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Springer eBooks

Free-Particle Wavepacket

James M. Feagin

Published Jan 1, 1994• 0 citations
Publisher
DOI
10.1007/978-1-4612-4328-1_4

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Evolution of a Gaussian Wavepacket

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Traditional quantum mechanics calculates a probability in space through W*(x)W(x) for a time-independent wavefunction W(x) = Sum over p a(p)exp(ip dot r). We argue that this probability is created at the interaction region and indicates a probability for outcomes. In particular, given a 2-slit problem, one calculates W(r) = exp( i p dot r1) + exp(i p dot r2), where r1 and r2 are vectors from the center of each slit to the same point on a screen far away. W(r)*W(r) then represents the probability for particles to arrive at various points on the screen. This may be realized by sending many particles/photons through the 2-slit apparatus, one at a time. Given that exp(ip dot r) is an interaction probability (we argue), its relevance holds in the interaction region which is presumably of about a wavelength long or so. We argue that exp(i p dot r) exp(-i p dot r) = 1 does not mean that one should interpret the particle/photon as being possibly present at any r in space. The particle still has a center-of-mass which follows x=vt for a free particle. The only issue is that exp(ipx) implies that the particle does not interact at its center-of-mass point necessarily, but can interact with probability exp(ipx) at different points within .5 hbar/p of the center of mass on either side. As a result, we argue that there is no issue concerning localization of either a free particle or photon. The object is localized and moving at x=vt and it has a precise p which may render impulse hits at points about the center-of-mass point in keeping with exp(ipx). As a result, Heisenberg’s uncertainty relation (standard deviation p) (standard deviation x) >= hbar/2 does not hold for exp(ipx) because standard deviation p =0. exp(ipx) applies to a free particle which is not interacting. When the particle interacts (which is the way it is measured), then exp(ipx) becomes Sum over p a(p)exp(i p dot r). In such a case, a standard deviation of p and x arise. The Heisenberg uncertainty principle holds for interactions which are the only way the particle can be measured and so applies to W(x) and not a single exp(ipx). Given the above ideas, we suggest there is no reason to create a free particle wavepacket. Such a wavepacket seems to be created in the literature (1) in order to localize a particle in space and to ensure that Heisenberg’s uncertainty relationship is enforced. We have argued that there is no reason for this relationship if the particle is not interacting. As for localization, the free particle is localized by its center-of-mass in x=vt. It is just that one is not measuring this point. Furthermore, the increase in a Gaussian wavepacket’s x standard deviation in time is unphysical. Given that there is no reason for a wavepacket, we then question the notion of the wavepacket as being associated with a Bohmian potential which creates the wavepacket so to speak, i.e. -1/2m (d/dx dd/x A(x)) / A(x), where W(x,t) = A(x) exp(i S(x,t) ).

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Time evolution of quantum systems is normally taught using the Schrödinger equation. Here, we use the Heisenberg approach to consider three problems that are exactly solvable in the Schrödinger picture when the initial wavepacket is Gaussian: the free particle, the simple harmonic oscillator and a free particle with a spin-orbit interaction. We show that complete solutions can be obtained for the free particle (with or without spin-orbit interaction). For the simple harmonic oscillator, however, the fact that the wavepacket remains a Gaussian as time progresses has to be inferred on the basis of the skewness and kurtosis as opposed to a study of all the relevant moments.

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WORLD SCIENTIFIC eBooks•Jun 1, 2011•0 citations
Traditional quantum mechanics calculates a probability in space through W*(x)W(x) for a time-independent wavefunction W(x) = Sum over p a(p)exp(ip dot r). We argue that this probability is created at the interaction region and indicates a probability for outcomes. In particular, given a 2-slit problem, one calculates W(r) = exp( i p dot r1) + exp(i p dot r2), where r1 and r2 are vectors from the center of each slit to the same point on a screen far away. W(r)*W(r) then represents the probability for particles to arrive at various points on the screen. This may be realized by sending many particles/photons through the 2-slit apparatus, one at a time. Given that exp(ip dot r) is an interaction probability (we argue), its relevance holds in the interaction region which is presumably of about a wavelength long or so. We argue that exp(i p dot r) exp(-i p dot r) = 1 does not mean that one should interpret the particle/photon as being possibly present at any r in space. The particle still has a center-of-mass which follows x=vt for a free particle. The only issue is that exp(ipx) implies that the particle does not interact at its center-of-mass point necessarily, but can interact with probability exp(ipx) at different points within .5 hbar/p of the center of mass on either side. As a result, we argue that there is no issue concerning localization of either a free particle or photon. The object is localized and moving at x=vt and it has a precise p which may render impulse hits at points about the center-of-mass point in keeping with exp(ipx). As a result, Heisenberg’s uncertainty relation (standard deviation p) (standard deviation x) >= hbar/2 does not hold for exp(ipx) because standard deviation p =0. exp(ipx) applies to a free particle which is not interacting. When the particle interacts (which is the way it is measured), then exp(ipx) becomes Sum over p a(p)exp(i p dot r). In such a case, a standard deviation of p and x arise. The Heisenberg uncertainty principle holds for interactions which are the only way the particle can be measured and so applies to W(x) and not a single exp(ipx). Given the above ideas, we suggest there is no reason to create a free particle wavepacket. Such a wavepacket seems to be created in the literature (1) in order to localize a particle in space and to ensure that Heisenberg’s uncertainty relationship is enforced. We have argued that there is no reason for this relationship if the particle is not interacting. As for localization, the free particle is localized by its center-of-mass in x=vt. It is just that one is not measuring this point. Furthermore, the increase in a Gaussian wavepacket’s x standard deviation in time is unphysical. Given that there is no reason for a wavepacket, we then question the notion of the wavepacket as being associated with a Bohmian potential which creates the wavepacket so to speak, i.e. -1/2m (d/dx dd/x A(x)) / A(x), where W(x,t) = A(x) exp(i S(x,t) ).
Time evolution of quantum systems is normally taught using the Schrödinger equation. Here, we use the Heisenberg approach to consider three problems that are exactly solvable in the Schrödinger picture when the initial wavepacket is Gaussian: the free particle, the simple harmonic oscillator and a free particle with a spin-orbit interaction. We show that complete solutions can be obtained for the free particle (with or without spin-orbit interaction). For the simple harmonic oscillator, however, the fact that the wavepacket remains a Gaussian as time progresses has to be inferred on the basis of the skewness and kurtosis as opposed to a study of all the relevant moments.