Momentum, Position and the Uncertainty Principle

8/10/2019 Momentum, Position and the Uncertainty Principle

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Momentum and the momentum operator

For momentumwe write an operator

We postulate this can be written as

with

where xo, y

o, and z

oare unit vectors

in the x, y, and z directions

ˆ p

ˆ p i

o o o x y z

x y z




With this postulated form we find that

and we have a correspondence between the

classical notion of the energy E

and the corresponding Hamiltonian operatoof the Schrödinger equation

ˆ p i 2 2

2ˆ2 2 p

m m

2

2 p

E V m

2 22 ˆˆ

2 2 p H V V

m m




Note that

This means the plane waves arethe eigenfunctions of the operator

with eigenvaluesWe can therefore say for these eigenstates that

the momentum isNote that p is a vector, with threecomponents with scalar values

not an operator

ˆ exp exp exp p i i i i k r k r k k r

exp i k rˆ p

k

p k



Position and the position operator

For the position operatorthe postulated operator is almost trivial

when we are working with functions ofpositionIt is simply the position vector, r , itself

At least when we are working in arepresentation that is in terms of position

we therefore typically do not writethough rigorously we shouldThe operator for the z-component of position

would, for example, also simply be z itself

r̂



The uncertainty principle

Here we illustrate the position-momentumuncertainty principle by example

We have looked at a Gaussian wavepacket beforWe could write this as a sum over waves of

different k-values, with Gaussian weightsor we could take the limit of that process by

using an integration

2

, exp exp2

Gk

k k z t i k t k

k




We could rewrite

at time t = 0 as

where

2

, exp exp2Gk

k k z t i k t kk

,0 expk k

z k ikz dk

2

exp2k

k k k

k




In

is the representation of the wavefunction in k

is the probability P k

strictly, the probability densitythat if we measured the momentum of the particle

actually the z component of momentumit would be found to have value

k k

2

k k

k

2

exp2

k

k k k

k




With

then this probability (density) of finding a valfor the momentum would be

This Gaussian corresponds to the statistical

Gaussian probability distributionwith standard deviation

2

exp2k

k k k

k

22

2exp2

k k

k k P k

k

k




Note also that

is the Fourier transform of and, as is well knownthe Fourier transform of a Gaussian is

a Gaussianspecifically here

,0 expk k

z k ikz dk

k k

2 2,0 exp z k z




If we want to rewrite

in the standard form

where the parameterwould now be the standard deviation

in the probability distribution for z

then

2 2 2,0 exp 2 z k z

22

2,0 exp 2

z z z

z

1/ 2k z




Fromif we now multiply by to get the standard deviati

we would measure in momentumwe have

which is the relation between the standarddeviations we would see in

measurements of position andmeasurements of momentum

1/ 2k z

2 p z




This relation

is as good as we can get for a Gaussian

For examplea Gaussian pulse will broaden in space as

it propagateseven though the range of k valuesremains the same

2 p z



It also turns out that the Gaussian shapeis the one with the minimum possible

product of andSo quite generally

which is the uncertainty principlefor position and momentum inone direction


2 p z

p z



The uncertainty principle in Fourier analysis

Uncertainty principles are well known in Fourier analyOne cannot simultaneously have both

a well defined frequency anda well defined time

If a signal is a short pulseit is necessarily made up out of a range of

frequencies

The shorter the pulse isthe larger the range of frequencies

12

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Momentum, Position and the Uncertainty Principle

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