Harmonic Wave Functions and the Schrodinger Equation

Before we start, I want to say that to a person well acquainted with the principles and nuances of non-relativistic quantum mechanics, this note will seem redundant, which is fine because that is not the purpose of this note. The goal here is to introduce those interested to the Schrodinger Equation, which can be a very confusing relation for some.

The Schrodinger Equation is, in its most general form: \(\hat{E}\Psi=\hat{H}\Psi\)

Where \(\hat{E}\) is the energy operator: \(\hat{E}=i\hbar\frac{\partial}{\partial t}\)

And \(\hat{H}\) is the Hamiltonian of the system the equation is being applied to. The Hamiltonian may take many different forms, but always describes the total energy of the system, that is, the kinetic and the potential energies. \(\hat{H}=\hat{T}+\hat{V}\)

So as you can probably see, the Schrodinger equation, though much more ambiguous than its classical counterpart, essentially says total energy equals kinetic energy plus potential energy. We won't go into detail about kinetic and potential energy operators. Instead, we will focus on the most common version of this equation, the time-dependent Schrodinger equation for a single non-relativistic particle. The equation reads:

\(i\hbar\frac{\partial\psi}{\partial t}=[\frac{-\hbar^2}{2m}\nabla^2 +V(\vec{r},t)]\psi\)

Where: \(\nabla^2\) is the Laplacian operator (the divergence of the gradient)

\(\hbar\) is the reduced Planck constant (\(\hbar=\frac{h}{2\pi}\))

\(i\) is the imaginary unit

\(m\) is the mass of the particle

\(V(\vec{r},t)\) is some potential function and

\(\psi\) is the wave function.

Now, this is a linear partial differential equation. As is the case with all partial differential equations, it cannot be solved without boundary/initial conditions. However, let's assume \(\psi\) is a harmonic function. By definition: \(\nabla^2 \psi=0\)

Then the Schrodinger equation becomes, for harmonic \(\psi\):

\(i\hbar\frac{\partial\psi}{\partial t}=V(\vec{r},t)\psi\)

This equation now reads: "Total energy equals potential energy," which of course implies that the particle in question has no kinetic energy. Then we can conclude with confidence that if a particle has a harmonic wave function, then it is static. Furthermore, if \(V(\vec{r},t)\) exists, we may conclude that the particle is static and bound.

Let's suppose that \(V(\vec{r},t)\) exists. We now have two partial differential equations that must be satisfied by \(\psi\):

\(\nabla^2 \psi=0\)

\(i\hbar\frac{\partial\psi}{\partial t}=V(\vec{r},t)\psi\)

This may seem somewhat restrictive, and in a sense it is, but notice that this may actually give us a bit more information to work with. If we don't know \(V(\vec{r},t)\) but we have ample initial conditions and a general idea of the behavior of our particle in its environment, we may use the first equation (Laplace's Equation) as an aid to finding both \(\psi\) and \(V(\vec{r},t)\). Now, this is of course exceptionally ambitious, as in reality the situation is rarely so simple.

Note by Ethan Robinett
3 years, 10 months ago

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Does that form of the kinetic energy mean that the quantum mechanical expression momentum is equal to \(\frac{\hbar}{i}\frac{\partial}{\partial x}\)? How does one prove that is the case?

Josh Silverman Staff - 3 years, 10 months ago

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The proof for that comes from plane wave solutions to the Schrodinger Equation. You can read about it in more detail here. However, it begs the question that if a wave function is not a plane wave, is the momentum operator different than that used for a plane wave? If it were, then the entire equation would essentially be pointless, because if we don't know what type of wave function we have (which we wouldn't, assuming we haven't actually solved the equation), then we can't choose our operators appropriately and hence can't even construct the equation. So essentially the proof that \( \hat{p} = -i\hbar \nabla\) is derived from a case in which the wave function takes a very specific form. I'm glad you asked this question, really got me thinking.

If you want to directly derive the kinetic energy expression, you have:


\(\hat{p} = -i\hbar \nabla \) \(\Rightarrow\) \(E_k = \frac{(-i\hbar\nabla)^2}{2m}\)

\(\Rightarrow\) \(E_k = \frac{-\hbar^2\nabla^2}{2m}\)

Ethan Robinett - 3 years, 10 months ago

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Ok. The momentum operator takes the above form because any wave can be mathematically constructed as a linear combination of plane waves. Hence, the derivation for this form of the operator is universal, due to this fact. That's my reasoning though, I could be wrong.

Ethan Robinett - 3 years, 10 months ago

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@Ethan Robinett This is correct. More accurately, consider the Fourier Transform of the equation w.r.t. \(t\)

Abhishek Sinha - 3 years, 10 months ago

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please tell me the actual meaning of laplacian operator & fourier transform?

Krv Ramanan - 3 years, 10 months ago

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