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Consider a two state system (2-state atom for instance) where each state corresponds to a different energy: $\left|\psi\right> = \frac{1}{\sqrt{2}}\left|0\right> + \frac{1}{\sqrt{2}}\left|1\right>$. The states $\left|0\right>$ and $\left|1\right>$ are the eigenstates of the Hamilton operator $\hat{\mathcal{H}}$. The corresponding eigenvalues are $E_0$ and $E_1$. Thus a measurement can have only 2 outcomes $E_0$ or $E_1$. For the first case the following collapse will occur: $\left|\psi\right> \to \left|0\right>$ and for the second case - $\left|\psi\right> \to \left|0\right>$.

As result after the measurement the system will be in a state with predefined energy i.e. without any additional experiment we can definitely say what the energy of the system is.

From the other side Energy-time Heisenberg inequality says that we can not define energy with 100% accuracy during a limited time frame.

How are the 2 statements are related each others?

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  • $\begingroup$ I wonder whether it is possible, in a world where |0> and |1> are the only existing state of that system, to perform a measurement. If it's not the case, then the state of the system is described in a more larger eigenstates space, and the energy is still not well defined due to the interaction of the measurement with the other degrees of freedom. $\endgroup$
    – JPG
    Apr 3, 2017 at 8:46
  • $\begingroup$ We can consider as many states as we want for instance look at the following one: $\left|\psi\right> = \sum_i c_i \left|i\right>$ with $\{E_i\}$ as eigenvalues. The result will be the same. I just chose the simplest model for my example. The only case that will not work is then $\hat{\mathcal{H}}$ has continuous spectra $\endgroup$
    – Ivan
    Apr 3, 2017 at 9:01
  • $\begingroup$ No; for example if you have two states of z-spin and 2 states of y-spin, they are incompatibles. If you measure z-spin with for example a Stern-Gerlach experiment, the y-spin becomes undetermined. Since you used a gradient of B along z axis, the magnetic field across y axis cannot be null. Then the energy coupling By and y-spin is undetermined. $\endgroup$
    – JPG
    Apr 3, 2017 at 9:11
  • $\begingroup$ I've got the idea. It makes sense $\endgroup$
    – Ivan
    Apr 3, 2017 at 9:17

1 Answer 1

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Your system is in an eigenstate of the Hamiltonian, so the state is not changing over time.

Therefore if $O$ is any observable, the probability distribution of possible outcomes of $O$ is not changing over time, and in particular the expectation value of $O$ is not changing over time.

Therefore the time until that expectation value changes by any given amount is infinite. So in the expression $$\Delta E\Delta t\ge \hbar/2$$ we have $\Delta E=0$ but $\Delta t=\infty$, so we are okay.

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    $\begingroup$ Not quite right $\hat{\mathcal{H}}\left|\psi\right> = \frac{E_0}{\sqrt{2}}\left|0\right> + \frac{E_1}{\sqrt{2}}\left|1\right> \ne E\left(\frac{1}{\sqrt{2}}\left|0\right> + \frac{1}{\sqrt{2}}\left|1\right>\right) = E \left|\psi\right>$. So the system is not in an eigenstate of the Hamiltonian $\endgroup$
    – Ivan
    Apr 3, 2017 at 12:49
  • $\begingroup$ @IvanMurashko: I don't quite see the relevance of your comment. You've made a measurement, so the system is in a state $\psi$ which is equal either to $0\rangle$ or $1\rangle$. $\endgroup$
    – WillO
    Apr 3, 2017 at 12:53
  • $\begingroup$ The $|\psi>$ you use in $\hat{H}|\psi>$ is not an eigenstate of the Hamiltionian, because it's taken before the wave function collapse, so $\hat{H}|\psi>=E|\psi>$ is wrong. Only after the measurement, the considered system is an eigenstate of the Hamiltionian. $\endgroup$ Apr 3, 2017 at 12:55
  • $\begingroup$ The $|\psi>$ you use in $\hat{H}|\psi>$ is not an eigenstate of the Hamiltionian, because it's taken before the wave function collapse, so $\hat{H}|\psi>=E|\psi>$ is wrong. Only after the measurement, the considered system is in an eigenstate of the Hamiltionian.After the measurement $|\psi>=|0>$ or $|\psi>=|1>$, so then, $\hat{H}|\psi>=\hat{H}|0>=E_0|0>$ or $\hat{H}|\psi>=\hat{H}|1>=E_1|1>$. The "weird" thing about the collapse of the wave function is, that the system changes completely, i.e. the resulting wave function has nothing anymore to do with the original wave function. $\endgroup$ Apr 3, 2017 at 13:01
  • $\begingroup$ @Willo My idea is that the system initially prepared in the state with undefined energy that is in the good relation with Heisenberg inequality (as soon as it is not an eigenstate of Hamiltonian). But after the measurement it transferred into the state with defined energy (that is not allowed accordingly Heisenberg inequality). $\endgroup$
    – Ivan
    Apr 3, 2017 at 13:13

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