I faced a weird question:what is the requirement of having a wave equation for the QM systems? I think unless we have it,we cannot predict the probabilistic time and space evolution of the wave function subjected to potential constraints.(dynamicity of the wave function will be lost). However,this does not help us to see the effect of performing measurements on the wave function... Any point missing???
Another thing: The essential features of this equations should be: (i) Should be consistent with statistical interpretation of wave function (ii) \(\psi\) should be normalized all the way (iii)The equation should be consistent with correspodence principle Any suggestion???
Well, first of all, the Schroedinger equation is a HEAT equation, not a wave equation. But otherwise, I don't understand the question.
In other words the question is why do we require a differential equation in QM and what are the features of the equation?
I hope it goes without saying, I have no idea what I'm talking about. But I was once told by a physics pal that nobody knows how to quantize non-periodic systems. Is this relevant? Is it right? Ah, this, from an old college chemistry text: the Schroedinger wave equation is an eigenfunction of the form \(\mathcal{H}\psi = \epsilon \psi\), \(\mathcal{H}\) is the Hamiltonian operator, and \(\epsilon\) is an eigenvalue for the operator \(\mathcal{H}\). I also read somewhere, that in the QM world, it is taken as axiomatic that any observable is the action of an operator, and the eigenvalues for that operator are the allowed possible outcomes of the measurements on the said observable. Is this gibberish?
Quark--- The Schroedinger equation is a HEAT equation Please Register or Log in to view the hidden image! It only has one derivative on time. As to quantizing non-periodic systems, you may be right. I will have to think of a counter-example! I have been turning neelakash's quesiton over in my head. Quantum mechanics needs a differential equaiton, but so does EVERY physical system you can think of! So I don't really know what the question is getting at.
Ben - sorry to appear argumentative, but my text states quite specifically that the S. equation is time-independent. Is it wrong? Like, the only derivatives it offers (in 3-space) are as \(\nabla^2\). Anyway, the S. equation most certainly is a wave equation. So there!
Well, ok. I guess it depends on which picute you're working in---there are Schroedinger picture (time dependant states) and Heisenberg pictuer (time dependant operators). Typically, you may see \(\mathcal{H} \psi = E_n \psi\) But this is for a time independant psi. If psi depends on time, either explicitly or implicitly, then you have \(i \hbar \frac{\partial}{\partial t}\Psi(x,t) = \left(-\frac{\hbar^2}{2m}\nabla^2 +V(x)\right)\Psi(x,t)\). This is most certainly a heat equation Please Register or Log in to view the hidden image! It describes diffusion---the way that the wavefunction (more aptly, the probability density) diffuses over a given time period.
The left hand side of my equation can be understood by taking the typical separation of variables for psi. You quickly find out (or you know already) that the time behavior goes like \(e^{-i\hbar t}\).
Jim Al-Khalili's book Quantum: A Guide for the Perplexed has a very good demonstration of this diffusion aspect. As shown, with enough given time, the probability distribution becomes roughly uniform over the entire bounded volume. The images in Al-Khalili's book even look like the kind of thing one would achieve by modeling a pressureless gas with the Navier-Stokes equations. Nice explanation Ben!
The Schrodinger equation is like a heat equation, but it has a totally different behavior because of the imaginary unit in front of the time derivative. As for the question, it might be related to the unitarity of the evolution?
There is time-dependant equation and i think a non-dependant one as well... Saying it is primarilly a heat equation wouldn't be entirely true. It is more than that. It describes the probabilistic nature of the wavelike nature of any physical system. It can describe more than just heat in short.
''However,this does not help us to see the effect of performing measurements on the wave function...'' If an observation is made on a particle which is in a wave-form, the wave function will be reduced to a single value.
But mathematically it is a diffusion equation. temur mentioned that the constants are imaginary, so there may be some issues there, but the wave-function itself is not directly observable---one can only measure its' modulous squared. So in this case the time dependance (which always goes like \(e^{-iEt}\)) will drop out. I guess understanding what the Schroedinger equation actually is is the aim of people who try to understand quantum mechanics on a deeper level.
It's just a separation of variables. If you ever took a class on PDE's, all you do is say that the full equation can be written as \(\Psi(x,t) = \sum_{n = 0}^{\infty} f_n(t) \psi_n(x)\) The full partial differential equation separates into two ordinary differential equations. One of those is the equation that QuarkHead quoted earlier \(\mathcal{H} \psi_n(x) = E_n \psi_n(x)\) Quite fun, actually Please Register or Log in to view the hidden image!
