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cragar

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- Thread starter cragar
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- #1

cragar

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- #2

mfb

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Nonrelativistic quantum mechanics will give wrong results when relativistic velocities are involved, but that should not be surprising.

- #3

zje2009

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The answer is a little complex.The theory of relativity and the quantum mechanics are always being used in the same situations.Relativity effects are obvious when we observe the cosmic ray or the particle in particle accelerators,while quantum mechanics principles,like uncertainty principle and the principle of superposition of states are also effective in those situations.What I really mean is relativity and quantum principle(not include Schrodinger equation) describe different sides of one object.There's no connection and constradiction between them.

- #4

Bill_K

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How, exactly?Relativistic quantum mechanics and quantum field theory take care about that.

- #5

mfb

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Sorry, I do not understand what exactly you are asking.

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Bill_K

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- #7

radioactive8

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- #8

TrickyDicky

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As opposed to x being...? Please clarify.Although this may be related to the issue of whether x is really the position operator.

- #9

Bill_K

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But it turns out to be impossible, using only positive energy modes, to construct a state which is entirely localized at a point (δ-function). Also, the operator in momentum space that one would normally call the position, i

Newton and Wigner defined a different position operator

- #10

mfb

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Which paradox do you mean?

- #11

TrickyDicky

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Hmmm...this alternative position operator

But it turns out to be impossible, using only positive energy modes, to construct a state which is entirely localized at a point (δ-function). Also, the operator in momentum space that one would normally call the position, i∇_{p}, is not even Hermitian. So for both these reasons, to answer the pertinent question, "what is the probability of finding the particle at a particular point", you can't usexto specify the point.

Newton and Wigner defined a different position operatorXto be the Hermitian part of i∇_{p}, namelyX= i∇_{p}-ip/(p^{2}+ (mc)^{2}).Xobeys the right commutation relations, [X, p] = iħ. And its eigenstates are little Hankel-function-shaped things about a Compton wavelength across.

- #12

Bill_K

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I know it's backwards from the usual situation where p is represented by a differential operator and x is not. But it works either way, and in quantum field theory the momentum representation is used all the time and for many purposes is a lot simpler.

For the standard

For the Newton-Wigner

- #13

zje2009

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x= iħ∇_{p}!

I know it's backwards from the usual situation where p is represented by a differential operator and x is not. But it works either way, and in quantum field theory the momentum representation is used all the time and for many purposes is a lot simpler.

For the standardx, the eigenvalue equation is iħ∇_{p}φ =xφ, with solution φ(_{0}p) = e^{-(i/ħ)p·x0}. To get the corresponding eigenfunction ψ(x) in coordinate space, Fourier transform it. The Fourier transform of a plane wave of course is δ(x-x), a wavefunction completely localized at_{0}x._{0}

For the Newton-WignerX, the same procedure leads to a φ(p) which is just slightly different: φ(p)= √p_{0}e^{-(i/ħ)p·x0}. But the Fourier transform of this back toxspace is no longer a delta function, it's slightly spread out.

I think you probably misunderstand the host's question.Your answers may be a little stray from the point and trap in complexity.The host has doubt that whether uncertainty principle should be modified in relativistic situation.The answer is absolutely no.

- #14

cragar

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- #15

zje2009

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You needn't.If you moving fast and watch an object,you would still find its velocity(or say momentum) and location satisfy the uncertainty principle.Because generally there's no exact value of velocity as well as location to describe the particle,the relativistic transformations are powerlessness.In some case,we regard the quantal particle as classic one,and relativistic transformations work.For example, the particles in the accelerator.

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- #16

cragar

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- #17

zje2009

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in that frame it couldn't radiate it doesn't have any energy to lose at least classically. If In another frame it does radiate, this radiation would affect its momentum. how would HUP take care of that.

I think your argument confuse something.You follow the particle and don't see any radiation,sound like different from the result in another frame.But these different phenomenons

are still unconcerned with HUP.The particle stays in the ground state and has no extra kinetic energy,but it doesn't matter.Because the radiation comes from the potential energy.We already understood this business in classic views.

- #18

radioactive8

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i mean that in the two slit experiment .. the electrons first behave like a wave .. but when scientist put an detector to see form which slit each electrons go (or both) the electron behaved like a particle ...Which paradox do you mean?

how did they explain this... ?

- #19

mfb

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With wave functions of the particle, and one of the many interpretations of quantum physics.i mean that in the two slit experiment .. the electrons first behave like a wave .. but when scientist put an detector to see form which slit each electrons go (or both) the electron behaved like a particle ...

how did they explain this... ?

- the wave function collapses when a measurement is done

- you get multiple worlds with different measurement results

- you have particles with a certain position and velocity, but their movement is given by wave functions

- some other interpretations

This is not really related to the uncertainty principle.

- #20

cragar

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[itex] \delta x [/itex] is small so the momentum becomes uncertain and starts to spread out in all directions.

- #21

radioactive8

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let me explain

i thought that electrons behave like waves when they don't have any mass..

e=mc

my explanation is this ... the only way for scientist to measure them is by trying to make them impact with other particles so they can do the math ... but when the two waves (i asuume that form the detector the other particle don't have enough enrgy to peform like a particle) meet each other there is enough energy for the electrons to behave like a particle ...

- #22

mfb

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While mass influences the wavelength<->energy relation, electrons can behave like waves with mass, too. You can perform double-slit experiments with massive electrons (in fact, all electrons are massive).i thought that electrons behave like waves when they don't have any mass..

If you want to detect electrons, they (usually) behave like particles, if you want to calculate their undetected movement, they (usually) behave like waves. It really depends on your setup which description is appropriate where.

- #23

radioactive8

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aaa ok thanks ...

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