# Writing Hermitian conjugate of quantum operator in integral form

I know that the Hermitian conjugate of a quantum operator $$\hat{Q}$$ can be represented as:

$$\displaystyle{\left\langle\phi_1,\hat{Q}\phi_2\right\rangle= \left\langle\hat{Q}^\dagger \phi_1,\phi_2\right\rangle}.$$

In integral form, how would I write this as an integral with $$\phi_1$$, $$\phi_2$$, and $$\hat{Q}$$, as well as the form with $$\hat{Q}^\dagger$$? Taking the conjugate of a Hermitian is what is confusing me.

• I can give you a hint: the LHS is $\int_{\mathbb{R}} ~ dx ~ \phi^{*}_{1}(x) Q \phi_2 (x)$. – DanielC Nov 3 at 20:12
• Now, I have $\int{dx (Q^\dagger \phi_1)^{*} \phi_2}$. Are there any other ways to write this? I'm curious as to how the take the conjugate of the Hermitian conjugate? – theta Nov 3 at 20:51
• I am not sure if OP’s use of the phrase “quantum operator” means a Hermitian operator. For example, the annihilation operator is not Hermitian, but it is an important operator in quantum mechanics. If you do mean a Hermitian operator, please edit your question wording. – Leo L. Nov 3 at 22:02

Recall that a Hermitian operator is its own conjugate, that is, $$Q^{\dagger} = Q$$. Then, we have: $$\displaystyle \int \phi^{*}_{1}(x) Q \phi_2 (x) dx = \int Q^{\dagger} \phi^*_1(x)\phi_2(x)= \int Q \phi^*_1(x)\phi_2(x)$$.
When you write $$\left( Q^{\dagger}\phi_1\right)^*$$ the order of operations matters. First, one has to get image of $$\phi_1$$, through the operator $$Q^{\dagger}$$, which is a function in a Hilbert space of complex functions, and then the complex conjugate is obtained by complex conjugation of the function $$Q^{\dagger}\phi_1$$.