Last modified on 21 November 2012, at 07:38

Quantum Mechanics/Time Independent Schrödinger

Consider a particle confined to a one-dimensional box with impenetrable walls. When you solve the Schrödinger equation for the wavefunctions you get two sets of solutions: those of positive parity, and those of negative parity:

\Psi_{P=1} = A \cos \left[\frac{(2n+1) \pi x}{a}\right] and \Psi_{P=-1} = A \cos \left(\frac{2n \pi x}{a}\right),

where n is any positive integer and A is a normalisation constant. Now, we can have all of these infinite states and if you've ever studied Fourier Analysis you may have noticed, with these states you can form any function you wish---that is, the wavefunctions are complete. So what have we learned? Well, a lot actually: we have discovered the eigenstates of the Hamiltonian which can be used to determine the particle's time dependance.

Derivation of the Time-Independent Schrödinger EquationEdit

We start with the general Schrödinger Equation, and use separation of variables. We have

 H \Psi = \hat \epsilon \Psi

We separate \Psi into two functions:

\Psi ( x , t ) = T ( t ) X ( x )

So now the Schrödinger Equation is

 H T X = \hat \epsilon T X

We know from earlier that the "interesting" part of the energy operator \hat \epsilon is a partial derivative with respect to time, and the "interesting" part of the Hamiltonian H is a partial derivative with respect to position. As T does not depend on position, it is not affected by H. Similarly, X is not affected by \hat \epsilon.

So we have:

 T H X = X \hat \epsilon T

We can multiply on the left by  T^{-1} X^{-1} to obtain

 X^{-1} H X = T^{-1} \hat \epsilon T

Note that the left side only depends on x, on the right side only depends on t. We have two functions which are totally indenpendent, but are somehow equal to each other. This is only possible if both functions are equal to a consant, which we call E.

i.e.

 X^{-1} H X = E T^{-1} \hat \epsilon T = E

Naturally this implies

 H X = E X

and

 \hat \epsilon T = E T

We can then expand H and \hat \epsilon and solve this equation.