- #36
PeterDonis
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vanhees71 said:the position is well defined as a self-adjoint operator
But the particle is not in an eigenstate of this operator, so it has no definite position.
vanhees71 said:the position is well defined as a self-adjoint operator
BvU said:My estimate is that the poster is hung on a mixup of p, x and n space in a Schroedinger picture and can't move on.
I'm always careful enough to say a self-adjoint operator is representing an observable in the formalism of QT, it's not the observable itself, but that's semantics.BvU said:@vanhees71: I have always learned that the position may well be an operator, but that you can not point at a position in the box and say: that's where the particle is located right now. All there is, is a probability density. My estimate is that the poster is hung on a mixup of p, x and n space in a Schroedinger picture and can't move on.
It cannot be in an eigenstate of the position operator, because the position eigenstates are distributions, not square-integrable functions. That's not different from the infinite-volume case.PeterDonis said:But the particle is not in an eigenstate of this operator, so it has no definite position.
BvU said:Ah, maybe I get it: For a given direction of pepep_e you have the number of states in the x-direction = 2Lxpe,xh2Lxpe,xh\displaystyle{2L_x p_{e,x}\over h}
and in the y-direction = 2Lype,yh2Lype,yh\displaystyle{2L_y p_{e,y}\over h} .
So in n-space you get ne=√n2x+n2y=2Lepehne=nx2+ny2=2Lepehn_e = \sqrt{n_x^2+n_y^2} = \displaystyle{2L_e p_e\over h }.
vanhees71 said:I still have no clue what you are after! Again: There is no momentum for the rigid-boundary box and thus it's nonsensical to look for momentum-level densities. For the periodic-boundary box it's an obvious and very important finding that the number of single-particle momentum states in a phase-space volume is given by
$$\mathrm{d}^3 p \frac{V}{(2 \pi \hbar)^3}.$$
For more details, see
https://th.physik.uni-frankfurt.de/~hees/publ/kolkata.pdf
Sect. 1.2 and Sect. 1.8.
vanhees71 said:Yes, the reason to introduce a finite volume and periodic boundary conditions often is to have a calculational tool to make sense of troublesome features of continuous eigenvalues of unbound operators in Hilbert space. It's a kind of regularization procedure. A highly non-trivial example is Haag's theorem in relativistic QFT, which is only due to using the continuous momentum spectrum (or "infinite-volume limit"). Then the trick with the finite volume and periodic boundary conditions to keep well-defined momenta helps a lot.
Even in an infinitesimally small ##dp##?BvU said:Because there are normally such an incredibly high number of states that are occupied.
BvU said:It's the density of states as a function of E (or ##|\vec p|##) that is of interest here. Not how grainy it is for infinitesimal ##dp##.
Sorry, I'm not sure which link and example you're talking about. Can't find any example 2.3 or figure 2.4 in the link I gave in post #45.BvU said:Look at the numbers in your link, e.g. in example 2.3 and figure 2.4
BvU said:Post #4, way back when. It works out your whole conundrum ...
BvU said:From (2.4.3) with ##p= \hbar k##: $$
N = 2 \times{1\over 8} \times \left(L\over \pi\right)^3 \times {4\over 3}\pi k^3 ={{4\over 3}\pi V p^3\over h^3 }\Rightarrow dN ={4\pi V p^2\over h^3 }\ dp
$$ With ##\quad E= \displaystyle {{ p^2\over 2m} \Rightarrow dE = { p\over m} dp \Rightarrow { dp\over dE } = { m\over p} }\quad## you use ##\quad\displaystyle {{dN\over dE } = { dN\over dp} {dp\over dE}}\quad ## to get $$
dN ={4\pi V p^2\over h^3 }\ {m\over p}\ dE = {4\pi\; V \sqrt{2mE} \over h^3 } m \ dE $$
BvU said:I'd say yes.
That is not exponentially but quadratically. And the 'delta-thickness' is constant.JohnnyGui said:the derivative ##\frac{dN}{dp}## actually shows that the number of quantum states per ##dp## increases exponentially as ##p## gets larger. How can an exponentially increasing number of quantum states per ##dp## fit into an n-shell with decreasing thickness per ##dp##
Apologies, I indeed meant quadratically the whole time. And I expected the thickness should be constant but there's a problem, please see below.BvU said:That is not exponentially but quadratically. And the 'delta-thickness' is constant.
BvU said:You know how to differentiate $$R = \frac{2Lp}{h}\Rightarrow dR = \frac{ 2L}{h} \;dp$$
BvU said:I supose you meant to place brackets around $$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$like this$$dR=\frac{p+dp}{p}\cdot\left ( n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}} \right ) $$which means $$
dR=\frac{p+dp}{p}\cdot 0 \quad ?$$In short: you forgot to work out ## n(|p+dp|)## for the ##n_i## in the first term.
So according to that, although the following is correct:BvU said:##n## depends on ##p##
No it does not:$$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}} = \frac{dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$and in #66 your first equation shows that this is equal to $$dR=\frac{2L}{h} dp$$I repeat: bottom line of #60. There's much more interesting stuff ahead.JohnnyGui said:But when I simply rewrite this equation in terms of the corresponding n-sphere radii...
$$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$
...then it shows that the thickness decreases with higher ##p## values.
BvU said:No it does not:$$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}} = \frac{dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$and in #66 your first equation shows that this is equal to $$dR=\frac{2L}{h} dp$$I repeat: bottom line of #60. There's much more interesting stuff ahead.
The collisions themselves !JohnnyGui said:What other factors than elasticity, mass and temperature can change the kinetic energy of a colliding particle?