Help with Schoedinger´s equation when V(x)=0

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In summary, Demystifier has confirmed that the function ##\Psi(x,t) = e^{i(x - t)/\hbar} ## is a solution of the Schroedinger's equation. Surprisingly the phase velocity of this wave function is not ##2\, m/s## but ##1\, m/s##.
  • #1
DaTario
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Hi All,

I have tried to solve Schroedinger´s equation for the case where ##V(x) = 0## for a particle with ##m=0.5## and ##v_0 = 2##. This yields ##E = 1## and ##p=1##.
From the fundamental relations of De Broglie and Planck, ##k = 1/ \hbar ## and ##\omega = 1/\hbar ##.
Thus, I have confirmed that the function ##\Psi(x,t) = e^{i(x - t)/\hbar} ## is a solution of the Schroedinger's equation. Surprisingly the phase velocity of this wave function is not ##2\, m/s## but ##1\, m/s##.
What have gone wrong ?

Best wishes,
DaTario
 
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  • #2
Nothing is wrong. For waves one can introduce two different velocities, namely phase velocity
$$v_p=\frac{\omega}{k}$$
and group velocity
$$v_g=\frac{\partial\omega}{\partial k}$$
For ##V(x)=0## one has ##\omega=\hbar k^2/2m##, so in your case ##v_g=2## and ##v_p=1##. To check this, be careful to compute ##\partial\omega/\partial k## before you put in a specific value of ##k##.
 
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  • #3
Ok. First, thank you. But two comments follows:

1) would it be ## \omega = \hbar^2 k^2/2m ## ?

2) I had some difficulty in thinking of group velocity having only one "Fourier" component. How this works with only one frequency present?
 
  • #4
DaTario said:
1) would it be ## \omega = \hbar^2 k^2/2m ## ?
No, because ## E = \hbar^2 k^2/2m ## and ##\omega=E/\hbar##.

DaTario said:
2) I had some difficulty in thinking of group velocity having only one "Fourier" component. How this works with only one frequency present?
I'm not sure what confuses you, but perhaps your question is answered by the fact that here we consider only one space dimension with coordinate ##x##.
 
  • #5
Ok, sorry with 1).

But wrt 2), this situation still resembles that one where we have a wave function (from the wave equation context) like ##\sin(kx - \omega t)##. Do we have here (in the context of a wave on a string) also a group velocity? It is equal to the phase velocity in this case, as it seems?
Is it correct to say that ##ω=ℏk^2/2m## is the dispersion relation and it is different from the one which is present in solving the wave equation (##\omega = k v##)?
 
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  • #6
DaTario said:
But wrt 2), this situation still resembles that one where we have a wave function (from the wave equation context) like ##\sin(kx - \omega t)##. Do we have here (in the context of a wave on a string) also a group velocity?
Of course.

DaTario said:
It is equal to the phase velocity in this case, as it seems?
For the simplest classical string dispersion relation is of the form ##\omega=vk##, so group velocity is equal to the phase velocity.

DaTario said:
Is it correct to say that ##ω=ℏk^2/2m## is the dispersion relation and it is different from the one which is present in solving the wave equation (##\omega = k v##)?
Yes.
 
  • #7
Note also that the group velocity formula
$$v_g=\frac{\partial\omega}{\partial k}$$
is closely related to the formula for velocity of a particle in classical Hamiltonian mechanics
$$\dot{x}=\frac{\partial H}{\partial p}$$
Hence, loosely speaking, it can be said that group velocity describes "particle-like" properties, while phase velocity describes "wave-like" properties. (Heuristically this makes sense in the context of quantum mechanics, but should not be taken too literally.)
 
  • #8
Demystifier said:
Note also that the group velocity formula
$$v_g=\frac{\partial\omega}{\partial k}$$
is closely related to the formula for velocity of a particle in classical Hamiltonian mechanics
$$\dot{x}=\frac{\partial H}{\partial p}$$
Hence, loosely speaking, it can be said that group velocity describes "particle-like" properties, while phase velocity describes "wave-like" properties. (Heuristically this makes sense in the context of quantum mechanics, but should not be taken too literally.)

Beautiful connection, in deed.

Thank you very much, Demystifier. I think now the entropy of the concepts inside my brain has decreased considerably.
 
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  • #9
DaTario said:
I think now the entropy of the concepts inside my brain has decreased considerably.
Which, by the second law of thermodynamics, considerably increased entropy in your brain's environment. :biggrin:

For further decrease of your brain's entropy you might be interested in my paper https://arxiv.org/abs/quant-ph/0609163
 
  • #10
If the thermodynamical consequence is only heat produced in the surroundings of my brain, attending to a nice explanation session, like this was, in a cold day may represent winning in both sides.

Thank you for sending me the link of your paper.
Best wishes.
 
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1. What is the Schoedinger's equation?

The Schoedinger's equation, also known as the quantum mechanical wave equation, is a mathematical equation that describes the behavior of a quantum mechanical system.

2. What does it mean when V(x)=0 in the Schoedinger's equation?

When V(x)=0, it means that the potential energy of the system is equal to zero. This could represent a free particle or a system in a region with no potential energy barriers or wells.

3. How does V(x)=0 affect the solutions of the Schoedinger's equation?

When V(x)=0, the solutions of the Schoedinger's equation will depend solely on the kinetic energy of the particle. This results in a simpler form of the equation and can be used to solve for the energy eigenvalues and eigenfunctions of the system.

4. Can the Schoedinger's equation be solved analytically when V(x)=0?

Yes, the Schoedinger's equation can be solved analytically when V(x)=0. This is because the potential energy term is simplified to zero, making the equation easier to solve.

5. What is the physical significance of V(x)=0 in the Schoedinger's equation?

V(x)=0 represents a system with no potential energy. This could correspond to a free particle or a system in a region with no potential energy barriers or wells. It allows for the simplification of the Schoedinger's equation and can provide insight into the behavior of the system.

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