## Continuous Measurement in Quantum Mechanics

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Let $\mathcal{P}(S^{\infty})$ denote the set of probability measures on the unit sphere $S^{\infty} \subset \mathcal{H}$ in the Hilbert space of states of a quantum mechanical system. Measurement of an observable $\Omega$ corresponds to orthogonal projection, sending $\delta_{|\psi \rangle}$ to a particular measure supported on the eigenvectors of $\Omega$, thus inducing a map $T_{\Omega}: \mathcal{P}(S^{\infty}) \rightarrow \mathcal{P}(S^{\infty})$. If we think of $T_{\Omega}$ as the transition matrix of a Markov chain, we can say that a continuum $\lbrace \Omega_t \rbrace_{t \in \mathbb{R} \geq 0}$ of observables induces a stochastic process on $S^{\infty}$.

• If we let $\Omega_t = H(t)$, the time-dependent Hamiltonian of our system, is the associated stochastic process the deterministic one described by the Schrödinger equation?

• What does this construction have to do with the time-energy uncertainty relation?

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This question is not well phrased--- what you can ask is "what is the effect of measuring an operator again and again, when the operator varies continuously in time". – Ron Maimon Aug 1 2011 at 18:39

## 2 Answers

The question is muddled regarding quantum mechanics: The projection operator T takes measures to measures, but it does not define a Markov chain, or if you like, the resulting Markov chain is not an interesting one, because it is deterministic most of the time. The reason is that T is idempotent, a second application of T does nothing. Even if omega varies continuously, doing T on a continuously varying omega is deterministic in the limit of continuous measurement (see below). The specific questions are not the right questions, but here is an answer:

1. There is no deterministic stochastic process associated with the Schrodinger equation, so this question is meaningless. There is a stochastic process associated with the analytic continuation of the Schrodinger equation to imaginary times, but this has nothing to do with measurement (or anything else in the physics in real time).

2. This construction has nothing to do with the time energy uncertainty principle.

Stripping away the pointless formalism, what you are asking in the question is: "What happens if you measure an observable again and again, so that the measurements become very dense?"

What happens is called the "quantum zeno effect". If you keep measuring a state that tries to change, you prevent the state from changing, instead you constrain it to stay the same Eigenvector of the observable you measure. But you have a continuous family of observables

But you are asking what happens if you measure an observable which varies with time. The result is that you follow the Eignevectors of the observable in a deterministic way. So the first operation will project you to one of the eigenvectors at random, then the remaining continuum of measurments will make the state change continuously to follow the changing eigenvectors of the operator. The reason there is no stochasticity is because if you measure after a time "epsilon", the probability of being found in a different eigenvector goes like "epsilon-squared", so that in the continuum limit, the process becomes completely deterministic, with 100% probability of following the corresponding eigenvector of H(t).

The only subtlety is when the eigenvectors collide (have the same eigenvalue at some time t), in which case, a continuous measurement will have to follow the eigenvector through the collision. So if to eigenvectors of H(t) coincide at time t, and afterwards come out in different direction, then there will be some stochasticity associated with the collision. The original direction of the eigenvector (assuming the generic case that only two eigenvectors collided) will have to be expanded in the directions of the two new vectors, and the square of the expansion parameters will tell you the probability of going off in different directions. You might be able to make a markov chain by colliding again and again, but this is not in the spirit of the original question.

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Thank you for your suggested steps towards a more subtle construction, and for pointing me to the "Quantum Zeno effect".  I'm reluctant to accept your answer to my second question above: quoting from the Wikipedia entry for the Quantum Zeno effect, "It is still an open question how closely one can approach the limit of an infinite number of interrogations due to the Heisenberg uncertainty involved in shorter measurement times. In 2006, Streed et al. at MIT observed the dependence of the Zeno effect on measurement pulse characteristics.[29]" – Alexander Moll Aug 2 2011 at 4:55
A measurement can be a failure of interaction. If you have an atom in its first excited state, it will decay to the ground state normally. You can apply an arbitrarily strong laser whose frequency is tuned to the energy difference between the ground state and the second excited state, and this measures when the atom transitioned to the ground state. This will prevent the atom from ever making the transition to the ground state, and there is no theoretical limit to the density of measurements from the uncertainty principle, because the different measurements are free photons. – Ron Maimon Aug 2 2011 at 14:06
It took me a little thinking to understand why people are saying that quantum zeno is limited by time-energy uncertainty, because it isn't at all. The wrong idea is that a measurement of energy to accuracy "delta" will have to take 1/delta seconds, so that the density of measurements can never exceed 1/delta. The first clause it true--- a measurement will have to take a long time, but the measurements can overlap, so that the density is unlimited. – Ron Maimon Aug 2 2011 at 14:48
Also, keep in mind that measurements can be pushed up to the level of psychology, the measurement can be thought to happen only at the last step, when you look at the measuring device. Then one can ask what physical reason allows a strong laser tuned to 0-2 transitions to prevent 1-0 transitions. The reason is that transition from 1-0 is accompanied by high frequency amplitudes for transitions to 2, which are entangled with incoming photon phases and therefore incoherent, so the down transition amplitudes get scrambled. The ground state in the presence of the photons is not stationary. – Ron Maimon Aug 2 2011 at 15:08

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You have two questions. The first one is if the induced stochastic process is described by the Schrodinger equation. The answer is no. To see this, notice that the stochastic process you describe takes a pure state $\psi$ to a mixed state $\rho$. The Shrodinger equation preserves purity.

What does this have to do with time-energy uncertainty relation. You need to go to the Heisenberg picture to see whether there is a connection. Notice that $\Omega_{t_0}$, say, is then going to evolve with time. So measuring $\Omega_{t_0}$ at time $t_1$ is quite different than measuring it at time $t_2$. However, they remain constant if $\Omega_t$ is independent of $t$ which is just a statement of the conservation of energy. However, even in such a case the time-energy uncertainty principle still applies (you can have uncertainty in energy of a state even if energy operators are constant). For this reason I don't see much of a connection here, though maybe I am wrong.

I'm afraid that's all I could think of. Hopefully it will be at least of some help :).

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Thanks, Sebastian - this certainly settles my first question, and you're right to point out the obvious difficulties with this construction in the Heisenberg (or interaction) picture.  I do think that the "Quantum Zeno effect" (see Ron's answer below) is what I was after, and I'd need to read a lot of what's out there before trying to get back on the horse. Thanks again! – Alexander Moll Aug 2 2011 at 5:08
This answer is not accurate--- the stochastic process, (on every step after the first (ignoring collisions) is entirely deterministic, and takes pure states to pure states. – Ron Maimon Aug 3 2011 at 19:10
I didn't see your comment earlier Ron. This is not usually deterministic. If $H_t$ is not constant then the spectrum may well change. Most importantly, if the eigenvectors change then even at a later step a pure state which was before an eigenvector of the Hamiltonian will be mapped to a mixed state due to it now not being an eigenvector anymore. That is if I am understanding the question correctly, and the distribution gets mapped to the mixed state corresponding to the distribution of outcomes rather than a particular outcome. – Sebastian Meznaric May 4 2012 at 16:18
@SebastianMeznaric: This is usually deterministic--- the "mixed state" stops being mixed as the time-steps become small, as the probability of deviating from the eigenvector path goes as "epsilon squared", while the number of timesteps only goes as 1/epsilon, so that in the continuous measurement limit, you have deterministic evolution. This only fails when the eigenvalues collide, which is measure zero, but at this point, you get a definite stochastic splitting depending only on the eigenvectors of the Hamiltonian just after and just before the collision. – Ron Maimon Jan 5 at 16:16
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