theoriginal169 Posted March 14, 2010 Share Posted March 14, 2010 can some 1 explain whhat is it. Link to comment Share on other sites More sharing options...

darkenlighten Posted March 14, 2010 Share Posted March 14, 2010 It's the time evolution of a system. Or how a system behaves given a certain hamiltonian (an equation describing its energy) I would suggest looking at http://en.wikipedia.org/wiki/Wave_function and asking questions from there. And so we know at least what knowledge level you are coming from. Link to comment Share on other sites More sharing options...

ajb Posted March 15, 2010 Share Posted March 15, 2010 You can think about the wave function as a classical field. It assigns a complex number to every point on your space under investigation, let say the real line. [math]\psi : \mathbb{R} \rightarrow \mathbb{C}[/math]. (More geometrically you can think of it as a section of a line bundle, this bundle may not be trivial in a more general setting. But this won't worry us now.) From this complex number you build a probability of the particle to be found at that point. We define a probability density as [math]|\psi(x)|^{2}[/math] (used the complex modulus here). Then probability of a measurement of the particle's position yielding a value in the interval [math][a,b][/math] is [math]P(a,b) = \int^{b}_{a}dx |\psi(x)|^{2}[/math], were we have assume the natural normalisation [math]1 = \int^{+\infty}_{- \infty}dx |\psi(x)|^{2}[/math]. Hope that is of some help. Link to comment Share on other sites More sharing options...

theoriginal169 Posted March 15, 2010 Author Share Posted March 15, 2010 is it a probibilty densty related to time? Link to comment Share on other sites More sharing options...

twistor59 Posted March 16, 2010 Share Posted March 16, 2010 is it a probibilty densty related to time? A bit better would be to think of it as a complex number you need to "square" (i.e. multiply by its conjugate) to get a probability density at a given point (x,t) Link to comment Share on other sites More sharing options...

Radical Edward Posted March 16, 2010 Share Posted March 16, 2010 is it a probibilty densty related to time? that depends entirely on the wave function you are talking about. Waves on water (obviously) have a wave function for example. In this case it is a surface amplitude related to distance and time. Light has a wave function which describes the electric and magnetic fields as a function of position and time. In essence it is a deviation from the average value (height, field strength or whatever) with distance and time. In the case of QM it is a deviation on the complex plane. Link to comment Share on other sites More sharing options...

ajb Posted March 16, 2010 Share Posted March 16, 2010 is it a probibilty densty related to time? In general yes it is a function of time (in the Schrodinger representation). However we can formally* separate it as [math]\Psi(t,x) = U(t) \psi(x)[/math] where the time-evolution operator is given by [math]U(t) = \exp \left( - \frac{i H t}{\hbar}\right)[/math], here [math]H[/math] is the Hamiltonian operator. The wave function [math]\Psi[/math] satisfies the time-dependant Schrodinger equation and [math]\psi[/math] satisfies the time-independent Schrodinger equation. Have a look at the Wiki article. * I will neglect any issues of convergence for unbounded operators in the exponential. So this separation will in general be understood very formally. Link to comment Share on other sites More sharing options...

theoriginal169 Posted March 17, 2010 Author Share Posted March 17, 2010 q(x,t) how does it works dynamical variable in wave function expectation value of p Merged post follows: Consecutive posts mergedIn general yes it is a function of time (in the Schrodinger representation). However we can formally* separate it as [math]\Psi(t,x) = U(t) \psi(x)[/math] where the time-evolution operator is given by [math]U(t) = \exp \left( - \frac{i H t}{\hbar}\right)[/math], here [math]H[/math] is the Hamiltonian operator. The wave function [math]\Psi[/math] satisfies the time-dependant Schrodinger equation and [math]\psi[/math] satisfies the time-independent Schrodinger equation. Have a look at the Wiki article. * I will neglect any issues of convergence for unbounded operators in the exponential. So this separation will in general be understood very formally. Hamiltonian operator what does it do derivatie and x -h2i Link to comment Share on other sites More sharing options...

ajb Posted March 17, 2010 Share Posted March 17, 2010 Hamiltonian operator what does it do derivatie and x -h2i In classical mechanics the Hamiltonian is the "energy function", it is the sum of the Kinetic and Potential energy. In quantum mechanics is it promoted to an operator on the Hilbert space of states. It's importance in both classical and quantum mechanics is that it describes the time evolution of the system. In classical mechanics we have Hamilton's equations which describe how canonical coordinates evolve (we have a flow on the phase space). In quantum mechanics we have Heisenberg equations which describe the evolution of the observables (or more general operators if we wish). Here we think of the states as being time independent and the operators as time dependant. As an aside the Stone--von Neumann theorem which states that for finite degrees of freedom the Schrodinger and Heisenberg representations of the canonical commutation relations are unitary equivalent. In essence, we have the operator [math]U[/math] which I defined earlier. So we can without any loss of generality consider either representation. There is also the "mixed" Dirac or interaction picture, which is useful for time-dependant Hamiltonians and scattering theory. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now