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 !


2 Answers 2


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.

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,

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

  • $\begingroup$ Sorry I did not quite get your explanation. Why is the Dirac notation of the Schroedinger equation not right ? As you said, the path the state traces out is just a map from a real number $t$ to a state vector $|\psi (t) \rangle$, do it does not obscure anything ? In other words, I don't see why you make the distinction between the function $\psi$ and the vector $| \psi \rangle$. Also, regarding your example of the function $f(x) = x^2$, its time derivative can certainly be zero. It depends on what $\frac{dx}{dt}$ is, right ? $\endgroup$ Jun 14, 2020 at 23:54
  • $\begingroup$ Lastly, what if you use the operator $A$ on a state vector at time $t_0$ and then afterwards you look at its time evolution ? This is actually how I interpreted the notation $A |\psi \rangle$, so you mean that this interpretation is false ? $\endgroup$ Jun 14, 2020 at 23:59
  • $\begingroup$ @Einsteinwasmyfather Dirac notation isn't wrong, it's just occasionally misleading if you aren't careful. My point about the function $f$ is that $f(x(t))$ does not obey the same time-evolution equation as $x(t)$ does; for the same reason, $A(\psi(t))$ does not obey they same time-evolution equation as $\psi(t)$ does. Lastly, the notation $A|\psi(t)\rangle$ means "the operator $A$ acting on the state $|\psi(t)\rangle$." It does not mean "the state obtained by evolving $A|\psi_0\rangle$ for a time $t$." $\endgroup$
    – J. Murray
    Jun 15, 2020 at 0:11
  • $\begingroup$ @Einsteinwasmyfather To clarify why I distinguish between the function and the state, the point I was making is that the time derivative operator acts on the function, but the Hamiltonian operator acts on the state the function spits out. This confusion leads people to ask questions like this one . $\endgroup$
    – J. Murray
    Jun 15, 2020 at 0:51

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.

  • 1
    $\begingroup$ The Schrodinger equation $i\hbar \psi'(t) = H\psi(t)$ does not assume that $\psi(t)$ is an eigenfunction of $H$, though. $\endgroup$
    – J. Murray
    Jun 19, 2020 at 15:01
  • $\begingroup$ @J.Murray Well, this equation by itself doesn't assume anything for $\psi(t)$,but when we rewrite the whole equation in the form of $H|\psi\rangle = E|\psi \rangle$ and another equation for time,we can see that $\psi_n$ (stationary state) is in fact an eigenfunction of hamiltonian.Because after all,H is a hermitian operator which acts on a function,and return that function with some eigenvalues.But if it doesn't sound good,I think my point still stands.$i\hbar \psi'(t) = H\psi(t)$ only works for $\psi(t)$ ,and $\psi(t)$ is unique,because we assume a special form ($\psi(x,t)=T(t)X(x)$) for it. $\endgroup$
    – Paradoxy
    Jun 19, 2020 at 15:21
  • $\begingroup$ @J.Murray , you can think of it like this $A|\psi \rangle$ does not return $a|\psi \rangle$, It changes $\psi$ to something else. that's why schrodinger equation doesn't work for $A|\psi \rangle$, it only works for $|\psi \rangle$ and I think it is obvious. $\endgroup$
    – Paradoxy
    Jun 19, 2020 at 15:26
  • $\begingroup$ But $H|\psi\rangle = E|\psi\rangle$ isn't the Schrodinger equation. You make the claim that the mistake is that $A\psi$ is no longer an eigenstate of $H$, but the problem has nothing to do with this. If we write $\psi(t) = \psi_1 e^{-iE_1 t/\hbar} + \psi_2 e^{-iE_2t/\hbar}$ where $\psi_1,\psi_2$ are eigenstates of the Hamiltonian, then $\psi(t)$ is not an eigenstate of the Hamiltonian but $i\hbar \psi'(t) = H\psi(t)$ nonetheless. $\endgroup$
    – J. Murray
    Jun 19, 2020 at 15:28
  • $\begingroup$ Even if $\psi$ were an eigenstate of $A$, the Schrodinger equation would not hold for $A\psi$ unless $\frac{\partial A}{\partial t}=0$. I agree with you that OP's manipulation is wrong - that's why I wrote my own answer - but I disagree with your assertion of why it is wrong. $\endgroup$
    – J. Murray
    Jun 19, 2020 at 15:35

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