# Schrodinger Equation in a representation

1. May 31, 2012

### youngurlee

The general evolution of a ket $|\psi\rangle$ is according to
$-i\hbar\frac{d}{dt}|\psi\rangle=H|\psi\rangle$
without specifying a representation.

From this equation, how can you simply get a equation in a certain representation $F$ as below:
$-i\hbar\frac{\partial}{\partial t}\langle f|\psi\rangle=\langle f|H|\psi\rangle$ ?

doesn't it need the validity of
$$\langle f|\frac{d}{dt}|\psi\rangle=\frac{∂}{∂t}\langle f|\psi\rangle$$
?

does this always hold for any ket and bra in a Hilbert space and its dual space?

2. Jun 1, 2012

### dextercioby

As long as f is time independent, yes. This is true in the Schroedinger picture for example, where f can be x or p_x and as operators these are time independent.

3. Jun 1, 2012

### youngurlee

thank you for your reply, but I really think some condition is required.

for example, look upon $\langle f|$ (independent of $t$) as a eigenbra of $F$ whose eigenbras are continuous in $f$. and $\langle f|$ acts on $|\psi\rangle$ as an linear functional.
that is:
$\langle f|\psi\rangle=\psi (f,t)=\int^{+\infty}_{-\infty}\phi^*(x,f)\psi (x,t)dx$

so, $\langle f|\frac{d}{dt}|\psi\rangle=\frac{∂}{∂t}\langle f|\psi\rangle$ in this case means:

$\int^{+\infty}_{-\infty}\phi^*(x,f)\frac{\partial}{\partial t}\psi (x,t)dx=\frac{\partial}{\partial t}\int^{+\infty}_{-\infty}\phi^*(x,f)\psi (x,t)dx$

I mean, doesn't this equation (exchange of derivative and improper integral) require some condition, such as the uniform convergence of the improper integral$\int^{+\infty}_{-\infty}$?

4. Jun 1, 2012

### strangerep

One way to deal with this more rigorously is to work in the context of so-called "rigged Hilbert space" (Gel'fand triples) -- if you're not familiar with these terms, think "generalized functions" or "distributions". Derivatives are then typically interpreted as some kind of "weak derivative".
http://en.wikipedia.org/wiki/Weak_derivative

5. Jun 8, 2012

### lugita15

The integral here is a Lebesgue integral, not a Riemann integral, so I think interchange of derivative and integral is allowed.

6. Jun 8, 2012

### youngurlee

lugita15, I know little about Lebesgue integral.
Be it a Lebesgue integral, interchange of derivative and integral is always allowed ?
I read somewhere that Lebesgue integral is a generalization of Riemann integral, then if the interchange in Riemann integral does not hold for a certain integral, will it hold in Lebesgue integral ?

7. Jun 9, 2012

8. Jun 9, 2012

thank you.