In section 6.3.1 of the following MIT Open Course Ware (PDF) (22-02, Introduction to Applied Nuclear Engineering), the author, Prof Paola Cappellaro, derives Heisenberg's equation using the definition of expectation values,
$$
\frac{d}{dt}\langle\hat{A}\rangle=\frac{d}{dt}\int d^3x\psi^*(x,t)\hat{A}\psi(x,t)\\
=\int d^3x\left(\frac{\partial\psi^*}{\partial t}\hat{A}\psi+\psi^*\frac{\partial\hat{A}}{\partial t}\psi+\psi^*\hat{A}\frac{\partial\psi}{\partial t}\right)
$$
and Schrodinger's equation,
$$
\frac{\partial\psi}{\partial t}=-\frac{i}{\hbar}\hat{H}\psi,\qquad
\frac{\partial\psi^*}{\partial t}=\frac{i}{\hbar}\left(\hat{H}\psi\right)^*
$$
Then the author uses $\left(\hat{H}\psi\right)^*=\psi^*\hat{H}^*=\psi^*\hat{H}$ to get
$$
\frac{d}{dt}\langle\hat{A}\rangle=\int d^3x\left(\frac{i}{\hbar}\psi^*\left[\hat{H}\hat{A}-\hat{A}\hat{H}\right]\psi+\psi^*\frac{\partial\hat{A}}{\partial t}\psi\right)\\
=\frac{i}{\hbar}\langle\left[\hat{H},\,\hat{A}\right]\rangle+\langle\frac{\partial\hat{A}}{\partial t}\rangle
$$
There are two things I don't understand in this derivation. The first is that the total time derivative $\frac{d}{dt}$ turns into a partial time derivative $\frac{\partial}{\partial t}$ when inside the integral.
The second is that the author states
$$(\hat{H}\psi)^*=\psi^*\hat{H}$$
This confuses me as $\hat H$ is the Hamiltonian operator, and contains an $x$ derivative. Thus the left hand side of the above equation is another state $\phi^*$ such that $\phi=\hat H\psi$, but the right hand side is an operator: the derivative is yet to be applied to anything. This equation by itself just makes no sense to me.
Any help with either of these two issues would be greatly appreciated!
Best Answer
Neither left-hand side nor right-hand side are properly operators that act on states and give states. They are a form of operators that act on states and give complex numbers. We call such a thing a bra (and a state is called a ket, so when you pair them you get a bra(c)ket).
The left-hand side acts like $$((H\psi)^*)(\varphi) = \int \psi^* H \varphi \, dx.$$ If this confuses you, think about the finite-dimensional case. Then $\psi$ is a column vector and $H$ is a square matrix. While there is a complex conjugation operation for vectors, what you really want is the Hermitian conjugate, which is transpose and complex conjugation, denoted with a ${}^\dagger$. Then $(H\psi)^\dagger$ is the transpose of a column vector, so it's a row vector, and $(H\psi)^\dagger = \psi^\dagger H$ and this can act on a column vector with matrix multiplication.
For wavefunctions, $(\psi^\dagger)(x) = \psi(x)^*$ where the star is complex conjugation, and matrix multiplication is of a row vector and a column vector is integration.