Can the Free Particle Wave Equation Be Solved for Periodic Boundary Conditions?

AI Thread Summary
The discussion centers on the solvability of the free particle wave equation under periodic boundary conditions. It highlights that the wave function can be expressed as a superposition of plane waves, with the relationship between the wave function and its Fourier transform being crucial. The conversation explores the implications of assuming a small delta k, which allows g(k) to behave like a delta function. There is a query about the general case of solving the wave equation for a defined periodic interval, specifically from 0 to 2πL. The overall consensus emphasizes that while the wave function can take various forms, the specific wave equation dictates how it evolves over time based on the conditions applied.
sniffer
Messages
112
Reaction score
0
for a free particle, the wave equation is a superposition of plane waves,

\psi(x,0)= \int_{-\infty}^{\infty}g(k)\exp(ikx)dk
and
g(k)= \int_{-\infty}^{\infty}\psi(0,0)\exp(-ikx)dx

one is the Fourier transform of the other. some cases to solve this is when we assume a small delta k, so g(k) behaves like a pulse "delta" function.

is there any more general case we can solve this?

i have been thinking hard if we have definite periodic x, say from 0 up to 2\pi L, is it solvable?

what would be the (periodic) eigen energy function (if it is)?
 
Physics news on Phys.org
sniffer said:
for a free particle, the wave equation is a superposition of plane waves,

\psi(x,0)= \int_{-\infty}^{\infty}g(k)\exp(ikx)dk
and
g(k)= \int_{-\infty}^{\infty}\psi(0,0)\exp(-ikx)dx

one is the Fourier transform of the other. some cases to solve this is when we assume a small delta k, so g(k) behaves like a pulse "delta" function.

The above pair is, as you say, a Fourier transform pair. Any "nice" function can be written that way, so the above is not a "wave equation" or something, it is a general way of writing a function.
The quantum state of a single scalar particle is described by just such a nice function, called the wave function. At any time, it can be (almost) any function. However, what the wave equation (not written here) gives you, is how this wavefunction AT A CERTAIN TIME t0 will change into the wavefunction at another time t1. This equation will be different according to the situation at hand (free particle, particle in a potential...).

cheers,
Patrick.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top