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I would like to know what is the derivative of an expectation value with respect to the molecular state

$$\frac{d}{d\psi}\langle\psi| \hat{\mathbf{O}} | \psi\rangle$$

Note that here $|\psi\rangle$ is a complex column vector of length $S$ where each of its components depend on the space $q$ and $\hat{\mathbf{O}}$ is a complex $S \times S$ matrix which also depends on the space.

I other words I want to know what is $$\frac{d}{d\mathbf{\Psi}(q)}\int dq \mathbf{\Psi}^{\dagger}(q)\hat{\mathbf{O}}(q) \mathbf{\Psi}(q)$$

Is the answer

$$\frac{d}{d\mathbf{\Psi}(q)}\int dq \mathbf{\Psi}^{\dagger}(q)\hat{\mathbf{O}}(q) \mathbf{\Psi}(q) = \int dq \mathbf{\Psi}^{\dagger}(q)\hat{\mathbf{O}}(q) $$

correct ?

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    $\begingroup$ It appears that you have overlooked the fact that the $q$ inside the integral is a dummy variable whereas the $q$ outside the integral is not. So, the final answer must depend on one independent variable, presumably through some delta functions. Your proposed final answer doesn't depend on any independent variable. $\endgroup$ – Dvij Mankad Aug 15 at 10:07
  • $\begingroup$ Okay I see, it made me realize that it is a functional derivative. $\endgroup$ – juls1612 Aug 15 at 10:44
  • $\begingroup$ Could you give a reference why this quantity is of interest? Thanks. $\endgroup$ – jim Aug 15 at 10:44
  • $\begingroup$ I don't have a reference. I am trying to solve a nonlinear Schrödinger equation where the Hamiltonian (more particularly the time-dependent part of the Hamiltonian) depends on the this expectation value. In order to solve this, I want to use Newton-Raphson and therefore need the Jacobian matrix of my nonlinear function. When computing the derivative with respect $\psi$ I end up with this derivative. $\endgroup$ – juls1612 Aug 15 at 10:49
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What you have is called a functional derivative. The notation is usually somewhat different. It is defined by the relationship $$ \frac{\delta \psi(q)}{\delta\psi(q')} = \delta(q-q') . $$

In your example, we have $$ \langle\psi|\hat{O}|\psi\rangle = \int \psi^*(q)O(q,q')\psi(q') dqdq' . $$ Note that I have changed the way the $q$'s appear to make the operator a more general kernel function. Now we can apply the functional derivative: $$ \frac{\delta}{\delta\psi(q_0)} \langle\psi|\hat{O}|\psi\rangle = \int \psi^*(q)O(q,q')\delta(q'-q_0) dqdq' = \int \psi^*(q)O(q,q_0) dq . $$ Here we have a different variable $q_0$ for the function with respect to which the functional derivative is evaluated.

Hope this answers your question.

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  • $\begingroup$ I see but now is it the same when $\psi$ is a vector and $O$ a matrix ? $\endgroup$ – juls1612 Aug 15 at 10:59
  • $\begingroup$ I have not yet encountered such a situation $\endgroup$ – flippiefanus Aug 16 at 4:13

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