Derivation of the uncertainty principle

Derivation of the quantum uncertainty principle

thewavefunction,Ψ,isequaltoA(e)i(kx−ϖt)(1),fromeuler,wehaveA(e)iϕ=A(cosϕ+isinϕ)(1.1),lettingkx−ϖt=ϕwehaveΨ(x,t)=cos(kx−ϖt)+isin(kx−ϖt)takingℜ(Ψ)weareleftwithΨ(x,t)=Acos(kx−ϖt)(1.2)nowwecandefinekas2πλ(1.3),whichisthewavenumberand,fromeinstein,wehaveE=hβ(1.4),wherecisthespeedoflight,β=cλandhisplanck‘sconstant.themomentumofaparticle,mv,canbeextendedtomasslessparticles(i.e.photons)byrelativityE2=(mc2)2+(pc)2(seerelativitypaper)sincem=0E=pchβ=pccβ=phorp=hβc(1.5)⇒λ=hp(1.5‘)andfinallyweletϖ=2πβ(1.6)whichistheangularfrequencysubstituting(1.6)and(1.3)into(1.2)wegetΨ(x,t)=Acos(2πxλ−2πβt)Ψ(x,t)=Acos(2πβxc−2πβtc)Ψ(x,t)=Acos(2πβc(x−ct)),β=EhΨ(x,t)=Acos(2πEhc(x−ct))h2π=ℏΨ(x,t)=Acos1ℏ(Exc−Et),E/c=psothisequalsAcos1ℏ(px−Et)letsthinkaboutthewaveformΨ,weknowbyfourier‘stransformsthatyocanrepresentΨas∫R\g(ϖ)cos(kx−ϖt)+Φ(ϖ)sin(kx−ϖt)dxthefourierconjugateofΨwillbespreadoutbecauseΨisalocalizedwaveformtomakeitsimpler,letslookatΨ(x)Ψ(x)=\F[ξ(η)]ifwethinkofξasaprobabilitydistribution,letsmakeξequaltothenormaldistributionΛn=μne−(n−n0)2/2δn2⇒ξ(n)=Λn∫RΛndn=∫Rξ(n)dn=1(theparticlehastobesomewhere)Λnistheenvelopeof∣Ψ∣∗∣Ψ∣,ΛnistheenvelopeofΨwecanfindthecoefficientμbynormalizingΛtheformforanormalizedcoefficientisμn=1σn2πsinceξandΨarefourierconjugatesμne−(n−n0)2/(4σ2n)=∫Rμxe−(x−x0)2/4σx2cosnxdx,ifweletx0=0RHS⇒2∫0∞e−(ax)2cosγxdx=π2ae−(b2a)2soweconcludethata=1σkandn=γsubstitutingthenewvalueswehaveμn=4πσ2x,cancellingthemoutandtakingthelnofeachsideweget−(k2σk)2=−(kσx)2whichsimplifiestoσk=12σxwhichisσkσx=12k=pℏΔxΔp=ℏ2butthisisfornormaldistributionsonlyifit‘snotanormaldistributionit‘smorethanh/4πFinalResultΔxΔp≥ℏ2 \\ \\ the\quad wave\quad function,\Psi ,is\quad equal\quad to\quad A({ e) }^{ i(kx-\varpi t) }\quad \quad (1),\quad from\quad euler,\quad we\quad have\quad A({ e) }^{ i\phi }=A(cos\phi +i\quad sin\phi )\quad (1.1)\quad \quad \quad ,\quad letting\quad kx-\varpi t=\phi \quad \\ \\ \\ we\quad have\quad \Psi (x,t)=cos(kx-\varpi t)+\quad i\quad sin(kx-\varpi t)\quad taking\quad \Re (\Psi )\quad we\quad are\quad left\quad with\quad \\ \\ \Psi (x,t)=A\quad cos(kx-\varpi t)\quad (1.2)\\ \\ now\quad we\quad can\quad define\quad k\quad as\quad \frac { 2\pi }{ \lambda } \quad (1.3)\quad ,\quad which\quad is\quad the\quad wave\quad number\quad and\quad ,from\quad einstein,\quad we\quad have\quad E=h\beta \quad (1.4)\quad ,where\quad c\quad is\quad \\ the\quad speed\quad of\quad light,\quad \beta =\frac { c }{ \lambda } \quad and\quad h\quad is\quad planck`s\quad constant.\\ \\ the\quad momentum\quad of\quad a\quad particle\quad ,mv,\quad can\quad be\quad extended\quad to\quad massless\quad particles(i.e.\quad photons)\quad by\quad relativity\quad \\ \\ { E }^{ 2 }={ { (mc }^{ 2 }) }^{ 2 }+{ (pc) }^{ 2 }\quad \quad \quad (see\quad relativity\quad paper)\\ \\ since\quad m=0\\ \\ E=pc\\ \\ h\beta =pc\\ \\ \frac { c }{ \beta } =\frac { p }{ h } \quad or\quad \\ \\ p=\frac { h\beta }{ c } \quad \quad \quad (1.5)\quad \Rightarrow \lambda =\frac { h }{ p } \quad (1.5`)\\ and\quad finally\quad we\quad let\quad \varpi =2\pi \beta \quad (1.6)\quad which\quad is\quad the\quad angular\quad frequency\\ \\ \\ substituting\quad (1.6)\quad and\quad (1.3)\quad into\quad (1.2)\quad we\quad get\quad \quad \\ \\ \Psi (x,t)=A\quad cos(\frac { 2\pi x }{ \lambda } -2\pi \beta t)\\ \\ \Psi (x,t)=A\quad cos(\frac { 2\pi \beta x }{ c } -\frac { 2\pi \beta t }{ c } )\\ \\ \\ \\ \Psi (x,t)=A\quad cos(\frac { 2\pi \beta }{ c } (x-ct))\quad \quad ,\beta =\frac { E }{ h } \\ \\ \Psi (x,t)=A\quad cos(\frac { 2\pi E }{ hc } (x-ct))\\ \\ \frac { h }{ 2\pi } =\hbar \\ \quad \\ \Psi (x,t)=A\quad cos\frac { 1 }{ \hbar } (\frac { Ex }{ c } -Et),\quad E/c=p\quad so\quad this\quad equalsA\quad cos\quad \frac { 1 }{ \hbar } (px-Et)\\ \\ \\ lets\quad think\quad about\quad the\quad waveform\quad \Psi ,\quad we\quad know\quad by\quad fourier`s\quad transforms\quad that\quad yo\quad can\quad represent\quad \Psi \quad as\\ \\ \int _{ R }^{ }{ \g (\varpi ) } cos(kx-\varpi t)+\Phi (\varpi )sin(kx-\varpi t)\quad dx\\ \\ the\quad fourier\quad conjugate\quad of\quad \Psi \quad will\quad be\quad spread\quad out\quad because\quad \Psi \quad is\quad a\quad localized\quad waveform\quad to\quad make\quad it\quad simpler,lets\quad look\quad at\quad \Psi (x)\\ \\ \Psi (x)=\F [\xi (\eta )]\\ \\ if\quad we\quad think\quad of\quad \xi \quad as\quad a\quad probability\quad distribution,\quad lets\quad make\quad \xi \quad equal\quad to\quad the\quad normal\quad distribution\\ \\ \\ { \Lambda }_{ n }={ \mu }_{ n }{ e }^{ -(n-{ n }_{ 0 })^{ 2 }/2{ { \delta }_{ n } }^{ 2 } }\quad \Rightarrow \xi (n)=\sqrt { { \Lambda }_{ n } } \\ \\ \int _{ R }^{ }{ { \Lambda }_{ n }\quad dn } =\int _{ R }^{ }{ \xi (n)\quad dn } =1\quad \quad (the\quad particle\quad has\quad to\quad be\quad somewhere)\\ \\ \\ { \Lambda }_{ n }\quad is\quad the\quad envelope\quad of\quad \left| \Psi \right| *\left| \Psi \right| ,\quad \sqrt { { \Lambda }_{ n } } is\quad the\quad envelope\quad of\quad \Psi \\ \\ \\ we\quad can\quad find\quad the\quad coefficient\quad \mu \quad by\quad normalizing\quad \Lambda \quad \quad \\ \\ the\quad form\quad for\quad a\quad normalized\quad coefficient\quad is\quad \\ \\ { \mu }_{ n }=\frac { 1 }{ { \sigma }_{ n }\sqrt { 2\pi } } \\ \\ since\quad \xi \quad and\quad \Psi \quad are\quad fourier\quad conjugates\\ \\ \sqrt { { \mu }_{ n } } { e }^{ -(n-{ n }_{ 0 })^{ 2 }/(4{ { \sigma }^{ 2 } }_{ n }) }=\int _{ R }^{ }{ \sqrt { { \mu }_{ x } } } { e }^{ -(x-{ x }_{ 0 })^{ 2 }/4{ { \sigma }_{ x } }^{ 2 } }cos\quad nx\quad dx,\quad if\quad we\quad let\quad { x }_{ 0 }=0\\ \\ \\ \\ \\ RHS\Rightarrow 2\int _{ 0 }^{ \infty }{ { e }^{ -{ (ax) }^{ 2 } } } cos\quad \gamma x\quad dx=\frac { \sqrt { \pi } }{ 2a } { e }^{ -({ \frac { b }{ 2a } ) }^{ 2 } }\\ \\ so\quad we\quad conclude\quad that\quad a\quad =\frac { 1 }{ { \sigma }_{ k } } \quad and\quad n=\gamma \\ \\ substituting\quad the\quad new\quad values\quad we\quad have\quad { \mu }_{ n }=4\pi { { \sigma }^{ 2 } }_{ x }\quad \quad ,\quad cancelling\quad them\quad out\quad and\quad taking\quad the\quad ln\quad of\quad each\quad side\quad we\quad get\\ \\ { -(\frac { k }{ 2{ \sigma }_{ k } } ) }^{ 2 }={ -(k{ \sigma }_{ x }) }^{ 2 }\quad which\quad simplifies\quad to\quad \\ \\ { \sigma }_{ k }\quad =\frac { 1 }{ 2{ \sigma }_{ x } } \\ \\ \\ \\ which\quad is\quad { \sigma }_{ k }{ \sigma }_{ x }=\frac { 1 }{ 2 } \\ \quad k=\frac { p }{ \hbar } \\ \\ \Delta x\Delta p=\frac { \hbar }{ 2 } \\ \\ but\quad this\quad is\quad for\quad normal\quad distributions\quad only\\ \\ if\quad it`s\quad not\quad a\quad normal\quad distribution\quad it`s\quad more\quad than\quad h/4\pi \\ \\ \\ Final\quad Result\\ \\ \Delta x\Delta p\ge \frac { \hbar }{ 2 } \\ \\ \\ \\ \\ \\

Are there any mistakes in my derivation?

Thanks.

#Mechanics

Note by Hamza A
5 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

From a quick scan this looks like a fine derivation for the plane wave particle. Can you derive it in the general case (any arbitrary wave function)?

I would just suggest using LaTeX directly on Brilliant, it's the easiest and best formatted option.

Josh Silverman Staff - 5 years, 5 months ago

this may be weird but i can`t find the backslash button on my laptop which is like really important in latex

and i will try to justify my assumption that if it is not normal then it too will have the uncertainties in position and momentum inversely proportional

i`ll let you know if i find anything

Thanks (:

Hamza A - 5 years, 4 months ago

It hurts my eyes to try to read this. Definitely do this in LaTeX.

I think I'll redo this in LaTeX first just so that I can read it.

Edit: Oh, thanks you've got it here.

Michael Mendrin - 5 years, 4 months ago

Log in to reply

Derivationofthequantumuncertaintyprinciplethewavefunction,Ψ,isequaltoA(e)i(kx−ϖt)(1),fromeuler,wehaveA(e)iϕ=A(cosϕ+isinϕ)(1.1),lettingkx−ϖt=ϕwehaveΨ(x,t)=cos(kx−ϖt)+isin(kx−ϖt)takingℜ(Ψ)weareleftwithΨ(x,t)=Acos(kx−ϖt)(1.2)nowwecandefinekas2πλ(1.3),whichisthewavenumberand,fromeinstein,wehaveE=hβ(1.4),wherecisthespeedoflight,β=cλandhisplanck‘sconstant.themomentumofaparticle,mv,canbeextendedtomasslessparticles(i.e.photons)byrelativityE2=(mc2)2+(pc)2(seerelativitypaper)sincem=0E=pchβ=pccβ=phorp=hβc(1.5)⇒λ=hp(1.5‘)andfinallyweletϖ=2πβ(1.6)whichistheangularfrequencysubstituting(1.6)and(1.3)into(1.2)wegetΨ(x,t)=Acos(2πxλ−2πβt)Ψ(x,t)=Acos(2πβxc−2πβtc)Ψ(x,t)=Acos(2πβc(x−ct)),β=EhΨ(x,t)=Acos(2πEhc(x−ct))h2π=ℏΨ(x,t)=Acos1ℏ(Exc−Et),E/c=psothisequalsAcos1ℏ(px−Et)letsthinkaboutthewaveformΨ,weknowbyfourier‘stransformsthatyocanrepresentΨas∫R\g(ϖ)cos(kx−ϖt)+Φ(ϖ)sin(kx−ϖt)dxthefourierconjugateofΨwillbespreadoutbecauseΨisalocalizedwaveformtomakeitsimpler,letslookatΨ(x)Ψ(x)=\F[ξ(η)]ifwethinkofξasaprobabilitydistribution,letsmakeξequaltothenormaldistributionΛn=μne−(n−n0)2/2δn2⇒ξ(n)=Λn∫RΛndn=∫Rξ(n)dn=1(theparticlehastobesomewhere)Λnistheenvelopeof∣Ψ∣∗∣Ψ∣,ΛnistheenvelopeofΨwecanfindthecoefficientμbynormalizingΛtheformforanormalizedcoefficientisμn=1σn2πsinceξandΨarefourierconjugatesμne−(n−n0)2/(4σ2n)=∫Rμxe−(x−x0)2/4σx2cosnxdx,ifweletx0=0RHS⇒2∫0∞e−(ax)2cosγxdx=π2ae−(b2a)2soweconcludethata=1σkandn=γsubstitutingthenewvalueswehaveμn=4πσ2x,cancellingthemoutandtakingthelnofeachsideweget−(k2σk)2=−(kσx)2whichsimplifiestoσk=12σxwhichisσkσx=12k=pℏΔxΔp=ℏ2butthisisfornormaldistributionsonlyifit‘snotanormaldistributionit‘smorethanh/4πFinalResultΔxΔp≥ℏ2\\ Derivation\quad of\quad the\quad quantum\quad uncertainty\quad principle\\ \\ \\ the\quad wave\quad function,\Psi ,is\quad equal\quad to\quad A({ e) }^{ i(kx-\varpi t) }\quad \quad (1),\quad from\quad euler,\quad we\quad have\quad A({ e) }^{ i\phi }=A(cos\phi +i\quad sin\phi )\quad (1.1)\quad \quad \quad ,\quad letting\quad kx-\varpi t=\phi \quad \\ \\ \\ we\quad have\quad \Psi (x,t)=cos(kx-\varpi t)+\quad i\quad sin(kx-\varpi t)\quad taking\quad \Re (\Psi )\quad we\quad are\quad left\quad with\quad \\ \\ \Psi (x,t)=A\quad cos(kx-\varpi t)\quad (1.2)\\ \\ now\quad we\quad can\quad define\quad k\quad as\quad \frac { 2\pi }{ \lambda } \quad (1.3)\quad ,\quad which\quad is\quad the\quad wave\quad number\quad and\quad ,from\quad einstein,\quad we\quad have\quad E=h\beta \quad (1.4)\quad ,where\quad c\quad is\quad \\ the\quad speed\quad of\quad light,\quad \beta =\frac { c }{ \lambda } \quad and\quad h\quad is\quad planck`s\quad constant.\\ \\ the\quad momentum\quad of\quad a\quad particle\quad ,mv,\quad can\quad be\quad extended\quad to\quad massless\quad particles(i.e.\quad photons)\quad by\quad relativity\quad \\ \\ { E }^{ 2 }={ { (mc }^{ 2 }) }^{ 2 }+{ (pc) }^{ 2 }\quad \quad \quad (see\quad relativity\quad paper)\\ \\ since\quad m=0\\ \\ E=pc\\ \\ h\beta =pc\\ \\ \frac { c }{ \beta } =\frac { p }{ h } \quad or\quad \\ \\ p=\frac { h\beta }{ c } \quad \quad \quad (1.5)\quad \Rightarrow \lambda =\frac { h }{ p } \quad (1.5`)\\ and\quad finally\quad we\quad let\quad \varpi =2\pi \beta \quad (1.6)\quad which\quad is\quad the\quad angular\quad frequency\\ \\ \\ substituting\quad (1.6)\quad and\quad (1.3)\quad into\quad (1.2)\quad we\quad get\quad \quad \\ \\ \Psi (x,t)=A\quad cos(\frac { 2\pi x }{ \lambda } -2\pi \beta t)\\ \\ \Psi (x,t)=A\quad cos(\frac { 2\pi \beta x }{ c } -\frac { 2\pi \beta t }{ c } )\\ \\ \\ \\ \Psi (x,t)=A\quad cos(\frac { 2\pi \beta }{ c } (x-ct))\quad \quad ,\beta =\frac { E }{ h } \\ \\ \Psi (x,t)=A\quad cos(\frac { 2\pi E }{ hc } (x-ct))\\ \\ \frac { h }{ 2\pi } =\hbar \\ \quad \\ \Psi (x,t)=A\quad cos\frac { 1 }{ \hbar } (\frac { Ex }{ c } -Et),\quad E/c=p\quad so\quad this\quad equalsA\quad cos\quad \frac { 1 }{ \hbar } (px-Et)\\ \\ \\ lets\quad think\quad about\quad the\quad waveform\quad \Psi ,\quad we\quad know\quad by\quad fourier`s\quad transforms\quad that\quad yo\quad can\quad represent\quad \Psi \quad as\\ \\ \int _{ R }^{ }{ \g (\varpi ) } cos(kx-\varpi t)+\Phi (\varpi )sin(kx-\varpi t)\quad dx\\ \\ the\quad fourier\quad conjugate\quad of\quad \Psi \quad will\quad be\quad spread\quad out\quad because\quad \Psi \quad is\quad a\quad localized\quad waveform\quad to\quad make\quad it\quad simpler,lets\quad look\quad at\quad \Psi (x)\\ \\ \Psi (x)=\F [\xi (\eta )]\\ \\ if\quad we\quad think\quad of\quad \xi \quad as\quad a\quad probability\quad distribution,\quad lets\quad make\quad \xi \quad equal\quad to\quad the\quad normal\quad distribution\\ \\ \\ { \Lambda }_{ n }={ \mu }_{ n }{ e }^{ -(n-{ n }_{ 0 })^{ 2 }/2{ { \delta }_{ n } }^{ 2 } }\quad \Rightarrow \xi (n)=\sqrt { { \Lambda }_{ n } } \\ \\ \int _{ R }^{ }{ { \Lambda }_{ n }\quad dn } =\int _{ R }^{ }{ \xi (n)\quad dn } =1\quad \quad (the\quad particle\quad has\quad to\quad be\quad somewhere)\\ \\ \\ { \Lambda }_{ n }\quad is\quad the\quad envelope\quad of\quad \left| \Psi \right| *\left| \Psi \right| ,\quad \sqrt { { \Lambda }_{ n } } is\quad the\quad envelope\quad of\quad \Psi \\ \\ \\ we\quad can\quad find\quad the\quad coefficient\quad \mu \quad by\quad normalizing\quad \Lambda \quad \quad \\ \\ the\quad form\quad for\quad a\quad normalized\quad coefficient\quad is\quad \\ \\ { \mu }_{ n }=\frac { 1 }{ { \sigma }_{ n }\sqrt { 2\pi } } \\ \\ since\quad \xi \quad and\quad \Psi \quad are\quad fourier\quad conjugates\\ \\ \sqrt { { \mu }_{ n } } { e }^{ -(n-{ n }_{ 0 })^{ 2 }/(4{ { \sigma }^{ 2 } }_{ n }) }=\int _{ R }^{ }{ \sqrt { { \mu }_{ x } } } { e }^{ -(x-{ x }_{ 0 })^{ 2 }/4{ { \sigma }_{ x } }^{ 2 } }cos\quad nx\quad dx,\quad if\quad we\quad let\quad { x }_{ 0 }=0\\ \\ \\ \\ \\ RHS\Rightarrow 2\int _{ 0 }^{ \infty }{ { e }^{ -{ (ax) }^{ 2 } } } cos\quad \gamma x\quad dx=\frac { \sqrt { \pi } }{ 2a } { e }^{ -({ \frac { b }{ 2a } ) }^{ 2 } }\\ \\ so\quad we\quad conclude\quad that\quad a\quad =\frac { 1 }{ { \sigma }_{ k } } \quad and\quad n=\gamma \\ \\ substituting\quad the\quad new\quad values\quad we\quad have\quad { \mu }_{ n }=4\pi { { \sigma }^{ 2 } }_{ x }\quad \quad ,\quad cancelling\quad them\quad out\quad and\quad taking\quad the\quad ln\quad of\quad each\quad side\quad we\quad get\\ \\ { -(\frac { k }{ 2{ \sigma }_{ k } } ) }^{ 2 }={ -(k{ \sigma }_{ x }) }^{ 2 }\quad which\quad simplifies\quad to\quad \\ \\ { \sigma }_{ k }\quad =\frac { 1 }{ 2{ \sigma }_{ x } } \\ \\ \\ \\ which\quad is\quad { \sigma }_{ k }{ \sigma }_{ x }=\frac { 1 }{ 2 } \\ \quad k=\frac { p }{ \hbar } \\ \\ \Delta x\Delta p=\frac { \hbar }{ 2 } \\ \\ but\quad this\quad is\quad for\quad normal\quad distributions\quad only\\ \\ if\quad it`s\quad not\quad a\quad normal\quad distribution\quad it`s\quad more\quad than\quad h/4\pi \\ \\ \\ Final\quad Result\\ \\ \Delta x\Delta p\ge \frac { \hbar }{ 2 } \\ \\ \\ \\ \\ \\

Hamza A - 5 years, 4 months ago

Log in to reply

so do you want the latex code?

Hamza A - 5 years, 4 months ago

Log in to reply

@Hamza A – Just hold on for a while.

Edit: Okay, it looks essentially correct, I don't see any outright mistakes in it, but it certainly needs to be cleaned up, and I think a short glossary or nomenclature of those Greek letters would be in order to make it an easier read.

Werner Heisenberg did not derive his famous uncertainty principle by this route, because he was grappling with the whole idea about just how reliable, concrete, or repeatable measurements could be made, i.e., limitations of physical measurements, before quantum wave mechanics was actually developed. His ideas not only lead to the development of quantum mechanics, the widespread applicability of his famous uncertainty principle is such that numerous other ways have been found to "prove" it, including the one you've posted here. Here's his original work:

Heisenberg's Original Derivation of the Uncertainty Principle

The proof you've posted here could be roughly summarized as follows: "A wide frequency spectrum transforms into a narrow wavelength spectrum, and vice-versa. The product of the half-widths of both cannot be less than a certain finite quantity". This is a well-known property in Fourier Transforms, so this is a mathematical fact that doesn't have to have a physical explanation for it. It's the other way around---the mathematical reality forces physical reality to follow.

As an aside, it does bring into mind the whole industry of "long baseline astronomical interferometry", which is where the use of several telescopes in different locations, particularly radio telescopes, have their data combined to improve their angular resolution, in the same way having a larger telescope aperture optically improves angular resolution (otherwise why keep building bigger and bigger telescopes?). The interesting point being raised here is that the wave interference formation the image is well understood for optical telescopes, but much harder to explain how that works for radio astronomical telescopes. The former uses a wave model, while the latter uses a statistical model. Yet, both obey similar laws and the same Heisenberg Uncertainty Principle! This is being pointed out because his principle holds true beyond just "wave physics". A deeper underlying physics is involved here--classical mechanics or electrodynamics offers no good explanation for this.

Michael Mendrin - 5 years, 4 months ago

Nice :) the derivation I have is the one for matrix

Nicole Tay - 5 years, 3 months ago

Log in to reply

thanks!:)

Hamza A - 5 years, 3 months ago
×

Problem Loading...

Note Loading...

Set Loading...