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It's the same. I don't know, what you are after.
vanhees71 said:It's the same. I don't know, what you are after.
vanhees71 said:Ok, then it must always been read as "arbitarily accurately" since there's no continuous quantity that can be measured exactly in this strict sense of the word. It came never to my mind that somebody could think that you can do such a thing.
vanhees71 said:The operators should be even self-adjoint ;-)).
Anyway, of course the measurement of a continuous observable always means that you measure it with a certain finite resolution. For any physicist that's self-evident, and it's not due to quantum theory but as valid within classical physics. So this is an empy debate.
Please learn first the language of quantum mechanics before dabbling in it. Only linear operators can haven eigenstates, but transition amplitudes are complex numbers, not operators.friend said:Do these transition amplitudes have momentum eigenstates
friend said:What is <x'|x> in quantum mechanics? I've seen it, but I don't know what it's suppose to mean in physical terms. It this the amplitude for an unspecified particle to go from x to x' ?
The article I link to here uses the symbols σx and σy in their gaussian distributions for the position and momentum. But I seem to be using √(-ħt/2mi) in place of σx. Can I still take the Fourier transform of [itex] < x'|U(t)|x > [/itex] and get a momentum?friend said:I was thinking more in terms of how position and momentum are Fourier transforms of each other in the calculation of Heisenberg's Uncertainty principle. See this pdf for details. There the wave-function in position space is a gaussian just as is [itex] < x'|U(t)|x > [/itex] when
[itex] < x'|U(t)|x > = {\left( {\frac{m}{{2\pi \hbar it}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}{e^{im{{(x' - x)}^2}/2\hbar t}}[/itex]. So I was asking what its conjugate momentum would look like. I think that's a gaussian as well, right?
So we can at least say thatfriend said:The article I link to here uses the symbols σx and σy in their gaussian distributions for the position and momentum. But I seem to be using √(-ħt/2mi) in place of σx. Can I still take the Fourier transform of [itex] < x'|U(t)|x > [/itex] and get a momentum?
It seems the Dirac delta in your post ruins the gaussian nature of ##\langle p|\hat{U}(t)|p' \rangle##. So I don't see how your equation is a gaussian which is the Fourier transform of a gaussian.vanhees71 said:Somehow everything is messed up now. For a free particle you have
$$\hat{U}(t)=\exp \left (-\frac{\mathrm{i} \hat{p}^2 t}{2m \hbar} \right),$$
and thus
$$\langle p|\hat{U}(t)|p' \rangle=\exp \left (-\frac{\mathrm{i} p^2 t}{2m \hbar} \right) \delta(p-p').$$
Now you can do a Fourier transform. You can regularize the integral by introducing a small imaginary part for ##t## ("infinitesimal Wick rotation"), i.e., you set ##t \rightarrow t-\mathrm{i} \epsilon## with ##\epsilon>0##. After the Fourier transform you can make ##\epsilon \rightarrow 0^+##.
In position space we havevanhees71 said:Do the calculation, and you'll find the Gaussian in position representation. I can't help it, the math is as it is!
What you seem to have shown is that you can Fourier transform the momentum to get the position. What I'm looking for is how to transform the position to get the momentum. I've tried to do that and got something that's starting to look like your ##\langle p|\hat{U}(t)|p' \rangle## equation in post #84. Give me a little while to write up the math and I'll show my progress, and maybe I can get some help finishing it. Thanks.vanhees71 said:I've given the answer in #84. In momentum space it's very simple to evaluate the propagator without much calculation, because the momentum eigenvectors are energy eigenvectors for a free particle. You can simply set ##t=0## in the equation and get ##\delta(p-p')## as it must be. I've also given you the hint, how to do the Fourier transformation from momentum to position space. Of course you can do the same in opposite direction. In any case you have to regularize the propagator before doing so. The reason is that it is not a function but a distribution (generalized function).
Wait! Are you saying that you've already seen the Fourier transform from position to momentum of <x|U(t)|x'>? Is this online somewhere?vanhees71 said:Sure! That's a good exercise. As I said, you have to regularize the integral before you can make sense of the Fourier transform. In this case you can set ##m \rightarrow m+\mathrm{i} \epsilon## with ##\epsilon>0## for ##t>0##. At the very end of the Fourier transform (still of a Gaussian with this regularization) you take the limit ##\epsilon \rightarrow 0^+##.
friend said:Wait! Are you saying that you've already seen the Fourier transform from position to momentum of <x|U(t)|x'>? Is this online somewhere?
stevendaryl said:If you let a plane wave [itex]e^{i k (x-x')}[/itex] evolve with time, it turns into:
[itex]e^{i k (x-x') - i \frac{k^2}{2m} t}[/itex] (letting [itex]\hbar = 1[/itex]).
friend said:Very interesting. Thank you for your response. I didn't quite get where you got the [itex] - i \frac{k^2}{2m} t[/itex] term to get time evolution.
I think my point was that ##{\psi _0}(x')## can always be put in the form ##\int_\mathbb{R} {dx''U({t_0},x',x'')} {\psi _{00}}(x'')## so that ultimately ##{\psi _{00}}(x'')## approaches (but never equal to) ## < x''|{e^{ - iH\varepsilon /\hbar }}|{x_{00}} >## , if you get my drift.vanhees71 said:No, ##U(t,x,x')## cannot represent a wave function, because it is not square integrable. It doesn't live in Hilbert space but in the dual of a smaller dense subspace that is the domain of the position operator. This dual space is larger than the Hilbert space and thus contains proper generalized functions like this propagator!
##{\psi _0}(x')## is supposed to be the initial wave-function at ##t=0##, or ##t_0##. But it occurs to me that in some circumstances, other events from a previous time could have lead to ##{\psi _0}(x')##. In that case we could just as easily write ##{\psi _0}(x')\,\, = \,\,\int_\mathbb{R} {dx''U({t_0},x',x'')} {\psi _{00}}(x'')##, where ##{\psi _{00}}(x'')## is the wave function from a previous time and ##U({t_0},x',x'')## propagates it from that previous time to ##t_0##. I have to wonder if perhaps this idea can be iterated back further in time yet again. And where can we say those iterations must stop. Must they stop where the iterated initial wave function begins to look like another one of those ##< {x_j}|{e^{ - iH\varepsilon /\hbar }}|{x_i} >## that are introduced by inserting the resolution of identity in yet another propagator as we iterate this process?vanhees71 said:I don't understand what you mean. What's ##\psi_{00}## supposed to be?
This is what I'm really interested in.friend said:Now I'm not so sure which of [itex]\langle p | U(t) | x' \rangle[/itex] or [itex]\langle p|U(t)|p' \rangle[/itex] I'm interested in. If [itex]\langle x|U(t)|x'\rangle[/itex] represents the transition amplitude of a particle from x' to x, then what is its conjugate momentum? Do I want [itex]\langle p | U(t) | x' \rangle[/itex] because that is its momentum after the transition? Or do I want [itex]\langle p|U(t)|p' \rangle[/itex] because [itex]\langle x|U(t)|x'\rangle[/itex] has an undetermined momentum at both x' and x? Does [itex]\langle x|U(t)|x'\rangle[/itex] have the momentum p' at x' and the momentum p at x ? Or maybe I'm looking for p-p'. Thanks.