Eigen Value and Eigen Function

[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.

 

[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.

 

[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 e^iωt 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" )​

 

[roshan2004]

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

 

[alxm]

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

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)