Does the Schrodinger equation obey the rule for differentiating a function if the function is in terms of the wavefunction?

Question

Does the Hamiltonian operator act like a derivative when acting on a functional in terms of wavefunctions? For example, does $$H\psi^2=2\psi H\psi$$ hold true? More generally, if the functional, $F(\cdot),$ is solely in terms of the wavefunction, $\psi$, then does $$HF(\psi)=F'(\psi)H\psi?$$ If not, then why not?

Answer 1

Before providing an answer, I first want to provide some clarifications on the use of terminology. If the Hamiltonian is an operators (as we find in quantum mechanics) then what it operates on is usually referred to as a state vectors (or just state) and it is commonly represented in terms of Dirac notation by $|\psi\rangle$. A $wave function$ is obtained from such a state vector with the aid of a specific set of basis vectors. For the position basis, we have $$ \psi(x) = \langle x|\psi\rangle , $$ and for the momentum basis (spatial Fourier basis) we have $$ \psi(k) = \langle k|\psi\rangle . $$ The Hamiltonian operator cannot operate on these wave functions, because they are just c-number quantities. Hence, I assume that the question meant to ask about (tensor) products of state vectors, rather than products of wave functions.

To address the question we first ask how do we treat the situation where the Hamiltonian operator is applied to a tensor product of two state vectors $$ |\psi\rangle = |\psi_1\rangle\otimes|\psi_2\rangle ? $$ Such a situation would for instance occur for two particles. One can express such a state as $$ |\psi\rangle = \hat{a}_1^{\dagger} \hat{a}_2^{\dagger} |vac\rangle , $$ where $$ \hat{a}_{1,2}^{\dagger} = \int \hat{a}^{\dagger}(k) \psi_{1,2}(k) dk , $$ with $\psi_{1,2}(k)$ being the Fourier domain wave functions. Since the Hamiltonian operators can also be expressed in terms of $\hat{a}^{\dagger}(k)$ one can now apply the Hamiltonian to tensor product state to see how it work. Indeed, we find that for a product of operators, the Hamiltonian operation follows a product rule, similar to the way it does for differentiations. This is because of the identity $$ [\hat{A},\hat{B}\hat{C}] = [\hat{A},\hat{B}]\hat{C} + \hat{B}[\hat{A},\hat{C}] . $$

Now we can consider the case where the state is represented by some holonomic function of the creation operators. Say $$ |\psi\rangle = F\left(\hat{a}_1^{\dagger}\right) |vac\rangle , $$ where $$ F(x) = \sum_m C_m x^m , $$ with $x\rightarrow \hat{a}_1^{\dagger}$. Such a situation is found in coherent states, for instance. When the Hamiltonian is now applied to this state, which we can evaluate for each term in the expansion, we find a result similar to the chain rule for differentiations.

Hope that answers your question.

Answer 2

Assuming a 1D hamiltionian of the form $H = p_x^2/ 2 m + V ( x )$ then $H \psi^2$ would look like $$ -\frac{\hbar^2}{2m} \frac{d^2}{d x^2} \psi^2 + V ( x ) \psi^2,$$ and you would have $$\frac{d}{d x} \psi^2 = 2 \psi \frac{d \psi}{d x}$$ so that $$\frac{d^2}{d x^2} \psi^2 = 2 \frac{d}{d x}\left( \psi \frac{d \psi}{d x} \right) = 2 \left( \frac{d \psi}{d x} \right)^2 + 2 \psi \frac{d^2 \psi}{d x^2}.$$ There is an extra term involving $\left( \frac{d \psi}{d x} \right)^2$ that messes up your initial suggestion.