How to Rigorously Find the Explicit Form of the Eigenvalues Equation?

Question

Even in the simplest case of a free particle, so subjected to the Hamiltonian: $$H=\frac{\hat{p}^2}{2m}$$ we often need to find the explicit form of the eigenvalue equation: $$H|E\rangle=E|E\rangle \ \ \ \ \ (1)$$ this explicit form is: $$-\frac{\hbar^2}{2m}\frac{\partial ^2 \psi _E(x)}{\partial x^2}=E\psi _E(x) \ \ \ \ \ (2)$$ To get from the obvious equation (1) to the less obvious equation (2) we first of all need to project into the base of the eigenvectors of the position operator of course: $$\langle x | H | E\rangle = E\langle x |E\rangle=E\psi _E (x)$$ so the right side is already ok, now we only need to show that: $$\langle x | H | E\rangle=-\frac{\hbar^2}{2m}\frac{\partial ^2 \psi _E(x)}{\partial x^2}$$ we can write the Hamiltonian operator explicitly: $$\langle x | H | E\rangle=\langle x | \frac{\hat{p}^2}{2m} | E\rangle=\frac{1}{2m}\langle x | \hat{p}^2|E\rangle$$ And now what? Of course we know that: $$\hat{p}^2|\psi\rangle=\hat{p}(\hat{p}|\psi\rangle)$$ and we also know that: $$\langle x | \hat{p}|\psi\rangle = -i\hbar \frac{\partial}{\partial x}\psi(x) \ \ \ \ \ (3)$$ But from here how we can rigorously derive equation (2)? Of course we can make arguments like: since we have the square of the operator then everything must be the "square" of equation (3), but this does not sound very rigorous at all. We can also think to state something like: $$\langle x | \hat{p}^2|\psi\rangle=\hat{p}^2\psi(x)$$ but this also does not seem really rigorous, as we defined $\hat{p}$ as acting on ket vectors and not on functions. So how can we get equation (2)? How can we deal rigorously with the square of an operator?

Answer 1

You have had this question answered multiple times in numerous questions you have asked in the last couple of weeks on the Dirac notation, and operator action on bras (not kets, as promised!), $$\langle x | \hat{p} = -i\hbar \partial_x \langle x | , \tag{3}$$ whence $$\langle x | \hat{p}^2 = (-i\hbar \partial_x)^2 \langle x | , \tag{4}$$ and therefore $$\langle x | \hat{p} ^2|\psi\rangle = (-i\hbar \partial_x )^2 \langle x |\psi\rangle =-\hbar^2\partial_x^2 \psi(x).$$

I insist you are doing yourself a huge disservice by using the same symbol, in your last equation, for the abstract operator $\hat p ^2$ and also the representation of it in the coordinate basis, $$ \hat p ^2(x)= -\hbar^2 \partial_x^2 ~. $$ They do the same thing, but in different spaces, and that is exactly why Dirac lavished so much detail on this in his legendary book.

In formal language, of course, $$ \hat p ^2 = -\hbar^2 \int \! dx ~|x\rangle~ \partial_x^2 ~\langle x|~, $$ which bypasses much confusion.

It has been stressed to you repeatedly that derivation of bras is not terribly less rigorous than derivation of functions: the genius of the notation is that it makes the two equivalent.

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NB in response to comment

Your (3) is of course the same as my (3). You have $$ \left (\langle x |\hat p +i\hbar \partial_x \langle x| \right ) ~ |\psi\rangle=0 , $$ crucially for any and all (x-independent) kets $|\psi\rangle$; this is an identity, not an equation! So the bra that is perpendicular to all kets $|\psi\rangle$ is the null bra, my (3).