# Schrodinger equation Eigenfunction problem

• joex444
In summary, Daniel is trying to find the potential energy and energy of an eigenfunction of a wave function using Schrodinger's equation. However, he has trouble understanding what the definition of hermitian is and gets lost in the math. He eventually solves the problem by realizing that the solutions are of the form e^{-x^m} where m is some integer (perhaps a real > 1??).

#### joex444

So, here's the question:

$$\psi(x) = A*(\frac{x}{x_{0}})^n*e^(\frac{-x}{x_{0}})$$
Where A, n, and X0 are constants.

Using Schrodinger's equation, find the potential U(x) and energy E such that the given wave function is an eigenfunction (we can assume that at x = infinity there is 0 potential).

So, in my futile attempt at beginning this I tried to use $$\hat p\psi(x) = p \psi(x)$$ but my partial of psi with respect to x left me with a difference, due to the product rule which left me with $$-i*\hbar(\frac{1}{x_{0}} - \frac{n}{x}) = p$$ which isn't an eigenvalue for momentum. Using an E operator makes no sense, because there's no t term so the derivative goes to 0 and it just doesn't make sense.

Forget the right track, is this the right station?

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you're looking for values of U(x) and E such that

$$\frac{\hat{p}^2}{2m}\psi+U(x)\psi = E\psi$$

So I'd begin by expanding this equation (i.e. do the derivatives involved and simplify, etc) and then start looking for U(x) and E sucht that make the equality true.

I don't know what "$$p\hat \psi(x) = p \psi(x)$$" means. If $\hat \psi$ is the complex conjugate of $\psi$, it isn't true, and you're mistaking what the defintion of hermitian is.

You know that:

$$\frac{\hat p^2}{2m} \psi +U(x) \psi = E \psi$$

So solve for U. There is a solution for every E, but I'm guessing they don't all go to zero at infinity.

StatusX said:
I don't know what "$$p\hat \psi(x) = p \psi(x)$$" means.

Oh, sorry, my tex was backwards its fixed now.

OK, so using the p operator squared, that equation becomes the time independent schrodinger equation.

Solving for the second derivative of psi I obtained $$\frac{\partial^2\psi(x)}{\partial x^2} = \psi(x)(\frac{1}{x_{0}^2} - \frac{2n}{x} + \frac{(n-1)n}{x^2})$$. I thought it was kind of useful I could pull psi out, but the rest of it is fairly ugly. I've checked the derivation three times, so I'm fairly sure it's right. I also tried putting it all in a common denominator of $$x^2x_{0}^2$$ but its worse.

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StatusX said:
I don't know what "$$p\hat \psi(x) = p \psi(x)$$" means. If $\hat \psi$ is the complex conjugate of $\psi$, it isn't true, and you're mistaking what the defintion of hermitian is.

You know that:

$$\frac{\hat p^2}{2m} \psi +U(x) \psi = E \psi$$

So solve for U. There is a solution for every E, but I'm guessing they don't all go to zero at infinity.

Not E, but the potential U(x) needs to go to zero at infinity. In this case, it does, since $n\geq 0$.

Daniel.

So here's how I'm doing this derivative, correctly this time.
$$u = A(\frac{x}{x_{0}})^n$$
$$du = Ac^{-n}nx^{n-1}$$
$$v = e^{-x/x_{0}}$$
$$dv = -\frac{e^{-x/x_{0}}}{x_{0}}$$
$$\frac{\partial\psi(x)}{\partialx} = udv + vdu$$
$$\frac{\partial\psi(x)}{\partialx} = \psi(x)(\frac{-1}{x_{0}}+\frac{n}{x})$$
$$\frac{\partial^2\psi(x)}{\partialx^2} = (\frac{-1}{x_{0}}+\frac{n}{x})(\frac{\partial\psi(x)}{\partialx}) + \psi(x)\frac{\partial\psi(x)}{\partialx}(\frac{-1}{x_{0}}+\frac{n}{x})$$
$$\frac{\partial^2\psi(x)}{\partialx^2} = \psi(x)(\frac{n^2}{x^2}+\frac{1}{x_{0}^2}-\frac{2n}{xx_{0}}-\frac{n}{x^2})$$

Now, what do I do?

I haven't had differential equations so I'm pretty useless at that if it's needed.

I have no idea what's going on here, because the way I see it I now have:

$$\frac{n^2}{x^2}+\frac{1}{x_{0}^2}-\frac{2n}{xx_{0}}-\frac{n}{x^2}+U(x)=E$$

So now the only restriction is that U(x) -> 0 as x -> infinity. Well, I'm not sure what I can do, it seems to me to be 2 unknowns and 1 equation so I get an infinite number of solutions.

If I let U(x) be 0, then E is found. If I let U(x) be e^-x then it satisfies the going to 0 thing and I have an E. If I let U(x) be $$e^{-x^2}$$ then I have yet another solution, and another E.

Perhaps I've hit upon something, that the solutions are of the form $$e^{-x^m}$$ where m is some integer (perhaps a real > 1??)...

E is a constant, so the LHS of the equation must also be a constant. Thus all the x dependence from the derivative terms must be canceled by the x dependence in U(x). Then E will be specified by the requirement that U(x) go to zero at infinity.

I have the same problem to solve, but I don't know how to start, can someone help me?
Thank you.

## 1. What is the Schrodinger equation Eigenfunction problem?

The Schrodinger equation Eigenfunction problem is a mathematical equation that describes the behavior of quantum mechanical systems, such as atoms and molecules. It is a fundamental equation in quantum mechanics and is used to calculate the probability of finding a particle in a particular position or energy state.

## 2. Who developed the Schrodinger equation?

The Schrodinger equation was developed by Austrian physicist Erwin Schrodinger in 1926. He was one of the pioneers of quantum mechanics and his equation revolutionized the field by providing a mathematical framework for understanding the behavior of particles at the atomic level.

## 3. What is an Eigenfunction?

An Eigenfunction is a function that, when acted upon by a linear operator, returns a multiple of itself. In the context of the Schrodinger equation, an Eigenfunction represents the wave function of a quantum mechanical system, which describes the probability of finding a particle at a specific position or energy state.

## 4. How is the Schrodinger equation solved?

The Schrodinger equation is a complex partial differential equation and can be solved using various mathematical techniques, such as separation of variables, perturbation theory, and numerical methods. The solution involves finding the Eigenfunctions and Eigenvalues of the system, which represent the energy levels and corresponding wave functions.

## 5. What is the significance of the Schrodinger equation Eigenfunction problem?

The Schrodinger equation Eigenfunction problem is significant because it provides a fundamental understanding of the behavior of quantum mechanical systems. It has been validated by numerous experiments and is used in various fields such as chemistry, physics, and engineering to predict the behavior of particles at the atomic level. It also forms the basis of many modern technologies, such as transistors and lasers.