Let's say I'm given a two state system that consists of base states $|1 \rangle$ and $|2 \rangle$, those being eigenstates of an hermitean operator $\hat{O}$ that commutes with the hamiltonian, and as well being eigenstates of the hamiltonian, with eigenvalues $o_1$ and $o_2$, or $h_1$ and $h_2$, respectively.

In the Schrödinger picture, the state is evoluted according to the Schrödinger equation. I begin with a state $|\Psi_0 \rangle$ that is evoluted to a state $|\Psi(t)\rangle$.

The time-independent operator $\hat{O}$ defines $|1\rangle$ and $|2\rangle$ as its eigenstates, and then I can write $|\Psi(t) \rangle = c_1(t) |1 \rangle + c_2 |2(t) \rangle$. So in the Schrödinger picture, we know the the relative phases of $|1\rangle$ and $|2\rangle$, and because $c_i(t) = e^{\hbar h_i t}$

In the Heisenberg picture, with our complete set of observables (that commutes with the hamiltonian), $|\Psi_0\rangle$ won't change during time evolution. $\hat{O}$ won't change as well, it commutes with the hamiltonian, and thus its eigenstates $|1(t)\rangle = |1\rangle$ and $|2(t)\rangle = |2\rangle$ won't change either.In this picture, and using only our "complete set of observables" to deduce the time evolution, we can't deduce the time evolution of the relative phases of $|\Psi_0 \rangle = c_1 |1\rangle + c_2 |2\rangle$.

Does that mean our complete set of observables doesn't contain everything that is knowable about the system? Or am I making a mistake somewhere?

  • $\begingroup$ Could you elaborate? Why in the last equation you write e.g. $|1(t)\rangle$?! $\endgroup$ Commented Sep 6, 2022 at 21:27
  • $\begingroup$ As a friendly comment, a more typical way to indicate the time dependence of a state would be to use a comma to separate the label of the state from the time variable, like $|1, t\rangle$, instead of $|1(t)\rangle$. (Since, like @JasonFunderberker is getting at, $1(t)$ is just perverse :-)). $\endgroup$
    – Andrew
    Commented Sep 6, 2022 at 22:58
  • 1
    $\begingroup$ Why aren't you including the Hamiltonian (energy) as one of the observables? $\endgroup$
    – Andrew
    Commented Sep 6, 2022 at 23:04
  • $\begingroup$ The states are in principle time dependent, in my example, they have constant time dependence. I thus removed the argument. $\endgroup$ Commented Sep 6, 2022 at 23:04
  • $\begingroup$ @Andrew the Hamiltonian also has constant time dependence, so it is not of use to deduce relative phases of the state $\endgroup$ Commented Sep 6, 2022 at 23:06

2 Answers 2


I'm not sure what the OP means by "complete set of observables". But what I can say is that (1) in the Heisenberg equations of motion you can in principle* determine the expectation value of any operator ($\sigma_x$, $\sigma_y$, $\sigma_z$, $\sigma_x^2$, $\sigma_x \sigma_y$, etc.) at any moment in time. Any repeated measurement will be related to some combination of such expectation values. (2) I'm pretty sure that knowledge of all of these expectations values is equivalent to knowledge of the wavefunction. That the wavefunction allows you to predict the expectation value of such operators is clear. That the wavefunction can be inferred from the expectation values is more challenging but I believe it is so. I believe it's related to a quantum generalization of Bochner's Theorem. See also this question of mine: Quantum Probability, what makes quantum characteristic functions quantum?

To answer the OPs question: Does knowing information about some set of observables that commutes with the Hamiltonian give us as much information about the state of the system as the wavefunction? No. You also need information about some observables that don't commute with the Hamiltonian for that.

*it may be computationally difficult

  • $\begingroup$ A complete set of commuting operators is a set of operators that, given an eigenvalue to any of the operators, uniquely specifies an eigenstate up to a complex phase. What worries me is that in the Schrödinger picture, I can calculate this wave function with just 2 operators. In the Heisenberg picture, I need (as you say) more operators, possibly "all the operators". Why is that? $\endgroup$ Commented Sep 7, 2022 at 10:24
  • $\begingroup$ @Quantumwhisp What do you mean “In the Schrodinger picture, I can calculate this wavefunction with just 2 operators”? What do you mean by calculate a wavefunction with operators? That sentence doesn’t make sense to me. $\endgroup$
    – Jagerber48
    Commented Sep 7, 2022 at 13:13
  • $\begingroup$ Given a start state, I can time evolute it using the Hamilton Operator. The operator $\hat{O}$ gives eigenstates, and the scalar products with those is what is usually called wave function (e.g. $\Psi (x) = \langle x | \Psi \rangle$) $\endgroup$ Commented Sep 7, 2022 at 14:44
  • $\begingroup$ @Quantumwhisp (1) The word "evolute" does not mean what you're using it to mean. It should e "time evolve" (2) In that case the "solving of the problem" was done by determining the time evolution operator. I wouldn't say that you "calculated the wavefunction using the operators". $\endgroup$
    – Jagerber48
    Commented Sep 7, 2022 at 19:16
  • $\begingroup$ What I will say is that if you know the wavefunction (for example maybe you can decompose the wavefunction into some eigenbasis and you know all the complex coefficients) then you know everything there is to know about the quantum system. Likewise, if you know the expectation values of all the operators mentioned in this answer over time then you also know everything there is to know about the quantum system. The Schrodinger picture allows you to calculate the former and the Heisenberg picture allows you to calculate the latter. $\endgroup$
    – Jagerber48
    Commented Sep 7, 2022 at 19:18

Your question is a bit confused but if I understand it well the phases $e^{-i E_j t/\hbar}$ do not enter in any calculation because your operator $\hat {\cal O}$ does not mix the eigenstates $\vert 1\rangle$ and $\vert 2\rangle$.

For instance, in the Schrödinger picture: $$ \langle \Psi(t)\vert \hat {\cal O}\vert \Psi(t)\rangle = \vert c_1(0) \vert^2 o_1 + \vert c_2(0) \vert^2 o_2 \tag{1} $$ as it should be because $\langle 1\vert 2\rangle=0$: no mixing appears.

In the Heisenberg picture: $$ \langle \Psi\vert e^{i\hat H t} \hat{\cal O} e^{-i\hat H t}\vert \Psi\rangle =\vert c_1(0) \vert^2 o_1 + \vert c_2(0) \vert^2 o_2 \tag{2} $$ with $\hat{\cal O}(t)=e^{i\hat H t} \hat{\cal O} e^{-i\hat H t}=\hat {\cal O}$ since $\hat H$ commutes with $\hat{\cal O}$.

So basically the problem is you chose an observable for which the relative phase induced by the evolution will never show up because the orthogonal eigenstates of $\hat H$ are also eigenstates of $\hat{\cal O}$.

  • $\begingroup$ I agree with any of your calculations. Additionally, in the Schrödinger picture, I know that the state evolutes by $e^{-i \hat{H} t}$, and so in this states time evolution, the component $c_i |i\rangle$ evolves to$ c_i e^{-i \hbar E_i t} |i\rangle$. I don't have this information accessible in the Heisenberg picture, without looking at the time evolution of more operators. $\endgroup$ Commented Sep 7, 2022 at 9:59
  • $\begingroup$ I do not understand your statement “I don’t have this information accessible”. You don’t have it in either picture since neither $\langle \hat{\cal O}\rangle$ nor any other calculation involving this observable, yield information about such a relative phase. $\endgroup$ Commented Sep 7, 2022 at 11:53
  • $\begingroup$ In the Schrödinger picture, I can calculate the relative phase, using just the operators mentioned. In the Heisenberg picture, this is not possible. From your comment, you seem to suggest that I have to be consistent in what qualifies as "accessible information", namely that "accessible information" is anything that I can obtain by an actual measurement only. Is that right? $\endgroup$ Commented Sep 7, 2022 at 12:05
  • $\begingroup$ I would not say you’re “calculating” this change of phase but wherever you stand on this you cannot measure it with this observable in either picture. You could “calculate” it wrong and wouldn’t know it with your model. $\endgroup$ Commented Sep 7, 2022 at 12:30

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