I've been dealing with annihilation operator recently where you can see related information


for definition.

Correction made by Valter Moretti and J.G.'s answer, there is no contradiction.

$$ \begin{align} i \hbar \frac{\mathrm{d}}{\mathrm{d}t} \left(a \left| \psi(n,t) \right\rangle \right) & ~=~ i \hbar \frac{\mathrm{d}}{\mathrm{d}t} \left(\sqrt{n} \left| \psi(n-1,t) \right\rangle \right) \\[5px] & ~=~ H \sqrt{n} \left| \psi(n-1,t) \right\rangle \\[5px] & ~=~ E_{n-1}\sqrt{n} \left| \psi(n-1,t) \right\rangle \end{align} \,.$$

and \begin{align} i\hbar\frac{d}{dt}(a|\psi(n,t)\rangle) & =i\hbar\frac{d}{dt}(a)|\psi(n,t)\rangle+a(i\hbar\frac{d}{dt}|\psi(n,t)\rangle) \\[5px] & =i\hbar \cdot iwa|\psi(n,t)\rangle+a(H|\psi(n,t)\rangle) \\[5px] & =-\hbar w a|\psi(n,t)\rangle+a(E_n|\psi(n,t)\rangle) \\[5px] & =(E_n-\hbar w)a|\psi(n,t)\rangle\\[5px] & =E_{n-1} a|\psi(n,t)\rangle\\[5px] & =E_{n-1} \sqrt{n}|\psi(n-1,t)\rangle \end{align}

Notice $\frac{d}{dt}(a)\neq 0$ despite the fact that $a$ in matrix represtation is a constant matrix.

  • 2
    $\begingroup$ Surely the error is assuming $\dot{a}=0$; comparing the two calculations lets you compute $\dot{a}$. You could try writing $a$ as an outer product as well. $\endgroup$
    – J.G.
    Commented Nov 4, 2018 at 7:29
  • $\begingroup$ The former is wrong. It is false that $a\psi_n(t) = \sqrt{n}\psi_{n-1}(t)$. $\endgroup$ Commented Nov 4, 2018 at 7:31
  • $\begingroup$ @ValterMoretti Could you explain a bit? $\endgroup$ Commented Nov 4, 2018 at 7:35
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    $\begingroup$ However, regarding your problem, the crucial observation is that $a$ does not commute with the evolutor operator since it does not commute with $H$. $\endgroup$ Commented Nov 4, 2018 at 8:02
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    $\begingroup$ @ValterMoretti can you put it in an answer? $\endgroup$ Commented Nov 4, 2018 at 8:37

2 Answers 2


The problem with your first answer is the identity (I prefer to modify notations) $$a \psi_n(t) = \sqrt{n} \psi_{n-1}(t)\:.\quad (false \:\:if \:\:t\neq 0)\tag{1}$$ This is false if $t\neq 0$ (it is trivially true for $n=0$, but I assume $n\neq 0$ henceforth).

Since $\psi(t) := e^{-itH/\hbar} \psi$, the identity above can be expanded as $$a e^{-itH/\hbar} \psi_n = \sqrt{n} e^{-itH/\hbar} \psi_{n-1} \quad(false \:\:if \:\:t\neq 0)\:.$$ In fact, both sides can be separately computed, $$a e^{-itH/\hbar} \psi_n = a e^{-itE_n/\hbar} \psi_n= e^{-itE_n/\hbar} a\psi_n= e^{-itE_n/\hbar} \sqrt{n}\psi_{n-1} = \sqrt{n}e^{-itE_n/\hbar} \psi_{n-1} $$ whereas $$\sqrt{n} e^{-itH/\hbar} \psi_{n-1} = \sqrt{n} e^{-itE_{n-1}/\hbar} \psi_{n-1}\:,$$ and $$\sqrt{n}e^{-itE_n/\hbar} \psi_{n-1} \neq \sqrt{n} e^{-itE_{n-1}/\hbar} \psi_{n-1}$$ because $$E_n\neq E_{n-1}\:.$$ The fundamental reason of the failure of (1) is that $a$ and $e^{-itH/\hbar}$ do not commute. It it were $$a e^{-itH/\hbar} = e^{-itH/\hbar} a\:,$$ then we would have $$\langle \psi_n |a e^{-itH/\hbar}\psi_m \rangle = \langle \psi_n | e^{-itH/\hbar} a \psi_m\rangle$$ i.e. $$\langle a^*\psi_n | e^{-itH/\hbar}\psi_m \rangle = \langle e^{itH/\hbar}\psi_n | a \psi_m\rangle\:.$$ taking the time derivative for $t=0$, $$\langle a^*\psi_n | H\psi_m \rangle = \langle H\psi_n | a \psi_m\rangle\:,$$ so that $$\langle \psi_n | aH\psi_m \rangle = \langle \psi_n | Ha \psi_m\rangle\:,$$ so that $$\langle \psi_n | (aH-Ha)\psi_m \rangle = 0$$ namely, since the Hermite functions $\psi_n$ span a dense space, $$aH\psi_m= Ha\psi_m\:.$$ By linearity, we would have $$aH= Ha\:.$$ at least restricting the domains to the span of Hermite functions $\psi_n$. The identity above is false in view of the commutation relations of $a$ and $a^*$ and $H = \hbar \omega (a^*a + I/2)$ over the said space.

Coming back to your question, the correct result is the second one.

  • $\begingroup$ Thank you. Up vote. However, still one little question. If the time derivative of the function returned $E_n$, does that mean $a|\psi(n,t)>$ is still in energy states $E_n$? because $i\hbar \frac{d}{dt}a|\psi(n,t)>=Ha|\psi(n,t)>=E_na|\psi(n,t)>$? So what's $a$'s effect to the states $a|\psi(n,t)>$? $\endgroup$ Commented Nov 4, 2018 at 15:22
  • $\begingroup$ No, $a\psi_n(t)$ has no longer energy $E_n$. Indeed, as a state, it has energy $E_{n-1}$ like $\psi_{n-1}$. The only difference is a factor: $a\psi_n(t) = e^{-i(E_{n-1}-E_n)t/\hbar}\sqrt{n}\psi_{n-1}$. So, there is an evident effect of the action of $a$, it changes the energy. $\endgroup$ Commented Nov 4, 2018 at 16:21
  • $\begingroup$ Can you clearfy a little? Because, if $i\hbar \frac{d}{dt}$ act on a states and return $E_n$ and that states, it's the same as $H$ act on that sates and obtained energy $E_n$ by the postulate right? So on one side I think I got your point, but on the other side your asnwer didn't seem to give the correct energy states. $\endgroup$ Commented Nov 4, 2018 at 16:31
  • $\begingroup$ Sorry I do not understand what you are saying. $d/dt$ acts on a function of $t$ taking values in the Hilbert space and produces a similar function. It is not an operator as $H$ is, which insteads works on a single vector and produces a vector. What it is clear it is what I wrote, $a\psi_n(t)$ is an eigenvector of $H$ with eigenvalue $E_{n-1}$ instead of $E_n$. $\endgroup$ Commented Nov 4, 2018 at 16:37
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    $\begingroup$ Sorry, I cannot understand. However, you wrote $i\hbar \frac{d}{dt}a \psi_n(t)= Ha \psi_n(t)$. But it is false. What it is true is $i\hbar \frac{d}{dt}a \psi_n(t)= aH \psi_n(t)$... $\endgroup$ Commented Nov 4, 2018 at 17:39

You have to be careful not to mix up implications specific to different pictures. For example, $\partial_t\langle m|a|n\rangle$ doesn't imply $\dot{a}=0$ in the Heisenberg picture because $|n,\,t\rangle=\exp -i\omega_n t|n,\,0\rangle,\,\omega_n:=\frac{E_n}{\hbar}$. While $\dot{a}=0$ in the Schrödinger picture, it can be shown $\dot{a}=-i\omega a$ in the Heisenberg picture, where the Hamiltonian is $-\frac{\hbar^2}{2m}\partial_x^2+\frac{m\omega^2}{2}x^2$. I'll give two proofs; there are plenty of others.

The first is one I mentioned in a comment: since $\partial_t|n\rangle=-i\omega_n|n\rangle$ and $a=\sum_{n\ge 1}\sqrt{n}|n-1\rangle\langle n|$, $\dot{a}=-i(\omega_n-\omega_{n-1})a=-i\omega a$.

Another approach uses $\dot{x}=\frac{p}{m},\,\dot{p}=-m\omega^2 x,\,a=\frac{m\omega x+ip}{\sqrt{2m\hbar\omega}}$.

  • $\begingroup$ $a$ does not depend on time here, we are dealing with Schroedinger picture here. What you wrote makes sense in Heisenberg picture. $\endgroup$ Commented Nov 4, 2018 at 11:38
  • $\begingroup$ @ValterMoretti I've added a picture discussion in case the OP had overlooked it too. $\endgroup$
    – J.G.
    Commented Nov 4, 2018 at 11:41
  • $\begingroup$ Ok, now what you wrote is correct, but I do not think that it answers OP's question. $\endgroup$ Commented Nov 4, 2018 at 11:49

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