Counting the states of a free particle (Periodic boundary conditions)

JK423
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Say you have a free particle, non relativistic, and you want to calculate the density of states (number of states with energy E-E+dE).
In doing that, textbooks apply periodic boundary conditions (PBC) in a box of length L, and they get L to infinity, and in this way the states become countable.
I have some questions and i'd really like your help. Thank you in advance!

1) Can someone explain to me what is the physical meaning of choosing PBC for the particle's wavefunction?

2)Why to choose boundary conditions at all for the free particle, since there are no boundaries..?

3) If PBC is just a mathematical trick that just gives the correct result, is there another more natural way of obtaining the density of states?

4) If we apply PBC, then the wavevector is quantized: k=2πn/L, where n=integer. When we take continuous limit L-->Infinity we get Δk-->dk because L is at the denominator and we can consider k to be a continuous variable. The problem is that when we make the integration, k's limits go from -Infinity to +Infinity and at these limits the approximation of treating k as a continuous variable is false, since 'n' is comparable in magnitude with L. Which means that at the limits (infinity) for large n, the integral should be replaced by sum.
So, how do we justify this? How can we prove that the error of keeping the integral all the way to infinity is negligible?
One idea that comes to my mind is the fact that the density of states is usually integrated with someother function f(k) which falls of rapidly when k-->Infinity, so the contribution of very large k is negligible. You think that this is correct?
 
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The problem is that when we make the integration, k's limits go from -Infinity to +Infinity and at these limits the approximation of treating k as a continuous variable is false, since 'n' is comparable in magnitude with L. Which means that at the limits (infinity) for large n, the integral should be replaced by sum.
No, the reason we can treat k as a continuous variable is that Δk = 2π/L, the distance between two neighboring values of k, is small, uniformly for all k. This is true whenever L is large. It's not the size of k that counts, it's the size of Δk.
 
Bill_K said:
No, the reason we can treat k as a continuous variable is that Δk = 2π/L, the distance between two neighboring values of k, is small, uniformly for all k. This is true whenever L is large. It's not the size of k that counts, it's the size of Δk.
Indeed.. Thanks!
 
Any help on the first 3 questions?
 
Is there anyone that can help please?
(I repost in order to update the thread..)
 
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