yn(x) = Ö(2/a) sin(npx/a) |
x, p = - i hbar d/dx , T = p2/(2m), V = mw2/2, H = T + V, ... |
En = n2p2 hbar2 / (2 m a2) |
If y = Sn anyn , <H> = Sn |an|2 En |
Need an = dnn' |
Discrete | Continuous | |
---|---|---|
Eigenvalue equation | Q|n>=ln|n> | Q|k>=lk|k> |
Orthonormality | <m|n>=dmn | <l|k>=d(l-k) |
Expansions | |y>=Sn cn|n> | |y>=òdk c(k)|k> |
Coefficients | cn=<n|y> | c(k)=<k|y> |
Probability of getting a particular eigenvalue | |cn|2 | |c(k)|2dk |
yn(x) = <x|n>
fn(p) = <p|n> = <p|x><x|n> = òdx y*p(x) yn(x)
you can expand y in terms of any complete states (i.e. basis): y = Sn cn yn to get the equation
Sn cn H |n> = E Sn cn |n>
Taking the inner product with <m| [in other words, multiplying by ym* and integrating]
Sn <m|H|n> Cn = E dmn Cn
which is a matrix eigenvalue equation. Hence solving the Schroedinger equation is equivalent to diagonalizing the Hamiltonian matrix in an arbitrary basis. This is why solving the Schroedinger equation is often referred to as "diagonalizing the Hamiltonian."
Last Revised 01/10/05 |