Schrodinger equation describes the time evolution of a (state) vector | \Psi \rangle in an abstract vector space \mathcal{H} (the Hilbert space) i \frac{\partial}{\partial t} | \Psi \rangle = H | \Psi \rangle . Written this way, it is clear that Schrodinger equation is independent of any representation spaces. And, it is exactly this property that makes QM so powerful and universal. To understand what I mean, let \{ | \alpha \rangle \} \in \mathcal{H} be complete orthonormal set of states with the variable \alpha takes on continuous, discrete or continuous and discrete values. Using the completeness property together with the inner-product in \mathcal{H}, we can transform the above Schrodinger equation to the following form i \frac{\partial}{\partial t} \langle \beta | \Psi \rangle = \sum_{\alpha} \int d \alpha \ \langle \beta | H | \alpha \rangle \langle \alpha | \Psi \rangle , \ \ \ (1) or i \frac{\partial}{\partial t} \Psi ( \beta ) = \sum_{\alpha} \int d \alpha \ \langle \beta | H | \alpha \rangle \ \Psi ( \alpha ) , \ \ \ (2) where \beta may or may not be in the set \{ | \alpha \rangle \}. I will suppress the time-dependence of the state vector, i.e, you should read \Psi ( \cdots ) as \Psi ( \cdots , t ).
(i) If we take, in (2), \beta = x and \alpha = y to be coordinates in the position space, we obtain the usual Schrodinger differential equation: i \frac{\partial}{\partial t} \Psi ( x ) = \int d^{3} y \ H ( y ) \ \delta^{3} ( x - y ) \ \Psi ( y ) = H ( x ) \Psi ( x ) . \ \ \ (3)
(ii) If you take \beta = p and \alpha = \bar{p} to be continuous momentum-space variables, then Schrodinger equation (2) becomes an integral equation (as it should be in momentum space) i \frac{\partial}{\partial t} \Psi ( p ) = \int d^{3} \bar{p} \ H ( p , \bar{p} ) \ \Psi ( \bar{p} ) . \ \ \ \ (4)
(iii) Now consider ( \beta , \alpha ) to be discrete energy levels ( E_{n} , E_{m} ) or spin variables ( \sigma_{n} , \sigma_{m} ), and call \Psi ( E_{n} ) \equiv C_{n} and \langle E_{n} | H | E_{m} \rangle \equiv H_{n m}. In this case, the Schrodinger equation becomes a matrix equation i \frac{\partial}{\partial t} C_{n} = \sum_{m} H_{n m} \ C_{m} . \ \ \ \ \ (5)
And finally, I leave you to work out the detail, take ( \beta , \alpha ) = ( x , p ) and deduce the Fourier transformation between momentum and position space wave functions \Psi ( x ) = \int d^{3} p \ \langle x | p \rangle \ \Psi ( p ) , with \langle x | p \rangle \sim \exp ( i p \cdot x ) serves as transformation matrix.
Sam
