# How is an operator applied to a wavefunction in quantum mechanics? [closed]

If you have the Hamiltonian operator written as such:

$$\hat H = -\frac{\hbar}{2m}\frac{1}{r}\frac{\partial^2}{\partial r^2}r \tag{1}$$

then to apply the Hamiltonian operator to a wavefunction, do you apply the separate operations in order from right to left, as in:

$$\hat H \Psi = -\frac{\hbar}{2m}(\frac{1}{r}(\frac{\partial^2}{\partial r^2}(r \Psi))) \tag{2}$$

and so on?

At first I thought you applied each operation to the wavefunction separately and then multiplied all the sections together but this seems rather clunky.

Forgive me for asking such a clueless question but I cannot find the answer anywhere!

## closed as unclear what you're asking by SRS, ZeroTheHero, Jon Custer, sammy gerbil, honeste_vivereAug 13 '17 at 13:12

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1. Yes, OP is right: We understand composition of operators $\hat{A}$ and $\hat{B}$ as $$(\hat{A}\circ \hat{B})(v)~:=~ \hat{A}(\hat{B}(v)), \tag{A}$$ where $v$ is a vector. Note that the composition symbol "$\circ$" and parenthesis "$()$" are often not written explicitly. This agrees with OP's eqs. (1) & (2). So far so good.
2. Things get surprisingly intricate, when we implement this rule (A) on the Dirac notation. Then we define $$\psi(x)~:=~\langle x | \psi \rangle ~=~| x \rangle^{\dagger}| \psi \rangle, \tag{B}$$ $$(\hat{A} (\psi))(x)~:=~ \langle x |\hat{A}| \psi \rangle ~=~\left(\hat{A}^{\dagger}| x \rangle\right)^{\dagger}| \psi \rangle, \tag{C}$$ $$((\hat{A}\circ \hat{B})(\psi))(x) ~:=~ \langle x |(\hat{A}\circ \hat{B})| \psi \rangle ~=~\left((\hat{B}^{\dagger}\circ \hat{A}^{\dagger})| x \rangle\right)^{\dagger}| \psi \rangle, \tag{D}$$ and so forth. Here $|x\rangle$ denotes the position ket state with eigenvalue $x$, $$\hat{x}|x\rangle ~=~x |x\rangle. \tag{E}$$
3. Let us for simplicity assume that the operators $\hat{A}$, $\hat{B}$, etc, are self-adjoint. If one has never seen the lhs. of eq. (D) before, one might worry that the operators $\hat{A}$ and $\hat{B}$ seem to be composed in the wrong order! It turns out that in the end, it works out correctly after all. See e.g. the next example.
4. Example: The convention (A) implies that $$\hat{p}\circ \hat{x}|x\rangle ~\stackrel{(A)+(E)}{=}~x \hat{p}|x\rangle \tag{F},$$ because $\hat{p}$ is a linear operator. Therefore, we calculate $$((\hat{x}\circ \hat{p})(\psi))(x) ~\stackrel{(D)}{=}~\left((\hat{p}\circ \hat{x})| x \rangle\right)^{\dagger}| \psi \rangle ~\stackrel{(F)}{=}~x\left(\hat{p}| x \rangle\right)^{\dagger}| \psi \rangle ~\stackrel{(C)}{=}~x(\hat{p}( \psi))(x),$$ as it should be. See also my related Phys.SE answer here.
That is, suppose I want to compute $\frac 12 \langle \hat X \hat P + \hat P \hat X\rangle,$ the closest Hermitian observable to the moment $\langle x p \rangle$ in classical mechanics. The relation that $[\hat X, \hat P] = i\hbar$ is usually taken to mean that in the position basis, $\hat X = (x\cdot)$ while $\hat P = -i\hbar \frac{\partial}{\partial x},$ so the first part of this integral is:$$\langle \hat X \hat P\rangle = -i\hbar \int_{-\infty}^\infty dx~\Psi^*(x)\cdot x\cdot\frac{\partial\Psi}{\partial x},$$ while the second part is $$\langle \hat P \hat X\rangle = -i\hbar\int_{-\infty}^\infty dx~\Psi^*(x)\cdot \frac{\partial}{\partial x}\big(x \cdot \Psi(x)\big).$$ The juxtaposition $\hat X~\hat P$ is really a sort of operator composition, just like how the matrix multiplication between two matrices forms a composition of the transforms that they describe.