Assuming that $A$ and $B$ are operators (not necessarily observables) which do not commute and that the quantum system in an arbitrary state $| \psi \rangle$, then ist it possible to get $\langle AB \rangle$ from $\langle A \rangle$ and $\langle B \rangle$, and vice versa ?

Take the example of a quantum harmonic oscillator where the operators involved are ladder operators. I'm using Ehrenfest theorem $$ \frac{d}{dt} \langle A \rangle = \frac{i}{\hbar}\langle [H,A] \rangle + \langle \frac{d}{dt} A \rangle $$ which yields ordinary differential equations for the expectation values of the operator $A$, no states are mentioned. So $A$ in the last equation above is to be $a^\dagger, a, a^\dagger a$ and $a^2$. So for example if one has that the initial values for $\langle Q \rangle$ and $\langle P \rangle$, it is easy to find the those for $\langle a \rangle$ and $\langle a^\dagger \rangle$. But is it possible from those initial conditions to get the corresponding initial expectation value $\langle T \rangle$, where $T = \frac{1}{2m}P^2$ is the kinetic energy, to get the time evolution of $\langle T \rangle$?

  • $\begingroup$ Could you clarify your question? As far as I understand, you want to know whether you can obtain $\langle a^\dagger a\rangle_\psi$ from the knowledge of $\langle a^\dagger\rangle_\psi$ and $\langle a\rangle_\psi$ for an arbitrary state $\psi$? $\endgroup$ Aug 23, 2021 at 9:11
  • $\begingroup$ Yes correct, that what I want $\endgroup$
    – Physor
    Aug 23, 2021 at 9:11
  • $\begingroup$ It may not be possible, I know $\endgroup$
    – Physor
    Aug 23, 2021 at 9:21
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    $\begingroup$ @Jakob That’s an answer. It’s better if you post it as such. $\endgroup$
    – joseph h
    Aug 23, 2021 at 9:22
  • $\begingroup$ From the second formula for covariance here, the negative answer to this question requires only statistics, not quantum mechanics. $\endgroup$
    – J.G.
    Aug 23, 2021 at 10:58

1 Answer 1


No, it is not possible for an arbitrary state: Let $\{|n\rangle\}_{n\in \mathbb N_0}$ denote the set of (normalized) eigenvectors of $N\equiv a^\dagger a$. From the ladder operator algebra we find that

$$\langle n|a|n\rangle =\langle n|a^\dagger|n\rangle = 0 \quad , $$

for all $n$. On the other hand, it holds that $$\langle n|N|n\rangle = n \quad . $$

Consequently, it is not possible to find the expectation value of $N$ from the knowledge of the expectation values of $a$ and $a^\dagger$ with respect to these states, since the latter are always zero, while the former can be any natural number.

However, note that here in this special case (of being in an eigenstate) the converse is true: Given the expectation value of $N$ with respect to its eigenstates, we trivially know the corresponding expectation values of $a$ and $a^\dagger$.

  • $\begingroup$ But that gives $$ \frac{d}{dt}\langle a\rangle = \frac{d}{dt}\langle a^\dagger\rangle = 0 \implies \langle a\rangle , \langle a^\dagger\rangle = const $$ and these are fixed by the initial $\langle Q\rangle$ and $\langle P\rangle$ $\endgroup$
    – Physor
    Aug 23, 2021 at 9:32
  • $\begingroup$ Okay - why does that make you worry? $\endgroup$ Aug 23, 2021 at 9:36
  • $\begingroup$ I thought that will enable us to find the initial $\langle N \rangle$, I meant that the expectation values of $a$ and $a^\dagger$ are constants not necessarily zeros, right ? and one can get them from those of $P$ and $Q$ $\endgroup$
    – Physor
    Aug 23, 2021 at 9:37
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    $\begingroup$ @Physor As I said, the expectation values of $a$ and $a^\dagger$ are zero for the eigenstates of $N$ (or equivalently $H$ in the case of the QHO); not for an arbitrary state. Hence, the expectation values of $Q$ and $P$ in theses states are zero, too. But this result you can also derive by calculating the expectation values in the position space representation. So for the QHO, we have that e.g. $\langle Q\rangle = 0 $ for eigenstates of $H$. Nevertheless $\langle Q^2\rangle \neq 0$. $\endgroup$ Aug 23, 2021 at 9:45
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    $\begingroup$ to add to @Jakob 's comment: in particular $\langle x\rangle$ and $\langle p\rangle$ are linear combinations of $\langle a\rangle$ and $\langle a^\dagger\rangle$, not products. $\endgroup$ Aug 23, 2021 at 14:01

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