# Eigen Value and Eigen Function

1. Jun 18, 2010

### roshan2004

I know that the eigen equation is given by $$A\psi =\lambda\psi$$ where A is an operator and $$\psi$$ is an eigen function and $$\lambda$$ is an eigen value.
For the particle in an infinite square well of width L the wavefunction is given by $$\psi (x)=\sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})$$ and the energy is given by $$E=\frac{n^2h^2}{8ml^2}$$
I want to know why $$\psi$$ is called eigen function and E is called the eigen value.

2. Jun 18, 2010

### alxm

From German eigen, meaning "distinct", "particular", "[one's] own".

It comes by analogy to linear algebra. If a vector $$\vec{v}$$ is such that when you multiply a matrix A, by it, you get back the same matrix multiplied by some scalar (lambda):
$$\mathbf{A}\vec{v} = \lambda\mathbf{A}$$
Then $$\vec{v}$$ is an eigenvector of A, and $$\lambda$$ the corresponding eigenvalue. Any linear algebra textbook will tell you all about it.

This concept can be generalized to linear operators (matrix multiplication being a linear operation). For a linear operator $$\hat{O}$$, and a function f, if:
$$\hat{O}f = \lambda f$$

Then f is an eigenfunction of $$\hat{O}$$, and $$\lambda$$ is again the corresponding eigenvalue. If you have a given operator and some boundary conditions, and wish to find the eigenfunctions and eigenvalues for it, then that's called an "eigenvalue problem".

For instance, the differential operator $$\hat{D}^n = \frac{d^n}{d x^n}$$ is a linear operator.
So differential equations, such as $$\frac{df}{d x} - \lambda f(x) = 0$$ are eigenvalue problems.

3. Jun 18, 2010

### tiny-tim

hi roshan2004!

(have a psi: ψ and a lambda: λ and a pi: π and a square-root: √ and a curly d: ∂ )

your wavefunction should also have an eiωt factor

the (energy) operator A is ih∂/∂t (where i'm writing h for h-bar = h/2π),

so the (energy) eigenvalue is hω

(which just happens to be h(nπ/L)2/2m = h2n2/8mL2

see http://en.wikipedia.org/wiki/Square_well" [Broken] )​

Last edited by a moderator: May 4, 2017
4. Jun 18, 2010

### roshan2004

Thanks, but in this case I couldnot find operator here, which is the operator in this case?

5. Jun 18, 2010

### alxm

The Hamiltonian. The time-independent Schrödinger equation is simply the eigenvalue problem for the Hamiltonian.
Also, since V = 0 and you only have one dimension in the infinite square well, the Hamiltonian in that case is just the second-order differential operator (multiplied by $$-\hbar/2m$$)