From bra & ket vectors to wave functions

Question

I have a hard time understanding how the transition happens between the two. Starting from Schrödinger eqaution for kets: $$i\hbar\frac{d}{dt}\left|\psi\left(t\right)\right\rangle =\hat{H}\left|\psi\left(t\right)\right\rangle \implies\left\langle x\right|i\hbar\frac{d}{dt}\left|\psi\left(t\right)\right\rangle =\left\langle x\left|H\right|\psi\left(t\right)\right\rangle \implies i\hbar\frac{d}{dt}\left\langle x|\psi\left(t\right)\right\rangle =\left\langle x\left|H\right|\psi\left(t\right)\right\rangle \implies i\hbar\frac{d}{dt}\psi\left(x,t\right)=\left\langle x\left|H\right|\psi\left(t\right)\right\rangle \underbrace{=}_{?}\hat{H}\psi\left(x,t\right)$$

Meaning, I don't get why should: $$\left\langle x\left|\hat{H}\right|\psi\left(t\right)\right\rangle =\hat{H}\left\langle x|\psi\left(t\right)\right\rangle $$ Is it generally true for any operator and in any case?

Answer 1

The notation here is confusing, because the same symbol $\hat{H}$ is used for two different things:

The operator $\hat{H}$ that you start with, i.e. the one you use in $\hat{H} |\psi(t) \rangle$, is the Hamiltonian as an abstract Hilbert space operator.

The operator $\hat{H}$ that appears in $\hat{H} \langle x | \psi(t) \rangle$ is the representation of the Hamiltonian in the position representation.

So it would be much cleaner to write something like:
$\hat{H}^{(pos.)} \langle x | \psi(t) \rangle = \langle x | \hat{H} | \psi(t) \rangle $

This last equation is actually a definition of operators in the position representation of the Hilbert space.

This can be easier understood by considering the momentum operator $\hat{p}$. In the abstract Hilbert space of kets, it is just an abstract operator given by $\hat{p} = \int \mathrm{d} p \ p |p \rangle \langle p | \ $.

Its representation in position space, i.e. $\hat{p}^{(pos.)} \langle x | \psi \rangle = \langle x | \hat{p} | \psi \rangle$, is the familiar: $ \ \hat{p}^{(pos.)} = - i \frac{\partial}{\partial x}$

In particular, it is a differential operator.

However, its representation in momentum space, i.e. $\hat{p}^{(mom.)} \langle p | \psi \rangle = \langle p | \hat{p} | \psi \rangle$ is given by $\hat{p}^{(mom.)} = p \ $. So in the momentum representation of the abstract Hilbert space, the momentum operator is represented by just a number. This is different from the differential operator we had in the position representation.

It is quite common in textbooks and papers to not have this explicit label $(mom.)$ or $(pos.)$, simply because people got used to representing the same physical operator in different ways.

To go sure, $\psi(x) = \langle x | \psi \rangle$ is the wavefunction associated with the state $|\psi \rangle$. The object $\widetilde{\psi}(p) := \langle p | \psi \rangle$ is the analogous object in momentum space.

Answer 2

The expression $$\langle x|H|\Psi(t)\rangle =H\langle x|\Psi(t)\rangle \ \ \ \ \ (\text{not true !!!})$$ You can insert an identity in between $$\langle x|H|\Psi(t)\rangle = \int dx'\ \langle x|H|x'\rangle \langle x'|\Psi(t)\rangle$$ That's it. This is how far you can go without putting $H=\frac{P^2}{2m}+V(X)$. $$ \int dx'\ \langle x|H|x'\rangle \langle x'|\Psi(t)\rangle=\frac{1}{2m}\int dx'\langle x|P^2|x'\rangle \psi(x',t)+\int dx'\langle x|V(X)|x'\rangle \psi(x',t)$$ Putting the matrix elements, leads to Shroedinger's equation.

Answer 3

Generally, $$ \langle x |\hat{H}|\psi\rangle = \int dx'\langle x|\hat{H}|x'\rangle \langle x'|\psi\rangle = \hat{H}\psi(x), $$ where the second and the third terms are just the two ways to express the same thing.