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Time Evolution of Quantum Harmonic Oscillator

Previously, we deried all the important results of the quantum harmonic oscillator. We will now use these results to study the time evolution of the quantum harmonic oscillator from multiple approaches. First, we will use the Heisenberg picture, and then we will use Feynman path integrals.

Table of Contents​

Heisenberg Picture​

Deriving from the Equations of Motion​

Recall that

We know that for the harmonic oscillator. Thus, the first equation becomes

There are a few ways to solve this system of equations. One enlightening way is to consider the phase space and use matrices to represent the equations. We will instead first rewrite the equations using the creation and annihilation operators.

First,

where we have used the equations of motion for and . Similarly, we have

The solutions to these equations are obvious; these are just complex exponentials:

This is to be expected. The quantum harmonic oscillator is still a harmonic oscillator, so it should somewhat resemble the classical harmonic oscillator (which has solutions of the form ). We simply need to convert these solutions back to the position and momentum operators to see the time evolution of the quantum harmonic oscillator. The two equations above can be rewritten as

Dividing both sides of the first equation by and applying Euler's formula gives

and can be solved by equating the real and imaginary parts of these equations. This gives us

These equations are the time evolution of the quantum harmonic oscillator in the Heisenberg picture. It is worth noting that these equations are the same as the classical equations of motion for a harmonic oscillator, except that the initial conditions are quantum operators instead of classical variables. Recall that the classical equations of motion for a harmonic oscillator are

Deriving it Directly​

We can also derive the time evolution of the quantum harmonic oscillator directly by the definition of the time evolution operator. Recall that we time-evolve a quantum state by applying the time evolution operator to the state. For instance, for the inner product , we have

In the Heisenberg picture, we group the three operators together and define it as . As such, we can directly evolve the position operator in time by applying the time evolution operator to it, which means

where is the Hamiltonian of the system. To evaluate this, we rely on the Campbell identity, a corollary of the Baker-Campbell-Hausdorff formula. The Campbell identity states that if and are two operators, then

Using , we have

We can now compute the commutators. The commutator of the Hamiltonian and the position operator is

Similarly, the commutator of the Hamiltonian and the momentum operator is

As such, the commutator is

The commutator is

The pattern goes on. Every other commutator will be proportional to , and every other commutator will be proportional to . They multiply by a factor of for every two subsequent commutators:

We can now substitute these commutators back into the equation for :

We can now factor out the and terms:

These series are the Taylor series for and , respectively. Thus, we have

which is the same result we derived earlier.

Feynman Path Integrals​

We can also derive the time evolution of the quantum harmonic oscillator using Feynman path integrals.

Classical Action of Simple Harmonic Oscillator​

First, we need to write down the classical action for the quantum harmonic oscillator, which is just

Splitting the integral into two parts, we have

The first term can be evaluated using integration by parts, where and . Using integration by parts, we have

Since ,

Substituting this back into the action, we have

To continue, we know that , where and are constants determined by the initial conditions. We can shift the time variable to and , where is the total time of the evolution. Plugging the boundary conditions into the action, we have

As such, is just

So

Similarly, is

At , we have . Thus, the action becomes

Path Integral​

With the classical action in hand, we can now write down the path integral for the quantum harmonic oscillator. Recall that the propagator can be written as

where and are the final and initial positions, respectively, and is the total time of the evolution. The path integral is taken over all paths that satisfy the boundary conditions and .

We now introduce a technique known as the stationary phase approximation. We can split it into two parts—one for the classical path (stationary action) and one for quantum fluctuations around the classical path. Then, , where is the classical path and is the quantum fluctuation. The action of the path is then

The path integral can then be written as

Why was this helpful? After all, we get another identical path integral just over . However, there is a key difference—there are different boundary conditions for . Namely, and . In other words, we have

This has a crucial role that we will see very soon.

Let's first expand the path integral over into

Let's use integration by parts again on the first term. This gives

This time, since and , we have

Thus, we can rewrite the path integral as

where we have defined the operator . The term on the right-hand side resembles an expression of the form , where . This allows us to use the trick which, if you remember, is

Plugging in , , and , we have

Let's find the determinant!

For a finite matrix, there is a standard procedure using the elements of the matrix. For an operator, we need to use another approach. Recall that the determinant of an operator is equal to the product of its eigenvalues. As such, instead of trying to find matrix elements, we can find the eigenfunctions of the operator.

First, we need to solve the eigenvalue equation for the operator :

where are the eigenvalues. This can be rewritten as

Since , we can write the solution as

where is a constant and is a constant that satisfies . is thus, by definition,

Since we must have and , we have . This means that , where is a positive integer. Thus, we have

The determinant is thus

and so the propagator is

The constant is a normalization constant that ensures that the propagator is properly normalized. To find it, let's take the limit as . This should yield the free particle propagator, which is a known result:

We can compare this to as . The product becomes

Next, we just multiply Equation by to get

We know that , so this is thus

Finally, we can substitute this back into Equation to get the full propagator for the quantum harmonic oscillator,

Equivalence to Heisenberg EOM​

We now show that the propagator yields the same equations of motion as derived using the Heisenberg picture. Recall that operator time-evolution is unique to the Heisenberg picture, so the equation will not be exactly the same. However, recall that expectation values of operators are independent of which picture we use. As such, we just need to show that both approaches yield

This is trivial to show for the Heisenberg picture—just use and , where is the state of the system.

For the path integral, there are two ways to show this. One is more straightforward but more tedious, while the other is more elegant.

The straightforward way is to directly use the propagator. Begin with an initial state at time . We can assume it to be a Gaussian wave packet given by

where is a normalization constant, is the initial position, and is the initial momentum. For a coherent state, . Then, at a time , the position-space wavefunction is given by

Plugging in the propagator and initial state will yield a Gaussian wave packet at time . The expected value of the position operator will then match the equation of motion derived in the Heisenberg picture. I will not go through the tedious details here, but you can try it out yourself.

The better way is to use the following argument. Recall that the propagator obeys Schrödinger's time-dependent wave equation

Therefore, any state that evolves according to the propagator will also obey Schrödinger's equation. The key point is that for any such state, Ehrenfest's theorem must hold. Therefore,

Taking the leftmost and rightmost terms, we have

This is exactly the same as a simple harmonic oscillator, which has the solution

This is the same equation of motion we derived using the Heisenberg picture.

Coherent States​

For a stationary state (i.e., an energy eigenket) , we have that the expected values of observables do not change in time. This means that for any energy eigenket, and . This is different from the classical case, where the position and momentum of a particle do change (oscillate) in time.

However, there is a special class of states called coherent states that do oscillate in time. These are superpositions of energy eigenstates that yield the same equations of motion as the classical harmonic oscillator. Such a state is defined using the eigenvalue equation

where is the annihilation operator, is a complex number, and is the coherent state.

Since it is a superposition, we can write it as

It turns out that , taken as a function of , is a Poisson distribution given by

The uncertainty relations for coherent states will always satisfy the minimum uncertainty relation. This means that the uncertainty in position and momentum will always satisfy

Lastly, coherent states are just translated ground states.