Is there a more direct way to show this identity for the projection operator?

Question

The following question I jotted down from an online quantum chemistry pdf.

Show that

$$i\hbar \frac{\partial P_j}{\partial t}= [H,P_j]$$

in which $P_j$ is the projection operator and $H$ is the Hamiltonian. We are given the usual time-dependent form of Schrodinger's equation.

What I did: I supposed that this was an application of Ehrenfest's theorem because it involved a commutator and derivatives. $P_j$ does not explicitly involve time and so Ehrenfest's theorem reduces to$^*$

$$\frac{d}{dt}\langle\psi|P_j|\psi\rangle=\frac{1}{i\hbar}\langle\psi|[H,P]|\psi\rangle $$

But assuming I think w.l.o.g. that $\psi$ is normalized we can pre-multiplyby $|\psi\rangle$ to get (using $|\psi\rangle\langle\psi|=1$)

$$i\hbar\frac{d}{dt}P_j|\psi\rangle~= [H,P_j]|\psi\rangle$$

If this works (I don't assume that), my question is whether there is a direct way to do this, using the definition of $P_j$, and if so whether it would not also be an appeal to this theorem?

$^*$ Using eq. (9) from this note.

Answer 1

Before I go into how this result can be derived, I want to point out a mistake in your attempt. $|\psi\rangle\langle\psi|\neq1$, but rather $\sum_i|i\rangle\langle i|=1$, where the sum of is over a complete set of states. Basically, this result only holds when you sum over all the basis vectors that span the space you are studying, not just for any single vector in the space. Its easiest to understand this in analogy to more conventional vectors. Assuming a column vector $v$ is normalized, than $v^\dagger v=1$, which is the inner/dot product of the vector with itself. This is analogous to $\langle\psi|\psi\rangle=1$ in bra-ket notation. If we reverse the order of the product, $vv^\dagger=\mathbf A$ where $\mathbf A$ is some matrix. This operation is called an outer product and is akin to $|\psi\rangle\langle\psi|$. As it turns out, for a complete orthonormal basis, the sum of these outer products has to equal the identity matrix $\mathbf1$. We can make sense of this by recognizing that the sum is a sum of projectors onto all the basis vectors, which in QM might be all of the possible eigenstates of the system. Your wavefunction has to be some linear combination of these basis vectors, so if your project your wavefunction onto all the eigenstates and then sum those projections, you should get your original wavefunction back, hence the sum of the all the projectors is the identity matrix.

For the actual derivation, the result proceeds directly from the time dependent Schrodinger equation (TDSE) $$i\hbar\frac{\partial\psi}{\partial t}=H\psi\to i\hbar\frac{\partial|{j}\rangle}{\partial t}=H|j\rangle$$

where in the second half of the equation, I have written the equation for the state $j$ that the projection operator references. Writing the projection operator explicitly as $P_j=|j\rangle\langle j|$, we can rewrite your original expression as $$i\hbar\frac{\partial(|{j}\rangle\langle j|)}{\partial t}=i\hbar\bigg[\frac{\partial|{j}\rangle}{\partial t}\langle j|+|j\rangle\frac{\partial\langle{j}|}{\partial t}\bigg]$$ where the equality is due to the product rule. The first term contains $i\hbar\frac{\partial|{j}\rangle}{\partial t}$ which we know from the TDSE equals $H|j\rangle$, so the first term becomes $H|j\rangle\langle j|=HP_j$. To find the other term, we take the complex conjugate transpose of the TDSE to obtain $-i\hbar\frac{\partial\langle j|}{\partial t}=\langle j|H$, which leads to the second term being $-P_jH$ and thus $$i\hbar\frac{\partial P_j}{dt}=[H,P_j]$$