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?

  • $\begingroup$ Sorry I updated my question to be more specific $\endgroup$
    – Abe
    Nov 4, 2022 at 2:28
  • 2
    $\begingroup$ 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

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.