Position and momentum spaces and Momentum space: Difference between pages

Content deleted Content added

VisualWikitext

Inline

Revision as of 21:24, 16 August 2012

Redirect to:

Position and momentum space

@@ Line 1: / Line 1: @@
+#REDIRECT [[Position and momentum space]]
-{{mergefrom|position space|discuss=Wikipedia talk:WikiProject Physics#Position space, position vector, momentum space, reciprocal lattice...|date=August 2012}}
-In [[physics]], the '''momentum space''' or '''k-space''' associated with a [[particle]] is the set of all [[wavevector]]s:
-:<math>\bold{k} = (k_1, k_2, k_3)</math>
-Since '''k''' is a vector, the set of all '''k'''-vectors forms a [[vector space]], in this case the "'''k'''-space". <ref>{{cite book |title=Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles|first1=R. |last1=Eisberg|first2=R.|last2= Resnick|edition=2nd | publisher=John Wiley & Sons|year=1985|isbn=978-0-471-873730}}</ref>
-The terms "[[momentum]]" (which has the symbol '''p''' and is also a vector) and "wavevector" are used interchangeably due to the [[de Broglie's relation]]:
-:<math>\bold{p} = \hbar \bold{k}</math>
-meaning they are equivalent, only [[up to]] [[Proportionality (mathematics)|proportionality]] (this is not true in a crystal, see below). The '''k'''-vector has [[dimensional analysis|dimension]]s of reciprocal [[length]], so '''k''' is the [[frequency]] analogue of the [[position vector]] '''r'''.
-Representing a problem in terms of the momenta of the particles involved, rather than in terms of their positions, can simplify some problems in [[physics]].{{citation needed|date=August 2012}}
-==Relation to quantum mechanics==
-{{See|Momentum operator}}
-In [[quantum physics]], a particle is described by a [[quantum state]]. This quantum state can be represented as a [[Quantum superposition|superposition]] (i.e. a [[linear combination]] as a [[weighted sum]]) of [[basis (linear algebra)|basis]] states. In principle one is free to choose the set of basis states, as long as they [[linear span|span]] the space. If one chooses the [[eigenfunction]]s of the [[position operator]] as a set of basis functions, one speaks of a state as a [[wave function]] ψ('''r''') in position space (our ordinary notion of [[space]] in terms of [[length]]). The familiar [[Schrödinger equation]] in terms of the position '''r''' is an example of quantum mechanics in the position representation.<ref>{{cite book |title=Quantum Mechanics|first1=Y. |last1=Peleg|first2=R.|last2= Pnini|first3=E.|last3= Zaarur|first4=E.|last4= Hecht|edition=2nd|publisher=McGraw Hill|year=2010|isbn=978-0-071-623582-2}}</ref>
-By choosing the eigenfunctions of a different operator as a set of basis functions, one can arrive at a number of different representations of the same state. If one picks the eigenfunctions of the [[momentum operator]] as a set of basis functions, the resulting wave function φ('''k''') is said to be the wave function in momentum space.<ref>{{cite book |title=Quantum Mechanics|first1=Y. |last1=Peleg|first2=R.|last2= Pnini|first3=E.|last3= Zaarur|first4=E.|last4= Hecht|edition=2nd|publisher=McGraw Hill|year=2010|isbn=978-0-071-623582-2}}</ref>
-==Relation to frequency domain==
-The momentum representation of a wave function is very closely related to the [[Fourier transform]] and the concept of [[frequency domain]]. Since a quantum mechanical particle has a frequency proportional to the momentum (de Broglie's equation given above), describing the particle as a sum of its momentum components is equivalent to describing it as a sum of frequency components (i.e. a Fourier transform). <ref>{{cite book |title=Quantum Mechanics|first1=E.|last1= Abers|edition=2nd|publisher=Addison Wesley, Prentice Hall Inc|year=2004|isbn=978-0-131-461000}}</ref> This becomes clear when we ask ourselves how we can transform from one representation to another.
-===Functions and operators in position space===
-Suppose we have a three-dimensional [[wave function]] in position space ψ('''r'''), then we can write this functions as a weighted sum of orthogonal basis functions ψ<sub>''j''</sub>('''r'''):
-:<math>\psi(\bold{r})=\sum_j \phi_j \psi_j(\bold{r})</math>
-or, in the continuous case, as an [[integral]]
-:<math>\psi(\bold{r})=\int_{\bold{k}{\rm-space}} \phi(\bold{k}) \psi_{\bold{k}}(\bold{r}) {\rm d}^3\bold{k}</math>
-It is clear that if we specify the set of functions ψ<sub>''j''</sub>('''r'''), say as the set of eigenfunctions of the momentum operator, the function φ('''k''') holds all the information necessary to reconstruct ψ('''r''') and is therefore an alternative description for the state ψ.
-In quantum mechanics, the [[momentum operator]] is given by
-:<math>\bold{\hat p} = -i \hbar\frac{\partial}{\partial \bold{r}}</math>
-(see [[matrix calculus#Scope|matrix calculus]] for the denominator notation) with eigenfunctions
-:<math>\psi_{\bold{k}}(\bold{r})=\frac{1}{(\sqrt{2\pi})^3} e^{i \bold{k}\cdot\bold{r}}</math>
-and [[eigenvalues]] ''ħ'''''k'''. So
-:<math>\psi(\bold{r})=\frac{1}{(\sqrt{2\pi})^3} \int_{\bold{k}{\rm-space}} \phi(\bold{k}) e^{i \bold{k}\cdot\bold{r}} {\rm d}^3\bold{k} </math>
-and we see that the momentum representation is related to the position representation by a Fourier transform.<ref>{{cite book |author=R. Penrose| title=[[The Road to Reality]]| publisher= Vintage books| year=2007 | isbn=0-679-77631-1}}</ref>
-===Functions and operators in momentum space===
-Conversely, a three-dimensional wave function in momentum space φ('''k''') as a weighted sum of orthogonal basis functions φ<sub>''j''</sub>('''k'''):
-:<math>\phi(\bold{k})=\sum_j \psi_j \phi_j(\bold{k})</math>
-or as an integral:
-:<math>\phi(\bold{k})=\int_{\bold{r}{\rm-space}} \psi(\bold{r}) \phi_{\bold{r}}(\bold{k}) {\rm d}^3\bold{r}</math>
-the position operator is given by
-:<math>\bold{\hat r} = i \hbar\frac{\partial}{\partial \bold p} = i\frac{\partial}{\partial \bold{k}}</math>
-with eigenfunctions
-:<math>\phi_{\bold{r}}(\bold{k})=\frac{1}{(\sqrt{2\pi})^3} e^{-i \bold{k}\cdot\bold{r}}</math>
-and [[eigenvalues]] '''r'''. So a similar decomposition of φ('''k''') can be made in terms of the eigenfunctions of this operator, which turns out to be the inverse Fourier transform:<ref>{{cite book |author=R. Penrose| title=[[The Road to Reality]]| publisher= Vintage books| year=2007 | isbn=0-679-77631-1}}</ref>
-:<math>\phi(\bold{k})=\frac{1}{(\sqrt{2\pi})^3} \int_{\bold{r}{\rm-space}} \psi(\bold{r}) e^{-i \bold{k}\cdot\bold{r}} {\rm d}^3\bold{r} </math>
-==In crystals==
-{{main|Reciprocal lattice}}
-For an electron (or other particle) in a crystal, its value of '''k''' relates almost always to its [[crystal momentum]], not its normal momentum. (Therefore '''k''' and '''p''' are not simply proportional but play different roles. See [[k·p perturbation theory]] for an example.) Crystal momentum is like a [[Envelope (waves)|wave envelope]] that describes how the wave varies from one [[unit cell]] to the next, but does ''not'' give any information about how the wave varies within each unit cell.
-When '''k''' relates to crystal momentum instead of true momentum, the concept of '''k'''-space is still meaningful and extremely useful, but it differs in several ways from the non-crystal k-space discussed above. For example, in a crystal's '''k'''-space, there is an infinite set of points called the [[reciprocal lattice]] which are "equivalent" to '''k'''=0 (this is analogous to [[aliasing]]). Likewise, the "[[first Brillouin zone]]" is a finite volume of '''k'''-space, such that every possible '''k''' is "equivalent" to exactly one point in this region.
-For more details see [[reciprocal lattice]].
-==References==
-{{reflist}}
-{{DEFAULTSORT:Momentum Space}}
-[[Category:Particle physics]]
-[[Category:Quantum mechanics]]
-[[de:Impulsraum]]