# Finding the potential function from the wavefunction

• Kaguro
In summary, the conversation discusses the process of differentiating a given equation and plugging it into the Schrödinger equation. The main issue is that the person does not have the necessary parameter, E, to complete the calculation. However, through further discussion and calculation, V(x) and E are eventually determined. The conversation also touches on the importance of the chain rule and the physical meaning of the problem.
Kaguro
Homework Statement
A particle is confined in one dimension by a potential V(x).
Given psi=[(x/a)^n]*exp(-x/a), find the potential.
Relevant Equations
H (psi)=E(psi)
H ={ -(h_bar)^2/2m +V}
I would differentiate this twice and plug it into the S.E, but for that I'll need E. Which I don't have. Please provide me some direction.

$$H\psi(x)=[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)]\psi(x)$$
$$\psi(x)=(\frac{x}{a})^n e^{-x/a}$$?

BvU
Sorry, I forgot typing the derivative.
Anyways, I can find the derivative and write the whole LHS, but for RHS, E(psi) , I need E. That's the main problem. What do I do with my lack of E?

What is the result of double differential of ##\psi?##

##\frac{d^2 \psi}{dx^2} = e^{-x/a}*[n(n-1)(x/a)^{n-2} -(2n+1)(x/a)^n + (x/a)^{n+2}]##

$$= \{n(n-1) (x/a)^{-2}+(x/a)^2-(2n+1)\} \psi(x) = \frac{2m}{\hbar^2}(V(x)-E)\psi(x)$$
So you find V(x) and E if your calculation is right. I am afraid not.

Last edited:
Kaguro
V(x) = ## \frac{\hbar ^2}{2m}((n)(n-1)(x/a)^{-2}+(x/a)^2)##
E = ## \frac{\hbar ^2(2n+1)}{2m}##

Thank You very much!

I was too worked up over not having E...

Check your math and do the same way.
$$\frac{d}{dx}e^{-x/a}=-\frac{1}{a} e^{-x/a}$$

Kaguro
Oh man! I forgot the chain rule! How can I forget the chain rule in final year undergrad?!

So, finally V(x) = ##\frac{\hbar^2}{2m}(\frac{n(n-1)}{x^2} -\frac{2n}{ax})##

You see ##|\psi|## increases to infinity for ##x→-\infty## though ##V \rightarrow 0##. Your teacher may explain physical meaning of the problem.

## 1. What is the potential function?

The potential function, also known as the potential energy, is a mathematical function that describes the energy of a system of particles based on their positions. It is an important concept in physics and is used to understand the behavior of particles in various physical systems.

## 2. How is the potential function related to the wavefunction?

The potential function is related to the wavefunction through the Schrödinger equation, which is a fundamental equation in quantum mechanics. The wavefunction describes the probability of finding a particle in a certain state, while the potential function describes the energy associated with that state.

## 3. How do you find the potential function from the wavefunction?

To find the potential function from the wavefunction, you need to solve the Schrödinger equation for the given system. This involves using mathematical techniques such as separation of variables and boundary conditions to determine the potential function that corresponds to the given wavefunction.

## 4. What are some common methods for finding the potential function?

Some common methods for finding the potential function include the variational method, the perturbation method, and the numerical method. These methods involve different mathematical approaches and are used depending on the complexity of the system and the accuracy required.

## 5. Why is finding the potential function important?

Finding the potential function is important because it allows us to understand the behavior of particles in various physical systems. It also helps us to predict the properties and interactions of particles, which is crucial in fields such as quantum mechanics, chemistry, and material science.

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