Levene's test: Difference between revisions

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In [[quantum mechanics]], the '''position operator''' is the [[operator (physics)|operator]] that corresponds to the position [[observable]] of a [[particle (physics)|particle]]. The [[eigenvalue]] of the operator is the [[position vector]] of the particle.<ref>{{cite book |title=Quanta: A handbook of concepts|first1=P.W.  |last1=Atkins|publisher=Oxford University Press|year=1974|isbn=0-19-855493-1}}</ref>
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==Introduction==
 
In one dimension, the wave function <math> \psi </math> represents the [[probability amplitude|probability density]] of finding the particle at position <math> x </math>.  Hence the [[expected value]] of a measurement of the position of the particle is
 
:<math> \langle x \rangle = \int_{-\infty}^{+\infty} x |\psi|^2 dx = \int_{-\infty}^{+\infty} \psi^* x \psi dx </math>
 
Accordingly, the quantum mechanical [[Operator (physics)#Operators_in_quantum_mechanics | operator]] corresponding to position is <math> \hat{x} </math>, where
 
:<math> (\hat{x} \psi)(x) = x\psi(x) </math>
 
==Eigenstates==
 
The eigenfunctions of the position operator, represented in position basis, are [[dirac delta functions]].
 
To show this, suppose <math> \psi </math> is an eigenstate of the position operator with eigenvalue <math> x_0 </math>. We write the eigenvalue equation in position coordinates,
 
:<math> \hat{x}\psi(x) = x  \psi(x) = x_0  \psi(x) </math>
 
recalling that <math> \hat{x} </math> simply multiplies the function by <math> x </math> in position representation. Since <math> x </math> is a variable while <math> x_0 </math> is a constant, <math> \psi </math> must be zero everywhere except at <math> x = x_0 </math>. The normalized solution to this is
 
:<math> \psi(x) = \delta(x - x_0) </math>
 
Although such a state is physically unrealizable and, strictly speaking, not a function, it can be thought of as an "ideal state" whose position is known exactly (any measurement of the position always returns the eigenvalue <math> x_0 </math>). Hence, by the [[uncertainty principle]], nothing is known about the momentum of such a state.
 
==Three dimensions==
 
The generalisation to three dimensions is straightforward. The wavefunction is now <math> \psi(\bold{r},t) </math> and the expectation value of the position is
 
:<math> \langle \bold{r} \rangle = \int \bold{r} |\psi|^2 d^3  \bold{r} </math>
 
where the integral is taken over all space. The position operator is
 
:<math>\bold{\hat{r}}\psi=\bold{r}\psi</math>
 
==Momentum space==
 
In [[momentum space]], the position operator in one dimension is
 
:<math> \hat{x} = i\hbar\frac{d}{dp} </math>
 
==Formalism==
 
Consider, for example, the case of a [[spin (physics)|spin]]less particle moving in one spatial dimension (i.e. in a line). The [[state space (physics)|state space]] for such a particle is [[Lp space|''L''<sup>2</sup>('''R''')]], the [[Hilbert space]] of [[complex number|complex-valued]] and [[Square-integrable function|square-integrable]] (with respect to the [[Lebesgue measure]]) [[Function (mathematics)|function]]s on the [[real line]]. The position operator, ''Q'', is then defined by:<ref>{{cite book |title=Quantum Mechanics Demystified|first1=D. |last1=McMahon|edition=2nd|publisher=Mc Graw Hill|year=2006|isbn=0 07 145546 9}}</ref><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=978-0071623582}}</ref>
 
:<math> Q (\psi)(x) = x  \psi (x) </math>
 
with domain
 
:<math>D(Q) = \{ \psi \in L^2({\mathbf R}) \,|\, Q \psi \in L^2({\mathbf R}) \}.</math>
 
Since all [[continuous function]]s with [[compact support]] lie in ''D(Q)'', ''Q'' is [[densely-defined operator|densely defined]]. ''Q'', being simply multiplication by ''x'', is a [[self adjoint operator]], thus satisfying the requirement of a quantum mechanical observable. Immediately from the definition we can deduce that the [[spectrum of an operator|spectrum]] consists of the entire [[real line]] and that ''Q'' has purely [[continuous spectrum]], therefore no discrete [[eigenvalues]]. The three dimensional case is defined analogously. We shall keep the one-dimensional assumption in the following discussion.
 
==Measurement==
As with any quantum mechanical [[observable]], in order to discuss [[measurement]], we need to calculate the spectral resolution of ''Q'':
 
:<math> Q = \int \lambda d \Omega_Q(\lambda).</math>
 
Since ''Q'' is just multiplication by ''x'', its spectral resolution is simple. For a [[Borel subset]] ''B'' of the real line, let <math>\chi _B</math> denote the [[indicator function]] of ''B''. We see that the [[projection-valued measure]] Ω<sub>''Q''</sub> is given by
 
:<math> \Omega_Q(B) \psi = \chi _B  \psi ,</math>
 
i.e. Ω<sub>''Q''</sub> is multiplication by the indicator function of ''B''. Therefore, if the [[system (physics)|system]] is prepared in state ''ψ'', then the [[probability]] of the measured position of the particle being in a [[Borel set]] ''B'' is
 
:<math> |\Omega_Q(B) \psi |^2  = | \chi _B  \psi |^2    = \int _B |\psi|^2  d \mu  ,</math>
 
where ''μ'' is the Lebesgue measure. After the measurement, the wave function [[wave function collapse|collapses]] to either
 
<math> \frac{\Omega_Q(B) \psi}{ \|\Omega_Q(B) \psi \|} </math>
 
or
 
<math> \frac{(1-\chi _B)  \psi}{ \|(1-\chi _B)  \psi \|} </math>, where <math>\|  \cdots \|</math> is the Hilbert space norm on ''L''<sup>2</sup>('''R''').
 
==See also==
 
* [[Position and momentum space]]
* [[Momentum operator]]
 
==References==
 
{{reflist}}
 
{{Physics operator}}
 
{{DEFAULTSORT:Position Operator}}
[[Category:Quantum mechanics]]

Latest revision as of 02:44, 17 July 2014

The author is known by the title of Numbers Lint. My working day occupation is a meter reader. He is truly fond of doing ceramics but he is having difficulties to discover time for it. North Dakota is exactly where me and my husband reside.

My webpage :: http://www.reachoutsociety.info/