1
$\begingroup$

Suppose that the Hilbert space of a quantum-mechanical system - which we will call the quantum door - is generated by two states, |open> and |closed>, forming an orthonormal basis. Suppose also that the system is prepared in the state

$ |\psi(x)> = \frac{1}{\sqrt{5}}(|OPEN> + 2|CLOSED>) $ We are given a device that measures whether the quantum door is open or closed.

(i)If we perform a measurement, which probability do we have to find the quantum door open?

(ii) Suppose the measurement returns that the quantum door is closed, and assume that the quantum Hamiltonian is identically 0 for this system at any future times. Does the door stay closed forever?

For part (i) I get

$P_{Open} = ( <open|\frac{1}{\sqrt{5}}(|OPEN> + 2|CLOSED>)^{2} $ = 1/5 ?

I also need help with part (ii), i am unure about this.

$\endgroup$
  • $\begingroup$ I'm no expert, but this seems underspecified: what if open and closed are both eigenfunctions of $H$ with the same eigenvalue? $\endgroup$ – Ian Jan 3 '17 at 21:46
  • $\begingroup$ @Ian It doesn't matter if the two states are degenerate (have same eigenvalues) or not. As long as the system is measured in one of the two states, then it will stay in that state unless other interaction happens. $\endgroup$ – Guangliang Jan 3 '17 at 22:37
  • $\begingroup$ @Guangliang Oh, I see what you mean; an eigenfunction is a stationary solution of the time-dependent Schrodinger equation. You've inserted the contextual assumption that the system is kept isolated, but that makes sense. $\endgroup$ – Ian Jan 3 '17 at 23:07
0
$\begingroup$

You got the part (i) correct. The probability of the door being measured as 'Open' is $1/5$.

On the part (ii), the door will stay as 'Closed' after being measured as in 'Closed' state. It will change only if the system is interacted by other operations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.