Question on wave function of free particle

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The discussion centers on the nature of the wave function for a free particle as described by the Schrödinger equation. It highlights the confusion surrounding the interpretation of a single plane wave versus a wave packet, emphasizing that while a single wave can be derived mathematically, it does not accurately represent a particle's behavior due to discrepancies between phase and group velocities. The correct representation of a particle is a wave packet formed by superposing multiple waves, which allows for a more accurate description of its properties. The conversation critiques the reliance on outdated theories and stresses the importance of understanding the probabilistic nature of quantum mechanics. Ultimately, a single plane wave can serve as an approximation under specific conditions, but it does not encompass the full complexity of a particle's wave function.
Shan K
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Hi,
I was studying the solution of Schrodinger equation with no potential and found that the wave function is just a single plane wave eikx for movement of the particle in positive x direction.
But when the phase velocity of a single wave is calculated it turns out to be less than the particle's velocity and from that argument is made that the de broglie wave, which is associated with a particle, is not a single wave but superposition of waves.
So my question is that how one single wave (which we have found by solving TISE for no potential) can describe a particle, when from the calculation of phase velocity and group velocity we know that a particle can be described by a group of waves?
Any kind of help will be highly appreciated.Thanks in advance
 
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Shan K said:
I was studying the solution of Schrodinger equation with no potential

Why do you want to analyse Schroedinger's equation in terms of a theory that was consigned to the dustbin when QM was fully developed?

You are caught up in one of the myths of QM - the wave particle duality - its neither wave or particle:
http://arxiv.org/pdf/quant-ph/0609163.pdf

One reason is taking the wave idea seriously you end up with the kind of issues you raise.

Thanks
Bill
 
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bhobba said:
Why do you want to analyse Schroedinger's equation in terms of a theory that was consigned to the dustbin when QM was fully developed?

You are caught up in one of the myths of QM - the wave particle duality - its neither wave or particle:
http://arxiv.org/pdf/quant-ph/0609163.pdf

One reason is taking the wave idea seriously you end up with the kind of issues you raise.

Thanks
Bill
But in the book by Resnick and Eisberg they have come to the wave function by considering the de broglie wave.In one hand they have said that a de Broglie wave cannot be a single wave because the phase and the particle velocity does not match and on the other hand they have said that the de Broglie wave is given by this wave function. That is what raising the difficulty for me to conceptualize .
 
Shan K said:
But in the book by Resnick and Eisberg they have come to the wave function by considering the de broglie wave.In one hand they have said that a de Broglie wave cannot be a single wave because the phase and the particle velocity does not match and on the other hand they have said that the de Broglie wave is given by this wave function. That is what raising the difficulty for me to conceptualize .

Then don't. The De-Broglie hypotheses is wrong as can be seen by what happens when you jump to a frame where the particle is at rest - what's its wavelength then?

The correct derivation of Schroedinger's equation is based on the invariance of probabilities between inertial frames - see chapter 3 Ballentine.

Its often done in beginning texts - they try to justify Schroedinger's equation by hand-wavey arguments something along the lines of this:


Its done so they don't have to say its pulled out of thin air - which is basically what Schroedinger did:
http://arxiv.org/pdf/1204.0653.pdf

The trouble is the correct derivation is complex - you probably know the math to follow it - but its a bit involved with each step slow going.

Thanks
Bill
 
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Shan K said:
So my question is that how one single wave (which we have found by solving TISE for no potential) can describe a particle, when from the calculation of phase velocity and group velocity we know that a particle can be described by a group of waves?

The correct description of a free particle in Schrödinger wave mechanics is a wave packet constructed by superposing waves ##Ae^{i(kx-\omega t)}## with various values of ##k = p/\hbar## and various amplitudes: $$\Psi(x,t) = \int_{-\infty}^{+\infty} {A(k) e^{i(kx-\omega t)} dk}$$ If the packet contains a very narrow range of momenta, i.e. the function A(k) is zero or almost zero almost everywhere, and very sharply peaked at some central value ##k_0##, then the packet has a large spread in position. In the central region of the packet, the single plane wave ##Ce^{i(k_0 x-\omega t)}## is a good approximation to the actual wave function. For some purposes we can represent the particle to a good approximation by using this plane wave.
 
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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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