# Expectation Values and Derivation of Heisenberg Equation?

Consider a system of particles with wave function $$\psi$$(x) (x can be understood to stand for all degrees of freedom of the system; so, if we have a system of two particles then x should represent {$$x_1; y_1; z_1; x_2; y_2; z_2$$}). The expectation value of an operator $$\hat{A}$$ that operates on is defined by : $$\langle\hat{A}\rangle = \int\psi^{*}\hat{A}\psi dx$$

Yup this makes sense to me and there's nothing new here.

If $$\psi$$ is an eigenfunction of $$\hat{A}$$ with eigenvalue $$a$$, then, assuming the wave function to be normalized, we have : $$⟨ \hat{A} ⟩ = a$$

This is where I want to confirm something.

$$\hat{A}\psi = a\psi$$

Hence, $$⟨ \hat{A} ⟩ =\int\psi^{*} a \psi dx$$

Since $$a$$ is a constant I can take it out :

$$\langle\hat{A}\rangle = a \int\psi^{*} \psi dx$$

We assumed that the wave function was normalized hence $$\int\psi^{*} \psi dx = 1$$

Leaving $$\langle\hat{A}\rangle = a$$

Now consider the rate of change of the expectation value of $$\langle\hat{A}\rangle$$:

$$\frac{d\langle\hat{A}\rangle}{dt} = \int{\frac{\partial}{\partial t}}(\psi^{*}\hat{A}\psi)dx$$

$$=\int{\frac{\partial \psi^{*}}{\partial t}\hat{A}\psi+\psi^{}\frac{\partial\hat{A}}{\partial t}\psi^{*}}+\frac{\partial \psi}{\partial t}\hat{A}\psi^{*} dx$$

$$=\int{\langle\frac{\partial\hat{A}}{\partial t}\rangle} +\frac{i}{\hbar}\int{[(\hat{H}\psi)^{*}\hat{A}\psi-\psi^{*}\hat{A}\hat{H}\psi]}dx$$

where we have used the Schrodinger equation :

$$i\hbar\frac{\partial \psi}{\partial t} = \hat{H}\psi$$

The second line is easily obtained via differentiation. The second term in the second line corresponds to the first term in the third line, correct ?

I do not see how this term was obtained. In particular where the $$\frac{i}{\hbar}$$ originates from :

$$\frac{i}{\hbar}\int{[(\hat{H}\psi)^{*}\hat{A}\psi-\psi^{*}\hat{A}\hat{H}\psi]}$$