Continuous Measurement in Quantum Mechanics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:36:52Z http://mathoverflow.net/feeds/question/70710 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70710/continuous-measurement-in-quantum-mechanics Continuous Measurement in Quantum Mechanics Alexander Moll 2011-07-19T05:12:40Z 2011-08-01T18:58:51Z <p>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 $S^{\infty}$.</p> <ul> <li><p>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?</p></li> <li><p>What does this construction have to do with the time-energy uncertainty relation?</p></li> </ul> http://mathoverflow.net/questions/70710/continuous-measurement-in-quantum-mechanics/71814#71814 Answer by Sebastian Meznaric for Continuous Measurement in Quantum Mechanics Sebastian Meznaric 2011-08-01T17:49:53Z 2011-08-01T17:49:53Z <p>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.</p> <p>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. </p> <p>I'm afraid that's all I could think of. Hopefully it will be at least of some help :). </p> http://mathoverflow.net/questions/70710/continuous-measurement-in-quantum-mechanics/71820#71820 Answer by Ron Maimon for Continuous Measurement in Quantum Mechanics Ron Maimon 2011-08-01T18:58:51Z 2011-08-01T18:58:51Z <p>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:</p> <ol> <li><p>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).</p></li> <li><p>This construction has nothing to do with the time energy uncertainty principle.</p></li> </ol> <p>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?"</p> <p>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 </p> <p>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).</p> <p>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.</p>