# Basic QFT: Operator/Bra/Ket Order

I am wondering about the order of terms from equation 34 this pdf by Hitoshi Murayama on second quantization. Specifically, why they are able to pull the H past the $\Psi(\vec x)$. i.e.

$$H|\psi(t)\rangle= \int dx\,H \Psi(\vec x,t)\phi^\dagger(\vec x)|0\rangle$$

makes sense, but how are they allowed to pull H in front of $\Psi(\vec x,t)$: $$\int dx\,H \Psi(\vec x,t)\phi^\dagger(\vec x)|0\rangle= \int dx\,\Psi(\vec x,t)H\phi^\dagger(\vec x)|0\rangle$$

However, here we have $|\Psi(t)\rangle$ already in the x basis, so I would posit it is because H is an operator and we can move $\Psi(\vec x,t)$ around. Yet, I find issue in this, because H frequently involves derivatives, since for a free particle it is $\frac{P^2}{2m}$, which involves a derivative in the x basis.

As a side note to help understand my question: For example, when inserting a complete set of states into an inner product, the order of operators, bras and kets are never changed, until we project the kets into a basis (i.e. the x basis):

$$\langle\psi|\hat A|\phi\rangle= \langle\psi|\mathbb I \, \hat A |\phi\rangle= \langle\psi|\Bigg(\int |x\rangle\langle x| \, dx\Bigg)\hat A |\phi\rangle= \int\langle\psi|x\rangle\langle x|\hat A |\phi\rangle \, dx= \int\psi(x)\phi_A(x)\, dx= \int\phi_A(x)\psi(x)\, dx$$

• Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Mar 11 '17 at 11:10

You are allowed to swap the positions, because $\Psi(x,t)$ is just a number, a coefficient that you expand the state $\Psi(t)$ into.
$| \Psi(t) \rangle = \int dx \Psi(x,t) |x \rangle$
Here $\Psi(x,t)$ is just a number, and $|x \rangle = \Phi^\dagger |0>$ are the states that your state $|\Psi(t)>$ is expanded in. You can switch the positions of $\Psi(x,t)$ and $H$, because $H$ is a linear operator.
You are right, when you choose a representation (for example, the position representation), then the operator $H$ contains derivatives that depend on the position. But you never choose a representation here, so everything is fine.
• Oh, can you explain why? With $\Psi(x,t) = \langle x | \Psi \rangle$ everything is fine, I'd say. – Quantumwhisp Nov 24 '17 at 0:06