Propagator for quantum linear potential

In summary, the conversation discusses a problem related to a quantum particle in free fall due to a constant force. The focus is on verifying the form of the position space propagator, which involves working in the momentum representation and using Fourier transforms. The initial solution obtained using separation of variables is verified and simplified using the Airy function and its Fourier transform. However, the final result does not match the expected form, leading to the suspicion of an error in the assumptions made. A simpler approach is suggested, using the Time-Dependent Schrodinger's Equation to verify the solution. The problem also mentions that the goal is to show the form of the propagator, rather than just verifying its functionality.
  • #1
Peeter
305
3

Homework Statement



Attempting a problem related to a quantum particle in free fall due to constant force (old exam question)

[tex]\begin{align*}H = \frac{1}{{2m}} P^2 + m g X\end{align*} [/tex]

The last part of the question asks to verify that the position space propagator has the following form

[tex]\begin{align*}\phi(x,t) &= \int_{-\infty}^\infty \phi(x', 0) G(x,x';t) dx' \\ G(x,x';t) &=\sqrt{\frac{m}{2 \pi \hbar i t}} \exp\left(\frac{im}{2 h t}\left( x - x' + gt^2/2\right)^2\right)\exp\left(-\frac{img}{h}\left( x t + gt^3/6\right)\right)\end{align*} [/tex]

(working in the momentum representation to get that far).


Homework Equations



One is able to show

[tex]\begin{align*}i \hbar \frac{\partial {}}{\partial {t}}\tilde{\phi}(p,t)= (mg) i \hbar \frac{\partial {}}{\partial {p}}\tilde{\phi}(p,t)+ \frac{p^2}{2m}\frac{\partial {}}{\partial {p}}\tilde{\phi}(p,t)\end{align*} [/tex]

I find as a solution, using separation of variables

[tex]\begin{align*}\tilde{\phi}(p,t) = C \exp \left( \frac{1}{{m g i \hbar}} \left( E p - \frac{p^3}{6m}\right) - \frac{i E t}{\hbar} \right)\end{align*} [/tex]

and verify that this has the form
[tex]\begin{align*}\tilde{\phi}(p,t) &= f( p + mg t ) \exp \left( - i \frac{ p^3} {6 m^2 \hbar g}\right) \\ f( p ) &=\tilde{\phi}(p,0) \exp \left( - i \frac{ p^3} {6 m^2 \hbar g}\right) \end{align*} [/tex]


The Attempt at a Solution



Relating the position and momentum representation with Fourier transforms

[tex]\begin{align*}\tilde{\phi}(p,t) &= \tilde{\phi}(p + m g t,0) \exp \left( - i \frac{ p^3 + (p + m g t)^3} {6 m^2 \hbar g}\right) \\ \phi(x,t) &= \frac{1}{{2 \pi \hbar}} \int dp\tilde{\phi}(p,t) e^{i p x/\hbar} \\ \tilde{\phi}(p + m g t,0) &=\frac{1}{{2 \pi \hbar}} \int dx'\phi(x', 0) e^{-i x'(p + mg t)/\hbar }\end{align*} [/tex]

Putting these all together and making some changes of variables I can reduce this to the propagator form:
[tex]\begin{align*}G(x,x'; t) = \frac{1}{{2 \pi}} e^{-i x' mg t/\hbar} \int dk \exp\left( i k (x-x') -\frac{i \hbar^2 }{6 m^2 g}\left( (k+ mgt/\hbar)^3 + k^3 \right)\right)\end{align*} [/tex]

But appear out of luck for any easy way to integrate this, so I suspect that I've messed this up in some fundamental way, and am looking to be pointed onto the right path.
 
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  • #2
With the help of the Airy function
[tex]\begin{align*}Ai(x) = \frac{1}{{2\pi}} \int_{-\infty}^\infty e^{i(u^3/3 + ux)} du\end{align*} [/tex]

and it's Fourier transform
[tex]\begin{align*}F(Ai(x)) = \frac{1}{{2\pi}} e^{i k^3/3}\end{align*} [/tex]

I was able to compute

[tex]\begin{align*}G(x,x'; t) = \frac{1}{{2 \pi}} \frac{6 m^2 g}{\hbar^2} e^{-i x' mg t/\hbar} \int_{-\infty}^\infty du \exp\left( i \left( u^3/3 + (x-x') u \right)\right) du\end{align*} [/tex]

(although I think I may have lost some factors of 3 along the way). This has some similarities to the expected form, but is not a match. We also haven't covered Airy functions in class, so I think there must be some simpler approach.

I'm wondering now if I've improperly assumed that [itex]\tilde{\phi}(p,t)[/itex] is related to [itex]\phi(x,t)[/itex] via Fourier transformation. Perhaps that only makes sense for the free particle case where [itex]\left\langle{{x}} \vert {{p}}\right\rangle \propto e^{i p x/\hbar}[/itex]?
 
  • #3
I haven't gone over your calculations, so I can't tell you where your error is, but I can suggest an easier way to do things... In order to verify a solution, all you really need to do is show that the solution satisfies the Time-Dependant Schrodinger's Equation... so why not just show that

[tex]i\hbar\frac{\partial}{\partial t} \phi(x,t) = H\phi(x,t)[/tex]

by calculating the appropriate derivatives?
 
  • #4
I was loose in my description of the problem, which asked to show that this is the value for G, and not just verify that it works. I suppose the verification you suggest would do that indirectly, but I think they really did want a derivation.
 
  • #5




Your approach seems to be on the right track, but there are a few mistakes in your calculations. First, when you write the solution in momentum space, you should have a term $e^{iEt/\hbar}$ instead of $e^{iEt}$. Also, in your calculation of $\tilde{\phi}(p,t)$, the exponential term should be $e^{iEpt/\hbar}$ instead of $e^{iEpt}$. Additionally, in your expression for $\tilde{\phi}(p,t)$, the factor $\exp(-i {p^3}/{6m^2 \hbar g})$ should be inside the integral instead of outside.

Furthermore, in your calculation of the propagator, you seem to have missed a factor of $1/\sqrt{2\pi\hbar}$ in the Fourier transform. Also, when changing variables, you should use $p'=p+mg t$ instead of $p'=p-mg t$. Finally, you can rewrite the exponential term in the propagator as $e^{i\hbar^2 k^2t/(2m)}$ to make the integral easier to solve.

Overall, your approach is correct but there are some minor mistakes in your calculations. I suggest reviewing your calculations and making the necessary corrections to arrive at the correct form of the propagator.
 

Related to Propagator for quantum linear potential

1. What is a propagator for quantum linear potential?

A propagator for quantum linear potential is a mathematical tool used in quantum mechanics to describe the evolution of a quantum system over time. It takes into account the potential energy of the system, which is linear in this case, and calculates the probability of the system transitioning from one state to another at a specific time.

2. How is a propagator for quantum linear potential derived?

A propagator for quantum linear potential can be derived using the Feynman path integral approach, which involves summing over all possible paths that the system could take between two given points in time. This results in a function that represents the probability amplitude for the system to transition from one state to another.

3. What is the significance of a propagator for quantum linear potential?

The propagator for quantum linear potential is an important tool for understanding and predicting the behavior of quantum systems. It allows us to calculate the probability of a system transitioning between different states, which is crucial for applications such as quantum computing and quantum information processing.

4. Can a propagator for quantum linear potential be generalized to other potential energy functions?

Yes, the concept of a propagator can be extended to other potential energy functions. However, the specific form of the propagator will depend on the type of potential energy being considered. For linear potential, the propagator is relatively simple to calculate, but for more complex potentials, it may involve more complicated mathematical techniques.

5. What are some limitations of using a propagator for quantum linear potential?

One limitation of using a propagator for quantum linear potential is that it assumes the potential energy is constant over time. This may not always be the case in real-world scenarios. Additionally, the propagator does not take into account the effects of external factors, such as interactions with other particles, which can also impact the behavior of a quantum system.

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