# What is $\frac{d}{d\psi}\langle\psi| \hat{O} | \psi\rangle$?

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

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

correct ?

• 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. Aug 15 '19 at 10:07
• Okay I see, it made me realize that it is a functional derivative. Aug 15 '19 at 10:44
• Could you give a reference why this quantity is of interest? Thanks.
– jim
Aug 15 '19 at 10:44
• 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. Aug 15 '19 at 10:49

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.
• I see but now is it the same when $\psi$ is a vector and $O$ a matrix ? Aug 15 '19 at 10:59