From bra & ket vectors to wave functions

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?

• The last line cannot be true (I know its notation, but...): On the left hand side is a $\mathbb{C}$ number, whereas on the right hand side its an operator (times a $\mathbb{C}$ number). But I guess its a common abuse of notation. It is the same issue with the momentum operator, where people write $\hat{p} \psi(x)$ etc... See for example this and this, I think these could help. May 21 at 8:36
• Hint: insert the resolution of identity between $H$ and the ket it acts on. Notation is indeed confusing but I assume on the r.h.s. we’re meant to act with $H$ on the wavefunction and take the value of the result at $x$. May 21 at 9:34

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.

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.

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.