Equivalence of Schrödinger's representation of state and Dirac's notation

Question

I am slightly confused regarding the equation $$\psi(x)=\langle x|\psi\rangle $$ Now, basically from my initial knowledge about Dirac's notation, I am able to see the expression $\langle x|\psi\rangle $ as some projection of the state vector on the position space. So, it is natural enough that it should be some function of the position variables which I can take as $\psi(x)$. But the problem is that, I like slightly more rigorous definitions with logic more than intuition. So, the confusion starts here when I consider the following two equations $$\hat H\psi(x,t)=i\hbar \frac{\partial \psi(x,t)}{\partial t}$$ $$\hat H|\psi(t)\rangle =i\hbar \frac{\partial |\psi(t)\rangle}{\partial t}$$ Suppose I solve the second equation, then I get a state vector which I can project onto the position space to get a wavefunction $\psi(x,t)$. But how can I ascertain that this map which has made on the position space exactly represents the solution of the first equation, considering that we solve both the equations separately?

Indirectly, the main question is that how can we justify on purely mathematical grounds that $\langle x|\psi\rangle =\psi(x)$ where LHS arises out of the second equation and the RHS arises from the first equation. Also, if we assume that it is a definition, how can we claim that it will not cause inconsistencies later?

Answer 1

Much of your confusion is sloppiness with symbols. This is a routine demonstration in elementary QM courses. $$\hat H|\psi(t)\rangle =i\hbar \frac{\partial |\psi(t)\rangle}{\partial t}\tag{1}$$ is the evolution equation of that ket, and let's henceforth skip t, as it always lives in ψ in all its forms, not in the $|x\rangle$s. Operate with the identity operator on $|\psi\rangle$, $$|\psi\rangle= \int\!\! dx ~ |x\rangle \langle x|\psi\rangle \equiv \int\!\! dx ~ |x\rangle \psi(x),$$ where we just defined the coefficients of the $|x\rangle$s as the function $\psi(x)$. (You recall this is also a function of t, which we skipped writing.)

You recall the x-representation of the operators $$ \hat x= \int\!\! dx ~ |x\rangle x \langle x|, \qquad \hat p= -i\hbar \int\!\! dx ~ |x\rangle \partial_x \langle x|,\leadsto \\ \langle y|\hat p|x\rangle= -i\hbar \partial_y\delta(x-y), ~~\leadsto ~~ \langle y|\hat p ^2|x\rangle= -\hbar^2 \partial_x^2\delta(x-y), \tag{2} $$ so you may work out the x-representation of the customary hamiltonian, e.g., $$ \hat H = \hat p^2/2m + V(\hat x), \leadsto \\ \langle y|\hat H|x\rangle =-\hbar^2 ~ \partial_x^2 \delta(x-y) ~/2m + V(x)\delta (x-y) \\ \equiv \hat H _x ~ \delta (x-y). $$ The x-dependent operator just defined, $\hat H _x $, is called "the x-representation of the abstract hamiltonian" of (1). It is not the same object you used the same symbol for.

Dot both sides of (1) by $\langle x| $ and insert a resolution of the identity after its hamiltonian, $$ \int\!\! dy ~ \langle x|\hat H|y\rangle \langle y|\psi\rangle =i\hbar \frac{\partial \langle x|\psi \rangle}{\partial t}~~\implies \\ \int\!\! dy ~ (\hat H _y \delta (y-x) )~~\psi (y) =i\hbar \frac{\partial \psi(x)}{\partial t} \implies \\ \hat H_x\psi(x )=i\hbar \frac{\partial \psi(x )}{\partial t}~~, \tag{3}$$ which is your x-representation TDSE you wished to justify: it is the endpoint, hardly the starting point, despite the perverse chronological order these things were arrived at. Dirac's masterful logic trumps all.