Time derivative of expectation value of observable is always zero (quantum mechanics)

Question

In my book about quantum mechanics it state that the time derivative of an arbitrary observable is: $$\frac{d}{dt}\langle A \rangle = \frac{1}{i\hbar} \langle [A,H] \rangle + \bigg{\langle }\frac{dA}{dt} \bigg{\rangle} $$ with $H$ being the Hamiltonian. They derived this equation by using the product rule of differentiation for the bra $\langle \psi|$ , the ket $|\psi\rangle$ and the operator $A$ and by using the Schrodinger equation (+ its conjugate form). However, when I used the product rule on only the bra $\langle \psi|$ and the ket $A|\psi\rangle$ I get the following: $$\frac{d}{dt}\langle A \rangle = \bigg{(}\frac{d}{dt} \langle \psi|\bigg{)} A|\psi\rangle + \langle \psi| \bigg{(}\frac{d}{dt} (A|\psi\rangle)\bigg{)} = -\frac{1}{i\hbar} \langle \psi|HA|\psi\rangle + \frac{1}{i\hbar} \langle \psi|HA|\psi\rangle = 0$$ Here, for the second term, I used the Schrodinger equation on the state $A|\psi\rangle$. What did I do wrong ?

Thanks in advance !

Answer 1

I think this is a nice question. It ultimately boils down to the following:

If $i\hbar\frac{d}{dt}|\psi\rangle = H|\psi\rangle$, then why does $i \hbar\frac{d}{dt}\big(A|\psi\rangle\big) \neq H\big(A|\psi\rangle\big)$, since $A|\psi\rangle$ is also a valid state vector?

The answer is a bit subtle. The time evolution of a quantum mechanical state takes the form of a path through the underlying Hilbert space - that is, a function $$\psi: \mathbb R\rightarrow \mathcal H$$ $$t \mapsto \psi(t)\in \mathcal H$$ The Schrodinger equation tells us that the physical paths through the Hilbert space are such that

$$i\hbar\psi'(t)= H\big(\psi(t)\big)$$ In particular, the time derivative acts on the function $\psi$, while the Hamiltonian operator acts on the state vector $\psi(t)$. The standard Dirac notation obscures this by writing $$i\frac{d}{dt}|\psi\rangle = H|\psi\rangle$$ from which it is easy to get the mistaken impression that it makes sense to differentiate a state vector with respect to time.

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Armed with this clarification, the answer is that $\psi(t)$ being a physical path does not guarantee that $A\big(\psi(t)\big)$ is a physical path. The latter is merely the image of a physical path under the action of the function (operator) $A$.

This concept is not reserved for quantum mechanics. Think about classical physics. Newton's law applied to a free particle yields $\frac{d^2}{dt^2} x = 0$. Does this imply that $\frac{d^2}{dt^2}f(x) = 0$ for some arbitrary function $f$? Certainly not - for example, consider $f(x)=x^2$.

If $\psi(t)$ is a physical path, then one has that $$\frac{d}{dt}(A\psi(t)) = \frac{\partial A}{\partial t} \psi(t) + A \psi'(t) = \frac{\partial A}{\partial t}\psi(t) + A\big(\frac{1}{i\hbar}H\psi(t)\big)$$

Inserting this into the expectation value then yields the correct result,

$$\begin{align}\frac{d}{dt}\langle \psi(t),A\psi(t)\rangle &= \langle \psi'(t),A\psi(t)\rangle + \langle \psi(t),\frac{\partial A}{\partial t}\psi(t)\rangle + \langle \psi(t),A\psi'(t)\rangle\\&=-\frac{1}{i\hbar}\langle H\psi,A\psi\rangle +\frac{1}{i\hbar}\langle \psi,AH\psi\rangle + \left\langle\frac{\partial A}{\partial t}\right\rangle\\&=-\frac{1}{i\hbar}\langle \psi,HA\psi\rangle +\frac{1}{i\hbar}\langle\psi,AH\psi\rangle + \left\langle\frac{\partial A}{\partial t}\right\rangle\\&=\frac{1}{i\hbar}\left\langle[A,H]\right\rangle + \left\langle\frac{\partial A}{\partial t}\right\rangle\end{align}$$

Answer 2

Well, I think there is a simpler explanation for that. First, note that we have

$$H|\psi\rangle = i\hbar\frac{d}{dt}|\psi\rangle ~~~(1)$$

That is, if Hamiltonian acts on its eigenstates, we would have that equality (This is Schrodinger equation after all, it gives eigenstates ($|\psi_n\rangle$) of Hamiltonian)(*see my edit at the bottom of answer). However, if we assume an arbitrary state, say $|\phi_m\rangle$ which is not an eigenstate of hamiltonian, we can't say anymore:

$$H|\phi_m\rangle = i\hbar\frac{d}{dt}|\phi_m\rangle$$ Rather we have to say

$$H|\phi_m\rangle = \sum_n i\hbar\frac{d}{dt}|\psi_n\rangle\langle \psi_n|\phi_m\rangle$$

That is, we have to expand $|\phi_n\rangle$ in term of $|\psi_n\rangle$.

Another important note, if an operator like A acts on a state, say $|m_n\rangle$ we don't have this equality: $$A|m_n\rangle = a|m_n\rangle$$ Unless of course we assume that $|m_n\rangle$ is an eigenstate of A. So generally speaking we have: $$A|m_n\rangle = a|Q_n\rangle$$ That is, an operator like A changes state $|m_n\rangle$ to something else, unless $|m_n\rangle$ happens to be an eigenstate of A.

Let's go back to your question. Note that $|\psi \rangle$ is NOT an eigenstate of A, It is a summation of eigenstates of Hamiltonian. So we have:

$$\frac{d}{dt}\langle \psi| A |\psi \rangle = \frac{d}{dt}(\langle \psi|)~~A |\psi \rangle + \langle \psi|~~\frac{d}{dt}(A |\psi \rangle)~~~(2)$$

$$\frac{d}{dt}(\langle \psi|)~~A |\psi \rangle = \frac{-1}{i\hbar}(\langle H\psi|)~~A |\psi \rangle = \frac{-1}{i\hbar}\langle \psi|HA |\psi \rangle ~~~(3)$$

So far so good, but this is where you did the math wrong. Note that in (3), $\frac{d}{dt}$ acts on $\langle \psi|$, so we can use conjugation of equation (1) with no problem. But for second term in (2), we can't do that. because $A$ changes $|\psi \rangle$ to something else.

Let's say $|f_n\rangle$ is eigenstate of A. So we can say:

$$\langle \psi|~~\frac{d}{dt}(A |\psi \rangle) = \langle \psi|~~ \sum_n \frac{d}{dt}(A|f_n\rangle \langle f_n| \psi \rangle)$$

I just expanded $|\psi \rangle$ in terms of eigenstates of A. also $A|f_n\rangle = a_n |f_n \rangle$ so:

$$\langle \psi|~~ \sum_n a_n\frac{d}{dt}(|f_n\rangle \langle f_n| \psi \rangle) = \langle \psi|~~ \sum_n a_n(|f_n'\rangle \langle f_n| \psi \rangle) + \langle \psi|~~ \sum_n a_n(|f_n\rangle \langle f_n'| \psi \rangle) + \langle \psi|~~ \sum_n a_n(|f_n\rangle \langle f_n| \psi '\rangle)~~(*)$$

Note that we can use (1) for third term of this equation, because after all $\frac{d}{dt}$ acts on $|\psi\rangle$ so

$$\langle \psi|~~ \sum_n a_n(|f_n\rangle \langle f_n| \psi '\rangle) = \frac{1}{i\hbar}\langle \psi|~~ \sum_n a_n(|f_n\rangle \langle f_n |H| \psi \rangle) = \frac{1}{i\hbar} \langle \psi|AH|\psi\rangle ~~~(4)$$ I simply re-compacted expansion. From summation of (4) and (3) we have:

$$\frac{1}{i\hbar} \langle \psi|AH|\psi\rangle - \frac{1}{i\hbar}\langle \psi|HA |\psi \rangle = \frac{1}{i\hbar} \langle [A,H] \rangle$$

remaining terms in (*) are $\langle \frac{\partial A}{\partial t}\rangle $.

*Edit: $|\psi(t) \rangle$ is not an eigenstate of hamiltonian, rather what I mean is since $$|\psi(t) \rangle = \sum_n |\psi_n \rangle exp(-iE_nt/\hbar)$$ we can write $$H|\psi(t) \rangle = \sum_n |H\psi_n \rangle exp(-iE_nt/\hbar)=\sum_n E_n|\psi_n \rangle exp(-iE_nt/\hbar)=i\hbar\frac{d}{dt}\sum_n |\psi_n \rangle exp(-iE_nt/\hbar) (**)$$ Thus we have: $$H|\psi\rangle = i\hbar\frac{d}{dt}|\psi\rangle$$ But if we start with $A|\psi \rangle$ instead of $|\psi \rangle$ we will have

$$HA|\psi(t) \rangle = H\sum_n \sum_m A |f_m \rangle \langle f_m||\psi_n \rangle exp(-iE_nt/\hbar) = \sum_n \sum_m aH|f_m \rangle \langle f_m||\psi_n \rangle exp(-iE_nt/\hbar)$$ But we know that $H |f_m \rangle \neq E|f_m \rangle$ since $|f_m \rangle$ is not an eigenstate of Hamiltonian, unless $[H,A] = 0$ which is not the case in general. So we can't assume $HA|\psi(t) \rangle = i\hbar\frac{d}{dt}(A|\psi(t) \rangle)$, because writing something like (**) for it, is not possible. As simple as that.