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For a given operator $A$ and states $|{\phi}\rangle$and $|{\psi}\rangle$ the following inner product in position ($x$) basis is:

$$\langle{\phi}|A|{\psi}\rangle=\int_{-\infty }^{\infty }\phi^{*}(x)A{\psi(x)}dx.$$

However, in dealing with discrete space I learned that the complex conjugate of the above in Dirac notation is $(\langle{\phi}|A|{\psi}\rangle)^{*}=(\langle{\psi}|A^{\dagger}|{\phi}\rangle)$. So now if I were to express in continuous spaces I would get the following:

$$\langle{\psi}|A^{\dagger}|{\phi}\rangle=\int_{-\infty }^{\infty }\psi^{*}(x)A^{\dagger}{\phi(x)}dx.$$

My question is, in this case, does $A^{\dagger}$ act on the right or left inside the integral? I.e, should the expression inside the integral be read as $\psi^{*}(x)(A^{\dagger}{\phi(x)})$ or $(A^{\dagger}\psi^{*}(x)){\phi(x)}$? I am asking this because I noticed that many books mix up operators inside the ket's symbol (e.g $A|{\phi}\rangle=|A{\phi}\rangle$) and it becomes very confusing when defining the conjugate of an operator for example here: https://quantummechanics.ucsd.edu/ph130a/130_notes/node133.html

How would one define the conjugate of an operator with Dirac notation as I wrote above?

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  • $\begingroup$ Sorry I updated my question to be more specific $\endgroup$
    – Abe
    Nov 4, 2022 at 2:28
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    $\begingroup$ IMHO if you want to understand the notion of adjoint, it is better to not use the bra-ket notation. Instead, write the inner product e.g. as $\langle \psi,\phi\rangle$ for $\psi,\phi\in H$. Now we have that $\langle \psi, A\phi\rangle = \langle A^\dagger \psi,\phi\rangle$, which holds (given some mathematical hypotheses in the infinite dimensional case) irrespective of the underlying complex Hilbert space. If $H=L^2$, then $\langle \psi, A\phi\rangle =\int \mathrm d x\, \psi^*(x)\,(A\phi)(x) = \int \mathrm d x\, (A^\dagger\psi)^*(x)\, \phi(x)$. $\endgroup$ Nov 4, 2022 at 6:53

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Strictly speaking, if you are converting the bras and kets into wave functions, then you should also convert the operator into a kernel function. The assumption is that you have a complete position basis, so that $$ \langle\phi|A|\psi\rangle = \int \langle\phi|x\rangle \langle x|A|y\rangle \langle y|\psi\rangle dx dy = \int \phi^*(x) A(x,y) \psi(y) dx dy . $$ Here, we inserted identities between A and the states on either side, respectively resolved in terms of the position bases, as represented by $x$ and $y$.

Now you can see what happens when you take the complex conjugate and interchange the order $$ (\langle\phi|A|\psi\rangle)^* = \int \psi^*(y) A^*(y,x) \phi(x) dx dy . $$

Hope that helps.

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