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- Thread starter alphaone
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- #2

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this is actually a wonderful way of looking at the HUP, it really takes the mystery out of it

the HUP, and the relationship that you point out above, are a consequence of the fact that in quantum mechanics particles are (a) described by waves and (b) quantized in energy

consider a gaussian wavepacket propogating through space..remember that the momentum and the wavelength are related by de Broglie's equation. The more accurately you can sample the positional "signal" of the wavepacket (i.e. make the digital "sampling interval" smaller and smaller) the worse you know the wavelength. Vice versa, in order to really know the wavelength well requires a greater sampling interval, ideally infinite, but this makes the positional interval large. The quantized energy of the waves that are being summed into the wavepacket place an upper bound on the standard error (i.e. <x^2> - <x>^2 and <p^2> - <p>^2) if you work it out exactly.

if you don't believe me, then work out <q|X> and <p|X> knowing the <x|p> and <p|x> relations, and you will get your fourier factors. See Dirac, 4th edition pg.97 for details.

these formulas can also be used to derive the Feynman path integral expression by inserting complete sets of states |q><q|, |p><p| into Tr{exp(-beta H)} and utilizing the Trotter identity.

- #3

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here's a concrete example.

[tex]\hat{p} <x|p> = <x|p|p> = p<x|p>[/tex]

but we also know that:

[tex]\hat{p} <x|p> = \frac{\hbar}{i} \frac{\partial}{\partial x} <x|p>[/tex]

consider the projection of the momentum into position space, <x|p>, as just some function f(x,p). Then we have the DE:

[tex]\frac{ip}{\hbar} f(x,p) = \frac{\partial}{\partial x} f(x,p)[/tex]

the solution of which (when normalized) is:

[tex]<x|p> = \frac{1}{\sqrt{2\pi \hbar}} e^{\frac{ipx}{\hbar}}[/tex]

and so it follows that

[tex]<p|x> = \frac{1}{\sqrt{2\pi \hbar}} e^{-\frac{ipx}{\hbar}}[/tex]

consider some arbitrary state projected into the momentum space:

[tex]<p|\psi >[/tex]

by the complete set of states, we have the following identity:

[tex]\int_{-\infty}^{\infty} dx |x><x| = \hat{1}[/tex]

and inserting this between the bra and ket:

[tex]<p|\psi> = <p |\left( \int_{-\infty}^{\infty} dx |x><x| \right) |\psi> = \int_{-\infty}^{\infty} dx <p|x><x|\psi> = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} dx e^{-\frac{ipx}{\hbar}} <x|\psi>[/tex]

and so the state projected into momentum space is given by the fourier transform of the state projected into position space.

[tex]\hat{p} <x|p> = <x|p|p> = p<x|p>[/tex]

but we also know that:

[tex]\hat{p} <x|p> = \frac{\hbar}{i} \frac{\partial}{\partial x} <x|p>[/tex]

consider the projection of the momentum into position space, <x|p>, as just some function f(x,p). Then we have the DE:

[tex]\frac{ip}{\hbar} f(x,p) = \frac{\partial}{\partial x} f(x,p)[/tex]

the solution of which (when normalized) is:

[tex]<x|p> = \frac{1}{\sqrt{2\pi \hbar}} e^{\frac{ipx}{\hbar}}[/tex]

and so it follows that

[tex]<p|x> = \frac{1}{\sqrt{2\pi \hbar}} e^{-\frac{ipx}{\hbar}}[/tex]

consider some arbitrary state projected into the momentum space:

[tex]<p|\psi >[/tex]

by the complete set of states, we have the following identity:

[tex]\int_{-\infty}^{\infty} dx |x><x| = \hat{1}[/tex]

and inserting this between the bra and ket:

[tex]<p|\psi> = <p |\left( \int_{-\infty}^{\infty} dx |x><x| \right) |\psi> = \int_{-\infty}^{\infty} dx <p|x><x|\psi> = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} dx e^{-\frac{ipx}{\hbar}} <x|\psi>[/tex]

and so the state projected into momentum space is given by the fourier transform of the state projected into position space.

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- #4

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Simple, the Fourier transform is the unitary application from the position coordinates Hilbert space [itex] L^{2}\left(R^{3},d^{3}r\right) [/itex] to the momentum coordinate Hilbert space [itex] L^{2}\left(R^{3},d^{3}p\right) [/itex]

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- #6

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more intuitive? your joking, right? are you familiar with fourier analysis of wave phenomena?

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- #8

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i do. please see post #2.

how do you typically transform from position space to wavelength space?

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Are you asking why a wave packet can be described by some equation

[itex]\psi (x,t) = \frac{1}{\sqrt{2\pi \hbar}} \int^{\,\infty}_{-\infty} dp\, \phi(p) e^{i(px-Et)/ \hbar}[/itex]

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- #11

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Can anyone explain how to derive the schrodinger equation in momentum space?

- #12

Avodyne

Science Advisor

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could somebody please explain to me why position and momentum space are related to one another by a fourier transform?

No, no one can. Seriously. This is just the way it turns out that the world works. It wouldn't work that way if the world had turned out to be governed by classical mechanics (as was believed to be true for about 200 years).

But classical mechanics turns out to be just a limit, approximately applicable in certain circumstances, to quantum mechanics, in which position and momentum are

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Ok i see... Incidently, how is the schrodinger equation in momentum space written?

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I think it's related to the fact that momentum generates translations.

- #15

samalkhaiat

Science Advisor

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Because the relevant dynamical equation ( Schrodinger's in your case) is LINEAR PDE and admits PLANE WAVE solutions. Physics (classical or quantum) has nothing to do with it.

Take, for example, the equation

[tex]\frac{\partial^{2}}{\partial t^{2}} u(r,t) = c^{2} \nabla^{2}u(r,t)[/tex]

Does u represent a massless boson or a sound wave? Who cares, the fact remains that the plane wave

[tex]\exp [i(\vec{k}.\vec{r} - c|k|t)][/tex]

solve our equation. Mathematics then tells you that

[tex]

u(r,t) = \int d^{3}k \ U(k) \exp [i(\vec{k}.\vec{r} - c|k|t)] = \int d^{3}k \ U(k,t) e^{i \vec{k}.\vec{r}}\ \ \ (F)

[/tex]

is a general solution. Now, why is it that coordinate and wave number spaces related by the above Fourier transformation? well, the one and only answer is; because eq(F) represents the most general SUPERPOSITION of PLANE WAVES.

regards

sam

- #16

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Now, why is it that coordinate and wave number spaces related by the above Fourier transformation? well, the one and only answer is; because eq(F) represents the most general SUPERPOSITION of PLANE WAVES.

regards

sam

Doesn't the interpretation of k as a wave-number depend on a dispersion relation between k_0 and k? If you instead put the time dependence into the Fourier coefficient, i.e., a(k,t)e^(ikx), then this is not a wave, and also a(k,k_0)e^(ikx)e^(ik_0t) is also not a wave - these are just general Fourier transforms. The differential equation will give you a relation between k_0 and k, but unless it's particularly simple, can it be called a wave?

And does a taut string undergoing only transverse motions have momentum in the longitudinal direction, and is it related to the wave-vector k?

- #17

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In quantum mechanics, as in classical mechanics, the physical system at hand obeys a ten dimensional symmetry group, the Galilean symmetry group. Three translations, three rotations, three reflections (?), time translation.

In quantum mechanics, as opposed to classical mechanics, we have vectors representing physical systems. When these vectors are infinite dimensional, for example a wavefunction in position space, the symmetry groups acting on these infinite dimensional vectors need to be in an infinite dimensional representation.

Say when you want to translate a wavefunction in space, you see that the generator of this group is id/dx (i so that the transformation is unitary). Or when you want to rotate, then id/dL. From that you can conclude, that they correspond to momentum respectively to angular momentum.

- #18

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Fourier transform fully preserves the information and so states can be described completely in such a transformed space.

As to why the fourier variable is interpreted as momentum...

It is because the eigenvector of the momentum operator is of the form [tex]e^{ipx}[/tex] and has the momentum eigenvalue, p, in place of the fourier variable.

- #19

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So given an arbitrary function f(x), the fourier transform allows you to express it as a sum of the basis functions e^(ipx/h) which are eigenstates of momentum.

This is just a hand-waiving argument I came up with, but even in classical mechanics translations generate momentum. The difference here is that the eigenvalues of operators represent the actual value of whatever it is your measuring (I think).

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