Your step $[1]$ is not OK. Just because $\hat H$ has eigenvectors whose eigenvalue is $E$ does not mean that any vector will be an eigenvector of $\hat H$ with eigenvalue $E$. In particular, the position eigenstates are emphatically not eigenstates of the hamiltonian.

To go beyond $$⟨\hat{H}⟩ = \int dx'\int dx⟨\psi|x'⟩⟨x'|\hat{H}|x⟩⟨x|\psi⟩$$ you need to specify further what you mean by $\hat H$ - if all you know is that it is some operator on $\mathcal H$ then you can't go any further, because there's no need for it to have any special relation to the position basis.

Usually, however, you are dealing with a non-relativistic single-particle hamiltonian of the form $H= p^2/2m +V(x)$. In this case, the correct way to handle $\hat H$ is to commute it with the bra to its left: $$ ⟨x'|\hat H=\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]⟨x'|. $$ Here the derivative acts on anything $x'$-dependent to the right of the derivative. (To see more about this, try this thread.)

To implement this in your expression for $⟨\hat H⟩$, you first start by completing the sandwich: \begin{align} ⟨x'|\hat H|x⟩ &= \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]⟨x'|x⟩ \\ &= \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]\delta(x-x'), \end{align} since the position states are orthogonal. You then perform the $x$ integral, to get \begin{align} \int dx⟨x'|\hat{H}|x⟩⟨x|\psi⟩ &= \int dx\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]\delta(x-x')⟨x|\psi⟩ \\&= \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]⟨x'|\psi⟩. \end{align} You can then do the integral over $x'$, to get \begin{align} ⟨\hat H⟩=\int dx'\int dx⟨\psi|x'⟩⟨x'|\hat{H}|x⟩⟨x|\psi⟩ &= \int dx'⟨\psi|x'⟩\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]⟨x'|\psi⟩ \\&= \int dx\psi(x)^*\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} +V(x)\right]\psi(x), \end{align} where I've relabelled $x'$ as $x$. This is the final result and there's nothing you can do to simplify it without further assumptions.

Your notation, $$\int_{-\infty}^\infty\mathrm{d}x\ \psi(x)^*E\psi(x) \tag{$*$}$$ is ambiguous and probably incorrect, but this depends on the context.

• If $E$ is a shorthand for the operator $-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2} +V(x)$, then this is formally correct but it is a misleading way to represent this. This is why we use hamiltonians and carets to denote operators, so that we keep clear tabs on what is a number and what is an operator.

• If $E$ is a $c$-number (i.e. not an operator) then the expression has very limited validity. More specifically, it is only valid if $\psi$ is an eigenstate of $\hat H$ with eigenvalue $E$, in which case you simply write $\hat H|\psi⟩=E|\psi⟩$ and therefore $⟨\hat H⟩=⟨\psi|\hat H|\psi⟩=E$.

There isn't really any case where the notation $(*)$ is both correct, useful, and non-misleading.

Your step $[1]$ is not OK. Just because $\hat H$ has eigenvectors whose eigenvalue is $E$ does not mean that any vector will be an eigenvector of $\hat H$ with eigenvalue $E$. In particular, the position eigenstates are emphatically not eigenstates of the hamiltonian.

To go beyond $$⟨\hat{H}⟩ = \int dx'\int dx⟨\psi|x'⟩⟨x'|\hat{H}|x⟩⟨x|\psi⟩$$ you need to specify further what you mean by $\hat H$ - if all you know is that it is some operator on $\mathcal H$ then you can't go any further, because there's no need for it to have any special relation to the position basis.

Usually, however, you are dealing with a non-relativistic single-particle hamiltonian of the form $H= p^2/2m +V(x)$. In this case, the correct way to handle $\hat H$ is to commute it with the bra to its left: $$ ⟨x'|\hat H=\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]⟨x'|. $$ Here the derivative acts on anything $x'$-dependent to the right of the derivative. (To see more about this, try this thread.)

To implement this in your expression for $⟨\hat H⟩$, you first start by completing the sandwich: \begin{align} ⟨x'|\hat H|x⟩ &= \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]⟨x'|x⟩ \\ &= \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]\delta(x-x'), \end{align} since the position states are orthogonal. You then perform the $x$ integral, to get \begin{align} \int dx⟨x'|\hat{H}|x⟩⟨x|\psi⟩ &= \int dx\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]\delta(x-x')⟨x|\psi⟩ \\&= \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]⟨x'|\psi⟩. \end{align} You can then do the integral over $x'$, to get \begin{align} ⟨\hat H⟩=\int dx'\int dx⟨\psi|x'⟩⟨x'|\hat{H}|x⟩⟨x|\psi⟩ &= \int dx'⟨\psi|x'⟩\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x'^2} +V(x')\right]⟨x'|\psi⟩ \\&= \int dx\psi(x)^*\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} +V(x)\right]\psi(x), \end{align} where I've relabelled $x'$ as $x$. This is the final result and there's nothing you can do to simplify it without further assumptions.

Your notation, $$\int_{-\infty}^\infty\mathrm{d}x\ \psi(x)^*E\psi(x) \tag{$*$}$$ is ambiguous and probably incorrect, but this depends on the context.

• If $E$ is a shorthand for the operator $-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2} +V(x)$, then this is formally correct but it is a misleading way to represent this. This is why we use hamiltonians and carets to denote operators, so that we keep clear tabs on what is a number and what is an operator.

• If $E$ is a $c$-number (i.e. not an operator) then the expression has very limited validity. More specifically, it is only valid if $\psi$ is an eigenstate of $\hat H$ with eigenvalue $E$, in which case you simply write $\hat H|\psi⟩=E|\psi⟩$ and therefore $⟨\hat H⟩=⟨\psi|\hat H|\psi⟩=E$.

There isn't really any case where the notation $(*)$ is both correct, useful, and non-misleading.