Understanding Wavepacket Spreading in QM

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Wavepacket spreading in quantum mechanics occurs due to the uncertainty principle, which dictates that precise knowledge of a particle's position leads to uncertainty in its momentum, causing the wavepacket to widen over time. This spreading is observable from both stationary and moving frames, though the extent may vary, necessitating a relativistic approach for accurate analysis. The flattening of the wavepacket over time reflects growing uncertainties in both position and momentum measurements. Additionally, the momentum-space wavefunction is not constant; it changes as the uncertainties evolve. Understanding quantum mechanics involves recognizing it as a framework for describing knowledge rather than an objective reality.
Saketh
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I'm working through the first few chapters of my QM textbook, so I am not yet familiar with the Schrodinger equation.

Consider a free particle, say an electron, moving through free space. I have done the calculations, and concluded that the wavepacket must spread -- that is, get wider. However, this does not make sense to me. How can the wavepacket spread with time? That is, why doesn't the wavepacket just translate?

Then I realized that if the wavepacket just translated, we could determine the particle's momentum from its translation.
My questions:
  1. What does the spreading of the wavepacket with time represent?
  2. Would spreading appear to both an observer at rest and a non-relativistic observer in motion?
  3. Why does the "hump" in the packet flatten out as time goes on? (I know the mathematics, but I'm not sure what it represents.)
  4. Am I right in assuming that the momentum wave function \phi(k) is constant with respect to time?

Thanks for your assistance.
 
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1. If you measured the particle's position, then there was some resolution to your postion-measuring apparatus. Due to the uncertainty relation, you will have a minimum uncertainty in the particle's momentum, determined by the precise form of your initial wavefunction and the resolution of your device, and this uncertainty will translate to a spreading of your uncertainty in its location, as time progresses.
2. Spreading would appear to both, though the exact amount would differ, in some way. However, in order to treat this properly you will need a relativistic theory, such as a quantum field theory, not just bare non-relativistic quantum mechanics. The exact field theory will depend on the particle. If it is an electron, then that will be quantum electrodynamics...
3. The flattening out represents an uncertainty in both the variable that you measured and in the variable conjugate to it. These uncertainties will grow with time according to your initial measurement precision.
4. The momentum-space wavefunction is not constant. Based on what I said regarding the statistical spreading and the uncertainty principle, think about why this should be so.
 
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degrees of knowledge

My suggestion is to think about quantum mechanics as a theory that describes what you know and how well you can know it, rather than some reality that is "out there", independent of experiments.
 
So the more accurately you know position or momentum at the beginning, the less accurately you know it later on.

That makes sense...thanks for your explanation!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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