Physics Help Forum Hamilton operator affecting observable

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 Mar 23rd 2017, 11:46 AM #1 Junior Member   Join Date: Mar 2017 Posts: 2 Hamilton operator affecting observable 'm working on this problem "Consider an experiment on a system that can be described using two basis functions. In this experiment, you begin in the ground state of Hamiltonian H0 at time t1. You have an apparatus that can change the Hamiltonian suddenly from H0 to H1. You turn this apparatus on at time t1. Then, at time tD > t1, you perform a measurement of an observable,D. In matrix notation, the Hamiltonians and the operator D are given below: H0 = [0 -4; -4 6], H1 = [1 0; 0 3]" b) If you perform many, many measurements, what will be the average observed value of D as a function of t1 and tD? d)You perform the experiment, but suspect your apparatus is malfunctioning and turning off at some systematic time t2 between t1 and tD. In other words, at some time t2, you suspect the Hamiltonian is reverting to H0. What qualitative effect would this have on your results from part b)? [Note: this only requires a qualitative description, not a full calculation.]" So I found that = (2/5)(exp(2i[tD-t1])+exp(-2i[tD-t1])+1) which would be the answer in b). I can't seem to figure out d). I was thinking that because H0 is the ground state, the energy would be lower than the one in b). Therefor D would be lower as well. But than again, the two eigenvalues of H0 is -2 an 8. Now -2 is lower than the eigenvalues of H1, but 8 is higher than both the eigenvalues of H1. How can I know which state we are in?
Mar 23rd 2017, 05:40 PM   #2

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 Originally Posted by UiOStud 'm working on this problem "Consider an experiment on a system that can be described using two basis functions. In this experiment, you begin in the ground state of Hamiltonian H0 at time t1. You have an apparatus that can change the Hamiltonian suddenly from H0 to H1. You turn this apparatus on at time t1. Then, at time tD > t1, you perform a measurement of an observable,D. In matrix notation, the Hamiltonians and the operator D are given below: H0 = [0 -4; -4 6], H1 = [1 0; 0 3]" b) If you perform many, many measurements, what will be the average observed value of D as a function of t1 and tD? d)You perform the experiment, but suspect your apparatus is malfunctioning and turning off at some systematic time t2 between t1 and tD. In other words, at some time t2, you suspect the Hamiltonian is reverting to H0. What qualitative effect would this have on your results from part b)? [Note: this only requires a qualitative description, not a full calculation.]" So I found that = (2/5)(exp(2i[tD-t1])+exp(-2i[tD-t1])+1) which would be the answer in b). I can't seem to figure out d). I was thinking that because H0 is the ground state, the energy would be lower than the one in b). Therefor D would be lower as well. But than again, the two eigenvalues of H0 is -2 an 8. Now -2 is lower than the eigenvalues of H1, but 8 is higher than both the eigenvalues of H1. How can I know which state we are in?
Two things. 1st it might help (though it probably doesn't matter much) if we had an inkling of what property D is measuring. The 2nd concerns the time average of D. As it turns out <D> is better noted to be
$\displaystyle <D> = \frac{4}{5} \left ( cos (2 [ t_D - t_1] \right ) + \left ( \frac{2}{5} \right )$.

Because this is real we know that D is Hermitian. This form should make it easier to consider what happens at $\displaystyle t_2$. I am also going to assume that the energy eigenvalues are also good quantum numbers.

So, armed with that and the fact that (I presume) the basis vectors for some $\displaystyle D_0$ is the ground state, and $\displaystyle D_1$ would be the larger eigenvalue. We can say that the wavefunction before $\displaystyle t_2$ is under time evolution. After $\displaystyle t_2$ the wavefunction is a mixed state of the $\displaystyle D_1$ state and the $\displaystyle D_0$ state. So what should happen in regard to the time evolution after $\displaystyle t_2$?

-Dan
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