# Commuting with the Hamiltonian operators are conserved quantities [closed]

In this question What does "commuting with the Hamiltonian" mean?. I've read that if an operator commutes with the Hamiltonian it is a conserved quantity, this means that the average value of that observable does not vary in the time namely: $$\frac{d \hat O}{dt} \ne 0$$.

The first question is: what does this mean? When I studied this subject, I read that in order to obtain the average value of an observable, we have to imagine this experiment: I prepare in the laboratory $$n$$ systems all equal and measure the same observable $$n$$ times obtaining $$n$$ results with a certain frequency. Then I make an average of the results obtained and I obtain the mean value of the observable on that system. But what does it mean to measure the mean value at different times? Do I do the operation I mentioned, let the eigenstates obtained evolve over time and repeat the operation?

The second question is: in general, if I have an operator that commutes with $$H$$, do its eigenstates not change over time? That is, if $$O$$ commutes with $$H$$ and I have an eigenstate of $$O$$ relative to a certain eigenvalue, as time passes, if I measure $$O$$ on this eigenstate, do I continue to obtain the same eigenvalue? What if the operator $$O$$ does not commute with $$H$$? In general, do the eigenstates of an operator that does not commute with $$H$$ evolve over time by changing? So that if I make a measurement of $$O$$ on one of its eigenstates at different times I might get different results?

To answer your first question let's fix what we mean by a time evolution (I'll use Shrodinger picture, but heisenberg one is equivalent.). A state of a quantum system is some vector $$\psi \in \mathcal H$$ of some Hilbert space. Quantum mechanics tells that if we start with a system in the state $$\psi$$ then we can know what state the system will be on any time $$t$$, namely $$\psi(t)$$. Now, what mean by the expected value is the average of measured values of a system in a given state (so there no mention of time here). Thus what it means for the expected value to be time invariant that the expected value of an operator $$A$$ that commutes with the Hamiltonian of a system in the state $$\psi_0$$ is the same as that of a system in the state $$\psi(t)$$.

To answer the second question we'll dive into the formalities. To any self-adjoint operator $$A$$ we can associate an one parameter group of unitary transformations $$U_A(t)=e^{itA}$$. For the Hamiltonian, we get $$e^{itH}$$ and this gives the time evolution of the system. That is, if in the time $$0$$ a system was in a state $$\psi_0 \in \mathcal H$$, the in a time $$t$$ the system will be on the state $$\psi(t)=e^{-iHt}\psi_0$$.

Now, suppose that $$\psi_0$$ is an eigenvector of $$A$$ and that $$A$$ commutes with $$H$$. We know that $$[A,e^{-iHt}]=0$$, so that $$A$$ commutes with $$e^{-iHt}$$ for all $$t$$. Now, note that

$$A(\psi(t))=A(e^{-iHt}\psi_0)=e^{iHt}A\psi_0= \lambda e^{-iHt}\psi_0=\lambda\psi(t).$$

Thus we conclude that the time evolution of eigenvectors of $$A$$ continue to be eigenvectors of A with the same associated eigenvalues. More generally, the probability of a measure of $$A$$ to lie on some subset $$E \subset \mathbb R$$ will not change over time.

• Just a reminder: note that in the standard case $\mathcal H = L^2(\mathbb R^n)$, so that a state is a function of $x \in \mathbb R^n$, i.e., $\psi=\psi(x)$. Now, we know that a state $\psi(x)$ changes to a function $\psi_t(x)$ over a time $t$, so we may set $\psi(x,t)=\psi_t(x)$ to get the tradtional case where a wave function is a function of both time and space. Feb 19, 2022 at 14:54
• I still don't understand how can we compare expected value of a state for $t=0$ and for $t=t_1$, do we have to measure an operator over a state, then let the state evolve in time and then make the same measure again? So that, if the operator commutes with the Hamiltonian, then the state after the first measure collapses into an eigenstate relative to an eigenvalue and after a time $t$, since the operator is a conserved quantity, the initial eigenstate is still an eigenstate in the eigenspace associated to the first eigenvalue so that the average value is the same? Feb 19, 2022 at 15:21
• Yes. A measure is "instantaneous", so that only state on the time of the measurement matters. The expected value of a system in the state, say, $\psi(t_1)$ is roughly the average of the values of several measures in the system ${in \ the \ state}$ $\psi(t_1)$. Thus, for each $t$ we have an expected value $E(t)$. What we mean then is that $d/dt E(t)=0$. Feb 19, 2022 at 15:35
• Sorry but I still can't get it. If we are in a laboratory and we want to prove this statement, what do we have to do? Do we have to prepare $n$ identical systems, make $n$ measure of an Hermitian operator which commutes with $H$ and take the average of the values we obtain? and after this we let the $n$ systems evolve with time and, at time $t_1$, repeat the same operation and take the new average and compare the two averages? Feb 19, 2022 at 15:41
• Yes! So, you would have to prepare a bunch of systems in the same initial state. And measure a bunch of them in a time $t_1$, a bunch of them in a time $t_2$, and so on, and then you would see that it doesn't matter how much time you wait to make the measurements, you would still have the same expected value. Feb 19, 2022 at 15:45