At this point we have already solved two quantum problems,
that of the particle in a box, and the quantum barrier potential. Now let’s tackle another,
which is called the quantum harmonic oscillator. If this term sounds confusing, don’t worry,
before we dive into all the math let’s just go over a few general concepts for context.
First we must understand that a harmonic oscillator model describes a system
with uniform periodicity over time. This is usually referring to a simple sinusoidal wave,
which indeed has a reliable periodicity, as opposed to systems that follow more complicated
patterns over time, which generally would not be referred to as harmonic oscillators.
If we can recall our study of Hooke’s law and simple harmonic motion from the classical physics
series, we can see an example of this type of behavior. This law describes the dynamics
of an object with mass m attached to a spring with the spring constant k. And specifically,
the law tells us that if a particle with mass m is subjected to a restoring force, such as
if it is held by a spring, then the mathematical expression for this restoring force is F = -kx,
where the spring constant k depends on the material and physical properties of the spring,
and x is the position. From this expression, we can determine what the potential looks like. To
do this, we can use a known relationship between a force and a potential that we learned earlier in
the series. This relationship states that a force equals the opposite of the spatial derivative of
the potential. What this means is that if we know the mathematical shape of a potential
V of x in the scalar form, meaning not as a vector, then we can find the mathematical shape
of the force that results from it as the spatial gradient of V of x. That will look like this.
To clarify, by spatial gradient, we mean a derivative with respect to spatial coordinates.
If we think of a three-dimensional system defined by vectors in this form,
then we need to take the partial derivatives with respect to the three spatial directions.
The negative sign is there because force equals the opposite of the gradient of a potential.
It’s like with a ball rolling up a hill. If the ball were in a flat region, the gradient,
or slope, would be zero, and thus there is no force. But going up the slope, the potential
energy increases, and the ball will feel a force in the opposite direction, down the slope. Without
the negative sign, it’s as though the force would cause the ball to accelerate up the hill, and thus
travel indefinitely to regions with higher energy instead of lower, which is an unphysical solution.
Now, we know that the gradient, represented by this symbol, is equal to the sum of these
partial differentials, but if we are dealing with a one-dimensional problem, as we are here,
we can just use d/dx. So applied to the potential, that will look like this, where F equals the
negative partial derivative of V of x with respect to x. Intuitively, this tells us that a force will
be larger where the gradient, or essentially the slope of the potential is larger, and where the
gradient is smaller, the force will be a smaller. So a flat potential would mean zero force,
and a large gradient or steep slope will mean a large force.
From here, we can substitute Hooke’s law in place of F, so F becomes negative kx. Let’s now multiply
both sides by negative one to get both sides positive, and then multiply by dx, so that we have
kx dx equals d V of x. Now let’s integrate both sides. Don’t get tripped up by d V of x, this is
not really any different than dx, and we know that integrating dx gives us x, so the integral of d V
of x will be V of x. Then integrating kx dx will give us one half kx squared.
Now, this happens to be the expression for a classical harmonic oscillator, V of x equals
one half k x squared. This expression will be very important moving forward, since it describes the
energy landscape that a particle encounters in a harmonic potential. What this means is that the
expression shows how the potential energy changes with respect to position. As the particle moves
in any direction, it will encounter an energy barrier, so to speak, that grows quadratically
with respect to x. Since we know that any system tends to move towards its lowest energy state,
low-energy particles will tend to remain at the region with minimum potential energy, which for
one half kx squared, is close to x equals zero. This is essentially like saying that a ball
on a spring will tend to remain at its equilibrium point, the point with the lowest potential energy.
Now, having defined the mathematical form of the harmonic potential,
we are ready to calculate the dynamics of a classical particle. To do this, we will use
Newton’s equations, which we are familiar with from the classical physics playlist. Solving the
Newtonian dynamics of this particle means finding how the particle’s position changes over time
when it experiences the force that stems from the spatial derivative of the harmonic potential.
So let’s go ahead and use the equation F = ma. This equation tells us that the force
applied to a particle equals the particle’s mass times the resulting acceleration,
and we can express acceleration as the second derivative of x with respect to time. Once again,
we also know that F = -kx, which is the force applied to the particle by the spring. So let’s
substitute this expression for F in our equation, and now we have this. This is a second-order
differential equation which we can now solve. First we have to rearrange things a bit.
Let’s get both of these terms on the same side, with their sum now equal to zero, and then let’s
divide through by m. To simplify a bit further let’s define a constant, beta, which is equal
to root k over m. That means we can place beta squared right here. Now we are ready to solve.
To speed up the solution of this second-order differential equation, let’s compare this equation
with the Schrödinger equation. Even though they represent completely different physical realities,
we may recognize that this equation looks like the Schrödinger equation with a potential that equals
zero, and where psi of x is substituted with simply x. To make things more clear, let’s set
V of x equal to zero, so we can get rid of this term, and then move the E term to the left side.
To go even further, in previous tutorials we have rearranged this to look more like this,
with the constants contained in this other constant k, which equals root 2mE over h bar,
and here we’ve also multiplied both sides by negative one to get everything positive.
And now in this form we can really see the similarity between our original
equation and the Schrödinger equation. We simply have x of t instead of psi of x.
Making this comparison is useful, because we have already solved this equation in earlier tutorials.
So we know the solution for x will be a linear combination of complex exponentials,
as we see here. And again, like with previous tutorials, we can solve this equation simply by
applying the appropriate boundary conditions. Remember that with the particle in a box and
with the quantum barrier potential, we had to make sure that the boundary conditions are fulfilled by
the solution for the Schrödinger equation, to yield continuous functions and either zero or
non-zero values as deemed necessary by the nature of the problem in question.
In these cases, we applied boundary conditions for the wavefunction in different regions,
and thereby found the constants that were multiplying the complex or real exponential terms.
In this situation, instead of a wavefunction in different regions, we are trying to find
what kind of shape, so to speak, the position and velocity of a particle must have as dictated by
Newton’s equations, more specifically this partial differential equation. So the main difference
is that for any quantum situation, we need the wavefunction to be continuous in space,
as psi of x depends on x, whereas with the classical particle we need continuity over time,
as x of t depends on t. And recall that the continuity condition requires that both
x of t and its derivative, d x of t over dt, must be continuous.
In addition, the interesting thing here is that unlike what we are used to with boundaries for x,
there are no time boundaries, except that we can not precede the time t equals zero,
so for this reason we will solve the time continuity equations only at t equals zero.
Now, with all of that understood, let’s say that our particle is at x equals zero at time t equals
zero, and that its velocity, given by v equals the partial derivative of x with respect to time,
is some constant velocity, v naught. Let’s apply these boundary conditions to the equation,
starting with velocity. So v zero will be equal to d x of t over dt, evaluated at zero,
which means we have to take the derivative of this expression with respect to t.
This just means bringing these constants out front for each term,
so we get i a beta e to the i beta t, minus i b beta e to the negative i beta t. Evaluating at t
equals zero makes both exponents zero, and anything to the zero power is one, so the e terms
essentially just go away, leaving us with i a beta minus i b beta. Again, we will call this v naught.
Then for the second condition, we know that x equals zero at time zero, so x of t becomes
x of zero, and we can set that equal to zero. This turns these exponents into zeros,
so these terms become equal to one, and we are left with a plus b equals zero.
So here is the resulting system of two equations. This is trivial to solve using basic algebra,
we can just multiply the second equation by i beta, and then we simply add them together.
The second terms cancel out, and we are left with two i beta a equals v naught.
Rearranging to solve for a we get this. Let’s take this and
plug it into the original second equation, which will allow us to then solve for b.
With values for a and b, we can now solve for x of t. We just plug these new expressions in for
a and b. To simplify let’s factor out this expression, leaving this parenthetical with
the exponential terms. Now we can get clever. We have used this Euler equation a few times by now,
and we can recognize that it has the same form as the equation we currently have, we just have beta
t in place of theta. We just need this expression to be over 2i in order to make the substitution,
so let’s multiply by 2i over 2i. These ones over here cancel out, leaving us with v naught
over beta, and then the other part we can now write as simply sine of beta t. Finally, recall
that beta represents root k over m, so let’s change the betas back to their original form.
With beta in the denominator here we simply invert the expression, so we can write root m over k, and
then root k over m can be written into the sine term. And there we have our expression for x of t.
The important thing to understand is that our expression for x of t contains a sine
function. That means the particle experiences an oscillatory behavior that looks just like a sine
wave. Its position depends on its mass, the spring constant k, and the initial velocity, v naught. We
can define the angular frequency of oscillation, omega, as being equal to the term within the sine
function, or root k over m. This will come into play a bit later. The frequency of oscillation is
simply the angular frequency divided by two pi, and this is much more straightforward,
measured in Hertz, and just represents the number of oscillations per unit time.
Let’s pause for a moment here. What we must understand is that mathematically speaking,
the particle can oscillate at any frequency, from zero, which means no oscillation at all,
or a particle at rest, up to any number you can think of. For example, if the mass is 200 grams
and k is 0.2 newtons per meter, then the particle would oscillate at a frequency of 0.16 hertz,
which is roughly one full oscillation every six seconds.
Now let’s have a look at the particle’s energy as it follows these oscillatory dynamics. The
particle’s kinetic energy is given by this expression, which we are familiar with from
classical physics. As we have been doing, let’s change velocity to the derivative of position
with respect to time. Now we just solved for x of t, so we should be able to take its derivative.
Because it is a sine function, and the derivative of sine is cosine, we will get the same expression
but with cosine, multiplied by what was inside the sine function because of the chain rule,
so root k over m comes out front. Conveniently, root m over k times root k
over m equals one, so we are left with v naught times this cosine term.
Now we can go back to our equation for kinetic energy and plug this expression
in where we see the derivative. Now this must be squared, so that just means we get v
naught squared times a cosine squared term. This is simply the well-known kinetic energy equation,
T equals one half m v naught squared, but now with an additonal oscillatory cosine squared term.
Potential energy is also quite trivial to solve for, since we defined it earlier as follows,
V of x equals one half k x squared, or now x of t squared. Let’s plug in our expression for x of t,
and then square it. The k’s out front will cancel out, and we get this, precisely the
same expression as for kinetic energy, just with a sine squared function instead of cosine squared.
Now the total energy of the system, E, will be equal to the kinetic energy, T, plus the potential
energy, V. So to get E, let’s just add these expressions together. We can factor one half m
v naught squared out, and we are left with this. If we recall any of our trigonometric identities,
probably the easiest one to remember is that sine squared theta plus cosine squared theta
equals one, which means that this entire parenthetical term simplifies to one.
This means the total energy of the particle is simply one half m v naught squared.
What does this result tell us? It tells us that the total energy of the classical harmonic
oscillator is a constant value that depends linearly on the initial velocity of the particle.
This means that a classical particle can have any energy from a continuous spectrum of energies. As
we will soon see, this is not the case for a quantum particle, and the quantum harmonic
oscillator. This should be no surprise by now, given the conclusions we have made with our
other quantum problems. Nevertheless, with this important classical context gathered,
let’s move forward and really dig into the quantum harmonic oscillator.