# Proof of the time-independent Schrödinger equation

I have a question regarding the proof of the time-independent Schrödinger equation. So if we have a time-Independent Hamiltonian, we can solve the Schrödinger equation by adopting separation of variables method: we write our general solution as $\psi(r,t) = \psi(r)*f(t)$ and we get to the two equation : one for the f(t) and one for the $\psi(r)$. $$f(t)=e^{-\frac{i}{\hbar}Ht}$$ and $$H\psi(r) = E\psi(r)$$ Books in general refer to the second equation as the TISE and it is seen as an eigenvalue problem for the hamiltonian, in order to find the stationary states. Now what i don't understand is why we see that equation as an eigenvalue problem for the hamiltonian, since we have a wave-function $\psi(x)$ which is supposed to be eigenstate for H. So if $\psi(x)$ is eigenstate for H, it means that H is diagonal in coordinate basis, but i know it's not true since H has the term $\frac{P^2}{2m}$ in it, which is not diagonal in coordinate basis. Where am i wrong?? Thank you very much

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$H$ isn't diagonal in coordinate basis, but in the $\psi(r)$ Eigenbasis you're computing... –  Christoph Jan 21 '14 at 10:27
ok thank you it deals with $\psi(r)$ and not the coordinates actually. But why are we sure that $\psi(r)$ is eigenfunction of $H$? When i separate the variables i just say i have a function $f(t)$ and a function $\psi(r)$, it can be any function to be determined –  Danny Jan 21 '14 at 10:48

$$\hat H \Psi(r, t) = i \hbar \partial_t \Psi (r, t).$$

Using the ansatz $\Psi(r,t) = \psi(r) f(t)$ yields

$$f(t) \hat H \psi(r) = i \hbar \psi(r) \partial_t f(t)$$

and, via the standard separation of variables trick,

$$i\hbar\frac{\dot f(t)}{f(t)} = \text{const} = \frac{\hat H\psi(r)}{\psi(r)};$$

the two sides are equal, but the LHS depends only on $t$ while the RHS depends only on $r$, so they each have to be constant. Let us call this constant $E$. Then, for the time-dependent part, we get

$$\dot f(t) = -i \frac{E}{\hbar} f(t)$$

Which is manifestly solved by

$$f(t) = \exp\left\{-i \frac{E}{\hbar} t\right\}.$$

For the spatial part, we find the time-independent Schrödinger equation

$$\hat H \psi(r) = E \psi(r),$$

which as you observe can be viewed as an eigenvalue equation for $\hat H$. This motivates the choice of name for $E$: Physically, the eigenvalue of the Hamiltonian is the energy.

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