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I am studying a Quantum Mechanics course and I have come across something that I am a little stuck on, mathematically. Physically it seems to make sense but I'm not sure which equations to use to justify the behaviour of the system.

Here is the question:

http://i.imgur.com/Jo7uFsa.png

Here is an excerpt of our notes:

enter image description here

The equation given (80) to find the probability of a quantum state seems to be for a one-dimensional system with time t and position x on the real number line. My problem has two dimensions: open and closed.

I'm unsure what approach to take, but I gave it a shot: I can only post two links, the imgur code is PFnqyTY.jpg

I expect that the probability of the door to be open is 1/5 but I'm not sure how to get to the answer without just jumping to it. I'd like to do it completely and respect the vectors and the notation involved. I'm slowly getting the hang of quantum systems and understanding all of the notation and their physical meanings, but I'm stuck on this one.

Any help with this would be greatly appreciated. Thank you!

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Your equation (80) does not mean that $x$ is a real number, $x$ is the state. For particular systems, it may correspond to a position, but that is clearly meaningless in this question. Also, your state does not depend on time, so you can just omit that. The probability of the door being open is thus

$$P_\text{open}(t) = |\langle \text{open} | \psi \rangle |^2$$ which is thus constant in time. We expect this since nothing depends on time here.

You were told that $|\text{open} \rangle$ and $|\text{closed} \rangle$ are orthonormal, so $$ \langle \text{open} | \text{closed} \rangle = 0, \qquad \langle\text{open}|\text{open} \rangle = \langle\text{closed}|\text{closed} \rangle = 1.$$

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  • $\begingroup$ Ahhh. That makes much more sense. Thank you so much! $\endgroup$ – Ben Gerry Mar 6 '15 at 19:38
  • $\begingroup$ Perhaps you could enlighten me on the final part? The only equation I can find with $\hat{H}$ is the Schrodinger equation, and setting $\hat{H} = 0$ there doesn't do much. I can't quite get my head around the idea of $\hat{H} = 0$ physically. Energy becomes zero, but what does that do to the system? $\endgroup$ – Ben Gerry Mar 6 '15 at 20:03
  • $\begingroup$ Your Hamiltonian is the thing which entirely determines the evolutionary dynamics of your system. If you look at the Schrodinger equation, setting $H=0$ forces $\frac\partial{\partial t} \langle x| \psi(t) \rangle =0,$; that is, the state is constant in time. $\endgroup$ – Tyler Holden Mar 6 '15 at 20:44
  • $\begingroup$ Ah. Thank you once again. I was getting confused, thinking that the state did not depend on time, but that is not necessarily true in this case. As $\hat{H} = 0$ => the state is constant in time, does that imply that the door will remain closed? Or does it just mean that it is still a 4/5 chance for all time? Would that imply a superposition? Sorry for all the questions. Thank you! $\endgroup$ – Ben Gerry Mar 6 '15 at 22:41
  • $\begingroup$ If the measurement results in the door being closed, then it is closed forever. Measuring the quantum system results in a collapse of the superposition and the system becomes classical. $\endgroup$ – Tyler Holden Mar 9 '15 at 16:27

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