The surprisingly simple linear potential solution

In summary, the linear potential in quantum mechanics can be solved using Airy functions, but it is not an easy or commonly discussed solution. An alternative solution involves a superposition of energy eigenstates with an arbitrary phase factor, but this is not a particularly useful wave function. The linear potential can also be solved using an accelerating plane wave, but this solution has a problem with normalization. However, the solution in terms of Airy functions is an exact solution and is much simpler. The solution is an eigenstate of momentum and satisfies the normalization condition.
  • #1
pellman
684
5
Once upon a time I was taking a quantum class and I asked the instructor why we did not cover the linear potential. What happened to F=ma in quantum mechanics? He said it was quite non-trivial and had to be done with Airy functions; not suitable for an introductory class.

Ok.

Given the Schrodinger equation

[tex]\{ -\frac{1}{2m}\frac{\partial^2}{\partial x^2}-Fx\}\Psi=i \frac{\partial\Psi}{\partial t} [/tex]

how about

[tex]\Psi\propto exp\{i[(k+Ft)x-\frac{k^2 t}{2m}-\frac{kFt^2}{2m}-\frac{F^2t^3}{6m}]\}[/tex]

where k is arbitrary?

Why isn't this well known? I spent some time digging through quantum texts in a science library and found only the barest mention of it, just a quick reference in a chapter exercise.

Seems like since it is so very simple it should get mentioned. I presume that it is because it is uninteresting. But why?
 
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  • #2
what is the normalization constant?
 
  • #3
The linear potential is easy to solve in momentum space, where the time-indepdent Schrodinger equation is (with [itex]\hbar=1[/itex])

[tex]{p^2\over 2m}\widetilde\psi_E(p) - iF{d\over dp}\widetilde\psi_E(p) = E\widetilde\psi_E(p)[/tex]

The (unnormalizable) solution is

[tex]\widetilde\psi_E(p) = \exp\!\left[{i\over F}\left(Ep-{p^3\over 6m\right)\right][/tex]

To get to position space, we have to Fourier transform from [itex]p[/itex] to [itex]x[/itex], and this yields an Airy function.

However, the general solution of the time-dependent Schrodinger equation (in momentum space) is given by

[tex]\widetilde\psi(p,t)=\int_{-\infty}^{+\infty}dE\;c(E)\,e^{-iEt}\;\widetilde\psi_E(p) [/tex]

where the coefficient [itex]c(E)[/itex] is an arbitary function of [itex]E[/itex]. If we choose [itex]c(E)=\exp(-ikE/F)[/itex], where [itex]k[/itex] is arbitrary, we get

[tex]\widetilde\psi(p,t)\propto \delta(p-k-Ft)\exp(-ip^3/6mF) [/tex]

The time-dependent wave function in position space is then

[tex]\psi(x,t) = {1\over\sqrt{2\pi}}\int_{-\infty}^{+\infty}dp\;e^{ipx}\;\widetilde\psi(p,t)[/tex]

which works out to be

[tex]\psi(x,t) \propto\exp[i(k+Ft)x-i(k+Ft)^3/6mF][/tex]

This is your solution with an extra constant phase factor of [itex]\exp(-ik^3/6mF)[/itex]. It is nice that it can be expressed exactly, but physically it corresponds to a superposition of energy eigenstates with equal probability for every energy (and a particular energy-dependent phase factor). This is not a particularly useful wave function to know.
 
  • #4
CPL.Luke said:
what is the normalization constant?

In 1-D, a normalization constant of [tex]\frac{1}{\sqrt{2\pi\hbar}}[/tex] gives [tex]\langle\Psi_k|\Psi_{k'}\rangle=\delta(k-k')[/tex] same as for free momentum states. Notice that this solution reduces to the plane wave for F=0.
 
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  • #5
Avodyne said:
It is nice that it can be expressed exactly, but physically it corresponds to a superposition of energy eigenstates with equal probability for every energy (and a particular energy-dependent phase factor). This is not a particularly useful wave function to know.

I expect you're right. I myself have no sense of what is useful or not.

However, you can also look at this solution as an accelerating plane wave. At any particular time t its shape is a pure plane wave corresponding to momentum k+Ft, that is, a plane wave whose wavelength shortens with time. So if you are dealing with a system in which momentum eigenstates are more important than energy eigenstates, wouldn't you prefer this solution?
 
  • #6
The solution for the linear potential is an Airy function. The details are found in many tests on solid state optics.
 
  • #7
Dr Transport said:
The solution for the linear potential is an Airy function. The details are found in many tests on solid state optics.

Yes, that is true. The solutions in terms of Airy functions are the energy eigenstates.

The expression in the OP is also an exact solution ... but one that is much simpler.
 
  • #8
I'm sorry pellman I was trying to point out that your solution has a problem with normalisation.

for one the nomalization depends on time, which means that a normalised solution would no longer be a solution to the schrodinger equation.


although your response puzzles me, are you trying to solve the eigenvalue problem for the linear potential (which are the airy functions) or are you trying to write the full solution. if its the latter that would mean your solution satisfies <Y||Y>=1
 
  • #9
Check again. Unless I'm mistaken, if

[tex]\Psi_k(x,t) = \frac{1}{\sqrt{2\pi}} exp{\{i[(k+Ft)x-\frac{k^2 t}{2m}-\frac{kFt^2}{2m}-\frac{F^2t^3}{6m}]\}}[/tex]

then

[tex]\int{\Psi_{k'}^*(x,t)\Psi_k(x,t)dx}=\delta(k-k')[/tex]

The solution is an eigenstate of momentum. The eigenvalue k+Ft depends on time, true, but since time is a parameter here and not a dynamical variable, I don't think that is a problem.
 
  • #10
my mistake, you are correct
 

Related to The surprisingly simple linear potential solution

1. What is the "surprisingly simple linear potential solution"?

The "surprisingly simple linear potential solution" is a mathematical model used to describe the behavior of particles in a one-dimensional potential field. It assumes that the force acting on a particle is proportional to its position, resulting in a linear potential function.

2. How is this solution different from other potential solutions?

This solution is unique in its simplicity. Many other potential solutions involve complex mathematical equations and require advanced knowledge to understand. The linear potential solution, on the other hand, is relatively easy to understand and apply.

3. What are the applications of the linear potential solution?

The linear potential solution has a wide range of applications in physics and engineering. It can be used to model the motion of particles in a variety of systems, including simple harmonic oscillators, electric fields, and gravitational fields. It is also used in the design of control systems and in the analysis of mechanical structures.

4. Is the linear potential solution an accurate representation of real-world systems?

While the linear potential solution is a simplified model, it can still provide valuable insights into the behavior of real-world systems. In many cases, it can accurately predict the motion and behavior of particles in a potential field. However, in more complex systems, additional factors may need to be considered for a more accurate representation.

5. How is the linear potential solution derived?

The linear potential solution is derived using the principles of classical mechanics and the concept of potential energy. By equating the force on a particle to the negative gradient of its potential energy, the resulting equation is a linear function of position. This solution can then be applied to a variety of systems to analyze their behavior.

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