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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.

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    $\begingroup$ I can give you a hint: the LHS is $\int_{\mathbb{R}} ~ dx ~ \phi^{*}_{1}(x) Q \phi_2 (x) $. $\endgroup$ – DanielC Nov 3 at 20:12
  • $\begingroup$ 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? $\endgroup$ – theta Nov 3 at 20:51
  • $\begingroup$ 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. $\endgroup$ – Leo L. Nov 3 at 22:02
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Recall that a Hermitian operator is its own conjugate, that is, $Q^{\dagger} = Q$. Then, we have: $\displaystyle \int \phi^{*}_{1}(x) Q \phi_2 (x) dx = \int Q^{\dagger} \phi^*_1(x)\phi_2(x)= \int Q \phi^*_1(x)\phi_2(x)$.

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  • $\begingroup$ You are using the term conjugate in a misleading way: when you say that a Hermitian operator is its own conjugate, conjugate is intended as hermitian conjugate. In the original question the conjugation was the usual complex conjugation. $\endgroup$ – GiorgioP Nov 3 at 22:03
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I think you have misunderstood the notation.

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 $Q^{\dagger}\phi_1$.

In no place conjugation of an operator is required.

Maybe a possible cause for such misunderstanding could be the reference to the case of a matrix representation of operator. However, in that case there is no integral representation anymore.

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