which are represented by these expressions.
Now we know that at the turning points, which can be represented as plus or minus XTP, the velocity
is zero. So let’s plug in XTP for x and zero for v. The kinetic energy term will then go away,
and we can solve for XTP, which will equal root 2E over k. Now let’s go back to the previous equation
for total energy and solve for v. We can subtract this potential energy term from both sides,
then multiply by two and divide by m, and we get v squared as being equal to this.
Distributing, we get 2E over m minus k over m times x squared. And since omega equals root k
over m, we can replace k over m with omega squared to get this. To manipulate further, let’s multiply
this first term by k over k, so we aren’t changing its value. Now with a k over m in this term,
we can replace that with an omega squared as well. And since we know that root 2E over k equals XTP,
then 2E over k can be replaced by XTP squared. Then finally we can factor omega squared
out of this difference. Remember that this whole time we had been solving for v squared, so let’s
now just take the square root, and v will be equal to omega times root XTP squared minus x squared.
Now we can plug this expression in for v in this earlier equation,
and we are left with this for the probability density for the classical scenario. But let’s
also recall that T equals two pi over omega, so let’s make that substitution as well, by placing
two pi in the denominator and omega in the numerator, since T was in the denominator.
Conveniently, the twos cancel and the omegas cancel, and we get one over pi times this radical.
Now it may be a little easier to conceive of this equation as a function of energy, so let’s
go ahead and convert XTP squared back into 2E over k. And that is where we finally end up.
Recalling that the quantum probability density will be determined by the probability density
function, as we have been doing for a number of tutorials by now, and with both of these
expressions we can plot the classical and quantum probability densities against one another.
We already calculated and plotted these P functions for several of the eigenstates for the
quantum harmonic oscillator just a bit earlier. Now let’s plot P of x and PCL of x together.
To do this, we will plug the eigenenergies from the quantum scenario into PCL of x.
Now we must be perfectly clear that the classical harmonic oscillator functions with an energy input
of any value, there is no quantization at play here. This is simply an exercise to compare
the classical and quantum situations. In both cases, we are looking at the probability density,
but in the quantum case, because of the De Broglie principle, we need to think of the
particle as a wave as well. Therefore, this comparison is not ideal, even if it is quite
enlightening, as we shall see momentarily. Let’s begin with the ground state. To do this,
we plug E naught equals one half h bar omega into the previous equation,
so we are getting P naught CL of x. This is the classical case, and will be plotted in magenta.
Then for the quantum case, again we have this probability density for P naught, which is
plotted in red. We have also clearly demarcated the turning points, plus or minus XTP, as these
vertical black lines. We will clearly notice that while a quantum particle tends to be found in
the central region of the harmonic oscillator, a classical particle tends to be found on the edges.
Now let’s look at other eigenenergies. First up, E one. We will notice that for the classical
particle things don’t change a whole lot. For the quantum particle, things change quite a bit,
as the particle is now unlikely to be found in the central region, but rather in one of these
two bumps. If we go to E two, again there is a high probability that the quantum particle
will be found at the center, while for the classical particle, still not much has changed.
If we continue to go higher in energy, looking at something like n equals 15, something stunning
begins to emerge. The quantum system is starting to more closely resemble the classical system,
with the probability density reaching its maximum closer to the edges,
and tapering down towards the center. We can easily predict the implication here,
that as energy increases, further and further, in the limit of modeling macroscopic objects,
classical behavior will emerge from this quantum description.
Yet another remarkable result is that we can clearly see a non-zero probability for the
quantum particle traversing the turning points plus and minus XTP. If we recall the lessons
we learned when examining the quantum barrier potential and the concept of quantum tunneling,
perhaps this is not too shocking, but it is nevertheless an interesting result.
In summary, we have performed calculations regarding both the classical motion
and quantum eigenenergies and eigenstates of a harmonic oscillator. In doing so,
we found clear differences between the classical and quantum scenarios. The eigenfunctions of
quantum particles are combinations of Hermite polynomials with bell-shaped functions, and their
eigenenergies take on the form of E sub n equals one half plus n times h bar omega, where n is the
eigenstate and omega is the harmonic oscillator frequency. It is the case that classical particles
can possess any energy, whereas quantum particles can take on only a discrete set of eigenenergies.
Once again, the energy of a quantum particle is quantized, which is one of the primary conclusions
of quantum mechanics that we are seeing in every single lesson throughout this series.
On average, classical particles are more likely to be found at the edges of their oscillatory motion,
which we call turning points. By contrast, low-energy quantum particles are more likely
to be found at the center of the harmonic oscillator. But in the most wondrous way,
as the energy of the quantum particle increases, its behavior converges with the classical results,
which is yet another stunning display of how classical physics emerges from the quantum world.