Does the non-linear Schrodinger equation satisfy quantum mechanics rules? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-17T16:37:20Z https://physics.stackexchange.com/feeds/question/444070 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/444070 -1 Does the non-linear Schrodinger equation satisfy quantum mechanics rules? zooby https://physics.stackexchange.com/users/84158 2018-11-29T16:11:09Z 2018-11-30T09:05:33Z <p>Thinking about the <strong>0+1 dimensional</strong> (time-only) non-linear Schrodinger equation: </p> <p><span class="math-container">$$i\frac{\partial}{\partial t} \psi(t) =\kappa |\psi(t)|^2 \psi(t).$$</span></p> <p>Treating <span class="math-container">$\psi$</span> as a wave function instead of a field does this satisfy the rules of quantum mechanics? Because it seems like probability would be conserved. But it is non-linear.</p> <p>Yes you could treat <span class="math-container">$\psi$</span> as a quantum field. Then you would have a wave function <span class="math-container">$\Psi[\psi]$</span>. But if you treated <span class="math-container">$\psi$</span> as a wave function by itself, would this work? (It would be a modification of quantum mechancs). If not why not?</p> <p>The rules of quantum mechanics as I can make out satisfy two things:</p> <ol> <li><p>Probabilities always add up to 1.</p></li> <li><p>States that are orthogonal stay orthogonal.</p></li> </ol> <p>Are these rules obeyed?</p> https://physics.stackexchange.com/questions/444070/-/444190#444190 2 Answer by Prof. Legolasov for Does the non-linear Schrodinger equation satisfy quantum mechanics rules? Prof. Legolasov https://physics.stackexchange.com/users/30833 2018-11-30T07:13:17Z 2018-11-30T09:05:33Z <p>In QM, the states are elements of a projective Hilbert space. Equivalently, <span class="math-container">$\left|\psi\right&gt;$</span> and <span class="math-container">$c \left|\psi\right&gt;$</span> refer to the same state for any <span class="math-container">$c \in \mathbb{C} \setminus \{0\}$</span>. Your equation doesn’t preserve this symmetry, so it doesn’t satisfy the axioms of QM.</p>