# Show that a wave function fits the Schrödinger's equation. (Harmonic oscillator)

• Asphyxion
In summary, the wave function \psi_0 (x) = A e^{- \dfrac{x^2}{2L^2}} represents the ground-state of a harmonic oscillator. The function \psi_1 (x) = L \dfrac{d}{dx} \psi_0 (x), with energy E_1 = 3/2 \hbar \omega, is also a solution of Schrödinger's equation and represents an excited state with one node.
Asphyxion

## Homework Statement

The wave function $$\psi_0 (x) = A e^{- \dfrac{x^2}{2L^2}}$$
represents the ground-state of a harmonic oscillator. (a) Show that $$\psi_1 (x) = L \dfrac{d}{dx} \psi_0 (x)$$ is also a solution of Schrödinger's equation. (b) What is the energy of this new state? (c) From a look at the nodes of this wave function, how would you classify this excited state?

## Homework Equations

$$E_n = (n + 1/2) \hbar \omega$$

SE:
$$-\frac{\hbar ^2}{2m} \dfrac{\partial}{\partial x} \psi + \frac{1}{2} m \omega ^2 \psi = E \psi$$

## The Attempt at a Solution

$$\dfrac{\partial}{\partial x} \psi_0 = -\dfrac{xA}{L^2} e^{- \dfrac{x^2}{2L^2}}$$

$$\dfrac{\partial}{\partial x} \psi_1 = - \dfrac{A}{L} e^{- \dfrac{x^2}{2L^2}} + \dfrac{x^2A}{L^3} e^{- \dfrac{x^2}{2L^2}}$$

$$\dfrac{\partial ^2}{\partial x^2} \psi_1 = (\dfrac{xA}{L^3} \dfrac{x^2}{2L^2}} + \dfrac{2xA}{L^3} \dfrac{x^2}{2L^2}} - \dfrac{x^3A}{L^5} \dfrac{x^2}{2L^2}})e^{- \dfrac{x^2}{2L^2}}$$

$$\dfrac{\partial ^2}{\partial x^2} \psi_1 = (\dfrac{3xA}{L^3} \dfrac{x^2}{2L^2}} - \dfrac{x^3A}{L^5} \dfrac{x^2}{2L^2}})e^{- \dfrac{x^2}{2L^2}}$$
Put into SE this gives me:
$$-\dfrac{\hbar ^2}{2m} (\dfrac{x^2}{L^4} - \dfrac{3}{L^2}) + \frac{1}{2} m \omega ^2 = E$$
This is where I'm not getting any further. I just can't see how this is supposed to match the energy state of a harmonic oscillator. Am I thinking completely wrong, have I done the maths wrong. Or both? Any pointers would be greatly appreciated! (First time trying to post my LaTeX work on a forum, so be forgiving!)

Last edited:
[STRIKE]Wait with reading this, this is total gibberish :)[/STRIKE]
Okay, I think my chain of thoughts should be understandable now.

Last edited:
Okay, two hints:

Hint #1:
$$-\dfrac{\hbar ^2}{2m} (\dfrac{x^2}{L^4} - \dfrac{3}{L^2}) + \frac{1}{2} m \omega ^2 = E$$
should actually be
$$-\dfrac{\hbar ^2}{2m} (\dfrac{x^2}{L^4} - \dfrac{3}{L^2}) + \frac{1}{2} m \omega ^2 x^2 = E$$

Hint #2
Should energy have an x dependence? In other words, you know this has to satisfy the SE, so what can you do that will make it do that?

Yeah, I've tried to set x=0 which gives me $$E_1= 3/2 \dfrac{\hbar ^2}{2mL^2}$$ which i don't find satisfy the $$E_1 = 3/2 \hbar \omega$$.
And on my paper i ofcourse had the x^2 part of the portential.

You can't just set x=0, you have to get rid of it in another way. Look at the text below if you get super stuck.

You have to collect the x^2 terms and use the constraint of the constants to take away the x^2 terms. You will be able to find out what L^2 is, and it will give you the E1 energy level.

I'm afraid I'm completely stuck even with the hint! I cannot find any example of anything similar to this in my book. And since you were hiding that hint, I can only assume it's supposed to be a rather short and easy step in the proof.

Okay here's how it goes:

$$-\dfrac{\hbar ^2}{2m} (\dfrac{x^2}{L^4} - \dfrac{3}{L^2}) + \frac{1}{2} m \omega ^2 x^2 = E$$

collect the x^2 terms

$$x^2(-\dfrac{\hbar ^2}{2mL^4}+\frac{1}{2} m \omega ^2) + \dfrac{3}{L^2}\dfrac{\hbar ^2}{2m} = E$$

set the constants multiplying the x^2 to 0 so that the x^2 term goes away
$$-\dfrac{\hbar ^2}{2mL^4}+\frac{1}{2} m \omega ^2 = 0$$

Solve for L^2 and plug it back into the new SE you will get the $3/2 \hbar \omega$ you are looking for.

Thanks a bunch! I see in retrospect that I would've never figured this out. I hope that's not a bad sign for me as a student of physics :)

## 1. What is the Schrödinger's equation?

The Schrödinger's equation is a fundamental equation in quantum mechanics that describes the time evolution of a quantum system. It is a partial differential equation that relates the wave function of a system to its energy and potential.

## 2. What is a wave function?

A wave function is a mathematical description of the quantum state of a system. It is a complex-valued function that contains all the information about the system, including its position, momentum, and energy.

## 3. What is the harmonic oscillator?

The harmonic oscillator is a quantum mechanical system that consists of a particle moving in a potential that is proportional to its displacement from a fixed point. It is a fundamental model used in many areas of physics, including quantum mechanics.

## 4. How do you show that a wave function fits the Schrödinger's equation for a harmonic oscillator?

To show that a wave function fits the Schrödinger's equation for a harmonic oscillator, we first need to express the wave function in terms of the position and momentum operators. Then, we substitute this expression into the Schrödinger's equation and solve for the energy eigenvalues and corresponding wave functions. If the resulting solutions satisfy the Schrödinger's equation, then we can conclude that the wave function fits the equation for a harmonic oscillator.

## 5. Why is it important to show that a wave function fits the Schrödinger's equation for a harmonic oscillator?

Showing that a wave function fits the Schrödinger's equation for a harmonic oscillator is important because it confirms that the wave function is a valid solution to the equation, and therefore accurately describes the quantum state of the system. This is crucial in understanding the behavior of quantum systems and making predictions about their properties and interactions.

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