# Quantum projection when measuring state population

• A
• BillKet
In 20ms, the state will be $$\left( \begin{array} a \sqrt{1.1}...b \sqrt{2.2} \\ -i\sqrt{0.1}...-j\sqrt{1.2} \end{array} \right)$$.f

#### BillKet

Hello! I have two energy levels of opposite parity close by (we can assume they are far from all the other levels in the system) and an off-diagonal term in the 2x2 matrix Hamiltonian that weakly couples them. I initially populate only one parity state, say the positive one, and after a while some population transfers to the other state. Finally, using a 2-step resonant ionization, such that the first step is parity selective (i.e. the ionization happens only from the initially unpopulated state - the negative parity one), I can test if the transition of population took place (in principle I do this many times and from there I extract the off-diagonal term). Let's assume for concreteness (the numbers are really arbitrary) that in 10 ms, I would transfer 10% of the population to the negative parity state and in 20 ms, I would transfer 20% of the population. I have 2 questions:

1. Let's say that my ionization probability (if I were to be 100% in the negative parity state), as well as my ion detection probability are 100%. Let's say that after 10 ms I send my lasers to try to ionize the system (I have 10% chance of doing so), and I don't detect an ion. I wait 10 ms more, and I try again. What will the probability of ionizing be after 20 ms? If I were to not try to ionize after 10 ms, the probability would be simply 20%, but as I tried after 10 ms, did that measurement act as a projection operator, thus reseting my system back to being 100% in the positive parity state, hence the probability after 20 ms would still be only 10%?

2. My questions is the same as the first one, but this time let's assume that the ionization probability is smaller than 100%. In this case, my measurement is not a fully quantum projection operator. What would be the probability of seeing an ion after 20 ms, if I don't see one after 10 ms?

Thank you!

What will the probability of ionizing be after 20 ms? If I were to not try to ionize after 10 ms, the probability would be simply 20%
Are you saying then that after 100ms the probability would be 100%? I don't understand.

Are you saying then that after 100ms the probability would be 100%? I don't understand.
I gave these numbers as an example, what is at 100ms doesn't really matter. For example you can think of an RF field driving the 2 levels. At 10 ms you have some population, at 20 ms you have more population, if the RF field is resonant with the transition after some time you will have 100% probability of finding the particle in the opposite parity state from where you started. But this is not really relevant for my question. I just want to know if the population gets reset or not after the failed attempt to ionize.

In most systems if you saturate the transition it will equally populate upper and lower states because of stimulated emission.

In most systems if you saturate the transition it will equally populate upper and lower states because of stimulated emission.
I am not sure I see how that is related to my question. In my setup I have only one atom (and we can ignore the lifetimes of these 2 states of interest), and indeed the Rabi oscillations will lead the population back and forth between the 2 levels of opposite parity i.e. cycles of stimulated excitation and stimulated emission and at a point it will reach 50% probability of being in either parity state (I assume this is what you mean by equally populated?). But my question is restricted to a time before that happens, which would be at a time t such that ##t\Omega = \pi/2##, where ##\Omega## is the Rabi frequency due to the driving field.

• Twigg
1. Yes, the ionization test will project the ion (assuming there is only 1 ion) back to the 100% positive parity state. The probability of ionization after 20ms will be still 10%, like you said.

2. That's a really good question. I think we can work it out from first principles. For simplicity, I'm going to assume that your 2x2 Hamiltonian and your ionization light are both resonant (##\delta=0##). After the first 10ms, the ion state now looks like $$\left( \begin{array} a \sqrt{0.9} \\ -i\sqrt{0.1} \end{array} \right)$$. Taking the ionization as a Rabi precession, your ionization pulse looks something like $$\left( \begin{array} a 1 & 0 & 0 \\ 0 & \sqrt{1 - \eta^2} & -i\sqrt{\eta} \\0 & -i\sqrt{\eta} & \sqrt{1 - \sqrt{\eta}^2} \\ \end{array} \right)$$ where ##\eta## is your ionization efficiency and the third state is the ionized state. That means, after ionization (but before detection) your state is $$\left( \begin{array} a \sqrt{0.9} \\ -i\sqrt{0.1 (1 - \eta^2)} \\ -\sqrt{0.1\eta} \\ \end{array} \right)$$ Then, when you detect whether any ions have been produced, if you don't observe an ion then you project the ion into the state $$k \left( \begin{array} a \sqrt{0.9} \\ -i\sqrt{0.1 (1 - \eta^2)} \\ 0 \\ \end{array} \right)$$ where k is just the normalization factor. To calculate the probability after another 10ms, multiply this projected state by the Rabi flop precession $$\left( \begin{array} a \sqrt{0.9} & -i\sqrt{0.1} \\ -i\sqrt{0.1} & \sqrt{0.9} \\ \end{array} \right)$$, then calculate the remaining probability of being in state 2. Then multiply that probability by ##\eta## (since you have to do another round of detection), and that's the final probability of detecting the state after 20ms.

1. Yes, the ionization test will project the ion (assuming there is only 1 ion) back to the 100% positive parity state. The probability of ionization after 20ms will be still 10%, like you said.

2. That's a really good question. I think we can work it out from first principles. For simplicity, I'm going to assume that your 2x2 Hamiltonian and your ionization light are both resonant (##\delta=0##). After the first 10ms, the ion state now looks like $$\left( \begin{array} a \sqrt{0.9} \\ -i\sqrt{0.1} \end{array} \right)$$. Taking the ionization as a Rabi precession, your ionization pulse looks something like $$\left( \begin{array} a 1 & 0 & 0 \\ 0 & \sqrt{1 - \eta^2} & -i\sqrt{\eta} \\0 & -i\sqrt{\eta} & \sqrt{1 - \sqrt{\eta}^2} \\ \end{array} \right)$$ where ##\eta## is your ionization efficiency and the third state is the ionized state. That means, after ionization (but before detection) your state is $$\left( \begin{array} a \sqrt{0.9} \\ -i\sqrt{0.1 (1 - \eta^2)} \\ -\sqrt{0.1\eta} \\ \end{array} \right)$$ Then, when you detect whether any ions have been produced, if you don't observe an ion then you project the ion into the state $$k \left( \begin{array} a \sqrt{0.9} \\ -i\sqrt{0.1 (1 - \eta^2)} \\ 0 \\ \end{array} \right)$$ where k is just the normalization factor. To calculate the probability after another 10ms, multiply this projected state by the Rabi flop precession $$\left( \begin{array} a \sqrt{0.9} & -i\sqrt{0.1} \\ -i\sqrt{0.1} & \sqrt{0.9} \\ \end{array} \right)$$, then calculate the remaining probability of being in state 2. Then multiply that probability by ##\eta## (since you have to do another round of detection), and that's the final probability of detecting the state after 20ms.
Thanks a lot! This was super useful! I have one more question (kind of an extension of the previous setup). Another thing we might try experimentally is, instead of ionization, to excite the system (it is a diatomic molecule) to an auto dissociative excited electronic state. After we send the laser to drive this transition, the state is as you mentioned:

$$\left( \begin{array} a \sqrt{0.9} \\ -i\sqrt{0.1 (1 - \eta^2)} \\ -\sqrt{0.1\eta} \\ \end{array} \right)$$

We can only detect an ion if the molecule dissociates (one of the products is an ion). Let's say that there is a probability p for it to dissociate from that state and 1-p to just decay to another state (let's assume there is just one extra such state and it is different from the original states). What would the state of the system be after this? Is it:

$$\left( \begin{array} a \sqrt{0.9} \\ -i\sqrt{0.1 (1 - \eta^2)} \\ 0 \\ -(1-p) \sqrt{0.1\eta} \end{array} \right)$$

where the 4th entry would be this extra state where the decay happened?

The headline to this post threw me for a second. I was wondering why someone needed quantum physics to conduct the U.S. Census. Duh! The body text, of course, clarified the issue.

• Twigg
The headline to this post threw me for a second. I was wondering why someone needed quantum physics to conduct the U.S. Census. Duh! The body text, of course, clarified the issue.
I'm sure the state population is entangled with whether or not the bureaucrats had their coffee Another thing we might try experimentally is, instead of ionization, to excite the system (it is a diatomic molecule) to an auto dissociative excited electronic state.
Ah, this is trickier because the spontaneous emission will also cause decoherence in addition to siphoning ions away from the ionized population. It's not as simple as a binomial process with paths p and (1-p). The math I gave above isn't really accurate, and you really need to work in a density matrix framework, using the solutions to the optical Bloch equations (in that linked text, the equations at the bottom of page 2 are what most people call the optical Bloch equations but the author seems to have a different naming convention, just fyi) to describe the evolution of the state during photodissociation/ionization.

My gut feeling is that if you include the effect of decoherence by spontaneous emission, you will find a greater chance of error (i.e. not detecting state 2) on the second 10ms round of precession+detection, but I couldn't tell you if it will end up being numerically significant or not. I think the probability of detection for the first 10ms round of precession+detection will be the same as if you did the math I gave above but with the substitution ##\eta \rightarrow \eta\times p##. This is because I believe the off-diagonal terms of the density matrix won't affect the first round of detection, but I think they will affect the second round.

• ohwilleke