Quantum mechanics basics - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-19T16:05:32Z http://mathoverflow.net/feeds/question/106721 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106721/quantum-mechanics-basics Quantum mechanics basics Ashley 2012-09-09T11:17:37Z 2012-09-09T15:20:05Z <p>Hello. I'm thinking about where does the basic quantum mechanics things comes from. I mean the forms of operators and a Shroedinger equation. The more intuitive explanation is better.</p> <p>To get forms of operators and Shroedinger eq., we can start from assumption that in our representation square of absolute value of wavefunction is probability density. Then it became clear that coordinate operator is just a multiplying by variable (obvious, looking for example to mean value expression). Next, Shroedinger equation anyway should be of the form $\frac{\partial \phi}{\partial t} = A \phi$ with some A. To see that $A$ is actually Hamiltonian multiplied by something, we can see that $\frac{\partial \phi}{\partial t}$ is an enegry multiplied by something. I have the only idea to explain this: assume additionally that in our representation defined momenta states are plane waves. Considering them, $i\frac{\partial \phi}{\partial t} = E$ is visible from De Broglie relation $E=\omega$ for plane waves (as well as $p=\frac{\partial \phi}{\partial x}$ visible from $p=k$). Then, the only thing i want explained deeper is De Broglie relations. Finally, now i also think that i want some explanation of introducing wavefunction as complex-valued. </p> <p>Or maybe there is an other way? I guess, it would be more general to start from commutation relations, but i'm afraid it would be hard and abstract. But i appreciate if you try to explain where are they come from.</p> http://mathoverflow.net/questions/106721/quantum-mechanics-basics/106723#106723 Answer by Alexander Chervov for Quantum mechanics basics Alexander Chervov 2012-09-09T11:41:16Z 2012-09-09T12:51:56Z <p>Probably the explanation via Heisenberg picture is more intuitive (at least for mathematically minded person like me).</p> <p>Equations of motion in classical mechanics can be described as Hamilton form as follows: </p> <p>d\dt f = {H, f}</p> <p>(here { } are Poisson brackets and these equations reduces to standard <a href="http://en.wikipedia.org/wiki/Hamiltonian_mechanics" rel="nofollow">Hamilton equations of motions</a>, if you take standard phase space R^2n p_i, q_i and standard Poisson bracket these equation will give d/dt p = -dH/dt ; d/dt q = dH/dt . However they make sense on arbitrary Poisson manifold.) </p> <p>Quantization in Heisenberg picture turns these equations into</p> <p>d/dt f = [H, f]</p> <p>(I have omitted (i/h). )</p> <p>That is more or less all, modula we need to explain what "[H, f]" is. It is commutator in non-commutative algebra which is <a href="http://en.wikipedia.org/wiki/Deformation_quantization" rel="nofollow">deformation quantization</a> of the Poisson algebra of classical observables. It should probably be pointed out here why deformation quantization might be considered as intuitively "clear". To explain this let us go in opposite direction: consider non-commutative algebras depending on parameter "h" such that for "h=0" algebra becomes commutative. The point is that Poisson bracket naturally arise in this step: define {f,g} = lim {h->0} (fg-gf) / h . Exercise - to check that such defined Poisson bracket satisfies the Jacobi identity (it follows from the associativity). </p> <p>So the moral is that <strong>Poisson bracket is "shadow of non-commutativity"</strong>, more precisely first order of the noncommutative product. The deformation quantization task is not construct the non-commutative algebra from the first this first order "shadow", it has been solved in certain generality by Kontsevich. Well, I am not sure how intuitive these reasons are, at least they are so for me.</p> <p>Another important remark is to relate Heisenber picture to the Shrodinger one. This is quite easy linear algebra. So Heisenberg picture is evolution of "operators (=matrices)" "f" according to equation d/dt f = [H,f]. The linear statement is the following: such evolution on matrices equivalent to Schrodinger like evolution on VECTORS: d/dt v = H v. You can make this "equivalent" in precise statement in various way e.g. lemma: consider f(t) which satisfy the "Heisenberg" equation d/dt f=[H,f], then vector v(t)= f(t)v_0 will satisfy the "Schrodinger" equation d/dt v =Hv, for any vector v_0.</p> <p>Another important remark is <a href="http://en.wikipedia.org/wiki/Stone-von_Neumann_theorem" rel="nofollow">Stone von Neumann</a> uniqueness theorem. It explains why usually operator corresponding to "p" is d/dq and corresponding to "q" is multiplication by "q".<br> Theorem says (up to details) that if you consider the algebra [p,q]=1 then it has the unique irreducible representation in the Hilbert space. The meaning of this theorem is the following: you can choose ANY other representation of operators corresponding to "p" and "q" ( not just "p"-> d/dq, q-> mult_q), but ALL these representation will be equivalent. So choose whatever you want and do not care. </p> http://mathoverflow.net/questions/106721/quantum-mechanics-basics/106734#106734 Answer by Ben Crowell for Quantum mechanics basics Ben Crowell 2012-09-09T15:20:05Z 2012-09-09T15:20:05Z <p>I'll try to help with intuition on why wavefunctions are complex-valued.</p> <p>First, it's not actually true that they have to be complex. When you quantize electromagnetic waves, the wavefunction is simply the electric and magnetic fields, which are real. The correct statement is that a spin-1/2 particle's wavefunction has to be complex.</p> <p>As a simple example of why this is, consider the case of two planar sine waves that are moving in antiparallel directions and then merge and superpose.</p> <p>If the wavefunction is a real scalar, then at a time when the two superposed waves are 180 degrees out of phase, their sum is identically zero. This violates conservation of energy and, perhaps more importantly, conservation of probability.</p> <p>If the wavefunction is a complex scalar, then one can prove that solutions of the Schrodinger equation always conserve probability. This is easy to check in an example like the superposition of $e^{i(kx+\omega t)}$ with $e^{i(-kx+\omega t)}$.</p> <p>To see that this argument doesn't prove that wavefunctions are always complex-valued, consider electromagnetic waves in the same situation of superposition of antiparallel plane waves. Because there is a right-handed relationship among $\mathbf{E}$, $\mathbf{B}$, and $\mathbf{k}$, you can't make both $\mathbf{E}$ and $\mathbf{B}$ cancel. For example, you could choose the polarization so that $\mathbf{E}$ would cancel, but then $\mathbf{B}$ wouldn't.</p> <p>To see that the fundamental issue is conservation of probability, so that this is really something specific to quantum mechanics rather than classical physics, consider the case of sound waves, which can be represented as real scalar functions $f$ measuring the pressure. Rerunning the same argument about superposing antiparallel plane waves, we find that it's possible for $f$ to cancel, but that's OK, because $f$ doesn't have a probability interpretation. We also still have conservation of energy, because the energy depends not just on $f$ (potential) but also on $\partial f/\partial t$ (kinetic), so $f$ can vanish without making the energy vanish.</p> <p>Complex numbers come up in a lot of places in quantum mechanics, not just in wavefunctions, and it's not always obvious when they're just a notational convenience. For example, Pauli basically reinvented the quaternions in 1924. His spin matrices $\sigma_1$, $\sigma_2$, and $\sigma_3$ are equivalent to the quaternions i, j, and k if you multiply them by i.</p> <p>Operators can be complex-valued, but expectation values are always supposed to be real, since they correspond to measurable quantities.</p>